-
@@ -569,7 +569,7 @@ Here follows a simple example where we set up an array of ten elements, all dete
n = 10
x = np.random.normal(size=n)
-print(x)
+print(x)
We defined a vector \( x \) with \( n=10 \) elements with its values given by the Normal distribution \( N(0,1) \). @@ -579,7 +579,7 @@ Another alternative is to declare a vector as follows
import numpy as np
x = np.array([1, 2, 3])
-print(x)
+print(x)
Here we have defined a vector with three elements, with \( x_0=1 \), \( x_1=2 \) and \( x_2=3 \). Note that both Python and C++ @@ -589,7 +589,7 @@ start numbering array elements from \( 0 \) and on. This means that a vector wit
import numpy as np
x = np.log(np.array([4, 7, 8]))
-print(x)
+print(x)
In the last example we used Numpy's unary function \( np.log \). This function is @@ -608,7 +608,7 @@ logarithms of a vector would be to write x = np.array([4, 7, 8]) for i in range(0, len(x)): x[i] = log(x[i]) -print(x) +print(x)
We note that our code is much longer already and we need to import the log function from the math module. @@ -618,7 +618,7 @@ The attentive reader will also notice that the output is \( [1, 1, 2] \). Python
import numpy as np
x = np.log(np.array([4, 7, 8], dtype = np.float64))
-print(x)
+print(x)
or simply write them as double precision numbers (Python uses 64 bits as default for floating point type variables), that is @@ -627,7 +627,7 @@ or simply write them as double precision numbers (Python uses 64 bits as default
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x)
+print(x)
To check the number of bytes (remember that one byte contains eight bits for double precision variables), you can use simple use the itemsize functionality (the array \( x \) is actually an object which inherits the functionalities defined in Numpy) as @@ -636,7 +636,7 @@ To check the number of bytes (remember that one byte contains eight bits for dou
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x.itemsize)
+print(x.itemsize)
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
-print(A)
+print(A)
If we use the shape function we would get \( (3, 3) \) as output, that is verifying that our matrix is a \( 3\times 3 \) matrix. We can slice the matrix and print for example the first column (Python organized matrix elements in a row-major order, see below) as @@ -661,7 +661,7 @@ If we use the shape function we would get \( (3, 3) \) as output, that is
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[:,0])
+print(A[:,0])
We can continue this was by printing out other columns or rows. The example here prints out the second column @@ -671,7 +671,7 @@ We can continue this was by printing out other columns or rows. The example here
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[1,:])
+print(A[1,:])
Numpy contains many other functionalities that allow us to slice, subdivide etc etc arrays. We strongly recommend that you look up the Numpy website for more details. Useful functions when defining a matrix are the np.zeros function which declares a matrix of a given dimension and sets all elements to zero @@ -682,7 +682,7 @@ Numpy contains many other functionalities that allow us to slice, subdivide etc n = 10 # define a matrix of dimension 10 x 10 and set all elements to zero A = np.zeros( (n, n) ) -print(A) +print(A)
or initializing all elements to @@ -693,7 +693,7 @@ or initializing all elements to n = 10 # define a matrix of dimension 10 x 10 and set all elements to one A = np.ones( (n, n) ) -print(A) +print(A)
or as unitarily distributed random numbers (see the material on random number generators in the statistics part) @@ -704,7 +704,7 @@ or as unitarily distributed random numbers (see the material on random number ge n = 10 # define a matrix of dimension 10 x 10 and set all elements to random numbers with x \in [0, 1] A = np.random.rand(n, n) -print(A) +print(A)
As we will see throughout these lectures, there are several extremely useful functionalities in Numpy. @@ -749,16 +749,16 @@ covariance matrix through the np.linalg.eig() function. n = 100 x = np.random.normal(size=n) -print(np.mean(x)) +print(np.mean(x)) y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) +print(np.mean(y)) z = x**3+np.random.normal(size=n) -print(np.mean(z)) +print(np.mean(z)) W = np.vstack((x, y, z)) Sigma = np.cov(W) -print(Sigma) +print(Sigma) Eigvals, Eigvecs = np.linalg.eig(Sigma) -print(Eigvals) +print(Eigvals)
@@ -767,9 +767,9 @@ Eigvals, Eigvecs = npimport matplotlib.pyplot as plt from scipy import sparse eye = np.eye(4) -print(eye) +print(eye) sparse_mtx = sparse.csr_matrix(eye) -print(sparse_mtx) +print(sparse_mtx) x = np.linspace(-10,10,100) y = np.sin(x) plt.plot(x,y,marker='x') @@ -852,8 +852,8 @@ cols = 5 a = np.random.randn(rows,cols) df = pd.DataFrame(a) display(df) -print(df.mean()) -print(df.std()) +print(df.mean()) +print(df.std()) display(df**2)
@@ -865,9 +865,9 @@ Thereafter we can select specific columns only and plot final results df.index = np.arange(10) display(df) -print(df['Second'].mean() ) -print(df.info()) -print(df.describe()) +print(df['Second'].mean() ) +print(df.info()) +print(df.describe()) from pylab import plt, mpl plt.style.use('seaborn') @@ -886,9 +886,9 @@ We can produce a \( 4\times 4 \) matrix
b = np.arange(16).reshape((4,4))
-print(b)
+print(b)
df1 = pd.DataFrame(b)
-print(df1)
+print(df1)
and many other operations. @@ -1100,7 +1100,7 @@ ypredict = linreg.plot(x, np.abs(ypredict-y)/abs(y), "ro") plt.axis([0,1.0,0.0, 0.5]) plt.xlabel(r'$x$') -plt.ylabel(r'$\epsilon_{\mathrm{relative}}$') +plt.ylabel(r'$\epsilon_{\mathrm{relative}}$') plt.title(r'Relative error') plt.show() @@ -1132,16 +1132,16 @@ y = 2.0+ linreg = LinearRegression() linreg.fit(x,y) ypredict = linreg.predict(x) -print('The intercept alpha: \n', linreg.intercept_) -print('Coefficient beta : \n', linreg.coef_) +print('The intercept alpha: \n', linreg.intercept_) +print('Coefficient beta : \n', linreg.coef_) # The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) +print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) # Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(y, ypredict)) +print('Variance score: %.2f' % r2_score(y, ypredict)) # Mean squared log error -print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) +print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) # Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) +print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) plt.plot(x, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0.0,1.0,1.5, 7.0]) @@ -1252,7 +1252,7 @@ plt.show() err=(y-yn)/yn return abs(np.sum(err))/len(err) -print (error(y)) +print (error(y))
The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). @@ -1498,18 +1498,18 @@ Now we can print measures of how our fit is doing, the coefficients from the fit
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
+print('Variance score: %.2f' % r2_score(Energies, fity))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
Masses['Eapprox'] = fity
# Generate a plot comparing the experimental with the fitted values values.
fig, ax = plt.subplots()
ax.set_xlabel(r'$A = N + Z$')
-ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
label='Ame2016')
ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
@@ -1554,8 +1554,8 @@ plt.title("
plt.legend()
save_fig("Masses2016Trees")
plt.show()
-print(Masses)
-print(np.mean( (Energies-y_1)**2))
+print(Masses)
+print(np.mean( (Energies-y_1)**2))
-
n = 10
x = np.random.normal(size=n)
-print(x)
+print(x)
We defined a vector \( x \) with \( n=10 \) elements with its values given by the Normal distribution \( N(0,1) \). @@ -580,7 +580,7 @@ Another alternative is to declare a vector as follows
import numpy as np
x = np.array([1, 2, 3])
-print(x)
+print(x)
Here we have defined a vector with three elements, with \( x_0=1 \), \( x_1=2 \) and \( x_2=3 \). Note that both Python and C++ @@ -590,7 +590,7 @@ start numbering array elements from \( 0 \) and on. This means that a vector wit
import numpy as np
x = np.log(np.array([4, 7, 8]))
-print(x)
+print(x)
In the last example we used Numpy's unary function \( np.log \). This function is @@ -609,7 +609,7 @@ logarithms of a vector would be to write x = np.array([4, 7, 8]) for i in range(0, len(x)): x[i] = log(x[i]) -print(x) +print(x)
We note that our code is much longer already and we need to import the log function from the math module. @@ -619,7 +619,7 @@ The attentive reader will also notice that the output is \( [1, 1, 2] \). Python
import numpy as np
x = np.log(np.array([4, 7, 8], dtype = np.float64))
-print(x)
+print(x)
or simply write them as double precision numbers (Python uses 64 bits as default for floating point type variables), that is @@ -628,7 +628,7 @@ or simply write them as double precision numbers (Python uses 64 bits as default
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x)
+print(x)
To check the number of bytes (remember that one byte contains eight bits for double precision variables), you can use simple use the itemsize functionality (the array \( x \) is actually an object which inherits the functionalities defined in Numpy) as @@ -637,7 +637,7 @@ To check the number of bytes (remember that one byte contains eight bits for dou
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x.itemsize)
+print(x.itemsize)
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
-print(A)
+print(A)
If we use the shape function we would get \( (3, 3) \) as output, that is verifying that our matrix is a \( 3\times 3 \) matrix. We can slice the matrix and print for example the first column (Python organized matrix elements in a row-major order, see below) as @@ -662,7 +662,7 @@ If we use the shape function we would get \( (3, 3) \) as output, that is
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[:,0])
+print(A[:,0])
We can continue this was by printing out other columns or rows. The example here prints out the second column @@ -672,7 +672,7 @@ We can continue this was by printing out other columns or rows. The example here
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[1,:])
+print(A[1,:])
Numpy contains many other functionalities that allow us to slice, subdivide etc etc arrays. We strongly recommend that you look up the Numpy website for more details. Useful functions when defining a matrix are the np.zeros function which declares a matrix of a given dimension and sets all elements to zero @@ -683,7 +683,7 @@ Numpy contains many other functionalities that allow us to slice, subdivide etc n = 10 # define a matrix of dimension 10 x 10 and set all elements to zero A = np.zeros( (n, n) ) -print(A) +print(A)
or initializing all elements to @@ -694,7 +694,7 @@ or initializing all elements to n = 10 # define a matrix of dimension 10 x 10 and set all elements to one A = np.ones( (n, n) ) -print(A) +print(A)
or as unitarily distributed random numbers (see the material on random number generators in the statistics part) @@ -705,7 +705,7 @@ or as unitarily distributed random numbers (see the material on random number ge n = 10 # define a matrix of dimension 10 x 10 and set all elements to random numbers with x \in [0, 1] A = np.random.rand(n, n) -print(A) +print(A)
As we will see throughout these lectures, there are several extremely useful functionalities in Numpy. @@ -756,16 +756,16 @@ covariance matrix through the np.linalg.eig() function. n = 100 x = np.random.normal(size=n) -print(np.mean(x)) +print(np.mean(x)) y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) +print(np.mean(y)) z = x**3+np.random.normal(size=n) -print(np.mean(z)) +print(np.mean(z)) W = np.vstack((x, y, z)) Sigma = np.cov(W) -print(Sigma) +print(Sigma) Eigvals, Eigvecs = np.linalg.eig(Sigma) -print(Eigvals) +print(Eigvals)
@@ -774,9 +774,9 @@ Eigvals, Eigvecs = np.linalg.eig(Sigma) import matplotlib.pyplot as plt from scipy import sparse eye = np.eye(4) -print(eye) +print(eye) sparse_mtx = sparse.csr_matrix(eye) -print(sparse_mtx) +print(sparse_mtx) x = np.linspace(-10,10,100) y = np.sin(x) plt.plot(x,y,marker='x') @@ -859,8 +859,8 @@ cols = 5 a = np.random.randn(rows,cols) df = pd.DataFrame(a) display(df) -print(df.mean()) -print(df.std()) +print(df.mean()) +print(df.std()) display(df**2)
@@ -872,9 +872,9 @@ Thereafter we can select specific columns only and plot final results df.index = np.arange(10) display(df) -print(df['Second'].mean() ) -print(df.info()) -print(df.describe()) +print(df['Second'].mean() ) +print(df.info()) +print(df.describe()) from pylab import plt, mpl plt.style.use('seaborn') @@ -893,9 +893,9 @@ We can produce a \( 4\times 4 \) matrix
b = np.arange(16).reshape((4,4))
-print(b)
+print(b)
df1 = pd.DataFrame(b)
-print(df1)
+print(df1)
and many other operations. @@ -1147,16 +1147,16 @@ y = 2.0+ 5print('The intercept alpha: \n', linreg.intercept_) -print('Coefficient beta : \n', linreg.coef_) +print('The intercept alpha: \n', linreg.intercept_) +print('Coefficient beta : \n', linreg.coef_) # The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) +print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) # Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(y, ypredict)) +print('Variance score: %.2f' % r2_score(y, ypredict)) # Mean squared log error -print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) +print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) # Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) +print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) plt.plot(x, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0.0,1.0,1.5, 7.0]) @@ -1279,7 +1279,7 @@ plt.show() err=(y-yn)/yn return abs(np.sum(err))/len(err) -print (error(y)) +print (error(y))
The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). @@ -1541,12 +1541,12 @@ Now we can print measures of how our fit is doing, the coefficients from the fit
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
+print('Variance score: %.2f' % r2_score(Energies, fity))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
Masses['Eapprox'] = fity
# Generate a plot comparing the experimental with the fitted values values.
@@ -1597,8 +1597,8 @@ plt.title("Decision Tree Regression""Masses2016Trees")
plt.show()
-print(Masses)
-print(np.mean( (Energies-y_1)**2))
+print(Masses)
+print(np.mean( (Energies-y_1)**2))
-
n = 10
x = np.random.normal(size=n)
-print(x)
+print(x)
We defined a vector \( x \) with \( n=10 \) elements with its values given by the Normal distribution \( N(0,1) \). @@ -539,7 +539,7 @@ Another alternative is to declare a vector as follows
import numpy as np
x = np.array([1, 2, 3])
-print(x)
+print(x)
Here we have defined a vector with three elements, with \( x_0=1 \), \( x_1=2 \) and \( x_2=3 \). Note that both Python and C++ @@ -549,7 +549,7 @@ start numbering array elements from \( 0 \) and on. This means that a vector wit
import numpy as np
x = np.log(np.array([4, 7, 8]))
-print(x)
+print(x)
In the last example we used Numpy's unary function \( np.log \). This function is @@ -568,7 +568,7 @@ logarithms of a vector would be to write x = np.array([4, 7, 8]) for i in range(0, len(x)): x[i] = log(x[i]) -print(x) +print(x)
We note that our code is much longer already and we need to import the log function from the math module. @@ -578,7 +578,7 @@ The attentive reader will also notice that the output is \( [1, 1, 2] \). Python
import numpy as np
x = np.log(np.array([4, 7, 8], dtype = np.float64))
-print(x)
+print(x)
or simply write them as double precision numbers (Python uses 64 bits as default for floating point type variables), that is @@ -587,7 +587,7 @@ or simply write them as double precision numbers (Python uses 64 bits as default
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x)
+print(x)
To check the number of bytes (remember that one byte contains eight bits for double precision variables), you can use simple use the itemsize functionality (the array \( x \) is actually an object which inherits the functionalities defined in Numpy) as @@ -596,7 +596,7 @@ To check the number of bytes (remember that one byte contains eight bits for dou
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x.itemsize)
+print(x.itemsize)
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
-print(A)
+print(A)
If we use the shape function we would get \( (3, 3) \) as output, that is verifying that our matrix is a \( 3\times 3 \) matrix. We can slice the matrix and print for example the first column (Python organized matrix elements in a row-major order, see below) as @@ -621,7 +621,7 @@ If we use the shape function we would get \( (3, 3) \) as output, that is
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[:,0])
+print(A[:,0])
We can continue this was by printing out other columns or rows. The example here prints out the second column @@ -631,7 +631,7 @@ We can continue this was by printing out other columns or rows. The example here
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[1,:])
+print(A[1,:])
Numpy contains many other functionalities that allow us to slice, subdivide etc etc arrays. We strongly recommend that you look up the Numpy website for more details. Useful functions when defining a matrix are the np.zeros function which declares a matrix of a given dimension and sets all elements to zero @@ -642,7 +642,7 @@ Numpy contains many other functionalities that allow us to slice, subdivide etc n = 10 # define a matrix of dimension 10 x 10 and set all elements to zero A = np.zeros( (n, n) ) -print(A) +print(A)
or initializing all elements to @@ -653,7 +653,7 @@ or initializing all elements to n = 10 # define a matrix of dimension 10 x 10 and set all elements to one A = np.ones( (n, n) ) -print(A) +print(A)
or as unitarily distributed random numbers (see the material on random number generators in the statistics part) @@ -664,7 +664,7 @@ or as unitarily distributed random numbers (see the material on random number ge n = 10 # define a matrix of dimension 10 x 10 and set all elements to random numbers with x \in [0, 1] A = np.random.rand(n, n) -print(A) +print(A)
As we will see throughout these lectures, there are several extremely useful functionalities in Numpy. @@ -709,16 +709,16 @@ covariance matrix through the np.linalg.eig() function. n = 100 x = np.random.normal(size=n) -print(np.mean(x)) +print(np.mean(x)) y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) +print(np.mean(y)) z = x**3+np.random.normal(size=n) -print(np.mean(z)) +print(np.mean(z)) W = np.vstack((x, y, z)) Sigma = np.cov(W) -print(Sigma) +print(Sigma) Eigvals, Eigvecs = np.linalg.eig(Sigma) -print(Eigvals) +print(Eigvals)
@@ -727,9 +727,9 @@ Eigvals, Eigvecs = np.linalg.eig(Sigma) import matplotlib.pyplot as plt from scipy import sparse eye = np.eye(4) -print(eye) +print(eye) sparse_mtx = sparse.csr_matrix(eye) -print(sparse_mtx) +print(sparse_mtx) x = np.linspace(-10,10,100) y = np.sin(x) plt.plot(x,y,marker='x') @@ -812,8 +812,8 @@ cols = 5 a = np.random.randn(rows,cols) df = pd.DataFrame(a) display(df) -print(df.mean()) -print(df.std()) +print(df.mean()) +print(df.std()) display(df**2)
@@ -825,9 +825,9 @@ Thereafter we can select specific columns only and plot final results df.index = np.arange(10) display(df) -print(df['Second'].mean() ) -print(df.info()) -print(df.describe()) +print(df['Second'].mean() ) +print(df.info()) +print(df.describe()) from pylab import plt, mpl plt.style.use('seaborn') @@ -846,9 +846,9 @@ We can produce a \( 4\times 4 \) matrix
b = np.arange(16).reshape((4,4))
-print(b)
+print(b)
df1 = pd.DataFrame(b)
-print(df1)
+print(df1)
and many other operations. @@ -1092,16 +1092,16 @@ y = 2.0+ 5print('The intercept alpha: \n', linreg.intercept_) -print('Coefficient beta : \n', linreg.coef_) +print('The intercept alpha: \n', linreg.intercept_) +print('Coefficient beta : \n', linreg.coef_) # The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) +print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) # Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(y, ypredict)) +print('Variance score: %.2f' % r2_score(y, ypredict)) # Mean squared log error -print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) +print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) # Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) +print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) plt.plot(x, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0.0,1.0,1.5, 7.0]) @@ -1212,7 +1212,7 @@ plt.show() err=(y-yn)/yn return abs(np.sum(err))/len(err) -print (error(y)) +print (error(y))
The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). @@ -1458,12 +1458,12 @@ Now we can print measures of how our fit is doing, the coefficients from the fit
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
+print('Variance score: %.2f' % r2_score(Energies, fity))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
Masses['Eapprox'] = fity
# Generate a plot comparing the experimental with the fitted values values.
@@ -1514,8 +1514,8 @@ plt.title("Decision Tree Regression""Masses2016Trees")
plt.show()
-print(Masses)
-print(np.mean( (Energies-y_1)**2))
+print(Masses)
+print(np.mean( (Energies-y_1)**2))
-
n = 10
x = np.random.normal(size=n)
-print(x)
+print(x)
We defined a vector \( x \) with \( n=10 \) elements with its values given by the Normal distribution \( N(0,1) \). @@ -544,7 +544,7 @@ Another alternative is to declare a vector as follows
import numpy as np
x = np.array([1, 2, 3])
-print(x)
+print(x)
Here we have defined a vector with three elements, with \( x_0=1 \), \( x_1=2 \) and \( x_2=3 \). Note that both Python and C++ @@ -554,7 +554,7 @@ start numbering array elements from \( 0 \) and on. This means that a vector wit
import numpy as np
x = np.log(np.array([4, 7, 8]))
-print(x)
+print(x)
In the last example we used Numpy's unary function \( np.log \). This function is @@ -573,7 +573,7 @@ logarithms of a vector would be to write x = np.array([4, 7, 8]) for i in range(0, len(x)): x[i] = log(x[i]) -print(x) +print(x)
We note that our code is much longer already and we need to import the log function from the math module. @@ -583,7 +583,7 @@ The attentive reader will also notice that the output is \( [1, 1, 2] \). Python
import numpy as np
x = np.log(np.array([4, 7, 8], dtype = np.float64))
-print(x)
+print(x)
or simply write them as double precision numbers (Python uses 64 bits as default for floating point type variables), that is @@ -592,7 +592,7 @@ or simply write them as double precision numbers (Python uses 64 bits as default
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x)
+print(x)
To check the number of bytes (remember that one byte contains eight bits for double precision variables), you can use simple use the itemsize functionality (the array \( x \) is actually an object which inherits the functionalities defined in Numpy) as @@ -601,7 +601,7 @@ To check the number of bytes (remember that one byte contains eight bits for dou
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
-print(x.itemsize)
+print(x.itemsize)
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
-print(A)
+print(A)
If we use the shape function we would get \( (3, 3) \) as output, that is verifying that our matrix is a \( 3\times 3 \) matrix. We can slice the matrix and print for example the first column (Python organized matrix elements in a row-major order, see below) as @@ -626,7 +626,7 @@ If we use the shape function we would get \( (3, 3) \) as output, that is
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[:,0])
+print(A[:,0])
We can continue this was by printing out other columns or rows. The example here prints out the second column @@ -636,7 +636,7 @@ We can continue this was by printing out other columns or rows. The example here
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
-print(A[1,:])
+print(A[1,:])
Numpy contains many other functionalities that allow us to slice, subdivide etc etc arrays. We strongly recommend that you look up the Numpy website for more details. Useful functions when defining a matrix are the np.zeros function which declares a matrix of a given dimension and sets all elements to zero @@ -647,7 +647,7 @@ Numpy contains many other functionalities that allow us to slice, subdivide etc n = 10 # define a matrix of dimension 10 x 10 and set all elements to zero A = np.zeros( (n, n) ) -print(A) +print(A)
or initializing all elements to @@ -658,7 +658,7 @@ or initializing all elements to n = 10 # define a matrix of dimension 10 x 10 and set all elements to one A = np.ones( (n, n) ) -print(A) +print(A)
or as unitarily distributed random numbers (see the material on random number generators in the statistics part) @@ -669,7 +669,7 @@ or as unitarily distributed random numbers (see the material on random number ge n = 10 # define a matrix of dimension 10 x 10 and set all elements to random numbers with x \in [0, 1] A = np.random.rand(n, n) -print(A) +print(A)
As we will see throughout these lectures, there are several extremely useful functionalities in Numpy. @@ -714,16 +714,16 @@ covariance matrix through the np.linalg.eig() function. n = 100 x = np.random.normal(size=n) -print(np.mean(x)) +print(np.mean(x)) y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) +print(np.mean(y)) z = x**3+np.random.normal(size=n) -print(np.mean(z)) +print(np.mean(z)) W = np.vstack((x, y, z)) Sigma = np.cov(W) -print(Sigma) +print(Sigma) Eigvals, Eigvecs = np.linalg.eig(Sigma) -print(Eigvals) +print(Eigvals)
@@ -732,9 +732,9 @@ Eigvals, Eigvecs = npimport matplotlib.pyplot as plt from scipy import sparse eye = np.eye(4) -print(eye) +print(eye) sparse_mtx = sparse.csr_matrix(eye) -print(sparse_mtx) +print(sparse_mtx) x = np.linspace(-10,10,100) y = np.sin(x) plt.plot(x,y,marker='x') @@ -817,8 +817,8 @@ cols = 5 a = np.random.randn(rows,cols) df = pd.DataFrame(a) display(df) -print(df.mean()) -print(df.std()) +print(df.mean()) +print(df.std()) display(df**2)
@@ -830,9 +830,9 @@ Thereafter we can select specific columns only and plot final results df.index = np.arange(10) display(df) -print(df['Second'].mean() ) -print(df.info()) -print(df.describe()) +print(df['Second'].mean() ) +print(df.info()) +print(df.describe()) from pylab import plt, mpl plt.style.use('seaborn') @@ -851,9 +851,9 @@ We can produce a \( 4\times 4 \) matrix
b = np.arange(16).reshape((4,4))
-print(b)
+print(b)
df1 = pd.DataFrame(b)
-print(df1)
+print(df1)
and many other operations. @@ -1065,7 +1065,7 @@ ypredict = linreg.plot(x, np.abs(ypredict-y)/abs(y), "ro") plt.axis([0,1.0,0.0, 0.5]) plt.xlabel(r'$x$') -plt.ylabel(r'$\epsilon_{\mathrm{relative}}$') +plt.ylabel(r'$\epsilon_{\mathrm{relative}}$') plt.title(r'Relative error') plt.show() @@ -1097,16 +1097,16 @@ y = 2.0+ linreg = LinearRegression() linreg.fit(x,y) ypredict = linreg.predict(x) -print('The intercept alpha: \n', linreg.intercept_) -print('Coefficient beta : \n', linreg.coef_) +print('The intercept alpha: \n', linreg.intercept_) +print('Coefficient beta : \n', linreg.coef_) # The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) +print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) # Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(y, ypredict)) +print('Variance score: %.2f' % r2_score(y, ypredict)) # Mean squared log error -print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) +print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) # Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) +print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) plt.plot(x, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0.0,1.0,1.5, 7.0]) @@ -1217,7 +1217,7 @@ plt.show() err=(y-yn)/yn return abs(np.sum(err))/len(err) -print (error(y)) +print (error(y))
The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). @@ -1463,18 +1463,18 @@ Now we can print measures of how our fit is doing, the coefficients from the fit
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
+print('Variance score: %.2f' % r2_score(Energies, fity))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
Masses['Eapprox'] = fity
# Generate a plot comparing the experimental with the fitted values values.
fig, ax = plt.subplots()
ax.set_xlabel(r'$A = N + Z$')
-ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
label='Ame2016')
ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
@@ -1519,8 +1519,8 @@ plt.title("
plt.legend()
save_fig("Masses2016Trees")
plt.show()
-print(Masses)
-print(np.mean( (Energies-y_1)**2))
+print(Masses)
+print(np.mean( (Energies-y_1)**2))
-
@@ -469,7 +469,7 @@ society.
-
-
-
-
@@ -449,7 +451,7 @@ MathJax.Hub.Config({
-Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \). +
diff --git a/doc/pub/Regression/html/._Regression-bs002.html b/doc/pub/Regression/html/._Regression-bs002.html index 2f239db32..7c9f1b69d 100644 --- a/doc/pub/Regression/html/._Regression-bs002.html +++ b/doc/pub/Regression/html/._Regression-bs002.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
+
-Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). -The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable. - -
-A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with +Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
@@ -447,7 +446,7 @@ A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\ver
-Consider an experiment in which \( p \) characteristics of \( n \) samples are -measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix -\( \mathbf{X} \). +Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). +The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
-The matrix \( \mathbf{X} \) is called the design -matrix. Additional information of the samples is available in the -form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is -generally referred to as the response variable. The aim of -regression analysis is to explain \( \boldsymbol{y} \) in terms of -\( \boldsymbol{X} \) through a functional relationship like \( y_i = -f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of -\( f(\cdot) \) is available, it is common to assume a linear relationship -between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to -the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots, -\beta_{p-1}]^{T} \) are the regression parameters. +A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with -
-Linear regression gives us a set of analytical equations for the parameters \( \beta_j \). +
+ The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
-In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \), -consider the model we discussed for describing nuclear binding energies.
-There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. -Assuming -$$ -BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, -$$ - -we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms. -This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a -\( p\times n \) matrix \( \boldsymbol{X} \). +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix +\( \mathbf{X} \).
-Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the -so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression. +The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \boldsymbol{y} \) in terms of +\( \boldsymbol{X} \) through a functional relationship like \( y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to +the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots, +\beta_{p-1}]^{T} \) are the regression parameters. + +
+Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
-Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function. +In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \), +consider the model we discussed for describing nuclear binding energies.
-Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is +There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. +Assuming $$ -y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, +BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, $$ -where \( \epsilon_i \) is the error in our approximation. +we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms. +This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a +\( p\times n \) matrix \( \boldsymbol{X} \). + +
+Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the +so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression.
-For every set of values \( y_i,x_i \) we have thus the corresponding set of equations +Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function. + +
+Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is $$ -\begin{align*} -y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ -y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ -y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ -\end{align*} +y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, $$ + +where \( \epsilon_i \) is the error in our approximation. + +
-Defining the vectors +For every set of values \( y_i,x_i \) we have thus the corresponding set of equations $$ -\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, +\begin{align*} +y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ +y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ +y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ - -and -$$ -\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, -$$ - -and -$$ -\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, -$$ - -and the design matrix -$$ -\boldsymbol{X}= -\begin{bmatrix} -1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ -1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ -1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ -\end{bmatrix} -$$ - -we can rewrite our equations as -$$ -\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. -$$ - -The above design matrix is called a Vandermonde matrix.
- -
-We are obviously not limited to the above polynomial expansions. We -could replace the various powers of \( x \) with elements of Fourier -series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j -x_i)} \), or time series or other orthogonal functions. For every set -of values \( y_i,x_i \) we can then generalize the equations to - +Defining the vectors $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$ -
-Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case! +and +$$ +\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, +$$ + +and +$$ +\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, +$$ + +and the design matrix +$$ +\boldsymbol{X}= +\begin{bmatrix} +1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ +1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ +1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ +\end{bmatrix} +$$ + +we can rewrite our equations as +$$ +\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. +$$ + +The above design matrix is called a Vandermonde matrix.
-We redefine in turn the matrix \( \boldsymbol{X} \) as + +
+We are obviously not limited to the above polynomial expansions. We +could replace the various powers of \( x \) with elements of Fourier +series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j +x_i)} \), or time series or other orthogonal functions. For every set +of values \( y_i,x_i \) we can then generalize the equations to + $$ -\boldsymbol{X}= -\begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ -\end{bmatrix} +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ -and without loss of generality we rewrite again our equations as -$$ -\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. -$$ - -The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values? +
+Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
-We have defined the matrix \( \boldsymbol{X} \) via the equations +We redefine in turn the matrix \( \boldsymbol{X} \) as $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\boldsymbol{X}= +\begin{bmatrix} +x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ +x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ +x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +\end{bmatrix} $$ -
-As we noted above, we stayed with a system with the design matrix - \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define -our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements. +and without loss of generality we rewrite again our equations as +$$ +\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. +$$ -
+The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
-In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code. - -
-We restate the parts of the code we are most interested in. -
- - -
# Common imports
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from IPython.display import display
-import os
-
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
-
-if not os.path.exists(PROJECT_ROOT_DIR):
- os.mkdir(PROJECT_ROOT_DIR)
-
-if not os.path.exists(FIGURE_ID):
- os.makedirs(FIGURE_ID)
-
-if not os.path.exists(DATA_ID):
- os.makedirs(DATA_ID)
-
-def image_path(fig_id):
- return os.path.join(FIGURE_ID, fig_id)
-
-def data_path(dat_id):
- return os.path.join(DATA_ID, dat_id)
-
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-infile = open(data_path("MassEval2016.dat"),'r')
-
-
-# Read the experimental data with Pandas
-Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
- names=('N', 'Z', 'A', 'Element', 'Ebinding'),
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- header=39,
- index_col=False)
-
-# Extrapolated values are indicated by '#' in place of the decimal place, so
-# the Ebinding column won't be numeric. Coerce to float and drop these entries.
-Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
-Masses = Masses.dropna()
-# Convert from keV to MeV.
-Masses['Ebinding'] /= 1000
-
-# Group the DataFrame by nucleon number, A.
-Masses = Masses.groupby('A')
-# Find the rows of the grouped DataFrame with the maximum binding energy.
-Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
-A = Masses['A']
-Z = Masses['Z']
-N = Masses['N']
-Element = Masses['Element']
-Energies = Masses['Ebinding']
-
-# Now we set up the design matrix X
-X = np.zeros((len(A),5))
-X[:,0] = 1
-X[:,1] = A
-X[:,2] = A**(2.0/3.0)
-X[:,3] = A**(-1.0/3.0)
-X[:,4] = A**(-1.0)
-# Then nice printout using pandas
-DesignMatrix = pd.DataFrame(X)
-DesignMatrix.index = A
-DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
-display(DesignMatrix)
--With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as +
+We have defined the matrix \( \boldsymbol{X} \) via the equations $$ -\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ -throughout these lectures. +
+As we noted above, we stayed with a system with the design matrix + \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define +our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements. + +
+
@@ -520,7 +461,7 @@ throughout these lectures.
-With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as +
+In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code. + +
+We restate the parts of the code we are most interested in. +
+ + +
# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from IPython.display import display
+import os
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("MassEval2016.dat"),'r')
+
+
+# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+
+# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+# Then nice printout using pandas
+DesignMatrix = pd.DataFrame(X)
+DesignMatrix.index = A
+DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
+display(DesignMatrix)
++With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as $$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$ -and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely -$$ -C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, -$$ - -or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as -$$ -C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)\right\}. -$$ - -This function is one possible way to define the so-called cost function. - -
-It is also common to define -the function \( Q \) as - -$$ -C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, -$$ - -since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out. -
@@ -466,7 +522,7 @@ since when taking the first derivative with respect to the unknown parameters \(
+With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as +$$ +\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, +$$ + +and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely +$$ +C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, +$$ + +or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as +$$ +C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)\right\}. +$$ + +This function is one possible way to define the so-called cost function.
-The function +It is also common to define +the function \( Q \) as + $$ -C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, +C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, $$ -can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value. -When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value -$$ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, -$$ - -
-where \( \langle y_i \rangle \) is the mean value. Keep in mind also that -till now we have treated \( y_i \) as the exact value. Normally, the -response (dependent or outcome) variable \( y_i \) the outcome of a -numerical experiment or another type of experiment and is thus only an -approximation to the true value. It is then always accompanied by an -error estimate, often limited to a statistical error estimate given by -the standard deviation discussed earlier. In the discussion here we -will treat \( y_i \) as our exact value for the response variable. - -
-In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. -$$ - -In practical terms it means we will require -$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, -$$ - -which results in -$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, -$$ - -or in a matrix-vector form as -$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). -$$ - -
+since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
-We can rewrite + +
+The function $$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), +C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, $$ -as +can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value. +When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value $$ -\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, -$$ - -and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution -$$ -\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. +y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, $$
-We note also that since our design matrix is defined as \( \boldsymbol{X}\in -{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in -{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \), -in our case \( p=5 \) meaning that we end up with inverting a small -\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional -matrices to invert. The methods discussed here and for many other -supervised learning algorithms like classification with logistic -regression or support vector machines, exhibit dimensionalities which -allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix -\( \boldsymbol{X}^T\boldsymbol{X} \). -
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +$$ + +In practical terms it means we will require +$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +$$ + +which results in +$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, +$$ + +or in a matrix-vector form as +$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). +$$
-
-Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+We can rewrite +$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), +$$ + +as +$$ +\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, +$$ + +and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution +$$ +\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. +$$
-The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and -matrices as upper case boldfaced letters. +We note also that since our design matrix is defined as \( \boldsymbol{X}\in +{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in +{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \), +in our case \( p=5 \) meaning that we end up with inverting a small +\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional +matrices to invert. The methods discussed here and for many other +supervised learning algorithms like classification with logistic +regression or support vector machines, exhibit dimensionalities which +allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix +\( \boldsymbol{X}^T\boldsymbol{X} \). +
+
+Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect? +
diff --git a/doc/pub/Regression/html/._Regression-bs016.html b/doc/pub/Regression/html/._Regression-bs016.html index 086e28144..d6a5afdbd 100644 --- a/doc/pub/Regression/html/._Regression-bs016.html +++ b/doc/pub/Regression/html/._Regression-bs016.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
-The residuals \( \boldsymbol{\epsilon} \) are in turn given by -$$ -\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}, -$$ - -and with -$$ -\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, -$$ - -we have -$$ -\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, -$$ - -meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. +
-
-Let us now return to our nuclear binding energies and simply code the above equations. +$$ +\frac{\partial (\boldsymbol{a}^T\boldsymbol{A}\boldsymbol{a})}{\partial \boldsymbol{a}} = (\boldsymbol{A}+\boldsymbol{A}^T)\boldsymbol{a}, +$$ + +$$ +\frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T, +$$ + +$$ +\frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T. +$$ -
+The residuals \( \boldsymbol{\epsilon} \) are in turn given by +$$ +\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}, +$$ + +and with +$$ +\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, +$$ + +we have +$$ +\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, +$$ + +meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
-It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to -write -
+
# matrix inversion to find beta
-beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
-# and then make the prediction
-ytilde = X @ beta
--Alternatively, you can use the least squares functionality in Numpy as -
- -
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
-ytildenp = np.dot(fit,X.T)
--And finally we plot our fit with and compare with data
+Let us now return to our nuclear binding energies and simply code the above equations. - -
Masses['Eapprox'] = ytilde
-# Generate a plot comparing the experimental with the fitted values values.
-fig, ax = plt.subplots()
-ax.set_xlabel(r'$A = N + Z$')
-ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
-ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
- label='Ame2016')
-ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
- label='Fit')
-ax.legend()
-save_fig("Masses2016OLS")
-plt.show()
-
@@ -471,7 +463,7 @@ plt.show()
-We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of _Scikit_Learn_ in the introductory slides. -Since we are not using _Scikit-Learn here we can define our own \( R2 \) function as +It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to +write
-
def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
+# and then make the prediction
+ytilde = X @ beta
-and we would be using it as
+Alternatively, you can use the least squares functionality in Numpy as
-
print(R2(Energies,ytilde))
+fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+ytildenp = np.dot(fit,X.T)
-We can easily add our MSE score as
+And finally we plot our fit with and compare with data
-
def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-print(MSE(Energies,ytilde))
+Masses['Eapprox'] = ytilde
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016OLS")
+plt.show()
-
-and finally the relative error as
-
-
-
-
def RelativeError(y_data,y_model):
- return abs((y_data-y_model)/y_data)
-print(RelativeError(Energies, ytilde))
-
-
-We could also add the so-called Huber norm, which we defined as
-$$
-H_{\delta}(a)={\begin{cases}{\frac {1}{2}}{a^{2}}&{\text{for }}|a|\leq \delta ,\\\delta (|a|-{\frac {1}{2}}\delta ),&{\text{otherwise.}}\end{cases}}},
-$$
-
-with \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
-
@@ -478,7 +473,7 @@ with \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
27
28
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- 111
+ 112
»
diff --git a/doc/pub/Regression/html/._Regression-bs019.html b/doc/pub/Regression/html/._Regression-bs019.html
index a2cd2a498..e249f0901 100644
--- a/doc/pub/Regression/html/._Regression-bs019.html
+++ b/doc/pub/Regression/html/._Regression-bs019.html
@@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source
@@ -281,116 +282,117 @@ MathJax.Hub.Config({
Contents
@@ -406,35 +408,51 @@ MathJax.Hub.Config({
-The \( \chi^2 \) function
-
-
-
+
Adding error analysis and training set up
-Normally, the response (dependent or outcome) variable \( y_i \) is the
-outcome of a numerical experiment or another type of experiment and is
-thus only an approximation to the true value. It is then always
-accompanied by an error estimate, often limited to a statistical error
-estimate given by the standard deviation discussed earlier. In the
-discussion here we will treat \( y_i \) as our exact value for the
-response variable.
-
+We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of _Scikit_Learn_ in the introductory slides.
+Since we are not using _Scikit-Learn here we can define our own \( R2 \) function as
-Introducing the standard deviation \( \sigma_i \) for each measurement
-\( y_i \), we define now the \( \chi^2 \) function (omitting the \( 1/n \) term)
-as
+
+
def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+
+
+and we would be using it as
+
+
+
+
print(R2(Energies,ytilde))
+
+
+We can easily add our MSE score as
+
+
+
+
def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+print(MSE(Energies,ytilde))
+
+
+and finally the relative error as
+
+
+
+
def RelativeError(y_data,y_model):
+ return abs((y_data-y_model)/y_data)
+print(RelativeError(Energies, ytilde))
+
+
+We could also add the so-called Huber norm, which we defined as
$$
-\chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+H_{\delta}(a)={\begin{cases}{\frac {1}{2}}{a^{2}}&{\text{for }}|a|\leq \delta ,\\\delta (|a|-{\frac {1}{2}}\delta ),&{\text{otherwise.}}\end{cases}}},
$$
-where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
-
-
-
-
-
+with \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
@@ -462,7 +480,7 @@ where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i
28
29
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+ 112
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diff --git a/doc/pub/Regression/html/._Regression-bs020.html b/doc/pub/Regression/html/._Regression-bs020.html
index 74eaadc1f..2d6b7a159 100644
--- a/doc/pub/Regression/html/._Regression-bs020.html
+++ b/doc/pub/Regression/html/._Regression-bs020.html
@@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source
@@ -281,116 +282,117 @@ MathJax.Hub.Config({
Contents
@@ -412,22 +414,26 @@ MathJax.Hub.Config({
-In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\beta}) \) by requiring
+Normally, the response (dependent or outcome) variable \( y_i \) is the
+outcome of a numerical experiment or another type of experiment and is
+thus only an approximation to the true value. It is then always
+accompanied by an error estimate, often limited to a statistical error
+estimate given by the standard deviation discussed earlier. In the
+discussion here we will treat \( y_i \) as our exact value for the
+response variable.
+
+
+Introducing the standard deviation \( \sigma_i \) for each measurement
+\( y_i \), we define now the \( \chi^2 \) function (omitting the \( 1/n \) term)
+as
+
$$
-\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
+\chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
$$
-which results in
-$$
-\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
-$$
+where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
-or in a matrix-vector form as
-$$
-\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right).
-$$
-
-where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).
+
@@ -458,7 +464,7 @@ where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\
-We can rewrite +In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\beta}) \) by requiring $$ -\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right), +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, $$ -as +which results in $$ -\boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta}, +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, $$ -and if the matrix \( \boldsymbol{A}^T\boldsymbol{A} \) is invertible we have the solution +or in a matrix-vector form as $$ -\boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}. +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right). $$ + +where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).
-If we then introduce the matrix +We can rewrite $$ -\boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}, +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right), $$ -we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \boldsymbol{H} \) are \( h_{ij} \)) +as $$ -\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} +\boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta}, $$ -We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) +and if the matrix \( \boldsymbol{A}^T\boldsymbol{A} \) is invertible we have the solution $$ -\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, -$$ - -resulting in -$$ -\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! +\boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}. $$
-The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write + +
+If we then introduce the matrix $$ -y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. +\boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}, $$ -By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by +we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \boldsymbol{H} \) are \( h_{ij} \)) $$ -\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, +\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} $$ -and +We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) $$ -\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. +\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, +$$ + +resulting in +$$ +\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! $$
- -
-For a linear fit (a first-order polynomial) we don't need to invert a matrix!! -Defining +The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write $$ -\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. $$ - +By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by $$ -\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, $$ - +and $$ -\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. $$ - - -$$ -\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, -$$ - - -$$ -\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, -$$ - -
-we obtain - -$$ -\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, -$$ - - -$$ -\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. -$$ - -
-This approach (different linear and non-linear regression) suffers -often from both being underdetermined and overdetermined in the -unknown coefficients \( \beta_i \). A better approach is to use the -Singular Value Decomposition (SVD) method discussed below. Or using -Lasso and Ridge regression. See below. - -
-Before we continue, let us introduce yet another example. We are going to fit the -nuclear equation of state using results from many-body calculations. -The equation of state we have made available here, as function of -density, has been derived using modern nucleon-nucleon potentials with -the addition of three-body -forces. This -time the file is presented as a standard csv file. +For a linear fit (a first-order polynomial) we don't need to invert a matrix!! +Defining +$$ +\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +$$ + + +$$ +\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +$$ + + +$$ +\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +$$ + + +$$ +\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, +$$ + + +$$ +\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, +$$
-The beginning of the Python code here is similar to what you have seen -before, with the same initializations and declarations. We use also -pandas again, rather extensively in order to organize our data. +we obtain + +$$ +\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, +$$ + + +$$ +\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. +$$
-The difference now is that we use Scikit-Learn's regression tools -instead of our own matrix inversion implementation. Furthermore, we -sneak in Ridge regression (to be discussed below) which includes a -hyperparameter \( \lambda \), also to be explained below. +This approach (different linear and non-linear regression) suffers +often from both being underdetermined and overdetermined in the +unknown coefficients \( \beta_i \). A better approach is to use the +Singular Value Decomposition (SVD) method discussed below. Or using +Lasso and Ridge regression. See below. + +
+
@@ -454,7 +490,7 @@ hyperparameter \( \lambda \), also to be explained below.
- - -
# Common imports
-import os
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-import matplotlib.pyplot as plt
-import sklearn.linear_model as skl
-from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
-
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
-
-if not os.path.exists(PROJECT_ROOT_DIR):
- os.mkdir(PROJECT_ROOT_DIR)
-
-if not os.path.exists(FIGURE_ID):
- os.makedirs(FIGURE_ID)
-
-if not os.path.exists(DATA_ID):
- os.makedirs(DATA_ID)
-
-def image_path(fig_id):
- return os.path.join(FIGURE_ID, fig_id)
-
-def data_path(dat_id):
- return os.path.join(DATA_ID, dat_id)
-
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-infile = open(data_path("EoS.csv"),'r')
-
-# Read the EoS data as csv file and organize the data into two arrays with density and energies
-EoS = pd.read_csv(infile, names=('Density', 'Energy'))
-EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
-EoS = EoS.dropna()
-Energies = EoS['Energy']
-Density = EoS['Density']
-# The design matrix now as function of various polytrops
-X = np.zeros((len(Density),4))
-X[:,3] = Density**(4.0/3.0)
-X[:,2] = Density
-X[:,1] = Density**(2.0/3.0)
-X[:,0] = 1
-
-# We use now Scikit-Learn's linear regressor and ridge regressor
-# OLS part
-clf = skl.LinearRegression().fit(X, Energies)
-ytilde = clf.predict(X)
-EoS['Eols'] = ytilde
-# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, ytilde))
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
-print(clf.coef_, clf.intercept_)
-
-# The Ridge regression with a hyperparameter lambda = 0.1
-_lambda = 0.1
-clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
-yridge = clf_ridge.predict(X)
-EoS['Eridge'] = yridge
-# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, yridge))
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
-print(clf_ridge.coef_, clf_ridge.intercept_)
-
-fig, ax = plt.subplots()
-ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
-ax.set_ylabel(r'Energy per particle')
-ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,
- label='Theoretical data')
-ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',
- label='OLS')
-ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',
- label='Ridge $\lambda = 0.1$')
-ax.legend()
-save_fig("EoSfitting")
-plt.show()
--The above simple polynomial in density \( \rho \) gives an excellent fit -to the data. +Before we continue, let us introduce yet another example. We are going to fit the +nuclear equation of state using results from many-body calculations. +The equation of state we have made available here, as function of +density, has been derived using modern nucleon-nucleon potentials with +the addition of three-body +forces. This +time the file is presented as a standard csv file.
-We note also that there is a small deviation between the -standard OLS and the Ridge regression at higher densities. We discuss this in more detail -below. +The beginning of the Python code here is similar to what you have seen +before, with the same initializations and declarations. We use also +pandas again, rather extensively in order to organize our data. + +
+The difference now is that we use Scikit-Learn's regression tools +instead of our own matrix inversion implementation. Furthermore, we +sneak in Ridge regression (to be discussed below) which includes a +hyperparameter \( \lambda \), also to be explained below.
@@ -532,7 +456,7 @@ below.
-It is normal in essentially all Machine Learning studies to split the -data in a training set and a test set (sometimes also an additional -validation set). Scikit-Learn has an own function for this. There -is no explicit recipe for how much data should be included as training -data and say test data. An accepted rule of thumb is to use -approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will -postpone a discussion of this splitting to the end of these notes and -our discussion of the so-called bias-variance tradeoff. Here we -limit ourselves to repeat the above equation of state fitting example -but now splitting the data into a training set and a test set. +
-
import os
+# Common imports
+import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
@@ -449,45 +443,71 @@ DATA_ID = "
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
-def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
-# Read the EoS data as csv file and organized into two arrays with density and energies
+# Read the EoS data as csv file and organize the data into two arrays with density and energies
EoS = pd.read_csv(infile, names=('Density', 'Energy'))
EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
EoS = EoS.dropna()
Energies = EoS['Energy']
Density = EoS['Density']
# The design matrix now as function of various polytrops
-X = np.zeros((len(Density),5))
-X[:,0] = 1
-X[:,1] = Density**(2.0/3.0)
-X[:,2] = Density
+X = np.zeros((len(Density),4))
X[:,3] = Density**(4.0/3.0)
-X[:,4] = Density**(5.0/3.0)
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
-# matrix inversion to find beta
-beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
-# and then make the prediction
-ytilde = X_train @ beta
-print("Training R2")
-print(R2(y_train,ytilde))
-print("Training MSE")
-print(MSE(y_train,ytilde))
-ypredict = X_test @ beta
-print("Test R2")
-print(R2(y_test,ypredict))
-print("Test MSE")
-print(MSE(y_test,ypredict))
+X[:,2] = Density
+X[:,1] = Density**(2.0/3.0)
+X[:,0] = 1
+
+# We use now Scikit-Learn's linear regressor and ridge regressor
+# OLS part
+clf = skl.LinearRegression().fit(X, Energies)
+ytilde = clf.predict(X)
+EoS['Eols'] = ytilde
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, ytilde))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
+print(clf.coef_, clf.intercept_)
+
+# The Ridge regression with a hyperparameter lambda = 0.1
+_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
+yridge = clf_ridge.predict(X)
+EoS['Eridge'] = yridge
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, yridge))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
+print(clf_ridge.coef_, clf_ridge.intercept_)
+
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
+ax.set_ylabel(r'Energy per particle')
+ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,
+ label='Theoretical data')
+ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',
+ label='OLS')
+ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',
+ label='Ridge $\lambda = 0.1$')
+ax.legend()
+save_fig("EoSfitting")
+plt.show()
+
+The above simple polynomial in density \( \rho \) gives an excellent fit
+to the data.
+
+
+We note also that there is a small deviation between the
+standard OLS and the Ridge regression at higher densities. We discuss this in more detail
+below.
+
@@ -514,7 +534,7 @@ ypredict = X_test @ beta
- + -
-The Boston housing -data set was originally a part of UCI Machine Learning Repository -and has been removed now. The data set is now included in Scikit-Learn's -library. There are 506 samples and 13 feature (predictor) variables -in this data set. The objective is to predict the value of prices of -the house using the features (predictors) listed here. +It is normal in essentially all Machine Learning studies to split the +data in a training set and a test set (sometimes also an additional +validation set). Scikit-Learn has an own function for this. There +is no explicit recipe for how much data should be included as training +data and say test data. An accepted rule of thumb is to use +approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will +postpone a discussion of this splitting to the end of these notes and +our discussion of the so-called bias-variance tradeoff. Here we +limit ourselves to repeat the above equation of state fitting example +but now splitting the data into a training set and a test set.
-The features/predictors are -
import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organized into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),5))
+X[:,0] = 1
+X[:,1] = Density**(2.0/3.0)
+X[:,2] = Density
+X[:,3] = Density**(4.0/3.0)
+X[:,4] = Density**(5.0/3.0)
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
+# and then make the prediction
+ytilde = X_train @ beta
+print("Training R2")
+print(R2(y_train,ytilde))
+print("Training MSE")
+print(MSE(y_train,ytilde))
+ypredict = X_test @ beta
+print("Test R2")
+print(R2(y_test,ypredict))
+print("Test MSE")
+print(MSE(y_test,ypredict))
+
diff --git a/doc/pub/Regression/html/._Regression-bs029.html b/doc/pub/Regression/html/._Regression-bs029.html index 4feec208c..e3aeef02f 100644 --- a/doc/pub/Regression/html/._Regression-bs029.html +++ b/doc/pub/Regression/html/._Regression-bs029.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
- + -
- - -
import numpy as np
-import matplotlib.pyplot as plt
-
-import pandas as pd
-import seaborn as sns
--and load the Boston Housing DataSet from Scikit-Learn +
- - -
from sklearn.datasets import load_boston
-
-boston_dataset = load_boston()
-
-# boston_dataset is a dictionary
-# let's check what it contains
-boston_dataset.keys()
--Then we invoke Pandas -
- - -
boston = pd.DataFrame(boston_dataset.data, columns=boston_dataset.feature_names)
-boston.head()
-boston['MEDV'] = boston_dataset.target
--and preprocess the data -
- - -
# check for missing values in all the columns
-boston.isnull().sum()
--We can then visualize the data -
- - -
# set the size of the figure
-sns.set(rc={'figure.figsize':(11.7,8.27)})
-
-# plot a histogram showing the distribution of the target values
-sns.distplot(boston['MEDV'], bins=30)
-plt.show()
--It is now useful to look at the correlation matrix -
- - -
# compute the pair wise correlation for all columns
-correlation_matrix = boston.corr().round(2)
-# use the heatmap function from seaborn to plot the correlation matrix
-# annot = True to print the values inside the square
-sns.heatmap(data=correlation_matrix, annot=True)
--From the above coorelation plot we can see that MEDV is strongly correlated to LSTAT and RM. We see also that RAD and TAX are stronly correlated, but we don't include this in our features together to avoid multi-colinearity +The Boston housing +data set was originally a part of UCI Machine Learning Repository +and has been removed now. The data set is now included in Scikit-Learn's +library. There are 506 samples and 13 feature (predictor) variables +in this data set. The objective is to predict the value of prices of +the house using the features (predictors) listed here.
+The features/predictors are - -
plt.figure(figsize=(20, 5))
+
+- CRIM: Per capita crime rate by town
+- ZN: Proportion of residential land zoned for lots over 25000 square feet
+- INDUS: Proportion of non-retail business acres per town
+- CHAS: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
+- NOX: Nitric oxide concentration (parts per 10 million)
+- RM: Average number of rooms per dwelling
+- AGE: Proportion of owner-occupied units built prior to 1940
+- DIS: Weighted distances to five Boston employment centers
+- RAD: Index of accessibility to radial highways
+- TAX: Full-value property tax rate per USD10000
+- B: \( 1000(Bk - 0.63)^2 \), where \( Bk \) is the proportion of [people of African American descent] by town
+- LSTAT: Percentage of lower status of the population
+- MEDV: Median value of owner-occupied homes in USD 1000s
+
-features = ['LSTAT', 'RM']
-target = boston['MEDV']
-
-for i, col in enumerate(features):
- plt.subplot(1, len(features) , i+1)
- x = boston[col]
- y = target
- plt.scatter(x, y, marker='o')
- plt.title(col)
- plt.xlabel(col)
- plt.ylabel('MEDV')
--Now we start training our model -
- - -
X = pd.DataFrame(np.c_[boston['LSTAT'], boston['RM']], columns = ['LSTAT','RM'])
-Y = boston['MEDV']
--We split the data into training and test sets - -
- - -
from sklearn.model_selection import train_test_split
-
-# splits the training and test data set in 80% : 20%
-# assign random_state to any value.This ensures consistency.
-X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state=5)
-print(X_train.shape)
-print(X_test.shape)
-print(Y_train.shape)
-print(Y_test.shape)
--Then we use the linear regression functionality from Scikit-Learn -
- - -
from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score
-
-lin_model = LinearRegression()
-lin_model.fit(X_train, Y_train)
-
-# model evaluation for training set
-
-y_train_predict = lin_model.predict(X_train)
-rmse = (np.sqrt(mean_squared_error(Y_train, y_train_predict)))
-r2 = r2_score(Y_train, y_train_predict)
-
-print("The model performance for training set")
-print("--------------------------------------")
-print('RMSE is {}'.format(rmse))
-print('R2 score is {}'.format(r2))
-print("\n")
-
-# model evaluation for testing set
-
-y_test_predict = lin_model.predict(X_test)
-# root mean square error of the model
-rmse = (np.sqrt(mean_squared_error(Y_test, y_test_predict)))
-
-# r-squared score of the model
-r2 = r2_score(Y_test, y_test_predict)
-
-print("The model performance for testing set")
-print("--------------------------------------")
-print('RMSE is {}'.format(rmse))
-print('R2 score is {}'.format(r2))
-- - -
# plotting the y_test vs y_pred
-# ideally should have been a straight line
-plt.scatter(Y_test, y_test_predict)
-plt.show()
-
diff --git a/doc/pub/Regression/html/._Regression-bs030.html b/doc/pub/Regression/html/._Regression-bs030.html index 06b1ac787..943548d0c 100644 --- a/doc/pub/Regression/html/._Regression-bs030.html +++ b/doc/pub/Regression/html/._Regression-bs030.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
+ + +
import numpy as np
+import matplotlib.pyplot as plt
+
+import pandas as pd
+import seaborn as sns
++and load the Boston Housing DataSet from Scikit-Learn
-
+ + +
from sklearn.datasets import load_boston
+
+boston_dataset = load_boston()
+
+# boston_dataset is a dictionary
+# let's check what it contains
+boston_dataset.keys()
++Then we invoke Pandas +
+ + +
boston = pd.DataFrame(boston_dataset.data, columns=boston_dataset.feature_names)
+boston.head()
+boston['MEDV'] = boston_dataset.target
++and preprocess the data +
+ + +
# check for missing values in all the columns
+boston.isnull().sum()
++We can then visualize the data +
+ + +
# set the size of the figure
+sns.set(rc={'figure.figsize':(11.7,8.27)})
+
+# plot a histogram showing the distribution of the target values
+sns.distplot(boston['MEDV'], bins=30)
+plt.show()
++It is now useful to look at the correlation matrix +
+ + +
# compute the pair wise correlation for all columns
+correlation_matrix = boston.corr().round(2)
+# use the heatmap function from seaborn to plot the correlation matrix
+# annot = True to print the values inside the square
+sns.heatmap(data=correlation_matrix, annot=True)
++From the above coorelation plot we can see that MEDV is strongly correlated to LSTAT and RM. We see also that RAD and TAX are stronly correlated, but we don't include this in our features together to avoid multi-colinearity
-The examples we have looked at so far are cases where we normally can -invert the matrix \( \boldsymbol{X}^T\boldsymbol{X} \). Using a polynomial expansion as we -did both for the masses and the fitting of the equation of state, -leads to row vectors of the design matrix which are essentially -orthogonal due to the polynomial character of our model. Obtaining the inverse of the design matrix is then often done via a so-called LU, QR or Cholesky decomposition. + + +
plt.figure(figsize=(20, 5))
+
+features = ['LSTAT', 'RM']
+target = boston['MEDV']
+
+for i, col in enumerate(features):
+ plt.subplot(1, len(features) , i+1)
+ x = boston[col]
+ y = target
+ plt.scatter(x, y, marker='o')
+ plt.title(col)
+ plt.xlabel(col)
+ plt.ylabel('MEDV')
++Now we start training our model +
+ + +
X = pd.DataFrame(np.c_[boston['LSTAT'], boston['RM']], columns = ['LSTAT','RM'])
+Y = boston['MEDV']
++We split the data into training and test sets
-This may -however not the be case in general and a standard matrix inversion -algorithm based on say LU, QR or Cholesky decomposition may lead to singularities. We will see examples of this below. + +
from sklearn.model_selection import train_test_split
+
+# splits the training and test data set in 80% : 20%
+# assign random_state to any value.This ensures consistency.
+X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state=5)
+print(X_train.shape)
+print(X_test.shape)
+print(Y_train.shape)
+print(Y_test.shape)
+-There is however a way to partially circumvent this problem and also gain some insight about the ordinary least squares approach. - +Then we use the linear regression functionality from Scikit-Learn
-This is given by the Singular Value Decomposition algorithm, perhaps -the most powerful linear algebra algorithm. Let us look at a -different example where we may have problems with the standard matrix -inversion algorithm. Thereafter we dive into the math of the SVD. + +
from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score
+
+lin_model = LinearRegression()
+lin_model.fit(X_train, Y_train)
+
+# model evaluation for training set
+
+y_train_predict = lin_model.predict(X_train)
+rmse = (np.sqrt(mean_squared_error(Y_train, y_train_predict)))
+r2 = r2_score(Y_train, y_train_predict)
+
+print("The model performance for training set")
+print("--------------------------------------")
+print('RMSE is {}'.format(rmse))
+print('R2 score is {}'.format(r2))
+print("\n")
+
+# model evaluation for testing set
+
+y_test_predict = lin_model.predict(X_test)
+# root mean square error of the model
+rmse = (np.sqrt(mean_squared_error(Y_test, y_test_predict)))
+
+# r-squared score of the model
+r2 = r2_score(Y_test, y_test_predict)
+
+print("The model performance for testing set")
+print("--------------------------------------")
+print('RMSE is {}'.format(rmse))
+print('R2 score is {}'.format(r2))
+-
# plotting the y_test vs y_pred
+# ideally should have been a straight line
+plt.scatter(Y_test, y_test_predict)
+plt.show()
+
@@ -465,7 +588,7 @@ inversion algorithm. Thereafter we dive into the math of the SVD.
-One of the typical problems we encounter with linear regression, in particular -when the matrix \( \boldsymbol{X} \) (our so-called design matrix) is high-dimensional, -are problems with near singular or singular matrices. The column vectors of \( \boldsymbol{X} \) -may be linearly dependent, normally referred to as super-collinearity. -This means that the matrix may be rank deficient and it is basically impossible to -to model the data using linear regression. As an example, consider the matrix -$$ -\begin{align*} -\mathbf{X} & = \left[ -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right] -\end{align*} -$$ +
-The columns of \( \boldsymbol{X} \) are linearly dependent. We see this easily since the -the first column is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of the space spanned by the -column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number -of linearly independent columns. In this particular case the matrix has rank 2. +The examples we have looked at so far are cases where we normally can +invert the matrix \( \boldsymbol{X}^T\boldsymbol{X} \). Using a polynomial expansion as we +did both for the masses and the fitting of the equation of state, +leads to row vectors of the design matrix which are essentially +orthogonal due to the polynomial character of our model. Obtaining the inverse of the design matrix is then often done via a so-called LU, QR or Cholesky decomposition.
-Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies -that the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) (the matrix we need to invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this -$$ -\begin{align*} -\boldsymbol{X} & = \left[ -\begin{array}{rr} -1 & -1 -\\ -1 & -1 -\end{array} \right]. -\end{align*} -$$ +This may +however not the be case in general and a standard matrix inversion +algorithm based on say LU, QR or Cholesky decomposition may lead to singularities. We will see examples of this below. + +
+There is however a way to partially circumvent this problem and also gain some insight about the ordinary least squares approach. + +
+This is given by the Singular Value Decomposition algorithm, perhaps +the most powerful linear algebra algorithm. Let us look at a +different example where we may have problems with the standard matrix +inversion algorithm. Thereafter we dive into the math of the SVD. + +
+
@@ -480,7 +467,7 @@ This is equivalent to saying that the matrix \( \boldsymbol{X} \) has at least a
-If our design matrix \( \boldsymbol{X} \) which enters the linear regression problem +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \boldsymbol{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \boldsymbol{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix $$ -\begin{align} -\boldsymbol{\beta} & = (\boldsymbol{X}^{T} \boldsymbol{X})^{-1} \boldsymbol{X}^{T} \boldsymbol{y}, -\tag{1} -\end{align} +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} $$ -has linearly dependent column vectors, we will not be able to compute the inverse -of \( \boldsymbol{X}^T\boldsymbol{X} \) and we cannot find the parameters (estimators) \( \beta_i \). -The estimators are only well-defined if \( (\boldsymbol{X}^{T}\boldsymbol{X})^{-1} \) exits. -This is more likely to happen when the matrix \( \boldsymbol{X} \) is high-dimensional. In this case it is likely to encounter a situation where -the regression parameters \( \beta_i \) cannot be estimated. -
-A cheap ad hoc approach is simply to add a small diagonal component to the matrix to invert, that is we change +The columns of \( \boldsymbol{X} \) are linearly dependent. We see this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +
+Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) (the matrix we need to invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this $$ -\boldsymbol{X}^{T} \boldsymbol{X} \rightarrow \boldsymbol{X}^{T} \boldsymbol{X}+\lambda \boldsymbol{I}, +\begin{align*} +\boldsymbol{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} $$ -where \( \boldsymbol{I} \) is the identity matrix. When we discuss Ridge regression this is actually what we end up evaluating. The parameter \( \lambda \) is called a hyperparameter. More about this later. +We see easily that \( \mbox{det}(\boldsymbol{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \boldsymbol{X} \) has at least an eigenvalue which is zero.
@@ -457,7 +482,7 @@ where \( \boldsymbol{I} \) is the identity matrix. When we discuss Ridge
-From standard linear algebra we know that a square matrix \( \boldsymbol{X} \) can be diagonalized if and only it is -a so-called normal matrix, that is if \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \) -we have \( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) or if \( \boldsymbol{X}\in {\mathbb{C}}^{n\times n} \) we have \( \boldsymbol{X}\boldsymbol{X}^{\dagger}=\boldsymbol{X}^{\dagger}\boldsymbol{X} \). -The matrix has then a set of eigenpairs - +If our design matrix \( \boldsymbol{X} \) which enters the linear regression problem $$ -(\lambda_1,\boldsymbol{u}_1),\dots, (\lambda_n,\boldsymbol{u}_n), +\begin{align} +\boldsymbol{\beta} & = (\boldsymbol{X}^{T} \boldsymbol{X})^{-1} \boldsymbol{X}^{T} \boldsymbol{y}, +\tag{1} +\end{align} $$ -and the eigenvalues are given by the diagonal matrix -$$ -\boldsymbol{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n). -$$ - -The matrix \( \boldsymbol{X} \) can be written in terms of an orthogonal/unitary transformation \( \boldsymbol{U} \) -$$ -\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, -$$ - -with \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) or \( \boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I} \). +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \boldsymbol{X}^T\boldsymbol{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\boldsymbol{X}^{T}\boldsymbol{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \boldsymbol{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated.
-Not all square matrices are diagonalizable. A matrix like the one discussed above +A cheap ad hoc approach is simply to add a small diagonal component to the matrix to invert, that is we change $$ -\boldsymbol{X} = \begin{bmatrix} -1& -1 \\ -1& -1\\ -\end{bmatrix} +\boldsymbol{X}^{T} \boldsymbol{X} \rightarrow \boldsymbol{X}^{T} \boldsymbol{X}+\lambda \boldsymbol{I}, $$ -is not diagonalizable, it is a so-called defective matrix. It is easy to see that the condition -\( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) is not fulfilled. +where \( \boldsymbol{I} \) is the identity matrix. When we discuss Ridge regression this is actually what we end up evaluating. The parameter \( \lambda \) is called a hyperparameter. More about this later.
@@ -468,7 +459,7 @@ is not diagonalizable, it is a so-called 42
-However, and this is the strength of the SVD algorithm, any general -matrix \( \boldsymbol{X} \) can be decomposed in terms of a diagonal matrix and -two orthogonal/unitary matrices. The Singular Value Decompostion -(SVD) theorem -states that a general \( m\times n \) matrix \( \boldsymbol{X} \) can be written in -terms of a diagonal matrix \( \boldsymbol{\Sigma} \) of dimensionality \( n\times n \) -and two orthognal matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \), where the first has -dimensionality \( m \times m \) and the last dimensionality \( n\times n \). -We have then +From standard linear algebra we know that a square matrix \( \boldsymbol{X} \) can be diagonalized if and only it is +a so-called normal matrix, that is if \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \) +we have \( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) or if \( \boldsymbol{X}\in {\mathbb{C}}^{n\times n} \) we have \( \boldsymbol{X}\boldsymbol{X}^{\dagger}=\boldsymbol{X}^{\dagger}\boldsymbol{X} \). +The matrix has then a set of eigenpairs -$$ -\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T $$ +(\lambda_1,\boldsymbol{u}_1),\dots, (\lambda_n,\boldsymbol{u}_n), +$$ + +and the eigenvalues are given by the diagonal matrix +$$ +\boldsymbol{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n). +$$ + +The matrix \( \boldsymbol{X} \) can be written in terms of an orthogonal/unitary transformation \( \boldsymbol{U} \) +$$ +\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, +$$ + +with \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) or \( \boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I} \).
-As an example, the above defective matrix can be decomposed as - +Not all square matrices are diagonalizable. A matrix like the one discussed above $$ -\boldsymbol{X} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1& 1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 0\\ \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -1 \\ 1& 1\\ \end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, +\boldsymbol{X} = \begin{bmatrix} +1& -1 \\ +1& -1\\ +\end{bmatrix} $$ -
-with eigenvalues \( \sigma_1=2 \) and \( \sigma_2=0 \). -The SVD exits always! +is not diagonalizable, it is a so-called defective matrix. It is easy to see that the condition +\( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) is not fulfilled.
@@ -460,7 +470,7 @@ The SVD exits always!
-Consider the following matrix which can be SVD decomposed as +However, and this is the strength of the SVD algorithm, any general +matrix \( \boldsymbol{X} \) can be decomposed in terms of a diagonal matrix and +two orthogonal/unitary matrices. The Singular Value Decompostion +(SVD) theorem +states that a general \( m\times n \) matrix \( \boldsymbol{X} \) can be written in +terms of a diagonal matrix \( \boldsymbol{\Sigma} \) of dimensionality \( n\times n \) +and two orthognal matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \), where the first has +dimensionality \( m \times m \) and the last dimensionality \( n\times n \). +We have then -$$ -\boldsymbol{X} = \frac{1}{15}\begin{bmatrix} 14 & 2\\ 4 & 22\\ 16 & 13\end{bmatrix}=\frac{1}{3}\begin{bmatrix} 1& 2 & 2 \\ 2& -1 & 1\\ 2 & 1& -2\end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 1\\ 0 & 0\end{bmatrix}\frac{1}{5}\begin{bmatrix} 3& 4 \\ 4& -3\end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. +$$ +\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T $$
-This is a \( 3\times 2 \) matrix which is decomposed in terms of a -\( 3\times 3 \) matrix \( \boldsymbol{U} \), and a \( 2\times 2 \) matrix \( \boldsymbol{V} \). It is easy to see -that \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal (how?). +As an example, the above defective matrix can be decomposed as + +$$ +\boldsymbol{X} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1& 1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 0\\ \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -1 \\ 1& 1\\ \end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, +$$
-And the SVD -decomposition (singular values) gives eigenvalues -\( \sigma_i\geq\sigma_{i+1} \) for all \( i \) and for dimensions larger than \( i=2 \), the -eigenvalues (singular values) are zero. - -
-In the general case, where our design matrix \( \boldsymbol{X} \) has dimension -\( n\times p \), the matrix is thus decomposed into an \( n\times n \) -orthogonal matrix \( \boldsymbol{U} \), a \( p\times p \) orthogonal matrix \( \boldsymbol{V} \) -and a diagonal matrix \( \boldsymbol{\Sigma} \) with \( r=\mathrm{min}(n,p) \) -singular values \( \sigma_i\geq 0 \) on the main diagonal and zeros filling -the rest of the matrix. There are at most \( p \) singular values -assuming that \( n > p \). In our regression examples for the nuclear -masses and the equation of state this is indeed the case, while for -the Ising model we have \( p > n \). These are often cases that lead to -near singular or singular matrices. - -
-The columns of \( \boldsymbol{U} \) are called the left singular vectors while the columns of \( \boldsymbol{V} \) are the right singular vectors. +with eigenvalues \( \sigma_1=2 \) and \( \sigma_2=0 \). +The SVD exits always!
@@ -467,7 +462,7 @@ The columns of \( \boldsymbol{U} \) are called the left singular vectors while t
-If we assume that \( n > p \), then our matrix \( \boldsymbol{U} \) has dimension \( n -\times n \). The last \( n-p \) columns of \( \boldsymbol{U} \) become however -irrelevant in our calculations since they are multiplied with the -zeros in \( \boldsymbol{\Sigma} \). +Consider the following matrix which can be SVD decomposed as + +$$ +\boldsymbol{X} = \frac{1}{15}\begin{bmatrix} 14 & 2\\ 4 & 22\\ 16 & 13\end{bmatrix}=\frac{1}{3}\begin{bmatrix} 1& 2 & 2 \\ 2& -1 & 1\\ 2 & 1& -2\end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 1\\ 0 & 0\end{bmatrix}\frac{1}{5}\begin{bmatrix} 3& 4 \\ 4& -3\end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. +$$
-The economy-size decomposition removes extra rows or columns of zeros -from the diagonal matrix of singular values, \( \boldsymbol{\Sigma} \), along with the columns -in either \( \boldsymbol{U} \) or \( \boldsymbol{V} \) that multiply those zeros in the expression. -Removing these zeros and columns can improve execution time -and reduce storage requirements without compromising the accuracy of -the decomposition. +This is a \( 3\times 2 \) matrix which is decomposed in terms of a +\( 3\times 3 \) matrix \( \boldsymbol{U} \), and a \( 2\times 2 \) matrix \( \boldsymbol{V} \). It is easy to see +that \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal (how?).
-If \( n > p \), we keep only the first \( p \) columns of \( \boldsymbol{U} \) and \( \boldsymbol{\Sigma} \) has dimension \( p\times p \). -If \( p > n \), then only the first \( n \) columns of \( \boldsymbol{V} \) are computed and \( \boldsymbol{\Sigma} \) has dimension \( n\times n \). -The \( n=p \) case is obvious, we retain the full SVD. -In general the economy-size SVD leads to less FLOPS and still conserving the desired accuracy. +And the SVD +decomposition (singular values) gives eigenvalues +\( \sigma_i\geq\sigma_{i+1} \) for all \( i \) and for dimensions larger than \( i=2 \), the +eigenvalues (singular values) are zero. + +
+In the general case, where our design matrix \( \boldsymbol{X} \) has dimension +\( n\times p \), the matrix is thus decomposed into an \( n\times n \) +orthogonal matrix \( \boldsymbol{U} \), a \( p\times p \) orthogonal matrix \( \boldsymbol{V} \) +and a diagonal matrix \( \boldsymbol{\Sigma} \) with \( r=\mathrm{min}(n,p) \) +singular values \( \sigma_i\geq 0 \) on the main diagonal and zeros filling +the rest of the matrix. There are at most \( p \) singular values +assuming that \( n > p \). In our regression examples for the nuclear +masses and the equation of state this is indeed the case, while for +the Ising model we have \( p > n \). These are often cases that lead to +near singular or singular matrices. + +
+The columns of \( \boldsymbol{U} \) are called the left singular vectors while the columns of \( \boldsymbol{V} \) are the right singular vectors.
@@ -454,7 +469,7 @@ In general the economy-size SVD leads to less FLOPS and still conserving the des
-There are several interesting mathematical properties which will be -relevant when we are going to discuss the differences between say -ordinary least squares (OLS) and Ridge regression. +If we assume that \( n > p \), then our matrix \( \boldsymbol{U} \) has dimension \( n +\times n \). The last \( n-p \) columns of \( \boldsymbol{U} \) become however +irrelevant in our calculations since they are multiplied with the +zeros in \( \boldsymbol{\Sigma} \).
-We have from OLS that the parameters of the linear approximation are given by -$$ -\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. -$$ +The economy-size decomposition removes extra rows or columns of zeros +from the diagonal matrix of singular values, \( \boldsymbol{\Sigma} \), along with the columns +in either \( \boldsymbol{U} \) or \( \boldsymbol{V} \) that multiply those zeros in the expression. +Removing these zeros and columns can improve execution time +and reduce storage requirements without compromising the accuracy of +the decomposition.
-The matrix to invert can be rewritten in terms of our SVD decomposition as - -$$ -\boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. -$$ - -Using the orthogonality properties of \( \boldsymbol{U} \) we have - -$$ -\boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T = \boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T, -$$ - -with \( \boldsymbol{D} \) being a diagonal matrix with values along the diagonal given by the singular values squared. - -
-This means that -$$ -(\boldsymbol{X}^T\boldsymbol{X})\boldsymbol{V} = \boldsymbol{V}\boldsymbol{D}, -$$ - -that is the eigenvectors of \( (\boldsymbol{X}^T\boldsymbol{X}) \) are given by the columns of the right singular matrix of \( \boldsymbol{X} \) and the eigenvalues are the squared singular values. It is easy to show (show this) that -$$ -(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}, -$$ - -that is, the eigenvectors of \( (\boldsymbol{X}\boldsymbol{X})^T \) are the columns of the left singular matrix and the eigenvalues are the same. - -
-Going back to our OLS equation we have -$$ -\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}. -$$ - -We will come back to this expression when we discuss Ridge regression. +If \( n > p \), we keep only the first \( p \) columns of \( \boldsymbol{U} \) and \( \boldsymbol{\Sigma} \) has dimension \( p\times p \). +If \( p > n \), then only the first \( n \) columns of \( \boldsymbol{V} \) are computed and \( \boldsymbol{\Sigma} \) has dimension \( n\times n \). +The \( n=p \) case is obvious, we retain the full SVD. +In general the economy-size SVD leads to less FLOPS and still conserving the desired accuracy.
@@ -481,7 +456,7 @@ We will come back to this expression when we discuss Ridge regression.
-Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is -our optimization problem is -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. -$$ +There are several interesting mathematical properties which will be +relevant when we are going to discuss the differences between say +ordinary least squares (OLS) and Ridge regression. -or we can state it as +
+We have from OLS that the parameters of the linear approximation are given by $$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2, -$$ - -where we have used the definition of a norm-2 vector, that is -$$ -\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. +\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$
-By minimizing the above equation with respect to the parameters -\( \boldsymbol{\beta} \) we could then obtain an analytical expression for the -parameters \( \boldsymbol{\beta} \). We can add a regularization parameter \( \lambda \) by -defining a new cost function to be optimized, that is +The matrix to invert can be rewritten in terms of our SVD decomposition as $$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2 +\boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. $$ +Using the orthogonality properties of \( \boldsymbol{U} \) we have + +$$ +\boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T = \boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T, +$$ + +with \( \boldsymbol{D} \) being a diagonal matrix with values along the diagonal given by the singular values squared. +
-which leads to the Ridge regression minimization problem where we -require that \( \vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t \), where \( t \) is -a finite number larger than zero. By defining +This means that +$$ +(\boldsymbol{X}^T\boldsymbol{X})\boldsymbol{V} = \boldsymbol{V}\boldsymbol{D}, +$$ +that is the eigenvectors of \( (\boldsymbol{X}^T\boldsymbol{X}) \) are given by the columns of the right singular matrix of \( \boldsymbol{X} \) and the eigenvalues are the squared singular values. It is easy to show (show this) that $$ -C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1, +(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}, $$ +that is, the eigenvectors of \( (\boldsymbol{X}\boldsymbol{X})^T \) are the columns of the left singular matrix and the eigenvalues are the same. +
-we have a new optimization equation +Going back to our OLS equation we have $$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1 +\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}. $$ -which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. - -
-Here we have defined the norm-1 as -$$ -\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. -$$ +We will come back to this expression when we discuss Ridge regression.
@@ -487,7 +483,7 @@ $$
-Using the matrix-vector expression for Ridge regression, - +Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is +our optimization problem is $$ -C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta}, +{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +$$ + +or we can state it as +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2, +$$ + +where we have used the definition of a norm-2 vector, that is +$$ +\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. $$
-by taking the derivatives with respect to \( \boldsymbol{\beta} \) we obtain then -a slightly modified matrix inversion problem which for finite values -of \( \lambda \) does not suffer from singularity problems. We obtain +By minimizing the above equation with respect to the parameters +\( \boldsymbol{\beta} \) we could then obtain an analytical expression for the +parameters \( \boldsymbol{\beta} \). We can add a regularization parameter \( \lambda \) by +defining a new cost function to be optimized, that is $$ -\boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}, +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2 $$
-with \( \boldsymbol{I} \) being a \( p\times p \) identity matrix with the constraint that +which leads to the Ridge regression minimization problem where we +require that \( \vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t \), where \( t \) is +a finite number larger than zero. By defining $$ -\sum_{i=0}^{p-1} \beta_i^2 \leq t, +C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1, $$
-with \( t \) a finite positive number. +we have a new optimization equation +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1 +$$ + +which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator.
-We see that Ridge regression is nothing but the standard -OLS with a modified diagonal term added to \( \boldsymbol{X}^T\boldsymbol{X} \). The -consequences, in particular for our discussion of the bias-variance tradeoff -are rather interesting. - -
-Furthermore, if we use the result above in terms of the SVD decomposition (our analysis was done for the OLS method), we had +Here we have defined the norm-1 as $$ -(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}. +\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. $$ -
-We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as -$$ -\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y} -$$ - -
-For Ridge regression this becomes - -$$ -\boldsymbol{X}\boldsymbol{\beta}^{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y}, -$$ - -
-with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \). -
@@ -488,7 +489,7 @@ with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \
-Since \( \lambda \geq 0 \), it means that compared to OLS, we have +Using the matrix-vector expression for Ridge regression, $$ -\frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1. +C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta}, $$
-Ridge regression finds the coordinates of \( \boldsymbol{y} \) with respect to the -orthonormal basis \( \boldsymbol{U} \), it then shrinks the coordinates by -\( \frac{\sigma_j^2}{\sigma_j^2+\lambda} \). Recall that the SVD has -eigenvalues ordered in a descending way, that is \( \sigma_i \geq -\sigma_{i+1} \). +by taking the derivatives with respect to \( \boldsymbol{\beta} \) we obtain then +a slightly modified matrix inversion problem which for finite values +of \( \lambda \) does not suffer from singularity problems. We obtain + +$$ +\boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}, +$$
-For small eigenvalues \( \sigma_i \) it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. -Actually, calculating the variance of \( \boldsymbol{X}\boldsymbol{v}_j \) shows that this quantity is equal to \( \sigma_j^2/n \). -With a parameter \( \lambda \) we can thus shrink the role of specific parameters. +with \( \boldsymbol{I} \) being a \( p\times p \) identity matrix with the constraint that + +$$ +\sum_{i=0}^{p-1} \beta_i^2 \leq t, +$$ + +
+with \( t \) a finite positive number. + +
+We see that Ridge regression is nothing but the standard +OLS with a modified diagonal term added to \( \boldsymbol{X}^T\boldsymbol{X} \). The +consequences, in particular for our discussion of the bias-variance tradeoff +are rather interesting. + +
+Furthermore, if we use the result above in terms of the SVD decomposition (our analysis was done for the OLS method), we had +$$ +(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}. +$$ + +
+We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as +$$ +\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y} +$$ + +
+For Ridge regression this becomes + +$$ +\boldsymbol{X}\boldsymbol{\beta}^{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y}, +$$ + +
+with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \).
@@ -453,7 +490,7 @@ With a parameter \( \lambda \) we can thus shrink the role of specific parameter
-For the sake of simplicity, let us assume that the design matrix is orthonormal, that is +Since \( \lambda \geq 0 \), it means that compared to OLS, we have $$ -\boldsymbol{X}^T\boldsymbol{X}=(\boldsymbol{X}^T\boldsymbol{X})^{-1} =\boldsymbol{I}. +\frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1. $$
-In this case the standard OLS results in -$$ -\boldsymbol{\beta}^{\mathrm{OLS}} = \boldsymbol{X}^T\boldsymbol{y}=\sum_{i=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\boldsymbol{y}, -$$ +Ridge regression finds the coordinates of \( \boldsymbol{y} \) with respect to the +orthonormal basis \( \boldsymbol{U} \), it then shrinks the coordinates by +\( \frac{\sigma_j^2}{\sigma_j^2+\lambda} \). Recall that the SVD has +eigenvalues ordered in a descending way, that is \( \sigma_i \geq +\sigma_{i+1} \).
-and - -$$ -\boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{I}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\left(1+\lambda\right)^{-1}\boldsymbol{\beta}^{\mathrm{OLS}}, -$$ - -
-that is the Ridge estimator scales the OLS estimator by the inverse of a factor \( 1+\lambda \), and -the Ridge estimator converges to zero when the hyperparameter goes to -infinity. - -
-We will come back to more interpreations after we have gone through some of the statistical analysis part. - -
-For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. -Similarly, Mehta et al's article is also recommended. +For small eigenvalues \( \sigma_i \) it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. +Actually, calculating the variance of \( \boldsymbol{X}\boldsymbol{v}_j \) shows that this quantity is equal to \( \sigma_j^2/n \). +With a parameter \( \lambda \) we can thus shrink the role of specific parameters.
@@ -466,7 +455,7 @@ Similarly, Mehta et al
+For the sake of simplicity, let us assume that the design matrix is orthonormal, that is - -
import numpy as np
-# SVD inversion
-def SVDinv(A):
- ''' Takes as input a numpy matrix A and returns inv(A) based on singular value decomposition (SVD).
- SVD is numerically more stable than the inversion algorithms provided by
- numpy and scipy.linalg at the cost of being slower.
- '''
- U, s, VT = np.linalg.svd(A)
-# print('test U')
-# print( (np.transpose(U) @ U - U @np.transpose(U)))
-# print('test VT')
-# print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))
- print(U)
- print(s)
- print(VT)
+$$
+\boldsymbol{X}^T\boldsymbol{X}=(\boldsymbol{X}^T\boldsymbol{X})^{-1} =\boldsymbol{I}.
+$$
- D = np.zeros((len(U),len(VT)))
- for i in range(0,len(VT)):
- D[i,i]=s[i]
- UT = np.transpose(U); V = np.transpose(VT); invD = np.linalg.inv(D)
- return np.matmul(V,np.matmul(invD,UT))
-
-
-X = np.array([ [1.0, -1.0, 2.0], [1.0, 0.0, 1.0], [1.0, 2.0, -1.0], [1.0, 1.0, 0.0] ])
-print(X)
-A = np.transpose(X) @ X
-print(A)
-# Brute force inversion of super-collinear matrix
-#B = np.linalg.inv(A)
-#print(B)
-C = SVDinv(A)
-print(C)
--The matrix \( \boldsymbol{X} \) has columns that are linearly dependent. The first -column is the row-wise sum of the other two columns. The rank of a -matrix (the column rank) is the dimension of space spanned by the -column vectors. The rank of the matrix is the number of linearly -independent columns, in this case just \( 2 \). We see this from the -singular values when running the above code. Running the standard -inversion algorithm for matrix inversion with \( \boldsymbol{X}^T\boldsymbol{X} \) results -in the program terminating due to a singular matrix. +In this case the standard OLS results in +$$ +\boldsymbol{\beta}^{\mathrm{OLS}} = \boldsymbol{X}^T\boldsymbol{y}=\sum_{i=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\boldsymbol{y}, +$$ + +
+and + +$$ +\boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{I}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\left(1+\lambda\right)^{-1}\boldsymbol{\beta}^{\mathrm{OLS}}, +$$ + +
+that is the Ridge estimator scales the OLS estimator by the inverse of a factor \( 1+\lambda \), and +the Ridge estimator converges to zero when the hyperparameter goes to +infinity. + +
+We will come back to more interpreations after we have gone through some of the statistical analysis part. + +
+For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. +Similarly, Mehta et al's article is also recommended.
@@ -480,7 +468,7 @@ in the program terminating due to a singular matrix.
- + -
-The parameter \( \lambda \) that we have introduced in the Ridge (and -Lasso as well) regression is often called a regularization parameter -or shrinkage parameter. It is common to call it a hyperparameter. What does it mean mathemtically? + +
import numpy as np
+# SVD inversion
+def SVDinv(A):
+ ''' Takes as input a numpy matrix A and returns inv(A) based on singular value decomposition (SVD).
+ SVD is numerically more stable than the inversion algorithms provided by
+ numpy and scipy.linalg at the cost of being slower.
+ '''
+ U, s, VT = np.linalg.svd(A)
+# print('test U')
+# print( (np.transpose(U) @ U - U @np.transpose(U)))
+# print('test VT')
+# print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))
+ print(U)
+ print(s)
+ print(VT)
+
+ D = np.zeros((len(U),len(VT)))
+ for i in range(0,len(VT)):
+ D[i,i]=s[i]
+ UT = np.transpose(U); V = np.transpose(VT); invD = np.linalg.inv(D)
+ return np.matmul(V,np.matmul(invD,UT))
+
+
+X = np.array([ [1.0, -1.0, 2.0], [1.0, 0.0, 1.0], [1.0, 2.0, -1.0], [1.0, 1.0, 0.0] ])
+print(X)
+A = np.transpose(X) @ X
+print(A)
+# Brute force inversion of super-collinear matrix
+#B = np.linalg.inv(A)
+#print(B)
+C = SVDinv(A)
+print(C)
+-Here we will first look at how to analyze the difference between the -standard OLS equations and the Ridge expressions in terms of a linear -algebra analysis using the SVD algorithm. Thereafter, we will link -(see the material on the bias-variance tradeoff below) these -observation to the statisical analysis of the results. In particular -we consider how the variance of the parameters \( \boldsymbol{\beta} \) is -affected by changing the parameter \( \lambda \). +The matrix \( \boldsymbol{X} \) has columns that are linearly dependent. The first +column is the row-wise sum of the other two columns. The rank of a +matrix (the column rank) is the dimension of space spanned by the +column vectors. The rank of the matrix is the number of linearly +independent columns, in this case just \( 2 \). We see this from the +singular values when running the above code. Running the standard +inversion algorithm for matrix inversion with \( \boldsymbol{X}^T\boldsymbol{X} \) results +in the program terminating due to a singular matrix.
@@ -448,7 +482,7 @@ affected by changing the parameter \( \lambda \).
- + -
-We have our design matrix - \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). With the SVD we decompose it as - -$$ -\boldsymbol{X} = \boldsymbol{U\Sigma V^T}, -$$ +The parameter \( \lambda \) that we have introduced in the Ridge (and +Lasso as well) regression is often called a regularization parameter +or shrinkage parameter. It is common to call it a hyperparameter. What does it mean mathemtically?
-with \( \boldsymbol{U}\in {\mathbb{R}}^{n\times n} \), \( \boldsymbol{\Sigma}\in {\mathbb{R}}^{n\times p} \) -and \( \boldsymbol{V}\in {\mathbb{R}}^{p\times p} \). - -
-The matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are unitary/orthonormal matrices, that is in case the matrices are real we have \( \boldsymbol{U}^T\boldsymbol{U}=\boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) and \( \boldsymbol{V}^T\boldsymbol{V}=\boldsymbol{V}\boldsymbol{V}^T=\boldsymbol{I} \). +Here we will first look at how to analyze the difference between the +standard OLS equations and the Ridge expressions in terms of a linear +algebra analysis using the SVD algorithm. Thereafter, we will link +(see the material on the bias-variance tradeoff below) these +observation to the statisical analysis of the results. In particular +we consider how the variance of the parameters \( \boldsymbol{\beta} \) is +affected by changing the parameter \( \lambda \).
@@ -449,7 +450,7 @@ The matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are unitary/orthonorm
-More material to be added here +We have our design matrix + \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). With the SVD we decompose it as + +$$ +\boldsymbol{X} = \boldsymbol{U\Sigma V^T}, +$$ + +
+with \( \boldsymbol{U}\in {\mathbb{R}}^{n\times n} \), \( \boldsymbol{\Sigma}\in {\mathbb{R}}^{n\times p} \) +and \( \boldsymbol{V}\in {\mathbb{R}}^{p\times p} \). + +
+The matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are unitary/orthonormal matrices, that is in case the matrices are real we have \( \boldsymbol{U}^T\boldsymbol{U}=\boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) and \( \boldsymbol{V}^T\boldsymbol{V}=\boldsymbol{V}\boldsymbol{V}^T=\boldsymbol{I} \).
@@ -437,7 +451,7 @@ More material to be added here
-Before we proceed, we need to rethink what we have been doing. In our -eager to fit the data, we have omitted several important elements in -our regression analysis. In what follows we will - -
@@ -446,7 +439,7 @@ This will allow us to link the standard linear algebra methods we have discussed
-Resampling methods are an indispensable tool in modern -statistics. They involve repeatedly drawing samples from a training -set and refitting a model of interest on each sample in order to -obtain additional information about the fitted model. For example, in -order to estimate the variability of a linear regression fit, we can -repeatedly draw different samples from the training data, fit a linear -regression to each new sample, and then examine the extent to which -the resulting fits differ. Such an approach may allow us to obtain -information that would not be available from fitting the model only -once using the original training sample. +
-Two resampling methods are often used in Machine Learning analyses, +Before we proceed, we need to rethink what we have been doing. In our +eager to fit the data, we have omitted several important elements in +our regression analysis. In what follows we will
-
@@ -463,7 +448,7 @@ cross-validation and the bootstrap method.
+Resampling methods are an indispensable tool in modern +statistics. They involve repeatedly drawing samples from a training +set and refitting a model of interest on each sample in order to +obtain additional information about the fitted model. For example, in +order to estimate the variability of a linear regression fit, we can +repeatedly draw different samples from the training data, fit a linear +regression to each new sample, and then examine the extent to which +the resulting fits differ. Such an approach may allow us to obtain +information that would not be available from fitting the model only +once using the original training sample.
-Resampling approaches can be computationally expensive, because they -involve fitting the same statistical method multiple times using -different subsets of the training data. However, due to recent -advances in computing power, the computational requirements of -resampling methods generally are not prohibitive. In this chapter, we -discuss two of the most commonly used resampling methods, -cross-validation and the bootstrap. Both methods are important tools -in the practical application of many statistical learning -procedures. For example, cross-validation can be used to estimate the -test error associated with a given statistical learning method in -order to evaluate its performance, or to select the appropriate level -of flexibility. The process of evaluating a model’s performance is -known as model assessment, whereas the process of selecting the proper -level of flexibility for a model is known as model selection. The -bootstrap is widely used. +Two resampling methods are often used in Machine Learning analyses, + +
-
+Resampling approaches can be computationally expensive, because they +involve fitting the same statistical method multiple times using +different subsets of the training data. However, due to recent +advances in computing power, the computational requirements of +resampling methods generally are not prohibitive. In this chapter, we +discuss two of the most commonly used resampling methods, +cross-validation and the bootstrap. Both methods are important tools +in the practical application of many statistical learning +procedures. For example, cross-validation can be used to estimate the +test error associated with a given statistical learning method in +order to evaluate its performance, or to select the appropriate level +of flexibility. The process of evaluating a model’s performance is +known as model assessment, whereas the process of selecting the proper +level of flexibility for a model is known as model selection. The +bootstrap is widely used. + +
-The probability distribution function (PDF) is a function -\( p(x) \) on the domain which, in the discrete case, gives us the -probability or relative frequency with which these values of \( X \) occur: -$$ -p(x) = \mathrm{prob}(X=x) -$$ -In the continuous case, the PDF does not directly depict the -actual probability. Instead we define the probability for the -stochastic variable to assume any value on an infinitesimal interval -around \( x \) to be \( p(x)dx \). The continuous function \( p(x) \) then gives us -the density of the probability rather than the probability -itself. The probability for a stochastic variable to assume any value -on a non-infinitesimal interval \( [a,\,b] \) is then just the integral: -$$ -\mathrm{prob}(a\leq X\leq b) = \int_a^b p(x)dx -$$ +
-A particularly useful class of special expectation values are the -moments. The \( n \)-th moment of the PDF \( p \) is defined as -follows: +The probability distribution function (PDF) is a function +\( p(x) \) on the domain which, in the discrete case, gives us the +probability or relative frequency with which these values of \( X \) occur: $$ -\langle x^n\rangle \equiv \int\! x^n p(x)\,dx +p(x) = \mathrm{prob}(X=x) $$ -The zero-th moment \( \langle 1\rangle \) is just the normalization condition of -\( p \). The first moment, \( \langle x\rangle \), is called the mean of \( p \) -and often denoted by the letter \( \mu \): +In the continuous case, the PDF does not directly depict the +actual probability. Instead we define the probability for the +stochastic variable to assume any value on an infinitesimal interval +around \( x \) to be \( p(x)dx \). The continuous function \( p(x) \) then gives us +the density of the probability rather than the probability +itself. The probability for a stochastic variable to assume any value +on a non-infinitesimal interval \( [a,\,b] \) is then just the integral: $$ -\langle x\rangle = \mu \equiv \int\! x p(x)\,dx +\mathrm{prob}(a\leq X\leq b) = \int_a^b p(x)dx $$ + +Qualitatively speaking, a stochastic variable represents the values of +numbers chosen as if by chance from some specified PDF so that the +selection of a large set of these numbers reproduces this PDF.
-A special version of the moments is the set of central moments, -the n-th central moment defined as: +A particularly useful class of special expectation values are the +moments. The \( n \)-th moment of the PDF \( p \) is defined as +follows: $$ -\langle (x-\langle x \rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx +\langle x^n\rangle \equiv \int\! x^n p(x)\,dx $$ -The zero-th and first central moments are both trivial, equal \( 1 \) and -\( 0 \), respectively. But the second central moment, known as the -variance of \( p \), is of particular interest. For the stochastic -variable \( X \), the variance is denoted as \( \sigma^2_X \) or \( \mathrm{var}(X) \): +The zero-th moment \( \langle 1\rangle \) is just the normalization condition of +\( p \). The first moment, \( \langle x\rangle \), is called the mean of \( p \) +and often denoted by the letter \( \mu \): $$ -\begin{align} -\sigma^2_X\ \ =\ \ \mathrm{var}(X) & = \langle (x-\langle x\rangle)^2\rangle = -\int\! (x-\langle x\rangle)^2 p(x)\,dx -\tag{2}\\ -& = \int\! \left(x^2 - 2 x \langle x\rangle^{2} + - \langle x\rangle^2\right)p(x)\,dx -\tag{3}\\ -& = \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2 -\tag{4}\\ -& = \langle x^2\rangle - \langle x\rangle^2 -\tag{5} -\end{align} +\langle x\rangle = \mu \equiv \int\! x p(x)\,dx $$ - -The square root of the variance, \( \sigma =\sqrt{\langle (x-\langle x\rangle)^2\rangle} \) is called the standard deviation of \( p \). It is clearly just the RMS (root-mean-square) -value of the deviation of the PDF from its mean value, interpreted -qualitatively as the spread of \( p \) around its mean.
-Another important quantity is the so called covariance, a variant of -the above defined variance. Consider again the set \( \{X_i\} \) of \( n \) -stochastic variables (not necessarily uncorrelated) with the -multivariate PDF \( P(x_1,\dots,x_n) \). The covariance of two -of the stochastic variables, \( X_i \) and \( X_j \), is defined as follows: +A special version of the moments is the set of central moments, +the n-th central moment defined as: +$$ +\langle (x-\langle x \rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx +$$ + +The zero-th and first central moments are both trivial, equal \( 1 \) and +\( 0 \), respectively. But the second central moment, known as the +variance of \( p \), is of particular interest. For the stochastic +variable \( X \), the variance is denoted as \( \sigma^2_X \) or \( \mathrm{var}(X) \): $$ \begin{align} -\mathrm{cov}(X_i,\,X_j) &\equiv \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle -\nonumber\\ -&= -\int\!\cdots\!\int\!(x_i-\langle x_i \rangle)(x_j-\langle x_j \rangle)\, -P(x_1,\dots,x_n)\,dx_1\dots dx_n -\tag{6} +\sigma^2_X\ \ =\ \ \mathrm{var}(X) & = \langle (x-\langle x\rangle)^2\rangle = +\int\! (x-\langle x\rangle)^2 p(x)\,dx +\tag{2}\\ +& = \int\! \left(x^2 - 2 x \langle x\rangle^{2} + + \langle x\rangle^2\right)p(x)\,dx +\tag{3}\\ +& = \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2 +\tag{4}\\ +& = \langle x^2\rangle - \langle x\rangle^2 +\tag{5} \end{align} $$ -with -$$ -\langle x_i\rangle = -\int\!\cdots\!\int\!x_i\,P(x_1,\dots,x_n)\,dx_1\dots dx_n -$$ +The square root of the variance, \( \sigma =\sqrt{\langle (x-\langle x\rangle)^2\rangle} \) is called the standard deviation of \( p \). It is clearly just the RMS (root-mean-square) +value of the deviation of the PDF from its mean value, interpreted +qualitatively as the spread of \( p \) around its mean.
-If we consider the above covariance as a matrix \( C_{ij}=\mathrm{cov}(X_i,\,X_j) \), then the diagonal elements are just the familiar -variances, \( C_{ii} = \mathrm{cov}(X_i,\,X_i) = \mathrm{var}(X_i) \). It turns out that -all the off-diagonal elements are zero if the stochastic variables are -uncorrelated. This is easy to show, keeping in mind the linearity of -the expectation value. Consider the stochastic variables \( X_i \) and -\( X_j \), (\( i\neq j \)): +Another important quantity is the so called covariance, a variant of +the above defined variance. Consider again the set \( \{X_i\} \) of \( n \) +stochastic variables (not necessarily uncorrelated) with the +multivariate PDF \( P(x_1,\dots,x_n) \). The covariance of two +of the stochastic variables, \( X_i \) and \( X_j \), is defined as follows: $$ \begin{align} -\mathrm{cov}(X_i,\,X_j) &= \langle(x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle -\tag{7}\\ -&=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle -\tag{8}\\ -&=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j\rangle + -\langle \langle x_i\rangle\langle x_j\rangle\rangle -\tag{9}\\ -&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + -\langle x_i\rangle\langle x_j\rangle -\tag{10}\\ -&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle -\tag{11} +\mathrm{cov}(X_i,\,X_j) &\equiv \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle +\nonumber\\ +&= +\int\!\cdots\!\int\!(x_i-\langle x_i \rangle)(x_j-\langle x_j \rangle)\, +P(x_1,\dots,x_n)\,dx_1\dots dx_n +\tag{6} \end{align} $$ + +with +$$ +\langle x_i\rangle = +\int\!\cdots\!\int\!x_i\,P(x_1,\dots,x_n)\,dx_1\dots dx_n +$$
-Suppose we have defined three vectors \( \hat{x}, \hat{y}, \hat{z} \) with -\( n \) elements each. The covariance matrix is defined as - +
+If we consider the above covariance as a matrix \( C_{ij}=\mathrm{cov}(X_i,\,X_j) \), then the diagonal elements are just the familiar +variances, \( C_{ii} = \mathrm{cov}(X_i,\,X_i) = \mathrm{var}(X_i) \). It turns out that +all the off-diagonal elements are zero if the stochastic variables are +uncorrelated. This is easy to show, keeping in mind the linearity of +the expectation value. Consider the stochastic variables \( X_i \) and +\( X_j \), (\( i\neq j \)): $$ -\hat{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ - \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ - \sigma_{zx} & \sigma_{zy} & \sigma_{zz} - \end{bmatrix}, +\begin{align} +\mathrm{cov}(X_i,\,X_j) &= \langle(x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle +\tag{7}\\ +&=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle +\tag{8}\\ +&=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j\rangle + +\langle \langle x_i\rangle\langle x_j\rangle\rangle +\tag{9}\\ +&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + +\langle x_i\rangle\langle x_j\rangle +\tag{10}\\ +&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle +\tag{11} +\end{align} $$ +
-The Numpy function np.cov calculates the covariance elements using -the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have -the exact mean valu\ es. - -
-The following simple function uses the np.vstack function which -takes each vector of dimension \( 1\times n \) and produces a \( 3\times n \) -matrix \( \hat{W} \) - -$$ -\hat{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\ - x_1 & y_1 & z_1 \\ - x_2 & y_2 & z_2 \\ - \dots & \dots & \dots \\ - x_{n-2} & y_{n-2} & z_{n-2} \\ - x_{n-1} & y_{n-1} & z_{n-1} - \end{bmatrix}, -$$ - -
-which in turn is converted into into the \( 3\times 3 \) covariance matrix -\( \hat{\Sigma} \) via the Numpy function np.cov(). We note that we can -also calculate the mean value of each set of samples \( \hat{x} \) etc -using the Numpy function np.mean(x). We can also extract the -eigenvalues of the covariance matrix through the np.linalg.eig() -function.
@@ -478,7 +464,7 @@ function.
+Suppose we have defined three vectors \( \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z} \) with +\( n \) elements each. The covariance matrix is defined as - -
# Importing various packages
-import numpy as np
+$$
+\boldsymbol{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
+ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\
+ \sigma_{zx} & \sigma_{zy} & \sigma_{zz}
+ \end{bmatrix},
+$$
+
+where for example
+$$
+\sigma_{xy} =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}).
+$$
-n = 100
-x = np.random.normal(size=n)
-print(np.mean(x))
-y = 4+3*x+np.random.normal(size=n)
-print(np.mean(y))
-z = x**3+np.random.normal(size=n)
-print(np.mean(z))
-W = np.vstack((x, y, z))
-Sigma = np.cov(W)
-print(Sigma)
-Eigvals, Eigvecs = np.linalg.eig(Sigma)
-print(Eigvals)
-+The Numpy function np.cov calculates the covariance elements using +the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have +the exact mean valu\ es. + +
+The following simple function uses the np.vstack function which +takes each vector of dimension \( 1\times n \) and produces a \( 3\times n \) +matrix \( \boldsymbol{W} \) + +$$ +\boldsymbol{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\ + x_1 & y_1 & z_1 \\ + x_2 & y_2 & z_2 \\ + \dots & \dots & \dots \\ + x_{n-2} & y_{n-2} & z_{n-2} \\ + x_{n-1} & y_{n-1} & z_{n-1} + \end{bmatrix}, +$$ + +
+which in turn is converted into into the \( 3\times 3 \) covariance matrix +\( \boldsymbol{\Sigma} \) via the Numpy function np.cov(). We note that we can +also calculate the mean value of each set of samples \( \boldsymbol{x} \) etc +using the Numpy function np.mean(x). We can also extract the +eigenvalues of the covariance matrix through the np.linalg.eig() +function. - -
import numpy as np
-import matplotlib.pyplot as plt
-from scipy import sparse
-eye = np.eye(4)
-print(eye)
-sparse_mtx = sparse.csr_matrix(eye)
-print(sparse_mtx)
-x = np.linspace(-10,10,100)
-y = np.sin(x)
-plt.plot(x,y,marker='x')
-plt.show()
-
@@ -468,7 +480,7 @@ plt.show()
-If \( X_i \) and \( X_j \) are independent, we get -\( \langle x_i x_j\rangle =\langle x_i\rangle\langle x_j\rangle \), resulting in \( \mathrm{cov}(X_i, X_j) = 0\ \ (i\neq j) \). +
-Also useful for us is the covariance of linear combinations of -stochastic variables. Let \( \{X_i\} \) and \( \{Y_i\} \) be two sets of -stochastic variables. Let also \( \{a_i\} \) and \( \{b_i\} \) be two sets of -scalars. Consider the linear combination: -$$ -U = \sum_i a_i X_i \qquad V = \sum_j b_j Y_j -$$ -By the linearity of the expectation value -$$ -\mathrm{cov}(U, V) = \sum_{i,j}a_i b_j \mathrm{cov}(X_i, Y_j) -$$ -
# Importing various packages
+import numpy as np
+n = 100
+x = np.random.normal(size=n)
+print(np.mean(x))
+y = 4+3*x+np.random.normal(size=n)
+print(np.mean(y))
+z = x**3+np.random.normal(size=n)
+print(np.mean(z))
+W = np.vstack((x, y, z))
+Sigma = np.cov(W)
+print(Sigma)
+Eigvals, Eigvecs = np.linalg.eig(Sigma)
+print(Eigvals)
++ +
import numpy as np
+import matplotlib.pyplot as plt
+from scipy import sparse
+eye = np.eye(4)
+print(eye)
+sparse_mtx = sparse.csr_matrix(eye)
+print(sparse_mtx)
+x = np.linspace(-10,10,100)
+y = np.sin(x)
+plt.plot(x,y,marker='x')
+plt.show()
+
@@ -456,7 +470,7 @@ $$
-Now, since the variance is just \( \mathrm{var}(X_i) = \mathrm{cov}(X_i, X_i) \), we get -the variance of the linear combination \( U = \sum_i a_i X_i \): +If \( X_i \) and \( X_j \) are independent, we get +\( \langle x_i x_j\rangle =\langle x_i\rangle\langle x_j\rangle \), resulting in \( \mathrm{cov}(X_i, X_j) = 0\ \ (i\neq j) \). + +
+Also useful for us is the covariance of linear combinations of +stochastic variables. Let \( \{X_i\} \) and \( \{Y_i\} \) be two sets of +stochastic variables. Let also \( \{a_i\} \) and \( \{b_i\} \) be two sets of +scalars. Consider the linear combination: $$ -\begin{equation} -\mathrm{var}(U) = \sum_{i,j}a_i a_j \mathrm{cov}(X_i, X_j) -\tag{12} -\end{equation} +U = \sum_i a_i X_i \qquad V = \sum_j b_j Y_j $$ -And in the special case when the stochastic variables are -uncorrelated, the off-diagonal elements of the covariance are as we -know zero, resulting in: +By the linearity of the expectation value $$ -\mathrm{var}(U) = \sum_i a_i^2 \mathrm{cov}(X_i, X_i) = \sum_i a_i^2 \mathrm{var}(X_i) +\mathrm{cov}(U, V) = \sum_{i,j}a_i b_j \mathrm{cov}(X_i, Y_j) $$ - -$$ -\mathrm{var}(\sum_i a_i X_i) = \sum_i a_i^2 \mathrm{var}(X_i) -$$ - -which will become very useful in our study of the error in the mean -value of a set of measurements.
-A stochastic process is a process that produces sequentially a -chain of values: +Now, since the variance is just \( \mathrm{var}(X_i) = \mathrm{cov}(X_i, X_i) \), we get +the variance of the linear combination \( U = \sum_i a_i X_i \): $$ -\{x_1, x_2,\dots\,x_k,\dots\}. +\begin{equation} +\mathrm{var}(U) = \sum_{i,j}a_i a_j \mathrm{cov}(X_i, X_j) +\tag{12} +\end{equation} $$ -We will call these -values our measurements and the entire set as our measured -sample. The action of measuring all the elements of a sample -we will call a stochastic experiment since, operationally, -they are often associated with results of empirical observation of -some physical or mathematical phenomena; precisely an experiment. We -assume that these values are distributed according to some -PDF \( p_X^{\phantom X}(x) \), where \( X \) is just the formal symbol for the -stochastic variable whose PDF is \( p_X^{\phantom X}(x) \). Instead of -trying to determine the full distribution \( p \) we are often only -interested in finding the few lowest moments, like the mean -\( \mu_X^{\phantom X} \) and the variance \( \sigma_X^{\phantom X} \). +And in the special case when the stochastic variables are +uncorrelated, the off-diagonal elements of the covariance are as we +know zero, resulting in: +$$ +\mathrm{var}(U) = \sum_i a_i^2 \mathrm{cov}(X_i, X_i) = \sum_i a_i^2 \mathrm{var}(X_i) +$$ + +$$ +\mathrm{var}(\sum_i a_i X_i) = \sum_i a_i^2 \mathrm{var}(X_i) +$$ + +which will become very useful in our study of the error in the mean +value of a set of measurements.
- + -
-In practical situations a sample is always of finite size. Let that -size be \( n \). The expectation value of a sample, the sample mean, is then defined as follows: +A stochastic process is a process that produces sequentially a +chain of values: $$ -\bar{x}_n \equiv \frac{1}{n}\sum_{k=1}^n x_k +\{x_1, x_2,\dots\,x_k,\dots\}. $$ -The sample variance is: -$$ -\mathrm{var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_k - \bar{x}_n)^2 -$$ - -its square root being the standard deviation of the sample. The -sample covariance is: -$$ -\mathrm{cov}(x)\equiv\frac{1}{n}\sum_{kl}(x_k - \bar{x}_n)(x_l - \bar{x}_n) -$$ +We will call these +values our measurements and the entire set as our measured +sample. The action of measuring all the elements of a sample +we will call a stochastic experiment since, operationally, +they are often associated with results of empirical observation of +some physical or mathematical phenomena; precisely an experiment. We +assume that these values are distributed according to some +PDF \( p_X^{\phantom X}(x) \), where \( X \) is just the formal symbol for the +stochastic variable whose PDF is \( p_X^{\phantom X}(x) \). Instead of +trying to determine the full distribution \( p \) we are often only +interested in finding the few lowest moments, like the mean +\( \mu_X^{\phantom X} \) and the variance \( \sigma_X^{\phantom X} \).
- + -
-Note that the sample variance is the sample covariance without the -cross terms. In a similar manner as the covariance in Eq. (6) is a measure of the correlation between -two stochastic variables, the above defined sample covariance is a -measure of the sequential correlation between succeeding measurements -of a sample. +In practical situations a sample is always of finite size. Let that +size be \( n \). The expectation value of a sample, the sample mean, is then defined as follows: +$$ +\bar{x}_n \equiv \frac{1}{n}\sum_{k=1}^n x_k +$$ -
-These quantities, being known experimental values, differ -significantly from and must not be confused with the similarly named -quantities for stochastic variables, mean \( \mu_X \), variance \( \mathrm{var}(X) \) -and covariance \( \mathrm{cov}(X,Y) \). +The sample variance is: +$$ +\mathrm{var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_k - \bar{x}_n)^2 +$$ + +its square root being the standard deviation of the sample. The +sample covariance is: +$$ +\mathrm{cov}(x)\equiv\frac{1}{n}\sum_{kl}(x_k - \bar{x}_n)(x_l - \bar{x}_n) +$$
-The law of large numbers -states that as the size of our sample grows to infinity, the sample -mean approaches the true mean \( \mu_X^{\phantom X} \) of the chosen PDF: -$$ -\lim_{n\to\infty}\bar{x}_n = \mu_X^{\phantom X} -$$ - -The sample mean \( \bar{x}_n \) works therefore as an estimate of the true -mean \( \mu_X^{\phantom X} \). +Note that the sample variance is the sample covariance without the +cross terms. In a similar manner as the covariance in Eq. (6) is a measure of the correlation between +two stochastic variables, the above defined sample covariance is a +measure of the sequential correlation between succeeding measurements +of a sample.
-What we need to find out is how good an approximation \( \bar{x}_n \) is to -\( \mu_X^{\phantom X} \). In any stochastic measurement, an estimated -mean is of no use to us without a measure of its error. A quantity -that tells us how well we can reproduce it in another experiment. We -are therefore interested in the PDF of the sample mean itself. Its -standard deviation will be a measure of the spread of sample means, -and we will simply call it the error of the sample mean, or -just sample error, and denote it by \( \mathrm{err}_X^{\phantom X} \). In -practice, we will only be able to produce an estimate of the -sample error since the exact value would require the knowledge of the -true PDFs behind, which we usually do not have. +These quantities, being known experimental values, differ +significantly from and must not be confused with the similarly named +quantities for stochastic variables, mean \( \mu_X \), variance \( \mathrm{var}(X) \) +and covariance \( \mathrm{cov}(X,Y) \).
-Let us first take a look at what happens to the sample error as the -size of the sample grows. In a sample, each of the measurements \( x_i \) -can be associated with its own stochastic variable \( X_i \). The -stochastic variable \( \overline X_n \) for the sample mean \( \bar{x}_n \) is -then just a linear combination, already familiar to us: +The law of large numbers +states that as the size of our sample grows to infinity, the sample +mean approaches the true mean \( \mu_X^{\phantom X} \) of the chosen PDF: $$ -\overline X_n = \frac{1}{n}\sum_{i=1}^n X_i +\lim_{n\to\infty}\bar{x}_n = \mu_X^{\phantom X} $$ -All the coefficients are just equal \( 1/n \). The PDF of \( \overline X_n \), -denoted by \( p_{\overline X_n}(x) \) is the desired PDF of the sample -means. +The sample mean \( \bar{x}_n \) works therefore as an estimate of the true +mean \( \mu_X^{\phantom X} \). + +
+What we need to find out is how good an approximation \( \bar{x}_n \) is to +\( \mu_X^{\phantom X} \). In any stochastic measurement, an estimated +mean is of no use to us without a measure of its error. A quantity +that tells us how well we can reproduce it in another experiment. We +are therefore interested in the PDF of the sample mean itself. Its +standard deviation will be a measure of the spread of sample means, +and we will simply call it the error of the sample mean, or +just sample error, and denote it by \( \mathrm{err}_X^{\phantom X} \). In +practice, we will only be able to produce an estimate of the +sample error since the exact value would require the knowledge of the +true PDFs behind, which we usually do not have.
-The probability density of obtaining a sample mean \( \bar x_n \) -is the product of probabilities of obtaining arbitrary values \( x_1, -x_2,\dots,x_n \) with the constraint that the mean of the set \( \{x_i\} \) -is \( \bar x_n \): +Let us first take a look at what happens to the sample error as the +size of the sample grows. In a sample, each of the measurements \( x_i \) +can be associated with its own stochastic variable \( X_i \). The +stochastic variable \( \overline X_n \) for the sample mean \( \bar{x}_n \) is +then just a linear combination, already familiar to us: $$ -p_{\overline X_n}(x) = \int p_X^{\phantom X}(x_1)\cdots -\int p_X^{\phantom X}(x_n)\ -\delta\!\left(x - \frac{x_1+x_2+\dots+x_n}{n}\right)dx_n \cdots dx_1 +\overline X_n = \frac{1}{n}\sum_{i=1}^n X_i $$ -And in particular we are interested in its variance \( \mathrm{var}(\overline X_n) \). +All the coefficients are just equal \( 1/n \). The PDF of \( \overline X_n \), +denoted by \( p_{\overline X_n}(x) \) is the desired PDF of the sample +means.
-It is generally not possible to express \( p_{\overline X_n}(x) \) in a -closed form given an arbitrary PDF \( p_X^{\phantom X} \) and a number -\( n \). But for the limit \( n\to\infty \) it is possible to make an -approximation. The very important result is called the central limit theorem. It tells us that as \( n \) goes to infinity, -\( p_{\overline X_n}(x) \) approaches a Gaussian distribution whose mean -and variance equal the true mean and variance, \( \mu_{X}^{\phantom X} \) -and \( \sigma_{X}^{2} \), respectively: +The probability density of obtaining a sample mean \( \bar x_n \) +is the product of probabilities of obtaining arbitrary values \( x_1, +x_2,\dots,x_n \) with the constraint that the mean of the set \( \{x_i\} \) +is \( \bar x_n \): $$ -\begin{equation} -\lim_{n\to\infty} p_{\overline X_n}(x) = -\left(\frac{n}{2\pi\mathrm{var}(X)}\right)^{1/2} -e^{-\frac{n(x-\bar x_n)^2}{2\mathrm{var}(X)}} -\tag{13} -\end{equation} +p_{\overline X_n}(x) = \int p_X^{\phantom X}(x_1)\cdots +\int p_X^{\phantom X}(x_n)\ +\delta\!\left(x - \frac{x_1+x_2+\dots+x_n}{n}\right)dx_n \cdots dx_1 $$ + +And in particular we are interested in its variance \( \mathrm{var}(\overline X_n) \).
-The desired variance -\( \mathrm{var}(\overline X_n) \), i.e. the sample error squared -\( \mathrm{err}_X^2 \), is given by: +It is generally not possible to express \( p_{\overline X_n}(x) \) in a +closed form given an arbitrary PDF \( p_X^{\phantom X} \) and a number +\( n \). But for the limit \( n\to\infty \) it is possible to make an +approximation. The very important result is called the central limit theorem. It tells us that as \( n \) goes to infinity, +\( p_{\overline X_n}(x) \) approaches a Gaussian distribution whose mean +and variance equal the true mean and variance, \( \mu_{X}^{\phantom X} \) +and \( \sigma_{X}^{2} \), respectively: $$ \begin{equation} -\mathrm{err}_X^2 = \mathrm{var}(\overline X_n) = \frac{1}{n^2} -\sum_{ij} \mathrm{cov}(X_i, X_j) -\tag{14} +\lim_{n\to\infty} p_{\overline X_n}(x) = +\left(\frac{n}{2\pi\mathrm{var}(X)}\right)^{1/2} +e^{-\frac{n(x-\bar x_n)^2}{2\mathrm{var}(X)}} +\tag{13} \end{equation} $$ - -We see now that in order to calculate the exact error of the sample -with the above expression, we would need the true means -\( \mu_{X_i}^{\phantom X} \) of the stochastic variables \( X_i \). To -calculate these requires that we know the true multivariate PDF of all -the \( X_i \). But this PDF is unknown to us, we have only got the measurements of -one sample. The best we can do is to let the sample itself be an -estimate of the PDF of each of the \( X_i \), estimating all properties of -\( X_i \) through the measurements of the sample.
-Our estimate of \( \mu_{X_i}^{\phantom X} \) is then the sample mean \( \bar x \) -itself, in accordance with the the central limit theorem: +The desired variance +\( \mathrm{var}(\overline X_n) \), i.e. the sample error squared +\( \mathrm{err}_X^2 \), is given by: $$ -\mu_{X_i}^{\phantom X} = \langle x_i\rangle \approx \frac{1}{n}\sum_{k=1}^n x_k = \bar x +\begin{equation} +\mathrm{err}_X^2 = \mathrm{var}(\overline X_n) = \frac{1}{n^2} +\sum_{ij} \mathrm{cov}(X_i, X_j) +\tag{14} +\end{equation} $$ -Using \( \bar x \) in place of \( \mu_{X_i}^{\phantom X} \) we can give an -estimate of the covariance in Eq. (14) -$$ -\mathrm{cov}(X_i, X_j) = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle -\approx\langle (x_i - \bar x)(x_j - \bar{x})\rangle, -$$ - -resulting in -$$ -\frac{1}{n} \sum_{l}^n \left(\frac{1}{n}\sum_{k}^n (x_k -\bar x_n)(x_l - \bar x_n)\right)=\frac{1}{n}\frac{1}{n} \sum_{kl} (x_k -\bar x_n)(x_l - \bar x_n)=\frac{1}{n}\mathrm{cov}(x) -$$ +We see now that in order to calculate the exact error of the sample +with the above expression, we would need the true means +\( \mu_{X_i}^{\phantom X} \) of the stochastic variables \( X_i \). To +calculate these requires that we know the true multivariate PDF of all +the \( X_i \). But this PDF is unknown to us, we have only got the measurements of +one sample. The best we can do is to let the sample itself be an +estimate of the PDF of each of the \( X_i \), estimating all properties of +\( X_i \) through the measurements of the sample.
-By the same procedure we can use the sample variance as an -estimate of the variance of any of the stochastic variables \( X_i \) +Our estimate of \( \mu_{X_i}^{\phantom X} \) is then the sample mean \( \bar x \) +itself, in accordance with the the central limit theorem: $$ -\mathrm{var}(X_i)=\langle x_i - \langle x_i\rangle\rangle \approx \langle x_i - \bar x_n\rangle\nonumber, +\mu_{X_i}^{\phantom X} = \langle x_i\rangle \approx \frac{1}{n}\sum_{k=1}^n x_k = \bar x $$ -which is approximated as +Using \( \bar x \) in place of \( \mu_{X_i}^{\phantom X} \) we can give an +estimate of the covariance in Eq. (14) $$ -\begin{equation} -\mathrm{var}(X_i)\approx \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)=\mathrm{var}(x) -\tag{15} -\end{equation} +\mathrm{cov}(X_i, X_j) = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle +\approx\langle (x_i - \bar x)(x_j - \bar{x})\rangle, $$ -
-Now we can calculate an estimate of the error -\( \mathrm{err}_X^{\phantom X} \) of the sample mean \( \bar x_n \): +resulting in +$$ +\frac{1}{n} \sum_{l}^n \left(\frac{1}{n}\sum_{k}^n (x_k -\bar x_n)(x_l - \bar x_n)\right)=\frac{1}{n}\frac{1}{n} \sum_{kl} (x_k -\bar x_n)(x_l - \bar x_n)=\frac{1}{n}\mathrm{cov}(x) $$ -\begin{align} -\mathrm{err}_X^2 -&=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) \nonumber \\ -&\approx&\frac{1}{n^2}\sum_{ij}\frac{1}{n}\mathrm{cov}(x) =\frac{1}{n^2}n^2\frac{1}{n}\mathrm{cov}(x)\nonumber\\ -&=\frac{1}{n}\mathrm{cov}(x) -\tag{16} -\end{align} -$$ - -which is nothing but the sample covariance divided by the number of -measurements in the sample.
- -
-In the special case that the measurements of the sample are -uncorrelated (equivalently the stochastic variables \( X_i \) are -uncorrelated) we have that the off-diagonal elements of the covariance -are zero. This gives the following estimate of the sample error: +By the same procedure we can use the sample variance as an +estimate of the variance of any of the stochastic variables \( X_i \) $$ -\mathrm{err}_X^2=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) = -\frac{1}{n^2} \sum_i \mathrm{var}(X_i), +\mathrm{var}(X_i)=\langle x_i - \langle x_i\rangle\rangle \approx \langle x_i - \bar x_n\rangle\nonumber, $$ -resulting in +which is approximated as $$ \begin{equation} -\mathrm{err}_X^2\approx \frac{1}{n^2} \sum_i \mathrm{var}(x)= \frac{1}{n}\mathrm{var}(x) -\tag{17} +\mathrm{var}(X_i)\approx \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)=\mathrm{var}(x) +\tag{15} \end{equation} $$ -where in the second step we have used Eq. (15). -The error of the sample is then just its standard deviation divided by -the square root of the number of measurements the sample contains. -This is a very useful formula which is easy to compute. It acts as a -first approximation to the error, but in numerical experiments, we -cannot overlook the always present correlations. +
+Now we can calculate an estimate of the error +\( \mathrm{err}_X^{\phantom X} \) of the sample mean \( \bar x_n \): +$$ +\begin{align} +\mathrm{err}_X^2 +&=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) \nonumber \\ +&\approx&\frac{1}{n^2}\sum_{ij}\frac{1}{n}\mathrm{cov}(x) =\frac{1}{n^2}n^2\frac{1}{n}\mathrm{cov}(x)\nonumber\\ +&=\frac{1}{n}\mathrm{cov}(x) +\tag{16} +\end{align} +$$ + +which is nothing but the sample covariance divided by the number of +measurements in the sample.
-For computational purposes one usually splits up the estimate of -\( \mathrm{err}_X^2 \), given by Eq. (16), into two -parts + +
+In the special case that the measurements of the sample are +uncorrelated (equivalently the stochastic variables \( X_i \) are +uncorrelated) we have that the off-diagonal elements of the covariance +are zero. This gives the following estimate of the sample error: $$ -\mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x) + \frac{1}{n}(\mathrm{cov}(x)-\mathrm{var}(x)), +\mathrm{err}_X^2=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) = +\frac{1}{n^2} \sum_i \mathrm{var}(X_i), $$ -which equals +resulting in $$ \begin{equation} -\frac{1}{n^2}\sum_{k=1}^n (x_k - \bar x_n)^2 +\frac{2}{n^2}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) -\tag{18} +\mathrm{err}_X^2\approx \frac{1}{n^2} \sum_i \mathrm{var}(x)= \frac{1}{n}\mathrm{var}(x) +\tag{17} \end{equation} $$ -The first term is the same as the error in the uncorrelated case, -Eq. (17). This means that the second -term accounts for the error correction due to correlation between the -measurements. For uncorrelated measurements this second term is zero. +where in the second step we have used Eq. (15). +The error of the sample is then just its standard deviation divided by +the square root of the number of measurements the sample contains. +This is a very useful formula which is easy to compute. It acts as a +first approximation to the error, but in numerical experiments, we +cannot overlook the always present correlations.
-Computationally the uncorrelated first term is much easier to treat -efficiently than the second. +For computational purposes one usually splits up the estimate of +\( \mathrm{err}_X^2 \), given by Eq. (16), into two +parts $$ -\mathrm{var}(x) = \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)^2 = -\left(\frac{1}{n}\sum_{k=1}^n x_k^2\right) - \bar x_n^2 +\mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x) + \frac{1}{n}(\mathrm{cov}(x)-\mathrm{var}(x)), $$ -We just accumulate separately the values \( x^2 \) and \( x \) for every -measurement \( x \) we receive. The correlation term, though, has to be -calculated at the end of the experiment since we need all the -measurements to calculate the cross terms. Therefore, all measurements -have to be stored throughout the experiment. +which equals +$$ +\begin{equation} +\frac{1}{n^2}\sum_{k=1}^n (x_k - \bar x_n)^2 +\frac{2}{n^2}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) +\tag{18} +\end{equation} +$$ + +The first term is the same as the error in the uncorrelated case, +Eq. (17). This means that the second +term accounts for the error correction due to correlation between the +measurements. For uncorrelated measurements this second term is zero.
-Let us analyze the problem by splitting up the correlation term into -partial sums of the form: +Computationally the uncorrelated first term is much easier to treat +efficiently than the second. $$ -f_d = \frac{1}{n-d}\sum_{k=1}^{n-d}(x_k - \bar x_n)(x_{k+d} - \bar x_n) +\mathrm{var}(x) = \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)^2 = +\left(\frac{1}{n}\sum_{k=1}^n x_k^2\right) - \bar x_n^2 $$ -The correlation term of the error can now be rewritten in terms of -\( f_d \) -$$ -\frac{2}{n}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) = -2\sum_{d=1}^{n-1} f_d -$$ - -The value of \( f_d \) reflects the correlation between measurements -separated by the distance \( d \) in the sample samples. Notice that for -\( d=0 \), \( f \) is just the sample variance, \( \mathrm{var}(x) \). If we divide \( f_d \) -by \( \mathrm{var}(x) \), we arrive at the so called autocorrelation function -$$ -\kappa_d = \frac{f_d}{\mathrm{var}(x)} -$$ - -which gives us a useful measure of pairwise correlations -starting always at \( 1 \) for \( d=0 \). +We just accumulate separately the values \( x^2 \) and \( x \) for every +measurement \( x \) we receive. The correlation term, though, has to be +calculated at the end of the experiment since we need all the +measurements to calculate the cross terms. Therefore, all measurements +have to be stored throughout the experiment.
-The sample error (see eq. (18)) can now be -written in terms of the autocorrelation function: +Let us analyze the problem by splitting up the correlation term into +partial sums of the form: $$ -\begin{align} -\mathrm{err}_X^2 &= -\frac{1}{n}\mathrm{var}(x)+\frac{2}{n}\cdot\mathrm{var}(x)\sum_{d=1}^{n-1} -\frac{f_d}{\mathrm{var}(x)}\nonumber\\ &=& -\left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\mathrm{var}(x)\nonumber\\ -&=\frac{\tau}{n}\cdot\mathrm{var}(x) -\tag{19} -\end{align} +f_d = \frac{1}{n-d}\sum_{k=1}^{n-d}(x_k - \bar x_n)(x_{k+d} - \bar x_n) $$ -and we see that \( \mathrm{err}_X \) can be expressed in terms the -uncorrelated sample variance times a correction factor \( \tau \) which -accounts for the correlation between measurements. We call this -correction factor the autocorrelation time: +The correlation term of the error can now be rewritten in terms of +\( f_d \) $$ -\begin{equation} -\tau = 1+2\sum_{d=1}^{n-1}\kappa_d -\tag{20} -\end{equation} +\frac{2}{n}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) = +2\sum_{d=1}^{n-1} f_d $$ + +The value of \( f_d \) reflects the correlation between measurements +separated by the distance \( d \) in the sample samples. Notice that for +\( d=0 \), \( f \) is just the sample variance, \( \mathrm{var}(x) \). If we divide \( f_d \) +by \( \mathrm{var}(x) \), we arrive at the so called autocorrelation function +$$ +\kappa_d = \frac{f_d}{\mathrm{var}(x)} +$$ + +which gives us a useful measure of pairwise correlations +starting always at \( 1 \) for \( d=0 \).
-For a correlation free experiment, \( \tau \) -equals 1. From the point of view of -eq. (19) we can interpret a sequential -correlation as an effective reduction of the number of measurements by -a factor \( \tau \). The effective number of measurements becomes: +The sample error (see eq. (18)) can now be +written in terms of the autocorrelation function: $$ -n_\mathrm{eff} = \frac{n}{\tau} +\begin{align} +\mathrm{err}_X^2 &= +\frac{1}{n}\mathrm{var}(x)+\frac{2}{n}\cdot\mathrm{var}(x)\sum_{d=1}^{n-1} +\frac{f_d}{\mathrm{var}(x)}\nonumber\\ &=& +\left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\mathrm{var}(x)\nonumber\\ +&=\frac{\tau}{n}\cdot\mathrm{var}(x) +\tag{19} +\end{align} $$ -To neglect the autocorrelation time \( \tau \) will always cause our -simple uncorrelated estimate of \( \mathrm{err}_X^2\approx \mathrm{var}(x)/n \) to -be less than the true sample error. The estimate of the error will be -too good. On the other hand, the calculation of the full -autocorrelation time poses an efficiency problem if the set of -measurements is very large. +and we see that \( \mathrm{err}_X \) can be expressed in terms the +uncorrelated sample variance times a correction factor \( \tau \) which +accounts for the correlation between measurements. We call this +correction factor the autocorrelation time: +$$ +\begin{equation} +\tau = 1+2\sum_{d=1}^{n-1}\kappa_d +\tag{20} +\end{equation} +$$
- + -
-Finally, we are going to discuss several statistical properties which can be obtained in terms of analytical expressions. -The -advantage of doing linear regression is that we actually end up with -analytical expressions for several statistical quantities. -Standard least squares and Ridge regression allow us to -derive quantities like the variance and other expectation values in a -rather straightforward way. - -
-It is assumed that \( \varepsilon_i -\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are -independent, i.e.: +
+For a correlation free experiment, \( \tau \) +equals 1. From the point of view of +eq. (19) we can interpret a sequential +correlation as an effective reduction of the number of measurements by +a factor \( \tau \). The effective number of measurements becomes: $$ -\begin{align*} -\mbox{Cov}(\varepsilon_{i_1}, -\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} -& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. -\end{align*} +n_\mathrm{eff} = \frac{n}{\tau} $$ -The randomness of \( \varepsilon_i \) implies that -\( \mathbf{y}_i \) is also a random variable. In particular, -\( \mathbf{y}_i \) is normally distributed, because \( \varepsilon_i \sim -\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \boldsymbol{\beta} \) is a -non-random scalar. To specify the parameters of the distribution of -\( \mathbf{y}_i \) we need to calculate its first two moments. +To neglect the autocorrelation time \( \tau \) will always cause our +simple uncorrelated estimate of \( \mathrm{err}_X^2\approx \mathrm{var}(x)/n \) to +be less than the true sample error. The estimate of the error will be +too good. On the other hand, the calculation of the full +autocorrelation time poses an efficiency problem if the set of +measurements is very large. +
-Recall that \( \boldsymbol{X} \) is a matrix of dimensionality \( n\times p \). The -notation above \( \mathbf{X}_{i,\ast} \) means that we are looking at the -row number \( i \) and perform a sum over all values \( p \).
@@ -467,7 +457,7 @@ row number \( i \) and perform a sum over all values \( p \).
- + -
-The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) -that there exists a function \( f(\boldsymbol{x}) \) and a normal distributed error \( \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) \) -which describe our data -$$ -\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon} -$$ +Finally, we are going to discuss several statistical properties which can be obtained in terms of analytical expressions. +The +advantage of doing linear regression is that we actually end up with +analytical expressions for several statistical quantities. +Standard least squares and Ridge regression allow us to +derive quantities like the variance and other expectation values in a +rather straightforward way.
-We approximate this function with our model from the solution of the linear regression equations, that is our -function \( f \) is approximated by \( \boldsymbol{\tilde{y}} \) where we want to minimize \( (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2 \), our MSE, with +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: $$ -\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}. +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} $$ +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{y}_i \) is also a random variable. In particular, +\( \mathbf{y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \boldsymbol{\beta} \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{y}_i \) we need to calculate its first two moments. + +
+Recall that \( \boldsymbol{X} \) is a matrix of dimensionality \( n\times p \). The +notation above \( \mathbf{X}_{i,\ast} \) means that we are looking at the +row number \( i \) and perform a sum over all values \( p \). +
@@ -449,7 +469,7 @@ $$
-We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \) +The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) +that there exists a function \( f(\boldsymbol{x}) \) and a normal distributed error \( \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) \) +which describe our data $$ -\begin{align*} -\mathbb{E}(y_i) & = -\mathbb{E}(\mathbf{X}_{i, \ast} \, \boldsymbol{\beta}) + \mathbb{E}(\varepsilon_i) -\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, -\end{align*} +\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon} $$ -while -its variance is +
+We approximate this function with our model from the solution of the linear regression equations, that is our +function \( f \) is approximated by \( \boldsymbol{\tilde{y}} \) where we want to minimize \( (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2 \), our MSE, with $$ -\begin{align*} \mbox{Var}(y_i) & = \mathbb{E} \{ [y_i -- \mathbb{E}(y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( y_i^2 ) - -[\mathbb{E}(y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, -\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & -= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \varepsilon_i -\mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i, -\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 -\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + -\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 -\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, -\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. -\end{align*} +\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}. $$ -Hence, \( y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2) \), that is \( \boldsymbol{y} \) follows a normal distribution with -mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (not be confused with the singular values of the SVD). -
@@ -464,7 +451,7 @@ mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (n
-With the OLS expressions for the parameters \( \boldsymbol{\beta} \) we can evaluate the expectation value +We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \) $$ -\mathbb{E}(\boldsymbol{\beta}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}. +\begin{align*} +\mathbb{E}(y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \boldsymbol{\beta}) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} $$ -This means that the estimator of the regression parameters is unbiased. - -
-We can also calculate the variance - -
-The variance of \( \boldsymbol{\beta} \) is +while +its variance is $$ -\begin{eqnarray*} -\mbox{Var}(\boldsymbol{\beta}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \} -\\ -& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \} -\\ -% & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -% & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -\\ -& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} -% \\ -% & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T -\\ -& = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1}, -\end{eqnarray*} +\begin{align*} \mbox{Var}(y_i) & = \mathbb{E} \{ [y_i +- \mathbb{E}(y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( y_i^2 ) - +[\mathbb{E}(y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} $$ -
-where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = -\mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + -\sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2 -\, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the -variance of the estimate of the \( j \)-th regression coefficient: -\( \hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ -[(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} } \). This may be used to -construct a confidence interval for the estimates. - -
-In a similar way, we can obtain analytical expressions for say the -expectation values of the parameters \( \boldsymbol{\beta} \) and their variance -when we employ Ridge regression, allowing us again to define a confidence interval. - -
-It is rather straightforward to show that -$$ -\mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}. -$$ - -We see clearly that -\( \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} \) for any \( \lambda > 0 \). We say then that the ridge estimator is biased. - -
-We can also compute the variance as - -$$ -\mbox{Var}[\boldsymbol{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, -$$ - -and it is easy to see that if the parameter \( \lambda \) goes to infinity then the variance of Ridge parameters \( \boldsymbol{\beta} \) goes to zero. - -
-With this, we can compute the difference - -$$ -\mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. -$$ - -The difference is non-negative definite since each component of the -matrix product is non-negative definite. -This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below. +Hence, \( y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2) \), that is \( \boldsymbol{y} \) follows a normal distribution with +mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (not be confused with the singular values of the SVD).
@@ -514,7 +466,7 @@ This means the variance we obtain with the standard OLS will always for \( \lamb
-With all these analytical equations for both the OLS and Ridge -regression, we will now outline how to assess a given model. This will -lead us to a discussion of the so-called bias-variance tradeoff (see -below) and so-called resampling methods. +With the OLS expressions for the parameters \( \boldsymbol{\beta} \) we can evaluate the expectation value +$$ +\mathbb{E}(\boldsymbol{\beta}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}. +$$ + +This means that the estimator of the regression parameters is unbiased.
-One of the quantities we have discussed as a way to measure errors is -the mean-squared error (MSE), mainly used for fitting of continuous -functions. Another choice is the absolute error. +We can also calculate the variance
-In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data, -we discuss the +The variance of \( \boldsymbol{\beta} \) is +$$ +\begin{eqnarray*} +\mbox{Var}(\boldsymbol{\beta}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \} +\\ +& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \} +\\ +% & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +% & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +\\ +& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} +% \\ +% & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T +\\ +& = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1}, +\end{eqnarray*} +$$ -
+where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = +\mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + +\sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2 +\, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the +variance of the estimate of the \( j \)-th regression coefficient: +\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 \sqrt{ +[(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} } \). This may be used to +construct a confidence interval for the estimates. -As our model becomes more and more complex, more of the training data tends to used. The training may thence adapt to more complicated structures in the data. This may lead to a decrease in the bias (see below for code example) and a slight increase of the variance for the test error. -For a certain level of complexity the test error will reach minimum, before starting to increase again. The -training error reaches a saturation. +
+In a similar way, we can obtain analytical expressions for say the +expectation values of the parameters \( \boldsymbol{\beta} \) and their variance +when we employ Ridge regression, allowing us again to define a confidence interval. + +
+It is rather straightforward to show that +$$ +\mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}. +$$ + +We see clearly that +\( \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} \) for any \( \lambda > 0 \). We say then that the ridge estimator is biased. + +
+We can also compute the variance as + +$$ +\mbox{Var}[\boldsymbol{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, +$$ + +and it is easy to see that if the parameter \( \lambda \) goes to infinity then the variance of Ridge parameters \( \boldsymbol{\beta} \) goes to zero. + +
+With this, we can compute the difference + +$$ +\mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. +$$ + +The difference is non-negative definite since each component of the +matrix product is non-negative definite. +This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.
@@ -458,7 +516,7 @@ training error reaches a saturation.
-Two famous -resampling methods are the independent bootstrap and the jackknife. +With all these analytical equations for both the OLS and Ridge +regression, we will now outline how to assess a given model. This will +lead us to a discussion of the so-called bias-variance tradeoff (see +below) and so-called resampling methods.
-The jackknife is a special case of the independent bootstrap. Still, the jackknife was made -popular prior to the independent bootstrap. And as the popularity of -the independent bootstrap soared, new variants, such as the dependent bootstrap. +One of the quantities we have discussed as a way to measure errors is +the mean-squared error (MSE), mainly used for fitting of continuous +functions. Another choice is the absolute error.
-The Jackknife and independent bootstrap work for -independent, identically distributed random variables. -If these conditions are not -satisfied, the methods will fail. Yet, it should be said that if the data are -independent, identically distributed, and we only want to estimate the -variance of \( \overline{X} \) (which often is the case), then there is no -need for bootstrapping. +In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data, +we discuss the + +
@@ -452,7 +460,7 @@ need for bootstrapping.
-The Jackknife works by making many replicas of the estimator \( \widehat{\theta} \). -The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values \( \boldsymbol{x} = (x_1,x_2,\cdots,X_n) \). -Let \( \boldsymbol{x}_i \) denote the vector -$$ -\boldsymbol{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n), -$$ +Two famous +resampling methods are the independent bootstrap and the jackknife.
-which equals the vector \( \boldsymbol{x} \) with the exception that observation -number \( i \) is left out. Using this notation, define -\( \widehat{\theta}_i \) to be the estimator -\( \widehat{\theta} \) computed using \( \vec{X}_i \). +The jackknife is a special case of the independent bootstrap. Still, the jackknife was made +popular prior to the independent bootstrap. And as the popularity of +the independent bootstrap soared, new variants, such as the dependent bootstrap. + +
+The Jackknife and independent bootstrap work for +independent, identically distributed random variables. +If these conditions are not +satisfied, the methods will fail. Yet, it should be said that if the data are +independent, identically distributed, and we only want to estimate the +variance of \( \overline{X} \) (which often is the case), then there is no +need for bootstrapping.
@@ -448,7 +454,7 @@ number \( i \) is left out. Using this notation, define
+The Jackknife works by making many replicas of the estimator \( \widehat{\theta} \). +The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values \( \boldsymbol{x} = (x_1,x_2,\cdots,X_n) \). +Let \( \boldsymbol{x}_i \) denote the vector +$$ +\boldsymbol{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n), +$$ - -
from numpy import *
-from numpy.random import randint, randn
-from time import time
+
+which equals the vector \( \boldsymbol{x} \) with the exception that observation
+number \( i \) is left out. Using this notation, define
+\( \widehat{\theta}_i \) to be the estimator
+\( \widehat{\theta} \) computed using \( \vec{X}_i \).
-def jackknife(data, stat):
- n = len(data);t = zeros(n); inds = arange(n); t0 = time()
- ## 'jackknifing' by leaving out an observation for each i
- for i in range(n):
- t[i] = stat(delete(data,i) )
-
- # analysis
- print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :")
- print("original bias std. error")
- print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))
-
- return t
-
-
-# Returns mean of data samples
-def stat(data):
- return mean(data)
-
-
-mu, sigma = 100, 15
-datapoints = 10000
-x = mu + sigma*random.randn(datapoints)
-# jackknife returns the data sample
-t = jackknife(x, stat)
-
@@ -465,7 +450,7 @@ t = jackknife(x, stat)
-Bootstrapping is a nonparametric approach to statistical inference -that substitutes computation for more traditional distributional -assumptions and asymptotic results. Bootstrapping offers a number of -advantages: +
-
from numpy import *
+from numpy.random import randint, randn
+from time import time
+
+def jackknife(data, stat):
+ n = len(data);t = zeros(n); inds = arange(n); t0 = time()
+ ## 'jackknifing' by leaving out an observation for each i
+ for i in range(n):
+ t[i] = stat(delete(data,i) )
+
+ # analysis
+ print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :")
+ print("original bias std. error")
+ print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))
+
+ return t
+# Returns mean of data samples
+def stat(data):
+ return mean(data)
+
+
+mu, sigma = 100, 15
+datapoints = 10000
+x = mu + sigma*random.randn(datapoints)
+# jackknife returns the data sample
+t = jackknife(x, stat)
+
@@ -451,7 +467,7 @@ advantages:
+Bootstrapping is a nonparametric approach to statistical inference +that substitutes computation for more traditional distributional +assumptions and asymptotic results. Bootstrapping offers a number of +advantages: + +
-Since \( \widehat{\theta} = \widehat{\theta}(\boldsymbol{X}) \) is a function of random variables, -\( \widehat{\theta} \) itself must be a random variable. Thus it has -a pdf, call this function \( p(\boldsymbol{t}) \). The aim of the bootstrap is to -estimate \( p(\boldsymbol{t}) \) by the relative frequency of -\( \widehat{\theta} \). You can think of this as using a histogram -in the place of \( p(\boldsymbol{t}) \). If the relative frequency closely -resembles \( p(\vec{t}) \), then using numerics, it is straight forward to -estimate all the interesting parameters of \( p(\boldsymbol{t}) \) using point -estimators.
@@ -445,7 +453,7 @@ estimators.
-In the case that \( \widehat{\theta} \) has -more than one component, and the components are independent, we use the -same estimator on each component separately. If the probability -density function of \( X_i \), \( p(x) \), had been known, then it would have -been straight forward to do this by: - -
@@ -451,7 +447,7 @@ idea is to use the relative frequency of \( \widehat{\theta}^* \)
-But -unless there is enough information available about the process that -generated \( X_1,X_2,\cdots,X_n \), \( p(x) \) is in general -unknown. Therefore, Efron in 1979 asked the -question: What if we replace \( p(x) \) by the relative frequency -of the observation \( X_i \); if we draw observations in accordance with -the relative frequency of the observations, will we obtain the same -result in some asymptotic sense? The answer is yes. +In the case that \( \widehat{\theta} \) has +more than one component, and the components are independent, we use the +same estimator on each component separately. If the probability +density function of \( X_i \), \( p(x) \), had been known, then it would have +been straight forward to do this by: -
-Instead of generating the histogram for the relative -frequency of the observation \( X_i \), just draw the values -\( (X_1^*,X_2^*,\cdots,X_n^*) \) with replacement from the vector -\( \boldsymbol{X} \). +
@@ -450,7 +453,7 @@ frequency of the observation \( X_i \), just draw the values
-The independent bootstrap works like this: +But +unless there is enough information available about the process that +generated \( X_1,X_2,\cdots,X_n \), \( p(x) \) is in general +unknown. Therefore, Efron in 1979 asked the +question: What if we replace \( p(x) \) by the relative frequency +of the observation \( X_i \); if we draw observations in accordance with +the relative frequency of the observations, will we obtain the same +result in some asymptotic sense? The answer is yes. -
+Instead of generating the histogram for the relative +frequency of the observation \( X_i \), just draw the values +\( (X_1^*,X_2^*,\cdots,X_n^*) \) with replacement from the vector +\( \boldsymbol{X} \).
@@ -454,7 +452,7 @@ example, if you are interested in estimating the variance of \( \widehat
-The following code starts with a Gaussian distribution with mean value -\( \mu =100 \) and variance \( \sigma=15 \). We use this to generate the data -used in the bootstrap analysis. The bootstrap analysis returns a data -set after a given number of bootstrap operations (as many as we have -data points). This data set consists of estimated mean values for each -bootstrap operation. The histogram generated by the bootstrap method -shows that the distribution for these mean values is also a Gaussian, -centered around the mean value \( \mu=100 \) but with standard deviation -\( \sigma/\sqrt{n} \), where \( n \) is the number of bootstrap samples (in -this case the same as the number of original data points). The value -of the standard deviation is what we expect from the central limit -theorem. +The independent bootstrap works like this: -
+
from numpy import *
-from numpy.random import randint, randn
-from time import time
-import matplotlib.mlab as mlab
-import matplotlib.pyplot as plt
+When you are done, you can draw a histogram of the relative frequency
+of \( \widehat \theta^* \). This is your estimate of the probability
+distribution \( p(t) \). Using this probability distribution you can
+estimate any statistics thereof. In principle you never draw the
+histogram of the relative frequency of \( \widehat{\theta}^* \). Instead
+you use the estimators corresponding to the statistic of interest. For
+example, if you are interested in estimating the variance of \( \widehat
+\theta \), apply the etsimator \( \widehat \sigma^2 \) to the values
+\( \widehat \theta ^* \).
-# Returns mean of bootstrap samples
-def stat(data):
- return mean(data)
-
-# Bootstrap algorithm
-def bootstrap(data, statistic, R):
- t = zeros(R); n = len(data); inds = arange(n); t0 = time()
- # non-parametric bootstrap
- for i in range(R):
- t[i] = statistic(data[randint(0,n,n)])
-
- # analysis
- print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :")
- print("original bias std. error")
- print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t)))
- return t
-
-
-mu, sigma = 100, 15
-datapoints = 10000
-x = mu + sigma*random.randn(datapoints)
-# bootstrap returns the data sample
-t = bootstrap(x, stat, datapoints)
-# the histogram of the bootstrapped data
-n, binsboot, patches = plt.hist(t, 50, normed=1, facecolor='red', alpha=0.75)
-
-# add a 'best fit' line
-y = mlab.normpdf( binsboot, mean(t), std(t))
-lt = plt.plot(binsboot, y, 'r--', linewidth=1)
-plt.xlabel('Smarts')
-plt.ylabel('Probability')
-plt.axis([99.5, 100.6, 0, 3.0])
-plt.grid(True)
-
-plt.show()
-
@@ -493,7 +456,7 @@ plt.show()
- + -
-When the repetitive splitting of the data set is done randomly, -samples may accidently end up in a fast majority of the splits in -either training or test set. Such samples may have an unbalanced -influence on either model building or prediction evaluation. To avoid -this \( k \)-fold cross-validation structures the data splitting. The -samples are divided into \( k \) more or less equally sized exhaustive and -mutually exclusive subsets. In turn (at each split) one of these -subsets plays the role of the test set while the union of the -remaining subsets constitutes the training set. Such a splitting -warrants a balanced representation of each sample in both training and -test set over the splits. Still the division into the \( k \) subsets -involves a degree of randomness. This may be fully excluded when -choosing \( k=n \). This particular case is referred to as leave-one-out -cross-validation (LOOCV). +The following code starts with a Gaussian distribution with mean value +\( \mu =100 \) and variance \( \sigma=15 \). We use this to generate the data +used in the bootstrap analysis. The bootstrap analysis returns a data +set after a given number of bootstrap operations (as many as we have +data points). This data set consists of estimated mean values for each +bootstrap operation. The histogram generated by the bootstrap method +shows that the distribution for these mean values is also a Gaussian, +centered around the mean value \( \mu=100 \) but with standard deviation +\( \sigma/\sqrt{n} \), where \( n \) is the number of bootstrap samples (in +this case the same as the number of original data points). The value +of the standard deviation is what we expect from the central limit +theorem. +
+ + +
from numpy import *
+from numpy.random import randint, randn
+from time import time
+import matplotlib.mlab as mlab
+import matplotlib.pyplot as plt
+
+# Returns mean of bootstrap samples
+def stat(data):
+ return mean(data)
+
+# Bootstrap algorithm
+def bootstrap(data, statistic, R):
+ t = zeros(R); n = len(data); inds = arange(n); t0 = time()
+ # non-parametric bootstrap
+ for i in range(R):
+ t[i] = statistic(data[randint(0,n,n)])
+
+ # analysis
+ print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :")
+ print("original bias std. error")
+ print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t)))
+ return t
+
+
+mu, sigma = 100, 15
+datapoints = 10000
+x = mu + sigma*random.randn(datapoints)
+# bootstrap returns the data sample
+t = bootstrap(x, stat, datapoints)
+# the histogram of the bootstrapped data
+n, binsboot, patches = plt.hist(t, 50, normed=1, facecolor='red', alpha=0.75)
+
+# add a 'best fit' line
+y = mlab.normpdf( binsboot, mean(t), std(t))
+lt = plt.plot(binsboot, y, 'r--', linewidth=1)
+plt.xlabel('Smarts')
+plt.ylabel('Probability')
+plt.axis([99.5, 100.6, 0, 3.0])
+plt.grid(True)
+
+plt.show()
+
@@ -450,7 +495,7 @@ cross-validation (LOOCV).
+When the repetitive splitting of the data set is done randomly, +samples may accidently end up in a fast majority of the splits in +either training or test set. Such samples may have an unbalanced +influence on either model building or prediction evaluation. To avoid +this \( k \)-fold cross-validation structures the data splitting. The +samples are divided into \( k \) more or less equally sized exhaustive and +mutually exclusive subsets. In turn (at each split) one of these +subsets plays the role of the test set while the union of the +remaining subsets constitutes the training set. Such a splitting +warrants a balanced representation of each sample in both training and +test set over the splits. Still the division into the \( k \) subsets +involves a degree of randomness. This may be fully excluded when +choosing \( k=n \). This particular case is referred to as leave-one-out +cross-validation (LOOCV).
@@ -461,7 +452,7 @@ $$
- + -
-For the various values of \( k \) - -
diff --git a/doc/pub/Regression/html/._Regression-bs093.html b/doc/pub/Regression/html/._Regression-bs093.html index 2ee6416bb..4dec8d54e 100644 --- a/doc/pub/Regression/html/._Regression-bs093.html +++ b/doc/pub/Regression/html/._Regression-bs093.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
-The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial. -
+For the various values of \( k \) - -
import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.model_selection import KFold
-from sklearn.linear_model import Ridge
-from sklearn.model_selection import cross_val_score
-from sklearn.preprocessing import PolynomialFeatures
+
+- shuffle the dataset randomly.
+- Split the dataset into \( k \) groups.
+- For each unique group:
-# A seed just to ensure that the random numbers are the same for every run.
-# Useful for eventual debugging.
-np.random.seed(3155)
+
+- Decide which group to use as set for test data
+- Take the remaining groups as a training data set
+- Fit a model on the training set and evaluate it on the test set
+- Retain the evaluation score and discard the model
+
-# Generate the data.
-nsamples = 100
-x = np.random.randn(nsamples)
-y = 3*x**2 + np.random.randn(nsamples)
+
diff --git a/doc/pub/Regression/html/._Regression-bs094.html b/doc/pub/Regression/html/._Regression-bs094.html index 4d66729b5..81458d449 100644 --- a/doc/pub/Regression/html/._Regression-bs094.html +++ b/doc/pub/Regression/html/._Regression-bs094.html @@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source @@ -281,116 +282,117 @@ MathJax.Hub.Config({
-We will discuss the bias-variance tradeoff in the context of -continuous predictions such as regression. However, many of the -intuitions and ideas discussed here also carry over to classification -tasks. Consider a dataset \( \mathcal{L} \) consisting of the data -\( \mathbf{X}_\mathcal{L}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\} \). - +The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial.
-Let us assume that the true data is generated from a noisy model -$$ -\boldsymbol{y}=f(\boldsymbol{x}) + \boldsymbol{\epsilon} -$$ + +
import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.model_selection import KFold
+from sklearn.linear_model import Ridge
+from sklearn.model_selection import cross_val_score
+from sklearn.preprocessing import PolynomialFeatures
-
-where \( \epsilon \) is normally distributed with mean zero and standard deviation \( \sigma^2 \).
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(3155)
-
-In our derivation of the ordinary least squares method we defined then
-an approximation to the function \( f \) in terms of the parameters
-\( \boldsymbol{\beta} \) and the design matrix \( \boldsymbol{X} \) which embody our model,
-that is \( \boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\beta} \).
+# Generate the data.
+nsamples = 100
+x = np.random.randn(nsamples)
+y = 3*x**2 + np.random.randn(nsamples)
-
-Thereafter we found the parameters \( \boldsymbol{\beta} \) by optimizing the means squared error via the so-called cost function
-$$
-C(\boldsymbol{X},\boldsymbol{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right].
-$$
+## Cross-validation on Ridge regression using KFold only
-
-We can rewrite this as
-$$
-\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\sigma^2.
-$$
+# Decide degree on polynomial to fit
+poly = PolynomialFeatures(degree = 6)
-
-The three terms represent the square of the bias of the learning
-method, which can be thought of as the error caused by the simplifying
-assumptions built into the method. The second term represents the
-variance of the chosen model and finally the last terms is variance of
-the error \( \boldsymbol{\epsilon} \).
+# Decide which values of lambda to use
+nlambdas = 500
+lambdas = np.logspace(-3, 5, nlambdas)
-
-To derive this equation, we need to recall that the variance of \( \boldsymbol{y} \) and \( \boldsymbol{\epsilon} \) are both equal to \( \sigma^2 \). The mean value of \( \boldsymbol{\epsilon} \) is by definition equal to zero. Furthermore, the function \( f \) is not a stochastics variable, idem for \( \boldsymbol{\tilde{y}} \).
-We use a more compact notation in terms of the expectation value
-$$
-\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}})^2\right],
-$$
+# Initialize a KFold instance
+k = 5
+kfold = KFold(n_splits = k)
-and adding and subtracting \( \mathbb{E}\left[\boldsymbol{\tilde{y}}\right] \) we get
-$$
-\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}}+\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right],
-$$
+# Perform the cross-validation to estimate MSE
+scores_KFold = np.zeros((nlambdas, k))
-which, using the abovementioned expectation values can be rewritten as
-$$
-\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\boldsymbol{\tilde{y}}\right]+\sigma^2,
-$$
+i = 0
+for lmb in lambdas:
+ ridge = Ridge(alpha = lmb)
+ j = 0
+ for train_inds, test_inds in kfold.split(x):
+ xtrain = x[train_inds]
+ ytrain = y[train_inds]
-that is the rewriting in terms of the so-called bias, the variance of the model \( \boldsymbol{\tilde{y}} \) and the variance of \( \boldsymbol{\epsilon} \).
+ xtest = x[test_inds]
+ ytest = y[test_inds]
+ Xtrain = poly.fit_transform(xtrain[:, np.newaxis])
+ ridge.fit(Xtrain, ytrain[:, np.newaxis])
+
+ Xtest = poly.fit_transform(xtest[:, np.newaxis])
+ ypred = ridge.predict(Xtest)
+
+ scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)
+
+ j += 1
+ i += 1
+
+
+estimated_mse_KFold = np.mean(scores_KFold, axis = 1)
+
+## Cross-validation using cross_val_score from sklearn along with KFold
+
+# kfold is an instance initialized above as:
+# kfold = KFold(n_splits = k)
+
+estimated_mse_sklearn = np.zeros(nlambdas)
+i = 0
+for lmb in lambdas:
+ ridge = Ridge(alpha = lmb)
+
+ X = poly.fit_transform(x[:, np.newaxis])
+ estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)
+
+ # cross_val_score return an array containing the estimated negative mse for every fold.
+ # we have to the the mean of every array in order to get an estimate of the mse of the model
+ estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
+
+ i += 1
+
+## Plot and compare the slightly different ways to perform cross-validation
+
+plt.figure()
+
+plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
+plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')
+
+plt.xlabel('log10(lambda)')
+plt.ylabel('mse')
+
+plt.legend()
+
+plt.show()
+
@@ -495,7 +531,7 @@ that is the rewriting in terms of the so-called bias, the variance of the model
+We will discuss the bias-variance tradeoff in the context of +continuous predictions such as regression. However, many of the +intuitions and ideas discussed here also carry over to classification +tasks. Consider a dataset \( \mathcal{L} \) consisting of the data +\( \mathbf{X}_\mathcal{L}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\} \). - -
import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.linear_model import LinearRegression, Ridge, Lasso
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.model_selection import train_test_split
-from sklearn.pipeline import make_pipeline
-from sklearn.utils import resample
+
+Let us assume that the true data is generated from a noisy model
-np.random.seed(2018)
+$$
+\boldsymbol{y}=f(\boldsymbol{x}) + \boldsymbol{\epsilon}
+$$
-n = 500
-n_boostraps = 100
-degree = 18 # A quite high value, just to show.
-noise = 0.1
+
+where \( \epsilon \) is normally distributed with mean zero and standard deviation \( \sigma^2 \).
-# Make data set.
-x = np.linspace(-1, 3, n).reshape(-1, 1)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + np.random.normal(0, 0.1, x.shape)
+
+In our derivation of the ordinary least squares method we defined then
+an approximation to the function \( f \) in terms of the parameters
+\( \boldsymbol{\beta} \) and the design matrix \( \boldsymbol{X} \) which embody our model,
+that is \( \boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\beta} \).
-# Hold out some test data that is never used in training.
-x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
+
+Thereafter we found the parameters \( \boldsymbol{\beta} \) by optimizing the means squared error via the so-called cost function
+$$
+C(\boldsymbol{X},\boldsymbol{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right].
+$$
-# Combine x transformation and model into one operation.
-# Not neccesary, but convenient.
-model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
+
+We can rewrite this as
+$$
+\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\sigma^2.
+$$
-# The following (m x n_bootstraps) matrix holds the column vectors y_pred
-# for each bootstrap iteration.
-y_pred = np.empty((y_test.shape[0], n_boostraps))
-for i in range(n_boostraps):
- x_, y_ = resample(x_train, y_train)
+
+The three terms represent the square of the bias of the learning
+method, which can be thought of as the error caused by the simplifying
+assumptions built into the method. The second term represents the
+variance of the chosen model and finally the last terms is variance of
+the error \( \boldsymbol{\epsilon} \).
- # Evaluate the new model on the same test data each time.
- y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
+
+To derive this equation, we need to recall that the variance of \( \boldsymbol{y} \) and \( \boldsymbol{\epsilon} \) are both equal to \( \sigma^2 \). The mean value of \( \boldsymbol{\epsilon} \) is by definition equal to zero. Furthermore, the function \( f \) is not a stochastics variable, idem for \( \boldsymbol{\tilde{y}} \).
+We use a more compact notation in terms of the expectation value
+$$
+\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}})^2\right],
+$$
-# Note: Expectations and variances taken w.r.t. different training
-# data sets, hence the axis=1. Subsequent means are taken across the test data
-# set in order to obtain a total value, but before this we have error/bias/variance
-# calculated per data point in the test set.
-# Note 2: The use of keepdims=True is important in the calculation of bias as this
-# maintains the column vector form. Dropping this yields very unexpected results.
-error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
-bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
-variance = np.mean( np.var(y_pred, axis=1, keepdims=True) )
-print('Error:', error)
-print('Bias^2:', bias)
-print('Var:', variance)
-print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance))
+and adding and subtracting \( \mathbb{E}\left[\boldsymbol{\tilde{y}}\right] \) we get
+$$
+\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}}+\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right],
+$$
+
+which, using the abovementioned expectation values can be rewritten as
+$$
+\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\boldsymbol{\tilde{y}}\right]+\sigma^2,
+$$
+
+that is the rewriting in terms of the so-called bias, the variance of the model \( \boldsymbol{\tilde{y}} \) and the variance of \( \boldsymbol{\epsilon} \).
-plt.plot(x[::5, :], y[::5, :], label='f(x)')
-plt.scatter(x_test, y_test, label='Data points')
-plt.scatter(x_test, np.mean(y_pred, axis=1), label='Pred')
-plt.legend()
-plt.show()
-
@@ -491,7 +497,7 @@ plt.show()
@@ -420,40 +422,48 @@ MathJax.Hub.Config({ np.random.seed(2018) -n = 40 +n = 500 n_boostraps = 100 -maxdegree = 14 - +degree = 18 # A quite high value, just to show. +noise = 0.1 # Make data set. -x = np.linspace(-3, 3, n).reshape(-1, 1) -y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape) -error = np.zeros(maxdegree) -bias = np.zeros(maxdegree) -variance = np.zeros(maxdegree) -polydegree = np.zeros(maxdegree) +x = np.linspace(-1, 3, n).reshape(-1, 1) +y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + np.random.normal(0, 0.1, x.shape) + +# Hold out some test data that is never used in training. x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2) -for degree in range(maxdegree): - model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False)) - y_pred = np.empty((y_test.shape[0], n_boostraps)) - for i in range(n_boostraps): - x_, y_ = resample(x_train, y_train) - y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel() +# Combine x transformation and model into one operation. +# Not neccesary, but convenient. +model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False)) - polydegree[degree] = degree - error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) ) - bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 ) - variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) ) - print('Polynomial degree:', degree) - print('Error:', error[degree]) - print('Bias^2:', bias[degree]) - print('Var:', variance[degree]) - print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree])) +# The following (m x n_bootstraps) matrix holds the column vectors y_pred +# for each bootstrap iteration. +y_pred = np.empty((y_test.shape[0], n_boostraps)) +for i in range(n_boostraps): + x_, y_ = resample(x_train, y_train) -plt.plot(polydegree, error, label='Error') -plt.plot(polydegree, bias, label='bias') -plt.plot(polydegree, variance, label='Variance') + # Evaluate the new model on the same test data each time. + y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel() + +# Note: Expectations and variances taken w.r.t. different training +# data sets, hence the axis=1. Subsequent means are taken across the test data +# set in order to obtain a total value, but before this we have error/bias/variance +# calculated per data point in the test set. +# Note 2: The use of keepdims=True is important in the calculation of bias as this +# maintains the column vector form. Dropping this yields very unexpected results. +error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) ) +bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 ) +variance = np.mean( np.var(y_pred, axis=1, keepdims=True) ) +print('Error:', error) +print('Bias^2:', bias) +print('Var:', variance) +print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance)) + +plt.plot(x[::5, :], y[::5, :], label='f(x)') +plt.scatter(x_test, y_test, label='Data points') +plt.scatter(x_test, np.mean(y_pred, axis=1), label='Pred') plt.legend() plt.show()
- - -
-The bias-variance tradeoff summarizes the fundamental tension in -machine learning, particularly supervised learning, between the -complexity of a model and the amount of training data needed to train -it. Since data is often limited, in practice it is often useful to -use a less-complex model with higher bias, that is a model whose asymptotic -performance is worse than another model because it is easier to -train and less sensitive to sampling noise arising from having a -finite-sized training dataset (smaller variance). -
-The above equations tell us that in -order to minimize the expected test error, we need to select a -statistical learning method that simultaneously achieves low variance -and low bias. Note that variance is inherently a nonnegative quantity, -and squared bias is also nonnegative. Hence, we see that the expected -test MSE can never lie below \( Var(\epsilon) \), the irreducible error. + +
import matplotlib.pyplot as plt
+import numpy as np
+from sklearn.linear_model import LinearRegression, Ridge, Lasso
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.model_selection import train_test_split
+from sklearn.pipeline import make_pipeline
+from sklearn.utils import resample
-
-What do we mean by the variance and bias of a statistical learning
-method? The variance refers to the amount by which our model would change if we
-estimated it using a different training data set. Since the training
-data are used to fit the statistical learning method, different
-training data sets will result in a different estimate. But ideally the
-estimate for our model should not vary too much between training
-sets. However, if a method has high variance then small changes in
-the training data can result in large changes in the model. In general, more
-flexible statistical methods have higher variance.
+np.random.seed(2018)
-
-You may also find this recent article of interest.
+n = 40
+n_boostraps = 100
+maxdegree = 14
+
+# Make data set.
+x = np.linspace(-3, 3, n).reshape(-1, 1)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
+error = np.zeros(maxdegree)
+bias = np.zeros(maxdegree)
+variance = np.zeros(maxdegree)
+polydegree = np.zeros(maxdegree)
+x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
+
+for degree in range(maxdegree):
+ model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
+ y_pred = np.empty((y_test.shape[0], n_boostraps))
+ for i in range(n_boostraps):
+ x_, y_ = resample(x_train, y_train)
+ y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
+
+ polydegree[degree] = degree
+ error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
+ bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
+ variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
+ print('Polynomial degree:', degree)
+ print('Error:', error[degree])
+ print('Bias^2:', bias[degree])
+ print('Var:', variance[degree])
+ print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
+
+plt.plot(polydegree, error, label='Error')
+plt.plot(polydegree, bias, label='bias')
+plt.plot(polydegree, variance, label='Variance')
+plt.legend()
+plt.show()
+
@@ -466,7 +485,7 @@ You may also find this recent 106
- + + +
+The bias-variance tradeoff summarizes the fundamental tension in +machine learning, particularly supervised learning, between the +complexity of a model and the amount of training data needed to train +it. Since data is often limited, in practice it is often useful to +use a less-complex model with higher bias, that is a model whose asymptotic +performance is worse than another model because it is easier to +train and less sensitive to sampling noise arising from having a +finite-sized training dataset (smaller variance). - -
"""
-============================
-Underfitting vs. Overfitting
-============================
+
+The above equations tell us that in
+order to minimize the expected test error, we need to select a
+statistical learning method that simultaneously achieves low variance
+and low bias. Note that variance is inherently a nonnegative quantity,
+and squared bias is also nonnegative. Hence, we see that the expected
+test MSE can never lie below \( Var(\epsilon) \), the irreducible error.
-This example demonstrates the problems of underfitting and overfitting and
-how we can use linear regression with polynomial features to approximate
-nonlinear functions. The plot shows the function that we want to approximate,
-which is a part of the cosine function. In addition, the samples from the
-real function and the approximations of different models are displayed. The
-models have polynomial features of different degrees. We can see that a
-linear function (polynomial with degree 1) is not sufficient to fit the
-training samples. This is called **underfitting**. A polynomial of degree 4
-approximates the true function almost perfectly. However, for higher degrees
-the model will **overfit** the training data, i.e. it learns the noise of the
-training data.
-We evaluate quantitatively **overfitting** / **underfitting** by using
-cross-validation. We calculate the mean squared error (MSE) on the validation
-set, the higher, the less likely the model generalizes correctly from the
-training data.
-"""
+
+What do we mean by the variance and bias of a statistical learning
+method? The variance refers to the amount by which our model would change if we
+estimated it using a different training data set. Since the training
+data are used to fit the statistical learning method, different
+training data sets will result in a different estimate. But ideally the
+estimate for our model should not vary too much between training
+sets. However, if a method has high variance then small changes in
+the training data can result in large changes in the model. In general, more
+flexible statistical methods have higher variance.
-print(__doc__)
+
+You may also find this recent article of interest.
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.pipeline import Pipeline
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.linear_model import LinearRegression
-from sklearn.model_selection import cross_val_score
-
-
-def true_fun(X):
- return np.cos(1.5 * np.pi * X)
-
-np.random.seed(0)
-
-n_samples = 30
-degrees = [1, 4, 15]
-
-X = np.sort(np.random.rand(n_samples))
-y = true_fun(X) + np.random.randn(n_samples) * 0.1
-
-plt.figure(figsize=(14, 5))
-for i in range(len(degrees)):
- ax = plt.subplot(1, len(degrees), i + 1)
- plt.setp(ax, xticks=(), yticks=())
-
- polynomial_features = PolynomialFeatures(degree=degrees[i],
- include_bias=False)
- linear_regression = LinearRegression()
- pipeline = Pipeline([("polynomial_features", polynomial_features),
- ("linear_regression", linear_regression)])
- pipeline.fit(X[:, np.newaxis], y)
-
- # Evaluate the models using crossvalidation
- scores = cross_val_score(pipeline, X[:, np.newaxis], y,
- scoring="neg_mean_squared_error", cv=10)
-
- X_test = np.linspace(0, 1, 100)
- plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label="Model")
- plt.plot(X_test, true_fun(X_test), label="True function")
- plt.scatter(X, y, edgecolor='b', s=20, label="Samples")
- plt.xlabel("x")
- plt.ylabel("y")
- plt.xlim((0, 1))
- plt.ylim((-2, 2))
- plt.legend(loc="best")
- plt.title("Degree {}\nMSE = {:.2e}(+/- {:.2e})".format(
- degrees[i], -scores.mean(), scores.std()))
-plt.show()
-
@@ -508,7 +468,7 @@ plt.show()
-
# Common imports
-import os
+"""
+============================
+Underfitting vs. Overfitting
+============================
+
+This example demonstrates the problems of underfitting and overfitting and
+how we can use linear regression with polynomial features to approximate
+nonlinear functions. The plot shows the function that we want to approximate,
+which is a part of the cosine function. In addition, the samples from the
+real function and the approximations of different models are displayed. The
+models have polynomial features of different degrees. We can see that a
+linear function (polynomial with degree 1) is not sufficient to fit the
+training samples. This is called **underfitting**. A polynomial of degree 4
+approximates the true function almost perfectly. However, for higher degrees
+the model will **overfit** the training data, i.e. it learns the noise of the
+training data.
+We evaluate quantitatively **overfitting** / **underfitting** by using
+cross-validation. We calculate the mean squared error (MSE) on the validation
+set, the higher, the less likely the model generalizes correctly from the
+training data.
+"""
+
+print(__doc__)
+
import numpy as np
-import pandas as pd
import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression, Ridge, Lasso
-from sklearn.model_selection import train_test_split
-from sklearn.utils import resample
-from sklearn.metrics import mean_squared_error
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
+from sklearn.pipeline import Pipeline
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.linear_model import LinearRegression
+from sklearn.model_selection import cross_val_score
-if not os.path.exists(PROJECT_ROOT_DIR):
- os.mkdir(PROJECT_ROOT_DIR)
-if not os.path.exists(FIGURE_ID):
- os.makedirs(FIGURE_ID)
+def true_fun(X):
+ return np.cos(1.5 * np.pi * X)
-if not os.path.exists(DATA_ID):
- os.makedirs(DATA_ID)
+np.random.seed(0)
-def image_path(fig_id):
- return os.path.join(FIGURE_ID, fig_id)
+n_samples = 30
+degrees = [1, 4, 15]
-def data_path(dat_id):
- return os.path.join(DATA_ID, dat_id)
+X = np.sort(np.random.rand(n_samples))
+y = true_fun(X) + np.random.randn(n_samples) * 0.1
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+plt.figure(figsize=(14, 5))
+for i in range(len(degrees)):
+ ax = plt.subplot(1, len(degrees), i + 1)
+ plt.setp(ax, xticks=(), yticks=())
-infile = open(data_path("EoS.csv"),'r')
+ polynomial_features = PolynomialFeatures(degree=degrees[i],
+ include_bias=False)
+ linear_regression = LinearRegression()
+ pipeline = Pipeline([("polynomial_features", polynomial_features),
+ ("linear_regression", linear_regression)])
+ pipeline.fit(X[:, np.newaxis], y)
-# Read the EoS data as csv file and organize the data into two arrays with density and energies
-EoS = pd.read_csv(infile, names=('Density', 'Energy'))
-EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
-EoS = EoS.dropna()
-Energies = EoS['Energy']
-Density = EoS['Density']
-# The design matrix now as function of various polytrops
+ # Evaluate the models using crossvalidation
+ scores = cross_val_score(pipeline, X[:, np.newaxis], y,
+ scoring="neg_mean_squared_error", cv=10)
-Maxpolydegree = 30
-X = np.zeros((len(Density),Maxpolydegree))
-X[:,0] = 1.0
-testerror = np.zeros(Maxpolydegree)
-trainingerror = np.zeros(Maxpolydegree)
-polynomial = np.zeros(Maxpolydegree)
-
-trials = 100
-for polydegree in range(1, Maxpolydegree):
- polynomial[polydegree] = polydegree
- for degree in range(polydegree):
- X[:,degree] = Density**(degree/3.0)
-
-# loop over trials in order to estimate the expectation value of the MSE
- testerror[polydegree] = 0.0
- trainingerror[polydegree] = 0.0
- for samples in range(trials):
- x_train, x_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
- model = LinearRegression(fit_intercept=True).fit(x_train, y_train)
- ypred = model.predict(x_train)
- ytilde = model.predict(x_test)
- testerror[polydegree] += mean_squared_error(y_test, ytilde)
- trainingerror[polydegree] += mean_squared_error(y_train, ypred)
-
- testerror[polydegree] /= trials
- trainingerror[polydegree] /= trials
- print("Degree of polynomial: %3d"% polynomial[polydegree])
- print("Mean squared error on training data: %.8f" % trainingerror[polydegree])
- print("Mean squared error on test data: %.8f" % testerror[polydegree])
-
-plt.plot(polynomial, np.log10(trainingerror), label='Training Error')
-plt.plot(polynomial, np.log10(testerror), label='Test Error')
-plt.xlabel('Polynomial degree')
-plt.ylabel('log10[MSE]')
-plt.legend()
+ X_test = np.linspace(0, 1, 100)
+ plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label="Model")
+ plt.plot(X_test, true_fun(X_test), label="True function")
+ plt.scatter(X, y, edgecolor='b', s=20, label="Samples")
+ plt.xlabel("x")
+ plt.ylabel("y")
+ plt.xlim((0, 1))
+ plt.ylim((-2, 2))
+ plt.legend(loc="best")
+ plt.title("Degree {}\nMSE = {:.2e}(+/- {:.2e})".format(
+ degrees[i], -scores.mean(), scores.std()))
plt.show()
@@ -516,7 +510,7 @@ plt.show()
- + -
@@ -417,11 +419,9 @@ MathJax.Hub.Config({
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression, Ridge, Lasso
+from sklearn.model_selection import train_test_split
+from sklearn.utils import resample
from sklearn.metrics import mean_squared_error
-from sklearn.model_selection import KFold
-from sklearn.model_selection import cross_val_score
-
-
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
@@ -443,7 +443,7 @@ DATA_ID = "
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
@@ -458,22 +458,35 @@ Density = EoS[&
Maxpolydegree = 30
X = np.zeros((len(Density),Maxpolydegree))
X[:,0] = 1.0
-estimated_mse_sklearn = np.zeros(Maxpolydegree)
+testerror = np.zeros(Maxpolydegree)
+trainingerror = np.zeros(Maxpolydegree)
polynomial = np.zeros(Maxpolydegree)
-k =5
-kfold = KFold(n_splits = k)
+trials = 100
for polydegree in range(1, Maxpolydegree):
polynomial[polydegree] = polydegree
for degree in range(polydegree):
X[:,degree] = Density**(degree/3.0)
- OLS = LinearRegression()
-# loop over trials in order to estimate the expectation value of the MSE
- estimated_mse_folds = cross_val_score(OLS, X, Energies, scoring='neg_mean_squared_error', cv=kfold)
-#[:, np.newaxis]
- estimated_mse_sklearn[polydegree] = np.mean(-estimated_mse_folds)
-plt.plot(polynomial, np.log10(estimated_mse_sklearn), label='Test Error')
+# loop over trials in order to estimate the expectation value of the MSE
+ testerror[polydegree] = 0.0
+ trainingerror[polydegree] = 0.0
+ for samples in range(trials):
+ x_train, x_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
+ model = LinearRegression(fit_intercept=True).fit(x_train, y_train)
+ ypred = model.predict(x_train)
+ ytilde = model.predict(x_test)
+ testerror[polydegree] += mean_squared_error(y_test, ytilde)
+ trainingerror[polydegree] += mean_squared_error(y_train, ypred)
+
+ testerror[polydegree] /= trials
+ trainingerror[polydegree] /= trials
+ print("Degree of polynomial: %3d"% polynomial[polydegree])
+ print("Mean squared error on training data: %.8f" % trainingerror[polydegree])
+ print("Mean squared error on test data: %.8f" % testerror[polydegree])
+
+plt.plot(polynomial, np.log10(trainingerror), label='Training Error')
+plt.plot(polynomial, np.log10(testerror), label='Test Error')
plt.xlabel('Polynomial degree')
plt.ylabel('log10[MSE]')
plt.legend()
@@ -505,7 +518,7 @@ plt.show()
-
-The one-dimensional Ising model with nearest neighbor interaction, no
-external field and a constant coupling constant \( J \) is given by
-
-$$
-\begin{align}
- H = -J \sum_{k}^L s_k s_{k + 1},
-\tag{21}
-\end{align}
-$$
-
-
-where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins
-in the system is determined by \( L \). For the one-dimensional system
-there is no phase transition.
-
-
-We will look at a system of \( L = 40 \) spins with a coupling constant of
-\( J = 1 \). To get enough training data we will generate 10000 states
-with their respective energies.
-
+
-Here we use ordinary least squares
-regression to predict the energy for the nearest neighbor
-one-dimensional Ising model on a ring, i.e., the endpoints wrap
-around. We will use linear regression to fit a value for
-the coupling constant to achieve this.
-
@@ -484,6 +472,7 @@ the coupling constant to achieve this.
-A more general form for the one-dimensional Ising model is
+The one-dimensional Ising model with nearest neighbor interaction, no
+external field and a constant coupling constant \( J \) is given by
$$
\begin{align}
- H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
-\tag{22}
+ H = -J \sum_{k}^L s_k s_{k + 1},
+\tag{21}
\end{align}
$$
-Here we allow for interactions beyond the nearest neighbors and a state dependent
-coupling constant. This latter expression can be formulated as
-a matrix-product
-$$
-\begin{align}
- \boldsymbol{H} = \boldsymbol{X} J,
-\tag{23}
-\end{align}
-$$
+where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins
+in the system is determined by \( L \). For the one-dimensional system
+there is no phase transition.
-where \( X_{jk} = s_j s_k \) and \( J \) is a matrix which consists of the
-elements \( -J_{jk} \). This form of writing the energy fits perfectly
-with the form utilized in linear regression, that is
-
-$$
-\begin{align}
- \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon},
-\tag{24}
-\end{align}
-$$
-
-
-We split the data in training and test data as discussed in the previous example
+We will look at a system of \( L = 40 \) spins with a coupling constant of
+\( J = 1 \). To get enough training data we will generate 10000 states
+with their respective energies.
-
+Here we use ordinary least squares
+regression to predict the energy for the nearest neighbor
+one-dimensional Ising model on a ring, i.e., the endpoints wrap
+around. We will use linear regression to fit a value for
+the coupling constant to achieve this.
+
@@ -476,6 +485,7 @@ X_train, X_test, y_train, y_test = train_tes
-In the ordinary least squares method we choose the cost function
+A more general form for the one-dimensional Ising model is
$$
\begin{align}
- C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}.
-\tag{25}
+ H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
+\tag{22}
\end{align}
$$
-We then find the extremal point of \( C \) by taking the derivative with respect to \( \boldsymbol{\beta} \) as discussed above.
-This yields the expression for \( \boldsymbol{\beta} \) to be
-
+Here we allow for interactions beyond the nearest neighbors and a state dependent
+coupling constant. This latter expression can be formulated as
+a matrix-product
$$
- \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}},
+\begin{align}
+ \boldsymbol{H} = \boldsymbol{X} J,
+\tag{23}
+\end{align}
$$
-which immediately imposes some requirements on \( \boldsymbol{X} \) as there must exist
-an inverse of \( \boldsymbol{X}^T \boldsymbol{X} \). If the expression we are modeling contains an
-intercept, i.e., a constant term, we must make sure that the
-first column of \( \boldsymbol{X} \) consists of \( 1 \). We do this here
+where \( X_{jk} = s_j s_k \) and \( J \) is a matrix which consists of the
+elements \( -J_{jk} \). This form of writing the energy fits perfectly
+with the form utilized in linear regression, that is
+
+$$
+\begin{align}
+ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon},
+\tag{24}
+\end{align}
+$$
+
+
+We split the data in training and test data as discussed in the previous example
-
-
-
-
@@ -473,6 +477,7 @@ beta = ols_inv(X_train_own, y_train)
-Doing the inversion directly turns out to be a bad idea since the matrix
-\( \boldsymbol{X}^T\boldsymbol{X} \) is singular. An alternative approach is to use the singular
-value decomposition. Using the definition of the Moore-Penrose
-pseudoinverse we can write the equation for \( \boldsymbol{\beta} \) as
+In the ordinary least squares method we choose the cost function
-$$
- \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y},
-$$
-
-
-where the pseudoinverse of \( \boldsymbol{X} \) is given by
-
-$$
- \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}.
-$$
-
-
-Using singular value decomposition we can decompose the matrix \( \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T \),
-where \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal(unitary) matrices and \( \boldsymbol{\Sigma} \) contains the singular values (more details below).
-where \( X^{+} = V\Sigma^{+} U^T \). This reduces the equation for
-\( \omega \) to
$$
\begin{align}
- \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}.
-\tag{26}
+ C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}.
+\tag{25}
\end{align}
$$
-Note that solving this equation by actually doing the pseudoinverse
-(which is what we will do) is not a good idea as this operation scales
-as \( \mathcal{O}(n^3) \), where \( n \) is the number of elements in a
-general matrix. Instead, doing \( QR \)-factorization and solving the
-linear system as an equation would reduce this down to
-\( \mathcal{O}(n^2) \) operations.
+We then find the extremal point of \( C \) by taking the derivative with respect to \( \boldsymbol{\beta} \) as discussed above.
+This yields the expression for \( \boldsymbol{\beta} \) to be
+
+$$
+ \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}},
+$$
+
+
+which immediately imposes some requirements on \( \boldsymbol{X} \) as there must exist
+an inverse of \( \boldsymbol{X}^T \boldsymbol{X} \). If the expression we are modeling contains an
+intercept, i.e., a constant term, we must make sure that the
+first column of \( \boldsymbol{X} \) consists of \( 1 \). We do this here
-
-
-When extracting the \( J \)-matrix we need to make sure that we remove the intercept, as is done here
-
-
-
-
-
-A way of looking at the coefficients in \( J \) is to plot the matrices as images.
-
-
-
-
-
-It is interesting to note that OLS
-considers both \( J_{j, j + 1} = -0.5 \) and \( J_{j, j - 1} = -0.5 \) as
-valid matrix elements for \( J \).
-In our discussion below on hyperparameters and Ridge and Lasso regression we will see that
-this problem can be removed, partly and only with Lasso regression.
-
-
-In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
-
@@ -511,6 +474,7 @@ In this case our matrix inversion was actually possible. The obvious question no
-Let us bring back the Ising model again, but now with an additional
-focus on Ridge and Lasso regression as well. We repeat some of the
-basic parts of the Ising model and the setup of the training and test
-data. The one-dimensional Ising model with nearest neighbor
-interaction, no external field and a constant coupling constant \( J \) is
-given by
+Doing the inversion directly turns out to be a bad idea since the matrix
+\( \boldsymbol{X}^T\boldsymbol{X} \) is singular. An alternative approach is to use the singular
+value decomposition. Using the definition of the Moore-Penrose
+pseudoinverse we can write the equation for \( \boldsymbol{\beta} \) as
+$$
+ \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y},
+$$
+
+
+where the pseudoinverse of \( \boldsymbol{X} \) is given by
+
+$$
+ \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}.
+$$
+
+
+Using singular value decomposition we can decompose the matrix \( \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T \),
+where \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal(unitary) matrices and \( \boldsymbol{\Sigma} \) contains the singular values (more details below).
+where \( X^{+} = V\Sigma^{+} U^T \). This reduces the equation for
+\( \omega \) to
$$
\begin{align}
- H = -J \sum_{k}^L s_k s_{k + 1},
-\tag{27}
-\end{align}
-$$
-
-where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
-
-
-We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
-
-
-
-
-
-A more general form for the one-dimensional Ising model is
-
-$$
-\begin{align}
- H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
-\tag{28}
+ \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}.
+\tag{26}
\end{align}
$$
-Here we allow for interactions beyond the nearest neighbors and a more
-adaptive coupling matrix. This latter expression can be formulated as
-a matrix-product on the form
-$$
-\begin{align}
- H = X J,
-\tag{29}
-\end{align}
-$$
-
-
-where \( X_{jk} = s_j s_k \) and \( J \) is the matrix consisting of the
-elements \( -J_{jk} \). This form of writing the energy fits perfectly
-with the form utilized in linear regression, viz.
-$$
-\begin{align}
- \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}.
-\tag{30}
-\end{align}
-$$
-
-We organize the data as we did above
-
-
-
-
-We will do all fitting with Scikit-Learn,
+Note that solving this equation by actually doing the pseudoinverse
+(which is what we will do) is not a good idea as this operation scales
+as \( \mathcal{O}(n^3) \), where \( n \) is the number of elements in a
+general matrix. Instead, doing \( QR \)-factorization and solving the
+linear system as an equation would reduce this down to
+\( \mathcal{O}(n^2) \) operations.
-
-When extracting the \( J \)-matrix we make sure to remove the intercept
-
-
-And then we plot the results
+When extracting the \( J \)-matrix we need to make sure that we remove the intercept, as is done here
+
+
+
+
+
+A way of looking at the coefficients in \( J \) is to plot the matrices as images.
+
-The results perfectly with our previous discussion where we used our own code.
+It is interesting to note that OLS
+considers both \( J_{j, j + 1} = -0.5 \) and \( J_{j, j - 1} = -0.5 \) as
+valid matrix elements for \( J \).
+In our discussion below on hyperparameters and Ridge and Lasso regression we will see that
+this problem can be removed, partly and only with Lasso regression.
+
+
+In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
@@ -557,6 +512,7 @@ The results perfectly with our previous discussion where we used our own code.
-Having explored the ordinary least squares we move on to ridge
-regression. In ridge regression we include a regularizer. This
-involves a new cost function which leads to a new estimate for the
-weights \( \boldsymbol{\beta} \). This results in a penalized regression problem. The
-cost function is given by
+Let us bring back the Ising model again, but now with an additional
+focus on Ridge and Lasso regression as well. We repeat some of the
+basic parts of the Ising model and the setup of the training and test
+data. The one-dimensional Ising model with nearest neighbor
+interaction, no external field and a constant coupling constant \( J \) is
+given by
$$
\begin{align}
- C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}.
-\tag{31}
+ H = -J \sum_{k}^L s_k s_{k + 1},
+\tag{27}
\end{align}
$$
+where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
+
+
+We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
+
-
+A more general form for the one-dimensional Ising model is
+
+$$
+\begin{align}
+ H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
+\tag{28}
+\end{align}
+$$
+
+
+Here we allow for interactions beyond the nearest neighbors and a more
+adaptive coupling matrix. This latter expression can be formulated as
+a matrix-product on the form
+$$
+\begin{align}
+ H = X J,
+\tag{29}
+\end{align}
+$$
+
+
+where \( X_{jk} = s_j s_k \) and \( J \) is the matrix consisting of the
+elements \( -J_{jk} \). This form of writing the energy fits perfectly
+with the form utilized in linear regression, viz.
+$$
+\begin{align}
+ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}.
+\tag{30}
+\end{align}
+$$
+
+We organize the data as we did above
+
+
+
+
+We will do all fitting with Scikit-Learn,
+
+
+
+
+
+When extracting the \( J \)-matrix we make sure to remove the intercept
+
+
+
+
+And then we plot the results
+
+
+
+
+The results perfectly with our previous discussion where we used our own code.
+
@@ -457,6 +558,7 @@ plt.show()
-In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function.
+Having explored the ordinary least squares we move on to ridge
+regression. In ridge regression we include a regularizer. This
+involves a new cost function which leads to a new estimate for the
+weights \( \boldsymbol{\beta} \). This results in a penalized regression problem. The
+cost function is given by
$$
\begin{align}
- C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \sqrt{\boldsymbol{\beta}^T\boldsymbol{\beta}}.
-\tag{32}
+ C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}.
+\tag{31}
\end{align}
$$
-
-Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from Scikit-Learn.
-
-
-It is quite striking how LASSO breaks the symmetry of the coupling
-constant as opposed to ridge and OLS. We get a sparse solution with
-\( J_{j, j + 1} = -1 \).
-
@@ -459,6 +458,7 @@ constant as opposed to ridge and OLS. We get a sparse solution with
-We see how the different models perform for a different set of values for \( \lambda \).
+In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function.
+
+$$
+\begin{align}
+ C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \sqrt{\boldsymbol{\beta}^T\boldsymbol{\beta}}.
+\tag{32}
+\end{align}
+$$
+
+
+Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from Scikit-Learn.
-
-We see that LASSO reaches a good solution for low
-values of \( \lambda \), but will "wither" when we increase \( \lambda \) too
-much. Ridge is more stable over a larger range of values for
-\( \lambda \), but eventually also fades away.
+It is quite striking how LASSO breaks the symmetry of the coupling
+constant as opposed to ridge and OLS. We get a sparse solution with
+\( J_{j, j + 1} = -1 \).
@@ -474,6 +460,7 @@ much. Ridge is more stable over a larger range of values for
-To determine which value of \( \lambda \) is best we plot the accuracy of
-the models when predicting the training and the testing set. We expect
-the accuracy of the training set to be quite good, but if the accuracy
-of the testing set is much lower this tells us that we might be
-subject to an overfit model. The ideal scenario is an accuracy on the
-testing set that is close to the accuracy of the training set.
+We see how the different models perform for a different set of values for \( \lambda \).
-
-From the above figure we can see that LASSO with \( \lambda = 10^{-2} \)
-achieves a very good accuracy on the test set. This by far surpasses the
-other models for all values of \( \lambda \).
+We see that LASSO reaches a good solution for low
+values of \( \lambda \), but will "wither" when we increase \( \lambda \) too
+much. Ridge is more stable over a larger range of values for
+\( \lambda \), but eventually also fades away.
-
-In order to judge which model performs best at varying values of \( \lambda \) (for ridge and LASSO) we compute \( R^2 \) which is given by
-
-$$
-\begin{align}
- R^2 = 1 - \frac{(y - \hat{y})^2}{(y - \bar{y})^2},
-\tag{45}
-\end{align}
-$$
-
-where \( y \) is a vector with the true values of the energy, \( \hat{y} \) is the predicted values of \( y \) from the models and \( \bar{y} \) is the mean of \( \hat{y} \).
+To determine which value of \( \lambda \) is best we plot the accuracy of
+the models when predicting the training and the testing set. We expect
+the accuracy of the training set to be quite good, but if the accuracy
+of the testing set is much lower this tells us that we might be
+subject to an overfit model. The ideal scenario is an accuracy on the
+testing set that is close to the accuracy of the training set.
-
-This is the same metric used by Scikit-learn for their regression models when scoring.
+From the above figure we can see that LASSO with \( \lambda = 10^{-2} \)
+achieves a very good accuracy on the test set. This by far surpasses the
+other models for all values of \( \lambda \).
+
-
-
-
@@ -449,7 +451,7 @@ MathJax.Hub.Config({
Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
@@ -183,7 +201,7 @@ Similarly, Mehta et a
@@ -206,7 +224,7 @@ A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\ver
@@ -236,7 +254,7 @@ Linear regression gives us a set of analytical equations for the parameters \( \
@@ -266,7 +284,7 @@ so-called General linear models
+
@@ -288,7 +306,7 @@ where \( \epsilon_i \) is the error in our approximation.
@@ -309,7 +327,7 @@ $$
@@ -361,7 +379,7 @@ The above design matrix is called a Generalizing the fitting procedure as a linear algebra problem
+
@@ -392,7 +410,7 @@ $$
@@ -423,7 +441,7 @@ The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\e
@@ -453,7 +471,7 @@ our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predict
In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
@@ -491,7 +509,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("MassEval2016.dat"),'r')
@@ -501,7 +519,7 @@ Masses = pd.read_fwf(infile, usecols=(2,
@@ -589,7 +607,7 @@ since when taking the first derivative with respect to the unknown parameters \(
@@ -654,7 +672,7 @@ $$
@@ -702,7 +720,7 @@ allow for the usage of direct linear algebra methods such as LU decomposi
The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and
@@ -735,7 +753,7 @@ $$
@@ -771,7 +789,7 @@ Let us now return to our nuclear binding energies and simply code the above equa
It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to
@@ -782,14 +800,14 @@ write
Alternatively, you can use the least squares functionality in Numpy as
-
@@ -814,7 +832,7 @@ plt.show()
We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of _Scikit_Learn_ in the introductory slides.
@@ -830,7 +848,7 @@ and we would be using it as
-
We can easily add our MSE score as
@@ -841,7 +859,7 @@ We can easily add our MSE score as
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
-print(MSE(Energies,ytilde))
+print(MSE(Energies,ytilde))
and finally the relative error as
@@ -850,7 +868,7 @@ and finally the relative error as
We could also add the so-called Huber norm, which we defined as
@@ -865,7 +883,7 @@ with \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
@@ -896,7 +914,7 @@ where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i
@@ -927,7 +945,7 @@ where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\
@@ -956,7 +974,7 @@ $$
@@ -992,7 +1010,7 @@ $$
@@ -1021,7 +1039,7 @@ $$
@@ -1085,7 +1103,7 @@ Lasso and Ridge regression. See below.
Before we continue, let us introduce yet another example. We are going to fit the
@@ -1110,7 +1128,7 @@ hyperparameter \( \lambda \), also to be explained below.
@@ -1145,7 +1163,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
@@ -1168,12 +1186,12 @@ clf = skl.LinearRegression().fit(X, Energies)
ytilde = clf.predict(X)
EoS['Eols'] = ytilde
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, ytilde))
+print('Variance score: %.2f' % r2_score(Energies, ytilde))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
-print(clf.coef_, clf.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
+print(clf.coef_, clf.intercept_)
# The Ridge regression with a hyperparameter lambda = 0.1
_lambda = 0.1
@@ -1181,12 +1199,12 @@ clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
yridge = clf_ridge.predict(X)
EoS['Eridge'] = yridge
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
+print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, yridge))
+print('Variance score: %.2f' % r2_score(Energies, yridge))
# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
-print(clf_ridge.coef_, clf_ridge.intercept_)
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
+print(clf_ridge.coef_, clf_ridge.intercept_)
fig, ax = plt.subplots()
ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
@@ -1213,7 +1231,7 @@ below.
It is normal in essentially all Machine Learning studies to split the
@@ -1256,7 +1274,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
def R2(y_data, y_model):
return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
@@ -1284,22 +1302,22 @@ X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=# matrix inversion to find beta
beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
# and then make the prediction
-ytilde = X_train @ beta
-print("Training R2")
-print(R2(y_train,ytilde))
-print("Training MSE")
-print(MSE(y_train,ytilde))
-ypredict = X_test @ beta
-print("Test R2")
-print(R2(y_test,ypredict))
-print("Test MSE")
-print(MSE(y_test,ypredict))
+ytilde = X_train @ beta
+print("Training R2")
+print(R2(y_train,ytilde))
+print("Training MSE")
+print(MSE(y_train,ytilde))
+ypredict = X_test @ beta
+print("Test R2")
+print(R2(y_test,ypredict))
+print("Test MSE")
+print(MSE(y_test,ypredict))
The Boston housing
@@ -1331,7 +1349,7 @@ The features/predictors are
@@ -1394,7 +1412,7 @@ It is now useful to look at the correlation matrix
correlation_matrix = boston.corr().round(2)
# use the heatmap function from seaborn to plot the correlation matrix
# annot = True to print the values inside the square
-sns.heatmap(data=correlation_matrix, annot=True)
+sns.heatmap(data=correlation_matrix, annot=True)
From the above coorelation plot we can see that MEDV is strongly correlated to LSTAT and RM. We see also that RAD and TAX are stronly correlated, but we don't include this in our features together to avoid multi-colinearity
@@ -1435,10 +1453,10 @@ We split the data into training and test sets
# splits the training and test data set in 80% : 20%
# assign random_state to any value.This ensures consistency.
X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state=5)
-print(X_train.shape)
-print(X_test.shape)
-print(Y_train.shape)
-print(Y_test.shape)
+print(X_train.shape)
+print(X_test.shape)
+print(Y_train.shape)
+print(Y_test.shape)
Then we use the linear regression functionality from Scikit-Learn
@@ -1457,11 +1475,11 @@ y_train_predict = lin_model.predict(X_train)
rmse = (np.sqrt(mean_squared_error(Y_train, y_train_predict)))
r2 = r2_score(Y_train, y_train_predict)
-print("The model performance for training set")
-print("--------------------------------------")
-print('RMSE is {}'.format(rmse))
-print('R2 score is {}'.format(r2))
-print("\n")
+print("The model performance for training set")
+print("--------------------------------------")
+print('RMSE is {}'.format(rmse))
+print('R2 score is {}'.format(r2))
+print("\n")
# model evaluation for testing set
@@ -1472,10 +1490,10 @@ rmse = (np.sqrt(mean_squared_error(Y_test, y_test_predict)))
# r-squared score of the model
r2 = r2_score(Y_test, y_test_predict)
-print("The model performance for testing set")
-print("--------------------------------------")
-print('RMSE is {}'.format(rmse))
-print('R2 score is {}'.format(r2))
+print("The model performance for testing set")
+print("--------------------------------------")
+print('RMSE is {}'.format(rmse))
+print('R2 score is {}'.format(r2))
@@ -1489,7 +1507,7 @@ plt.show()
One of the typical problems we encounter with linear regression, in particular
@@ -1576,7 +1594,7 @@ This is equivalent to saying that the matrix \( \boldsymbol{X} \) has at least a
If our design matrix \( \boldsymbol{X} \) which enters the linear regression problem
@@ -1608,7 +1626,7 @@ where \( \boldsymbol{I} \) is the identity matrix. When we discuss Ridge
From standard linear algebra we know that a square matrix \( \boldsymbol{X} \) can be diagonalized if and only it is
@@ -1655,7 +1673,7 @@ is not diagonalizable, it is a so-called The SVD, a Fantastic Algorithm
+
However, and this is the strength of the SVD algorithm, any general
@@ -1690,7 +1708,7 @@ The SVD exits always!
Consider the following matrix which can be SVD decomposed as
@@ -1730,7 +1748,7 @@ The columns of \( \boldsymbol{U} \) are called the left singular vectors while t
If we assume that \( n > p \), then our matrix \( \boldsymbol{U} \) has dimension \( n
@@ -1755,7 +1773,7 @@ In general the economy-size SVD leads to less FLOPS and still conserving the des
There are several interesting mathematical properties which will be
@@ -1819,7 +1837,7 @@ We will come back to this expression when we discuss Ridge regression.
Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is
@@ -1891,7 +1909,7 @@ $$
Using the matrix-vector expression for Ridge regression,
@@ -1962,7 +1980,7 @@ with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \
Since \( \lambda \geq 0 \), it means that compared to OLS, we have
@@ -1988,7 +2006,7 @@ With a parameter \( \lambda \) we can thus shrink the role of specific parameter
For the sake of simplicity, let us assume that the design matrix is orthonormal, that is
@@ -2031,7 +2049,7 @@ Similarly, Mehta et a
@@ -2048,9 +2066,9 @@ Similarly, Mehta et a
# print( (np.transpose(U) @ U - U @np.transpose(U)))
# print('test VT')
# print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))
- print(U)
- print(s)
- print(VT)
+ print(U)
+ print(s)
+ print(VT)
D = np.zeros((len(U),len(VT)))
for i in range(0,len(VT)):
@@ -2060,14 +2078,14 @@ Similarly, Mehta et a
X = np.array([ [1.0, -1.0, 2.0], [1.0, 0.0, 1.0], [1.0, 2.0, -1.0], [1.0, 1.0, 0.0] ])
-print(X)
-A = np.transpose(X) @ X
-print(A)
+print(X)
+A = np.transpose(X) @ X
+print(A)
# Brute force inversion of super-collinear matrix
#B = np.linalg.inv(A)
#print(B)
C = SVDinv(A)
-print(C)
+print(C)
The matrix \( \boldsymbol{X} \) has columns that are linearly dependent. The first
@@ -2082,7 +2100,7 @@ in the program terminating due to a singular matrix.
The parameter \( \lambda \) that we have introduced in the Ridge (and
@@ -2101,7 +2119,7 @@ affected by changing the parameter \( \lambda \).
We have our design matrix
@@ -2123,7 +2141,7 @@ The matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are unitary/orthonorm
More material to be added here
@@ -2131,7 +2149,7 @@ More material to be added here
Before we proceed, we need to rethink what we have been doing. In our
@@ -2149,7 +2167,7 @@ This will allow us to link the standard linear algebra methods we have discussed
@@ -2182,7 +2200,7 @@ cross-validation and the bootstrap method.
@@ -2208,7 +2226,7 @@ bootstrap is widely used.
@@ -2275,7 +2293,7 @@ selection of a large set of these numbers reproduces this PDF.
@@ -2301,7 +2319,7 @@ $$
@@ -2342,7 +2360,7 @@ qualitatively as the spread of \( p \) around its mean.
@@ -2376,7 +2394,7 @@ $$
@@ -2409,15 +2427,15 @@ $$
-Suppose we have defined three vectors \( \hat{x}, \hat{y}, \hat{z} \) with
+Suppose we have defined three vectors \( \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z} \) with
\( n \) elements each. The covariance matrix is defined as
The following simple function uses the np.vstack function which
takes each vector of dimension \( 1\times n \) and produces a \( 3\times n \)
-matrix \( \hat{W} \)
+matrix \( \boldsymbol{W} \)
which in turn is converted into into the \( 3\times 3 \) covariance matrix
-\( \hat{\Sigma} \) via the Numpy function np.cov(). We note that we can
-also calculate the mean value of each set of samples \( \hat{x} \) etc
+\( \boldsymbol{\Sigma} \) via the Numpy function np.cov(). We note that we can
+also calculate the mean value of each set of samples \( \boldsymbol{x} \) etc
using the Numpy function np.mean(x). We can also extract the
eigenvalues of the covariance matrix through the np.linalg.eig()
function.
@@ -2464,7 +2482,7 @@ function.
@@ -2474,16 +2492,16 @@ function.
n = 100
x = np.random.normal(size=n)
-print(np.mean(x))
+print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
-print(np.mean(y))
+print(np.mean(y))
z = x**3+np.random.normal(size=n)
-print(np.mean(z))
+print(np.mean(z))
W = np.vstack((x, y, z))
Sigma = np.cov(W)
-print(Sigma)
+print(Sigma)
Eigvals, Eigvecs = np.linalg.eig(Sigma)
-print(Eigvals)
+print(Eigvals)
@@ -2492,9 +2510,9 @@ Eigvals, Eigvecs = np.linalg.eig(Sigma)
import matplotlib.pyplot as plt
from scipy import sparse
eye = np.eye(4)
-print(eye)
+print(eye)
sparse_mtx = sparse.csr_matrix(eye)
-print(sparse_mtx)
+print(sparse_mtx)
x = np.linspace(-10,10,100)
y = np.sin(x)
plt.plot(x,y,marker='x')
@@ -2504,7 +2522,7 @@ plt.show()
@@ -2533,7 +2551,7 @@ $$
@@ -2570,7 +2588,7 @@ value of a set of measurements.
@@ -2599,7 +2617,7 @@ interested in finding the few lowest moments, like the mean
@@ -2630,7 +2648,7 @@ $$
@@ -2650,7 +2668,7 @@ and covariance \( \mathrm{cov}(X,Y) \).
@@ -2683,7 +2701,7 @@ true PDFs behind, which we usually do not have.
@@ -2706,7 +2724,7 @@ means.
@@ -2728,7 +2746,7 @@ And in particular we are interested in its variance \( \mathrm{var}(\overline X_
@@ -2754,7 +2772,7 @@ $$
@@ -2784,7 +2802,7 @@ estimate of the PDF of each of the \( X_i \), estimating all properties of
@@ -2816,7 +2834,7 @@ $$
@@ -2860,7 +2878,7 @@ measurements in the sample.
@@ -2896,7 +2914,7 @@ cannot overlook the always present correlations.
@@ -2928,7 +2946,7 @@ measurements. For uncorrelated measurements this second term is zero.
@@ -2951,7 +2969,7 @@ have to be stored throughout the experiment.
@@ -2989,7 +3007,7 @@ starting always at \( 1 \) for \( d=0 \).
@@ -3025,7 +3043,7 @@ $$
@@ -3051,7 +3069,7 @@ measurements is very large.
Finally, we are going to discuss several statistical properties which can be obtained in terms of analytical expressions.
@@ -3091,7 +3109,7 @@ row number \( i \) and perform a sum over all values \( p \).
The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off)
@@ -3115,7 +3133,7 @@ $$
We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \)
@@ -3154,7 +3172,7 @@ mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (n
With the OLS expressions for the parameters \( \boldsymbol{\beta} \) we can evaluate the expectation value
@@ -3202,7 +3220,7 @@ where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) =
\sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2
\, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the
variance of the estimate of the \( j \)-th regression coefficient:
-\( \hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{
+\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 \sqrt{
[(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} } \). This may be used to
construct a confidence interval for the estimates.
@@ -3249,7 +3267,7 @@ This means the variance we obtain with the standard OLS will always for \( \lamb
With all these analytical equations for both the OLS and Ridge
@@ -3279,7 +3297,7 @@ training error reaches a saturation.
Two famous
@@ -3302,7 +3320,7 @@ need for bootstrapping.
The Jackknife works by making many replicas of the estimator \( \widehat{\theta} \).
@@ -3323,7 +3341,7 @@ number \( i \) is left out. Using this notation, define
@@ -3338,9 +3356,9 @@ number \( i \) is left out. Using this notation, define
t[i] = stat(delete(data,i) )
# analysis
- print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :")
- print("original bias std. error")
- print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))
+ print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :")
+ print("original bias std. error")
+ print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))
return t
@@ -3360,7 +3378,7 @@ t = jackknife(x, stat)
@@ -3382,7 +3400,7 @@ advantages:
Since \( \widehat{\theta} = \widehat{\theta}(\boldsymbol{X}) \) is a function of random variables,
@@ -3398,7 +3416,7 @@ estimators.
In the case that \( \widehat{\theta} \) has
@@ -3421,7 +3439,7 @@ idea is to use the relative frequency of \( \widehat{\theta}^* \)
But
@@ -3442,7 +3460,7 @@ frequency of the observation \( X_i \), just draw the values
The independent bootstrap works like this:
@@ -3468,7 +3486,7 @@ example, if you are interested in estimating the variance of \( \widehat
The following code starts with a Gaussian distribution with mean value
@@ -3505,9 +3523,9 @@ theorem.
t[i] = statistic(data[randint(0,n,n)])
# analysis
- print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :")
- print("original bias std. error")
- print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t)))
+ print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :")
+ print("original bias std. error")
+ print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t)))
return t
@@ -3525,7 +3543,7 @@ lt = plt.plot(binsboot, y, 'r--', li
plt.xlabel('Smarts')
plt.ylabel('Probability')
plt.axis([99.5, 100.6, 0, 3.0])
-plt.grid(True)
+plt.grid(True)
plt.show()
When the repetitive splitting of the data set is done randomly,
@@ -3554,7 +3572,7 @@ cross-validation (LOOCV).
For the various values of \( k \)
@@ -3610,7 +3628,7 @@ For the various values of \( k \)
The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial.
@@ -3711,7 +3729,7 @@ plt.show()
We will discuss the bias-variance tradeoff in the context of
@@ -3789,7 +3807,7 @@ that is the rewriting in terms of the so-called bias, the variance of the model
@@ -3817,7 +3835,7 @@ x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=# Combine x transformation and model into one operation.
# Not neccesary, but convenient.
-model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
+model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
# The following (m x n_bootstraps) matrix holds the column vectors y_pred
# for each bootstrap iteration.
@@ -3834,13 +3852,13 @@ y_pred = np.empty((y_test.shape[0], n_boostr
# calculated per data point in the test set.
# Note 2: The use of keepdims=True is important in the calculation of bias as this
# maintains the column vector form. Dropping this yields very unexpected results.
-error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
-bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
-variance = np.mean( np.var(y_pred, axis=1, keepdims=True) )
-print('Error:', error)
-print('Bias^2:', bias)
-print('Var:', variance)
-print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance))
+error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
+bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
+variance = np.mean( np.var(y_pred, axis=1, keepdims=True) )
+print('Error:', error)
+print('Bias^2:', bias)
+print('Var:', variance)
+print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance))
plt.plot(x[::5, :], y[::5, :], label='f(x)')
plt.scatter(x_test, y_test, label='Data points')
@@ -3852,7 +3870,7 @@ plt.show()
@@ -3881,21 +3899,21 @@ polydegree = np.zeros(maxdegree)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
for degree in range(maxdegree):
- model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
+ model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
y_pred = np.empty((y_test.shape[0], n_boostraps))
for i in range(n_boostraps):
x_, y_ = resample(x_train, y_train)
y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
polydegree[degree] = degree
- error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
- bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
- variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
- print('Polynomial degree:', degree)
- print('Error:', error[degree])
- print('Bias^2:', bias[degree])
- print('Var:', variance[degree])
- print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
+ error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
+ bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
+ variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
+ print('Polynomial degree:', degree)
+ print('Error:', error[degree])
+ print('Bias^2:', bias[degree])
+ print('Var:', variance[degree])
+ print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
plt.plot(polydegree, error, label='Error')
plt.plot(polydegree, bias, label='bias')
@@ -3907,7 +3925,7 @@ plt.show()
The bias-variance tradeoff summarizes the fundamental tension in
@@ -3944,7 +3962,7 @@ You may also find this recent Another Example from Scikit-Learn's Repository
+
@@ -3970,7 +3988,7 @@ You may also find this recent training data.
"""
-print(__doc__)
+print(__doc__)
import numpy as np
import matplotlib.pyplot as plt
@@ -3997,7 +4015,7 @@ plt.figure(figsize=(14, False)
+ include_bias=False)
linear_regression = LinearRegression()
pipeline = Pipeline([("polynomial_features", polynomial_features),
("linear_regression", linear_regression)])
@@ -4024,7 +4042,7 @@ plt.show()
@@ -4059,7 +4077,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
@@ -4089,7 +4107,7 @@ trials = 100
trainingerror[polydegree] = 0.0
for samples in range(trials):
x_train, x_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
- model = LinearRegression(fit_intercept=True).fit(x_train, y_train)
+ model = LinearRegression(fit_intercept=True).fit(x_train, y_train)
ypred = model.predict(x_train)
ytilde = model.predict(x_test)
testerror[polydegree] += mean_squared_error(y_test, ytilde)
@@ -4097,9 +4115,9 @@ trials = 100
testerror[polydegree] /= trials
trainingerror[polydegree] /= trials
- print("Degree of polynomial: %3d"% polynomial[polydegree])
- print("Mean squared error on training data: %.8f" % trainingerror[polydegree])
- print("Mean squared error on test data: %.8f" % testerror[polydegree])
+ print("Degree of polynomial: %3d"% polynomial[polydegree])
+ print("Mean squared error on training data: %.8f" % trainingerror[polydegree])
+ print("Mean squared error on test data: %.8f" % testerror[polydegree])
plt.plot(polynomial, np.log10(trainingerror), label='Training Error')
plt.plot(polynomial, np.log10(testerror), label='Test Error')
@@ -4112,7 +4130,7 @@ plt.show()
@@ -4149,7 +4167,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
@@ -4189,7 +4207,7 @@ plt.show()
@@ -4233,7 +4251,7 @@ plt.show()
The one-dimensional Ising model with nearest neighbor interaction, no
@@ -4268,7 +4286,7 @@ with their respective energies.
import scipy.linalg as scl
from sklearn.model_selection import train_test_split
import tqdm
-sns.set(color_codes=True)
+sns.set(color_codes=True)
cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
L = 40
@@ -4292,7 +4310,7 @@ the coupling constant to achieve this.
A more general form for the one-dimensional Ising model is
@@ -4349,7 +4367,7 @@ X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=
-
In the ordinary least squares method we choose the cost function
@@ -4395,14 +4413,14 @@ X_test_own = np.concatenate(
Doing the inversion directly turns out to be a bad idea since the matrix
@@ -4452,7 +4470,7 @@ linear system as an equation would reduce this down to
@@ -4495,7 +4513,7 @@ In this case our matrix inversion was actually possible. The obvious question no
Let us bring back the Ising model again, but now with an additional
@@ -4530,7 +4548,7 @@ We will look at a system of \( L = 40 \) spins with a coupling constant of \( J
from sklearn.model_selection import train_test_split
import sklearn.linear_model as skl
import tqdm
-sns.set(color_codes=True)
+sns.set(color_codes=True)
cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
L = 40
@@ -4637,7 +4655,7 @@ The results perfectly with our previous discussion where we used our own code.
Having explored the ordinary least squares we move on to ridge
@@ -4675,7 +4693,7 @@ plt.show()
In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function.
@@ -4715,7 +4733,7 @@ constant as opposed to ridge and OLS. We get a sparse solution with
We see how the different models perform for a different set of values for \( \lambda \).
@@ -4769,7 +4787,7 @@ much. Ridge is more stable over a larger range of values for
To determine which value of \( \lambda \) is best we plot the accuracy of
diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html
index 648f5afa1..41cd94417 100644
--- a/doc/pub/Regression/html/Regression-solarized.html
+++ b/doc/pub/Regression/html/Regression-solarized.html
@@ -61,24 +61,21 @@ div { text-align: justify; text-justify: inter-word; }
-
Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
@@ -333,7 +351,7 @@ Similarly, Mehta et a
@@ -358,7 +376,7 @@ A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\ver
@@ -391,7 +409,7 @@ Linear regression gives us a set of analytical equations for the parameters \( \
@@ -420,7 +438,7 @@ so-called General linear models
+
@@ -441,7 +459,7 @@ where \( \epsilon_i \) is the error in our approximation.
@@ -461,7 +479,7 @@ $$
@@ -504,7 +522,7 @@ The above design matrix is called a Generalizing the fitting procedure as a linear algebra problem
+
@@ -536,7 +554,7 @@ $$
@@ -564,7 +582,7 @@ The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\e
@@ -593,7 +611,7 @@ our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predict
In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
@@ -631,7 +649,7 @@ DATA_ID = "DataFiles/"
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
+ plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("MassEval2016.dat"),'r')
@@ -641,7 +659,7 @@ Masses = pd.read_fwf(infile, usecols=(2,
@@ -720,7 +738,7 @@ since when taking the first derivative with respect to the unknown parameters \(
@@ -776,7 +794,7 @@ $$
@@ -820,7 +838,7 @@ allow for the usage of direct linear algebra methods such as LU decomposi
The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and
@@ -844,7 +862,7 @@ $$
@@ -875,7 +893,7 @@ Let us now return to our nuclear binding energies and simply code the above equa
It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to
@@ -886,14 +904,14 @@ write
Alternatively, you can use the least squares functionality in Numpy as
-
@@ -917,7 +935,7 @@ plt.show()
We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of _Scikit_Learn_ in the introductory slides.
@@ -933,7 +951,7 @@ and we would be using it as
-The same example but now with cross-validation
-Cross-validation with Ridge
import numpy as np
+
# Common imports
+import os
+import numpy as np
+import pandas as pd
import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression, Ridge, Lasso
+from sklearn.metrics import mean_squared_error
from sklearn.model_selection import KFold
-from sklearn.linear_model import Ridge
from sklearn.model_selection import cross_val_score
-from sklearn.preprocessing import PolynomialFeatures
-# A seed just to ensure that the random numbers are the same for every run.
-np.random.seed(3155)
-# Generate the data.
-n = 100
-x = np.linspace(-3, 3, n).reshape(-1, 1)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
-# Decide degree on polynomial to fit
-poly = PolynomialFeatures(degree = 10)
-# Decide which values of lambda to use
-nlambdas = 500
-lambdas = np.logspace(-3, 5, nlambdas)
-# Initialize a KFold instance
-k = 5
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organize the data into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+
+Maxpolydegree = 30
+X = np.zeros((len(Density),Maxpolydegree))
+X[:,0] = 1.0
+estimated_mse_sklearn = np.zeros(Maxpolydegree)
+polynomial = np.zeros(Maxpolydegree)
+k =5
kfold = KFold(n_splits = k)
-estimated_mse_sklearn = np.zeros(nlambdas)
-i = 0
-for lmb in lambdas:
- ridge = Ridge(alpha = lmb)
- estimated_mse_folds = cross_val_score(ridge, x, y, scoring='neg_mean_squared_error', cv=kfold)
- estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
- i += 1
-plt.figure()
-plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
-plt.xlabel('log10(lambda)')
-plt.ylabel('MSE')
+
+for polydegree in range(1, Maxpolydegree):
+ polynomial[polydegree] = polydegree
+ for degree in range(polydegree):
+ X[:,degree] = Density**(degree/3.0)
+ OLS = LinearRegression()
+# loop over trials in order to estimate the expectation value of the MSE
+ estimated_mse_folds = cross_val_score(OLS, X, Energies, scoring='neg_mean_squared_error', cv=kfold)
+#[:, np.newaxis]
+ estimated_mse_sklearn[polydegree] = np.mean(-estimated_mse_folds)
+
+plt.plot(polynomial, np.log10(estimated_mse_sklearn), label='Test Error')
+plt.xlabel('Polynomial degree')
+plt.ylabel('log10[MSE]')
plt.legend()
plt.show()
The Ising model
-
-Cross-validation with Ridge
import numpy as np
import matplotlib.pyplot as plt
-from mpl_toolkits.axes_grid1 import make_axes_locatable
-import seaborn as sns
-import scipy.linalg as scl
-from sklearn.model_selection import train_test_split
-import tqdm
-sns.set(color_codes=True)
-cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+from sklearn.model_selection import KFold
+from sklearn.linear_model import Ridge
+from sklearn.model_selection import cross_val_score
+from sklearn.preprocessing import PolynomialFeatures
-L = 40
-n = int(1e4)
+# A seed just to ensure that the random numbers are the same for every run.
+np.random.seed(3155)
+# Generate the data.
+n = 100
+x = np.linspace(-3, 3, n).reshape(-1, 1)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
+# Decide degree on polynomial to fit
+poly = PolynomialFeatures(degree = 10)
-spins = np.random.choice([-1, 1], size=(n, L))
-J = 1.0
-
-energies = np.zeros(n)
-
-for i in range(n):
- energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
+# Decide which values of lambda to use
+nlambdas = 500
+lambdas = np.logspace(-3, 5, nlambdas)
+# Initialize a KFold instance
+k = 5
+kfold = KFold(n_splits = k)
+estimated_mse_sklearn = np.zeros(nlambdas)
+i = 0
+for lmb in lambdas:
+ ridge = Ridge(alpha = lmb)
+ estimated_mse_folds = cross_val_score(ridge, x, y, scoring='neg_mean_squared_error', cv=kfold)
+ estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
+ i += 1
+plt.figure()
+plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
Reformulating the problem to suit regression
+The Ising model
X = np.zeros((n, L ** 2))
+
import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.axes_grid1 import make_axes_locatable
+import seaborn as sns
+import scipy.linalg as scl
+from sklearn.model_selection import train_test_split
+import tqdm
+sns.set(color_codes=True)
+cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+
+L = 40
+n = int(1e4)
+
+spins = np.random.choice([-1, 1], size=(n, L))
+J = 1.0
+
+energies = np.zeros(n)
+
for i in range(n):
- X[i] = np.outer(spins[i], spins[i]).ravel()
-y = energies
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+ energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
Linear regression
+Reformulating the problem to suit regression
X_train_own = np.concatenate(
- (np.ones(len(X_train))[:, np.newaxis], X_train),
- axis=1
-)
-X_test_own = np.concatenate(
- (np.ones(len(X_test))[:, np.newaxis], X_test),
- axis=1
-)
-
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
- return scl.inv(x.T @ x) @ (x.T @ y)
-beta = ols_inv(X_train_own, y_train)
+
X = np.zeros((n, L ** 2))
+for i in range(n):
+ X[i] = np.outer(spins[i], spins[i]).ravel()
+y = energies
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
Singular Value decomposition
+Linear regression
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
- u, s, v = scl.svd(x)
- return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
+
X_train_own = np.concatenate(
+ (np.ones(len(X_train))[:, np.newaxis], X_train),
+ axis=1
+)
+X_test_own = np.concatenate(
+ (np.ones(len(X_test))[:, np.newaxis], X_test),
+ axis=1
+)
beta = ols_svd(X_train_own,y_train)
+
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
+ return scl.inv(x.T @ x) @ (x.T @ y)
+beta = ols_inv(X_train_own, y_train)
J = beta[1:].reshape(L, L)
-
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J, **cmap_args)
-plt.title("OLS", fontsize=18)
-plt.xticks(fontsize=18)
-plt.yticks(fontsize=18)
-cb = fig.colorbar(im)
-cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
-plt.show()
-
The one-dimensional Ising model
+Singular Value decomposition
import numpy as np
-import matplotlib.pyplot as plt
-from mpl_toolkits.axes_grid1 import make_axes_locatable
-import seaborn as sns
-import scipy.linalg as scl
-from sklearn.model_selection import train_test_split
-import sklearn.linear_model as skl
-import tqdm
-sns.set(color_codes=True)
-cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
-
-L = 40
-n = int(1e4)
-
-spins = np.random.choice([-1, 1], size=(n, L))
-J = 1.0
-
-energies = np.zeros(n)
-
-for i in range(n):
- energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
-
X = np.zeros((n, L ** 2))
-for i in range(n):
- X[i] = np.outer(spins[i], spins[i]).ravel()
-y = energies
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
-
-X_train_own = np.concatenate(
- (np.ones(len(X_train))[:, np.newaxis], X_train),
- axis=1
-)
-
-X_test_own = np.concatenate(
- (np.ones(len(X_test))[:, np.newaxis], X_test),
- axis=1
-)
-
clf = skl.LinearRegression().fit(X_train, y_train)
+
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
+ u, s, v = scl.svd(x)
+ return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
J_sk = clf.coef_.reshape(L, L)
+
beta = ols_svd(X_train_own,y_train)
J = beta[1:].reshape(L, L)
+
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_sk, **cmap_args)
-plt.title("LinearRegression from Scikit-learn", fontsize=18)
+im = plt.imshow(J, **cmap_args)
+plt.title("OLS", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
@@ -535,7 +483,14 @@ cb.ax.se
plt.show()
Ridge regression
+The one-dimensional Ising model
_lambda = 0.1
-clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
-J_ridge_sk = clf_ridge.coef_.reshape(L, L)
-fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_ridge_sk, **cmap_args)
-plt.title("Ridge from Scikit-learn", fontsize=18)
+
import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.axes_grid1 import make_axes_locatable
+import seaborn as sns
+import scipy.linalg as scl
+from sklearn.model_selection import train_test_split
+import sklearn.linear_model as skl
+import tqdm
+sns.set(color_codes=True)
+cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+
+L = 40
+n = int(1e4)
+
+spins = np.random.choice([-1, 1], size=(n, L))
+J = 1.0
+
+energies = np.zeros(n)
+
+for i in range(n):
+ energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
+
X = np.zeros((n, L ** 2))
+for i in range(n):
+ X[i] = np.outer(spins[i], spins[i]).ravel()
+y = energies
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
+
+X_train_own = np.concatenate(
+ (np.ones(len(X_train))[:, np.newaxis], X_train),
+ axis=1
+)
+
+X_test_own = np.concatenate(
+ (np.ones(len(X_test))[:, np.newaxis], X_test),
+ axis=1
+)
+
clf = skl.LinearRegression().fit(X_train, y_train)
+
J_sk = clf.coef_.reshape(L, L)
+
fig = plt.figure(figsize=(20, 14))
+im = plt.imshow(J_sk, **cmap_args)
+plt.title("LinearRegression from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
-
plt.show()
LASSO regression
+Ridge regression
clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
-J_lasso_sk = clf_lasso.coef_.reshape(L, L)
+
_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
+J_ridge_sk = clf_ridge.coef_.reshape(L, L)
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_lasso_sk, **cmap_args)
-plt.title("Lasso from Scikit-learn", fontsize=18)
+im = plt.imshow(J_ridge_sk, **cmap_args)
+plt.title("Ridge from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
@@ -436,11 +440,6 @@ cb.ax.se
plt.show()
Performance as function of the regularization parameter
+LASSO regression
lambdas = np.logspace(-4, 5, 10)
-
-train_errors = {
- "ols_sk": np.zeros(lambdas.size),
- "ridge_sk": np.zeros(lambdas.size),
- "lasso_sk": np.zeros(lambdas.size)
-}
-
-test_errors = {
- "ols_sk": np.zeros(lambdas.size),
- "ridge_sk": np.zeros(lambdas.size),
- "lasso_sk": np.zeros(lambdas.size)
-}
-
-plot_counter = 1
-
-fig = plt.figure(figsize=(32, 54))
-
-for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
- for key, method in zip(
- ["ols_sk", "ridge_sk", "lasso_sk"],
- [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
- ):
- method = method.fit(X_train, y_train)
-
- train_errors[key][i] = method.score(X_train, y_train)
- test_errors[key][i] = method.score(X_test, y_test)
-
- omega = method.coef_.reshape(L, L)
-
- plt.subplot(10, 5, plot_counter)
- plt.imshow(omega, **cmap_args)
- plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
- plot_counter += 1
+
clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
+J_lasso_sk = clf_lasso.coef_.reshape(L, L)
+fig = plt.figure(figsize=(20, 14))
+im = plt.imshow(J_lasso_sk, **cmap_args)
+plt.title("Lasso from Scikit-learn", fontsize=18)
+plt.xticks(fontsize=18)
+plt.yticks(fontsize=18)
+cb = fig.colorbar(im)
+cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
Finding the optimal value of \( \lambda \)
+Performance as function of the regularization parameter
fig = plt.figure(figsize=(20, 14))
+
diff --git a/doc/pub/Regression/html/._Regression-bs111.html b/doc/pub/Regression/html/._Regression-bs111.html
index e9064994b..c218d38b8 100644
--- a/doc/pub/Regression/html/._Regression-bs111.html
+++ b/doc/pub/Regression/html/._Regression-bs111.html
@@ -6,6 +6,7 @@ Automatically generated HTML file from DocOnce source
+
lambdas = np.logspace(-4, 5, 10)
-colors = {
- "ols_sk": "r",
- "ridge_sk": "y",
- "lasso_sk": "c"
+train_errors = {
+ "ols_sk": np.zeros(lambdas.size),
+ "ridge_sk": np.zeros(lambdas.size),
+ "lasso_sk": np.zeros(lambdas.size)
}
-for key in train_errors:
- plt.semilogx(
- lambdas,
- train_errors[key],
- colors[key],
- label="Train {0}".format(key),
- linewidth=4.0
- )
+test_errors = {
+ "ols_sk": np.zeros(lambdas.size),
+ "ridge_sk": np.zeros(lambdas.size),
+ "lasso_sk": np.zeros(lambdas.size)
+}
+
+plot_counter = 1
+
+fig = plt.figure(figsize=(32, 54))
+
+for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
+ for key, method in zip(
+ ["ols_sk", "ridge_sk", "lasso_sk"],
+ [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
+ ):
+ method = method.fit(X_train, y_train)
+
+ train_errors[key][i] = method.score(X_train, y_train)
+ test_errors[key][i] = method.score(X_test, y_test)
+
+ omega = method.coef_.reshape(L, L)
+
+ plt.subplot(10, 5, plot_counter)
+ plt.imshow(omega, **cmap_args)
+ plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
+ plot_counter += 1
-for key in test_errors:
- plt.semilogx(
- lambdas,
- test_errors[key],
- colors[key] + "--",
- label="Test {0}".format(key),
- linewidth=4.0
- )
-plt.legend(loc="best", fontsize=18)
-plt.xlabel(r"$\lambda$", fontsize=18)
-plt.ylabel(r"$R^2$", fontsize=18)
-plt.tick_params(labelsize=18)
plt.show()
Performance of the different models
+Finding the optimal value of \( \lambda \)
def r_squared(y, y_hat):
- return 1 - np.sum((y - y_hat) ** 2) / np.sum((y - np.mean(y_hat)) ** 2)
+
diff --git a/doc/pub/Regression/html/Regression-bs.html b/doc/pub/Regression/html/Regression-bs.html
index 383bb6110..ee4ea89d0 100644
--- a/doc/pub/Regression/html/Regression-bs.html
+++ b/doc/pub/Regression/html/Regression-bs.html
@@ -41,24 +41,21 @@ Automatically generated HTML file from DocOnce source
@@ -281,116 +282,117 @@ MathJax.Hub.Config({
fig = plt.figure(figsize=(20, 14))
+
+colors = {
+ "ols_sk": "r",
+ "ridge_sk": "y",
+ "lasso_sk": "c"
+}
+
+for key in train_errors:
+ plt.semilogx(
+ lambdas,
+ train_errors[key],
+ colors[key],
+ label="Train {0}".format(key),
+ linewidth=4.0
+ )
+
+for key in test_errors:
+ plt.semilogx(
+ lambdas,
+ test_errors[key],
+ colors[key] + "--",
+ label="Test {0}".format(key),
+ linewidth=4.0
+ )
+plt.legend(loc="best", fontsize=18)
+plt.xlabel(r"$\lambda$", fontsize=18)
+plt.ylabel(r"$R^2$", fontsize=18)
+plt.tick_params(labelsize=18)
+plt.show()
y_hat = clf.predict(X_test)
-r_test = r_squared(y_test, y_hat)
-sk_r_test = clf.score(X_test, y_test)
-
-assert abs(r_test - sk_r_test) < 1e-2
-
Dec 24, 2019
Aug 19, 2020
-Dec 24, 2019
Aug 19, 2020
Why Linear Regression (aka Ordinary Least Squares and family)
+To do list
+
+
+
+Why Linear Regression (aka Ordinary Least Squares and family)
Regression analysis, overarching aims
+Regression analysis, overarching aims
Regression analysis, overarching aims II
+Regression analysis, overarching aims II
Examples
+Examples
General linear models
Rewriting the fitting procedure as a linear algebra problem
+Rewriting the fitting procedure as a linear algebra problem
Rewriting the fitting procedure as a linear algebra problem, more details
+Rewriting the fitting procedure as a linear algebra problem, more details
Generalizing the fitting procedure as a linear algebra problem
Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
Optimizing our parameters
+Optimizing our parameters
Our model for the nuclear binding energies
+Our model for the nuclear binding energies
Optimizing our parameters, more details
+Optimizing our parameters, more details
Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
Some useful matrix and vector expressions
+Some useful matrix and vector expressions
Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
Own code for Ordinary Least Squares
+Own code for Ordinary Least Squares
# matrix inversion to find beta
beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
# and then make the prediction
-ytilde = X @ beta
+ytilde = X @ beta
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
ytildenp = np.dot(fit,X.T)
Adding error analysis and training set up
+Adding error analysis and training set up
print(R2(Energies,ytilde))
+
print(R2(Energies,ytilde))
def RelativeError(y_data,y_model):
return abs((y_data-y_model)/y_data)
-print(RelativeError(Energies, ytilde))
+print(RelativeError(Energies, ytilde))
The \( \chi^2 \) function
+The \( \chi^2 \) function
The \( \chi^2 \) function
+The \( \chi^2 \) function
The \( \chi^2 \) function
+The \( \chi^2 \) function
The \( \chi^2 \) function
+The \( \chi^2 \) function
The \( \chi^2 \) function
+The \( \chi^2 \) function
The \( \chi^2 \) function
+The \( \chi^2 \) function
Fitting an Equation of State for Dense Nuclear Matter
+Fitting an Equation of State for Dense Nuclear Matter
The code
+The code
Splitting our Data in Training and Test data
+Splitting our Data in Training and Test data
The Boston housing data example
+The Boston housing data example
Housing data, the code
+Housing data, the code
We start by importing the libraries
The singular value decomposition
+The singular value decomposition
Linear Regression Problems
+Linear Regression Problems
Fixing the singularity
+Fixing the singularity
Basic math of the SVD
+Basic math of the SVD
The SVD, a Fantastic Algorithm
Another Example
+Another Example
Economy-size SVD
+Economy-size SVD
Mathematical Properties
+Mathematical Properties
Ridge and LASSO Regression
+Ridge and LASSO Regression
More on Ridge Regression
+More on Ridge Regression
Interpreting the Ridge results
+Interpreting the Ridge results
More interpretations
+More interpretations
Codes for the SVD
+Codes for the SVD
A better understanding of regularization
+A better understanding of regularization
Decomposing the OLS and Ridge expressions
+Decomposing the OLS and Ridge expressions
Spectral Decomposition of the OLS
+Spectral Decomposition of the OLS
Where are we going?
+Where are we going?
Resampling methods
+Resampling methods
Resampling approaches can be computationally expensive
+Resampling approaches can be computationally expensive
Why resampling methods ?
+Why resampling methods ?
@@ -2221,7 +2239,7 @@ bootstrap is widely used.
Statistical analysis
+Statistical analysis
@@ -2241,7 +2259,7 @@ bootstrap is widely used.
Statistics
+Statistics
Statistics, moments
+Statistics, moments
Statistics, central moments
+Statistics, central moments
Statistics, covariance
+Statistics, covariance
Statistics, more covariance
+Statistics, more covariance
Covariance example
+Covariance example
$$
-\hat{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
+\boldsymbol{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
\sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\
\sigma_{zx} & \sigma_{zy} & \sigma_{zz}
\end{bmatrix},
@@ -2439,11 +2457,11 @@ the exact mean valu\ es.
$$
-\hat{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\
+\boldsymbol{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
\dots & \dots & \dots \\
@@ -2455,8 +2473,8 @@ $$
Covariance in numpy
+Covariance in numpy
Statistics, independent variables
+Statistics, independent variables
Statistics, more variance
+Statistics, more variance
Statistics and stochastic processes
+Statistics and stochastic processes
Statistics and sample variables
+Statistics and sample variables
Statistics, sample variance and covariance
+Statistics, sample variance and covariance
Statistics, law of large numbers
+Statistics, law of large numbers
Statistics, more on sample error
+Statistics, more on sample error
Statistics
+Statistics
Statistics, central limit theorem
+Statistics, central limit theorem
Statistics, more technicalities
+Statistics, more technicalities
Statistics
+Statistics
Statistics and sample variance
+Statistics and sample variance
Statistics, uncorrelated results
+Statistics, uncorrelated results
Statistics, computations
+Statistics, computations
Statistics, more on computations of errors
+Statistics, more on computations of errors
Statistics, wrapping up 1
+Statistics, wrapping up 1
Statistics, final expression
+Statistics, final expression
Statistics, effective number of correlations
+Statistics, effective number of correlations
Linking the regression analysis with a statistical interpretation
+Linking the regression analysis with a statistical interpretation
Assumptions made
+Assumptions made
Expectation value and variance
+Expectation value and variance
Expectation value and variance for \( \boldsymbol{\beta} \)
+Expectation value and variance for \( \boldsymbol{\beta} \)
Resampling methods
+Resampling methods
Resampling methods: Jackknife and Bootstrap
+Resampling methods: Jackknife and Bootstrap
Resampling methods: Jackknife
+Resampling methods: Jackknife
Jackknife code example
+Jackknife code example
Resampling methods: Bootstrap
+Resampling methods: Bootstrap
Resampling methods: Bootstrap background
+Resampling methods: Bootstrap background
Resampling methods: More Bootstrap background
+Resampling methods: More Bootstrap background
Resampling methods: Bootstrap approach
+Resampling methods: Bootstrap approach
Resampling methods: Bootstrap steps
+Resampling methods: Bootstrap steps
Code example for the Bootstrap method
+Code example for the Bootstrap method
Various steps in cross-validation
+Various steps in cross-validation
How to set up the cross-validation for Ridge and/or Lasso
+How to set up the cross-validation for Ridge and/or Lasso
Cross-validation in brief
+Cross-validation in brief
Code Example for Cross-validation and \( k \)-fold Cross-validation
+Code Example for Cross-validation and \( k \)-fold Cross-validation
The bias-variance tradeoff
+The bias-variance tradeoff
Example code for Bias-Variance tradeoff
+Example code for Bias-Variance tradeoff
Understanding what happens
+Understanding what happens
Summing up
+Summing up
Another Example from Scikit-Learn's Repository
More examples on bootstrap and cross-validation and errors
+More examples on bootstrap and cross-validation and errors
The same example but now with cross-validation
+The same example but now with cross-validation
Cross-validation with Ridge
+Cross-validation with Ridge
The Ising model
+The Ising model
Reformulating the problem to suit regression
+Reformulating the problem to suit regression
Linear regression
+Linear regression
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
- return scl.inv(x.T @ x) @ (x.T @ y)
+ return scl.inv(x.T @ x) @ (x.T @ y)
beta = ols_inv(X_train_own, y_train)
Singular Value decomposition
+Singular Value decomposition
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
u, s, v = scl.svd(x)
- return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
+ return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
The one-dimensional Ising model
+The one-dimensional Ising model
Ridge regression
+Ridge regression
LASSO regression
+LASSO regression
Performance as function of the regularization parameter
+Performance as function of the regularization parameter
Finding the optimal value of \( \lambda \)
+Finding the optimal value of \( \lambda \)
Dec 24, 2019
Aug 19, 2020
-Why Linear Regression (aka Ordinary Least Squares and family)
+To do list
+
+
+
+
+
+
+Why Linear Regression (aka Ordinary Least Squares and family)
-Regression analysis, overarching aims
+Regression analysis, overarching aims
-Regression analysis, overarching aims II
+Regression analysis, overarching aims II
-Examples
+Examples
General linear models
-Rewriting the fitting procedure as a linear algebra problem
+Rewriting the fitting procedure as a linear algebra problem
-Rewriting the fitting procedure as a linear algebra problem, more details
+Rewriting the fitting procedure as a linear algebra problem, more details
Generalizing the fitting procedure as a linear algebra problem
-Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
-Optimizing our parameters
+Optimizing our parameters
-Our model for the nuclear binding energies
+Our model for the nuclear binding energies
-Optimizing our parameters, more details
+Optimizing our parameters, more details
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
-Some useful matrix and vector expressions
+Some useful matrix and vector expressions
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
-Own code for Ordinary Least Squares
+Own code for Ordinary Least Squares
# matrix inversion to find beta
beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
# and then make the prediction
-ytilde = X @ beta
+ytilde = X @ beta
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
ytildenp = np.dot(fit,X.T)
-Adding error analysis and training set up
+Adding error analysis and training set up
print(R2(Energies,ytilde))
+