@@ -409,6 +411,12 @@ on the other hand \( \boldsymbol{y} \) is a scalar, the Jacobian becomes a
\( 1\times n \) matrix.
+
When this matrix is a square matrix \( m=n \), its determinant is often referred to as the Jacobian
+determinant. Both the matrix and (if \( m=n \)) the determinant are
+often referred to simply as the Jacobian. The Jacobian matrix represents the differential of \( \boldsymbol{y} \) at every point where the
+vector is differentiable.
+
+
@@ -434,7 +442,7 @@ on the other hand \( \boldsymbol{y} \) is a scalar, the Jacobian becomes a
A very important matrix we will meet again and again in Machine
-Learning is the Hessian. It is given by the second derivative of the
-cost function with respect to the parameter \( \beta \). Using the above
-expression for derivatives of vectors and matrices, we find that the
-second derivative of the cost function is,
+
We define a scalar (our cast functions are in general also scalars, think of the mean squared error) as the result of some matrix vector multiplications
with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors.
+In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first
+\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then
For ordinary least squares, it is inversely proportional (derivation
-next week) with the variance of the optimal parameters
-\( \hat{\boldsymbol{\beta}} \). Furthermore, we will see later this week that is
-(beside \( 1/n \)) equal to the covariance matrix. It plays also a very
-important role in optmization algorithms and Principal Component
-Analysis as a way to reduce the dimensionality of a machine learning
-problem.
-
+
Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
Linear algebra question: Can we use the Hessian matrix to say something about properties of the cost function (our optmization problem)? (hint: think about convex or concave problems and how to relate these to a matrix!).
The residuals \( \boldsymbol{\epsilon} \) are in turn given by
+
Meet the Hessian Matrix
+
+
A very important matrix we will meet again and again in Machine
+Learning is the Hessian. It is given by the second derivative of the
+cost function with respect to the parameter \( \beta \). Using the above
+expression for derivatives of vectors and matrices, we find that the
+second derivative of the cost function is,
+
meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
-
-
+
For ordinary least squares, it is inversely proportional (derivation
+next week) with the variance of the optimal parameters
+\( \hat{\boldsymbol{\beta}} \). Furthermore, we will see later this week that is
+(beside \( 1/n \)) equal to the covariance matrix. It plays also a very
+important role in optmization algorithms and Principal Component
+Analysis as a way to reduce the dimensionality of a machine learning
+problem.
+
+
Linear algebra question: Can we use the Hessian matrix to say something about properties of the cost function (our optmization problem)? (hint: think about convex or concave problems and how to relate these to a matrix!).
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
-
meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
-
-
-
-
-
-
-
-
-
-
-
-
-
-
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
-
-
-
-
-
-
-
-
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()
-
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
@@ -403,8 +405,10 @@ Since we are not using Scikit-Learn here we can define our own \( R2 \) f
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.
+
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
importos
-importnumpyasnp
-importpandasaspd
-importmatplotlib.pyplotasplt
-fromsklearn.model_selectionimport train_test_split
-
-
-defR2(y_data, y_model):
- return1- np.sum((y_data - y_model) **2) / np.sum((y_data - np.mean(y_data)) **2)
-defMSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-x = np.random.rand(100)
-y =2.0+5*x*x+0.1*np.random.randn(100)
-
-
-# The design matrix now as function of a given polynomial
-X = np.zeros((len(x),3))
-X[:,0] =1.0
-X[:,1] = x
-X[:,2] = x**2
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-# matrix inversion to find beta
-beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
-print(beta)
-# 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))
-
+
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.
+
importos
+importnumpyasnp
+importpandasaspd
+importmatplotlib.pyplotasplt
+fromsklearn.model_selectionimport train_test_split
- np.random.shuffle(inputs_shuffled)
- np.random.shuffle(labels_shuffled)
- train_end =int(n_inputs*train_size)
- X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]
- Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]
+defR2(y_data, y_model):
+ return1- np.sum((y_data - y_model) **2) / np.sum((y_data - np.mean(y_data)) **2)
+defMSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
- return X_train, X_test, Y_train, Y_test
+x = np.random.rand(100)
+y =2.0+5*x*x+0.1*np.random.randn(100)
+
+
+# The design matrix now as function of a given polynomial
+X = np.zeros((len(x),3))
+X[:,0] =1.0
+X[:,1] = x
+X[:,2] = x**2
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
+print(beta)
+# 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))
@@ -429,10 +455,6 @@ MathJax.Hub.Config({
-
But since scikit-learn has its own function for doing this and since
-it interfaces easily with tensorflow and other libraries, we
-normally recommend using the latter functionality.
-
@@ -459,7 +481,7 @@ normally recommend using the latter functionality.
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.
+
+
+
But since scikit-learn has its own function for doing this and since
+it interfaces easily with tensorflow and other libraries, we
+normally recommend using the latter functionality.
-
The features/predictors are
-
-
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
-
@@ -442,7 +461,7 @@ the house using the features (predictors) listed here.
and load the Boston Housing DataSet from Scikit-Learn
-
-
-
-
-
-
-
-
-
fromsklearn.datasetsimport 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
-
-
-
-
-
-
-
-
-
plt.figure(figsize=(20, 5))
-
-features = ['LSTAT', 'RM']
-target = boston['MEDV']
-
-for i, col inenumerate(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')
-
fromsklearn.model_selectionimport 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
-
-
-
-
-
-
-
-
fromsklearn.linear_modelimport LinearRegression
-fromsklearn.metricsimport 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()
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
The Boston housing data example
+
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
+
+
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
Reducing the number of degrees of freedom, overarching view
-
-
-
+
Housing data, the code
+
We start by importing the libraries
-
Many Machine Learning problems involve thousands or even millions of
-features for each training instance. Not only does this make training
-extremely slow, it can also make it much harder to find a good
-solution, as we will see. This problem is often referred to as the
-curse of dimensionality. Fortunately, in real-world problems, it is
-often possible to reduce the number of features considerably, turning
-an intractable problem into a tractable one.
-
+
+
+
+
+
+
+
importnumpyasnp
+importmatplotlib.pyplotasplt
-
Later we will discuss some of the most popular dimensionality reduction
-techniques: the principal component analysis (PCA), Kernel PCA, and
-Locally Linear Embedding (LLE).
-
-
-
Principal component analysis and its various variants deal with the
-problem of fitting a low-dimensional affine
-subspace to a set of of
-data points in a high-dimensional space. With its family of methods it
-is one of the most used tools in data modeling, compression and
-visualization.
-
+importpandasaspd
+importseabornassns
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
and load the Boston Housing DataSet from Scikit-Learn
+
+
+
+
+
+
+
+
+
fromsklearn.datasetsimport 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
+
+
+
+
+
+
+
+
+
plt.figure(figsize=(20, 5))
+
+features = ['LSTAT', 'RM']
+target = boston['MEDV']
+
+for i, col inenumerate(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')
+
fromsklearn.model_selectionimport 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
+
+
+
+
+
+
+
+
fromsklearn.linear_modelimport LinearRegression
+fromsklearn.metricsimport 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()
+
Reducing the number of degrees of freedom, overarching view
-
Before we proceed however, we will discuss how to preprocess our
-data. Till now and in connection with our previous examples we have
-not met so many cases where we are too sensitive to the scaling of our
-data. Normally the data may need a rescaling and/or may be sensitive
-to extreme values. Scaling the data renders our inputs much more
-suitable for the algorithms we want to employ.
+
Many Machine Learning problems involve thousands or even millions of
+features for each training instance. Not only does this make training
+extremely slow, it can also make it much harder to find a good
+solution, as we will see. This problem is often referred to as the
+curse of dimensionality. Fortunately, in real-world problems, it is
+often possible to reduce the number of features considerably, turning
+an intractable problem into a tractable one.
-
For data sets gathered for real world applications, it is rather normal that
-different features have very different units and
-numerical scales. For example, a data set detailing health habits may include
-features such as age in the range \( 0-80 \), and caloric intake of order \( 2000 \).
-Many machine learning methods sensitive to the scales of the features and may perform poorly if they
-are very different scales. Therefore, it is typical to scale
-the features in a way to avoid such outlier values.
+
Later we will discuss some of the most popular dimensionality reduction
+techniques: the principal component analysis (PCA), Kernel PCA, and
+Locally Linear Embedding (LLE).
+
+
+
Principal component analysis and its various variants deal with the
+problem of fitting a low-dimensional affine
+subspace to a set of of
+data points in a high-dimensional space. With its family of methods it
+is one of the most used tools in data modeling, compression and
+visualization.
@@ -441,7 +448,7 @@ the features in a way to avoid such outlier values.
Scikit-Learn has several functions which allow us to rescale the
-data, normally resulting in much better results in terms of various
-accuracy scores. The StandardScaler function in Scikit-Learn
-ensures that for each feature/predictor we study the mean value is
-zero and the variance is one (every column in the design/feature
-matrix). This scaling has the drawback that it does not ensure that
-we have a particular maximum or minimum in our data set. Another
-function included in Scikit-Learn is the MinMaxScaler which
-ensures that all features are exactly between \( 0 \) and \( 1 \). The
+
Before we proceed however, we will discuss how to preprocess our
+data. Till now and in connection with our previous examples we have
+not met so many cases where we are too sensitive to the scaling of our
+data. Normally the data may need a rescaling and/or may be sensitive
+to extreme values. Scaling the data renders our inputs much more
+suitable for the algorithms we want to employ.
+
For data sets gathered for real world applications, it is rather normal that
+different features have very different units and
+numerical scales. For example, a data set detailing health habits may include
+features such as age in the range \( 0-80 \), and caloric intake of order \( 2000 \).
+Many machine learning methods sensitive to the scales of the features and may perform poorly if they
+are very different scales. Therefore, it is typical to scale
+the features in a way to avoid such outlier values.
+
+
+
+
+
@@ -429,7 +443,7 @@ ensures that all features are exactly between \( 0 \) and \( 1 \). The
The Normalizer scales each data
-point such that the feature vector has a euclidean length of one. In other words, it
-projects a data point on the circle (or sphere in the case of higher dimensions) with a
-radius of 1. This means every data point is scaled by a different number (by the
-inverse of it’s length).
-This normalization is often used when only the direction (or angle) of the data matters,
-not the length of the feature vector.
+
Scikit-Learn has several functions which allow us to rescale the
+data, normally resulting in much better results in terms of various
+accuracy scores. The StandardScaler function in Scikit-Learn
+ensures that for each feature/predictor we study the mean value is
+zero and the variance is one (every column in the design/feature
+matrix). This scaling has the drawback that it does not ensure that
+we have a particular maximum or minimum in our data set. Another
+function included in Scikit-Learn is the MinMaxScaler which
+ensures that all features are exactly between \( 0 \) and \( 1 \). The
-
The RobustScaler works similarly to the StandardScaler in that it
-ensures statistical properties for each feature that guarantee that
-they are on the same scale. However, the RobustScaler uses the median
-and quartiles, instead of mean and variance. This makes the
-RobustScaler ignore data points that are very different from the rest
-(like measurement errors). These odd data points are also called
-outliers, and might often lead to trouble for other scaling
-techniques.
-
Many features are often scaled using standardization to improve performance. In Scikit-Learn this is given by the StandardScaler function as discussed above. It is easy however to write your own.
-Mathematically, this involves subtracting the mean and divide by the standard deviation over the data set, for each feature:
+
+
+
+
The Normalizer scales each data
+point such that the feature vector has a euclidean length of one. In other words, it
+projects a data point on the circle (or sphere in the case of higher dimensions) with a
+radius of 1. This means every data point is scaled by a different number (by the
+inverse of it’s length).
+This normalization is often used when only the direction (or angle) of the data matters,
+not the length of the feature vector.
where \( \overline{x}_j \) and \( \sigma(x_j) \) are the mean and standard deviation, respectively, of the feature \( x_j \).
-This ensures that each feature has zero mean and unit standard deviation. For data sets where we do not have the standard deviation or don't wish to calculate it, it is then common to simply set it to one.
+
The RobustScaler works similarly to the StandardScaler in that it
+ensures statistical properties for each feature that guarantee that
+they are on the same scale. However, the RobustScaler uses the median
+and quartiles, instead of mean and variance. This makes the
+RobustScaler ignore data points that are very different from the rest
+(like measurement errors). These odd data points are also called
+outliers, and might often lead to trouble for other scaling
+techniques.
+
+
+
@@ -430,7 +445,7 @@ This ensures that each feature has zero mean and unit standard deviation. For d
Let us consider the following vanilla example where we use both
-Scikit-Learn and write our own function as well. We produce a
-simple test design matrix with random numbers. Each column could then
-represent a specific feature whose mean value is subracted.
+
Many features are often scaled using standardization to improve performance. In Scikit-Learn this is given by the StandardScaler function as discussed above. It is easy however to write your own.
+Mathematically, this involves subtracting the mean and divide by the standard deviation over the data set, for each feature:
importsklearn.linear_modelasskl
-fromsklearn.metricsimport mean_squared_error
-fromsklearn.model_selectionimport train_test_split
-fromsklearn.preprocessingimport MinMaxScaler, StandardScaler, Normalizer
-importnumpyasnp
-importpandasaspd
-fromIPython.displayimport display
-np.random.seed(100)
-# setting up a 10 x 5 matrix
-rows =10
-cols =5
-X = np.random.randn(rows,cols)
-XPandas = pd.DataFrame(X)
-display(XPandas)
-print(XPandas.mean())
-print(XPandas.std())
-XPandas = (XPandas -XPandas.mean())
-display(XPandas)
-# This option does not include the standard deviation
-scaler = StandardScaler(with_std=False)
-scaler.fit(X)
-Xscaled = scaler.transform(X)
-display(XPandas-Xscaled)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Small exercise: perform the standard scaling by including the standard deviation and compare with what Scikit-Learn gives.
+
where \( \overline{x}_j \) and \( \sigma(x_j) \) are the mean and standard deviation, respectively, of the feature \( x_j \).
+This ensures that each feature has zero mean and unit standard deviation. For data sets where we do not have the standard deviation or don't wish to calculate it, it is then common to simply set it to one.
+
Another commonly used scaling method is min-max scaling. This is very
-useful for when we want the features to lie in a certain interval. To
-scale the feature \( x_j \) to the interval \( [a, b] \), we can apply the
-transformation
+
Let us consider the following vanilla example where we use both
+Scikit-Learn and write our own function as well. We produce a
+simple test design matrix with random numbers. Each column could then
+represent a specific feature whose mean value is subracted.
Testing the Means Squared Error as function of Complexity
-
One of
-the aims is to reproduce Figure 2.11 of Hastie et al.
-We will also use Ridge and Lasso regression.
+
Min-Max Scaling
+
+
Another commonly used scaling method is min-max scaling. This is very
+useful for when we want the features to lie in a certain interval. To
+scale the feature \( x_j \) to the interval \( [a, b] \), we can apply the
+transformation
-
Our data is defined by \( x\in [-3,3] \) with a total of for example \( 100 \) data points.
Mathematical Interpretation of Ordinary Least Squares
-
-
What is presented here is a mathematical analysis of various regression algorithms (ordinary least squares, Ridge and Lasso Regression). The analysis is based on an important algorithm in linear algebra, the so-called Singular Value Decomposition (SVD).
-
-
We have shown that in ordinary least squares the optimal parameters \( \beta \) are given by
The matrix \( \boldsymbol{A} \) has the important property that \( \boldsymbol{A}^2=\boldsymbol{A} \). This is the definition of a projection matrix.
-We can then interpret our optimal model \( \tilde{\boldsymbol{y}} \) as being represented by an orthogonal projection of \( \boldsymbol{y} \) onto a space defined by the column vectors of \( \boldsymbol{X} \). In our case here the matrix \( \boldsymbol{A} \) is a square matrix. If it is a general rectangular matrix we have an oblique projection matrix.
-
+
Material for lecture Thursday, August 31
@@ -448,7 +420,7 @@ We can then interpret our optimal model \( \tilde{\boldsymbol{y}} \) as being re
Mathematical Interpretation of Ordinary Least Squares
+
+
What is presented here is a mathematical analysis of various regression algorithms (ordinary least squares, Ridge and Lasso Regression). The analysis is based on an important algorithm in linear algebra, the so-called Singular Value Decomposition (SVD).
+
+
We have shown that in ordinary least squares the optimal parameters \( \beta \) are given by
The residual errors are then the projections of \( \boldsymbol{y} \) onto the orthogonal component of the space defined by the column vectors of \( \boldsymbol{X} \).
+
The hat over \( \boldsymbol{\beta} \) means we have the optimal parameters after minimization of the cost function.
The matrix \( \boldsymbol{A} \) has the important property that \( \boldsymbol{A}^2=\boldsymbol{A} \). This is the definition of a projection matrix.
+We can then interpret our optimal model \( \tilde{\boldsymbol{y}} \) as being represented by an orthogonal projection of \( \boldsymbol{y} \) onto a space defined by the column vectors of \( \boldsymbol{X} \). In our case here the matrix \( \boldsymbol{A} \) is a square matrix. If it is a general rectangular matrix we have an oblique projection matrix.
+
The residual errors are then the projections of \( \boldsymbol{y} \) onto the orthogonal component of the space defined by the column vectors of \( \boldsymbol{X} \).
If the matrix \( \boldsymbol{X} \) is an orthogonal (or unitary in case of complex values) matrix, we have
-
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 where we fit of various functions 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.
-
As we will also see in the first project,
-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.
-
+
In this case the matrix \( \boldsymbol{A} \) becomes
There is however a way to circumvent this problem and also
-gain some insights about the ordinary least squares approach, and
-later shrinkage methods like Ridge and Lasso regressions.
-
-
-
This is given by the Singular Value Decomposition (SVD) algorithm,
-perhaps the most powerful linear algebra algorithm. The SVD provides
-a numerically stable matrix decomposition that is used in a large
-swath oc applications and the decomposition is always stable
-numerically.
-
-
-
In machine learning it plays a central role in dealing with for
-example design matrices that may be near singular or singular.
-Furthermore, as we will see here, the singular values can be related
-to the covariance matrix (and thereby the correlation matrix) and in
-turn the variance of a given quantity. It plays also an important role
-in the principal component analysis where high-dimensional data can be
-reduced to the statistically relevant features.
-
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
-
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 where we fit of various functions 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
+
As we will also see in the first project,
+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.
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.
+
There is however a way to circumvent this problem and also
+gain some insights about the ordinary least squares approach, and
+later shrinkage methods like Ridge and Lasso regressions.
+
This is given by the Singular Value Decomposition (SVD) algorithm,
+perhaps the most powerful linear algebra algorithm. The SVD provides
+a numerically stable matrix decomposition that is used in a large
+swath oc applications and the decomposition is always stable
+numerically.
+
+
+
In machine learning it plays a central role in dealing with for
+example design matrices that may be near singular or singular.
+Furthermore, as we will see here, the singular values can be related
+to the covariance matrix (and thereby the correlation matrix) and in
+turn the variance of a given quantity. It plays also an important role
+in the principal component analysis where high-dimensional data can be
+reduced to the statistically relevant features.
+
+
+
+
+
@@ -465,7 +462,7 @@ 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
+
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.
+
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.
-
A cheap ad hoc approach is simply to add a small diagonal component to the matrix to invert, that is we change
+
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
+
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.
+
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
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
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.
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 \( m\times n \)
-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
The SVD
-decomposition (singular values) gives eigenvalues
-\( \sigma_i\geq\sigma_{i+1} \) for all \( i \) and for dimensions larger than \( i=p \), the
-eigenvalues (singular values) are zero.
-
+
The matrix \( \boldsymbol{X} \) can be written in terms of an orthogonal/unitary transformation \( \boldsymbol{U} \)
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.
-
+
with \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) or \( \boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I} \).
-
The columns of \( \boldsymbol{U} \) are called the left singular vectors while the columns of \( \boldsymbol{V} \) are the right singular vectors.
+
Not all square matrices are diagonalizable. A matrix like the one discussed above
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.
+
@@ -463,7 +454,7 @@ near singular or singular matrices.
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 \( \boldsymbol{\Sigma} \).
+
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 \( m\times n \)
+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
-
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.
+$$
+\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T
+$$
+
+
As an example, the above defective matrix can be decomposed as
with eigenvalues \( \sigma_1=2 \) and \( \sigma_2=0 \).
+The SVD exits always!
-
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.
+
The SVD
+decomposition (singular values) gives eigenvalues
+\( \sigma_i\geq\sigma_{i+1} \) for all \( i \) and for dimensions larger than \( i=p \), 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.
+
@@ -438,7 +465,7 @@ In general the economy-size SVD leads to less FLOPS and still conserving the des
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 \( \boldsymbol{\Sigma} \).
+
-
-
-
-
-
-
-
importnumpyasnp
-# SVD inversion
-defSVD(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,full_matrices=True)
- 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)
+
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.
+
- D = np.zeros((len(U),len(VT)))
- for i inrange(0,len(VT)):
- D[i,i]=S[i]
- return U @ D @ VT
-
-
-X = np.array([ [1.0,-1.0], [1.0,-1.0]])
-#X = np.array([[1, 2], [3, 4], [5, 6]])
-
-print(X)
-C = SVD(X)
-# Print the difference between the original matrix and the SVD one
-print(C-X)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
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.
+
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.
@@ -479,7 +440,7 @@ in the program terminating due to a singular matrix.
The \( U \), \( S \), and \( V \) matrices returned from the svd() function
-cannot be multiplied directly.
-
-
As you can see from the code, the \( S \) vector must be converted into a
-diagonal matrix. This may cause a problem as the size of the matrices
-do not fit the rules of matrix multiplication, where the number of
-columns in a matrix must match the number of rows in the subsequent
-matrix.
-
+
+
+
+
+
+
+
importnumpyasnp
+# SVD inversion
+defSVD(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,full_matrices=True)
+ 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)
-
If you wish to include the zero singular values, you will need to
-resize the matrices and set up a diagonal matrix as done in the above
-example
+ D = np.zeros((len(U),len(VT)))
+ for i inrange(0,len(VT)):
+ D[i,i]=S[i]
+ return U @ D @ VT
+
+
+X = np.array([ [1.0,-1.0], [1.0,-1.0]])
+#X = np.array([[1, 2], [3, 4], [5, 6]])
+
+print(X)
+C = SVD(X)
+# Print the difference between the original matrix and the SVD one
+print(C-X)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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.
As you can see from the code, the \( S \) vector must be converted into a
+diagonal matrix. This may cause a problem as the size of the matrices
+do not fit the rules of matrix multiplication, where the number of
+columns in a matrix must match the number of rows in the subsequent
+matrix.
+
where \( \boldsymbol{U} \) is an orthogonal matrix of dimension \( n\times n \), meaning that \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{U}^T\boldsymbol{U}=\boldsymbol{I}_n \). Here \( \boldsymbol{I}_n \) is the unit matrix of dimension \( n \times n \).
-
-
Similarly, \( \boldsymbol{V} \) is an orthogonal matrix of dimension \( p\times p \), meaning that \( \boldsymbol{V}\boldsymbol{V}^T=\boldsymbol{V}^T\boldsymbol{V}=\boldsymbol{I}_p \). Here \( \boldsymbol{I}_n \) is the unit matrix of dimension \( p \times p \).
-
-
Finally \( \boldsymbol{\Sigma} \) contains the singular values \( \sigma_i \). This matrix has dimension \( n\times p \) and the singular values \( \sigma_i \) are all positive. The non-zero values are ordered in descending order, that is
where \( \boldsymbol{U} \) is an orthogonal matrix of dimension \( n\times n \), meaning that \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{U}^T\boldsymbol{U}=\boldsymbol{I}_n \). Here \( \boldsymbol{I}_n \) is the unit matrix of dimension \( n \times n \).
-
contains only the singular values. Note also (and we will use this below) that
+
Similarly, \( \boldsymbol{V} \) is an orthogonal matrix of dimension \( p\times p \), meaning that \( \boldsymbol{V}\boldsymbol{V}^T=\boldsymbol{V}^T\boldsymbol{V}=\boldsymbol{I}_p \). Here \( \boldsymbol{I}_n \) is the unit matrix of dimension \( p \times p \).
+
+
Finally \( \boldsymbol{\Sigma} \) contains the singular values \( \sigma_i \). This matrix has dimension \( n\times p \) and the singular values \( \sigma_i \) are all positive. The non-zero values are ordered in descending order, that is
is a \( 3\times 3 \) matrix. The last row and column of this last matrix
-contain only zeros. This will have important consequences for our SVD
-decomposition of the design matrix.
-
+
All values beyond \( p-1 \) are all zero.
@@ -473,7 +451,7 @@ decomposition of the design matrix.
We define \( \boldsymbol{\Sigma}^T\boldsymbol{\Sigma}=\tilde{\boldsymbol{\Sigma}}^2 \) which is a diagonal matrix containing only the singular values squared. It has dimensionality \( p \times p \).
-
-
We can now insert the result for the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) into our equation for ordinary least squares where
It means that the ordinary least square model (with the optimal
-parameters) \( \boldsymbol{\tilde{y}} \), corresponds to an orthogonal
-transformation of the output (or target) vector \( \boldsymbol{y} \) by the
-vectors of the matrix \( \boldsymbol{U} \). Note that the summation ends at \( p-1 \),
-that is \( \boldsymbol{\tilde{y}}\ne \boldsymbol{y} \).
+
is a \( 3\times 3 \) matrix. The last row and column of this last matrix
+contain only zeros. This will have important consequences for our SVD
+decomposition of the design matrix.
@@ -457,7 +475,7 @@ that is \( \boldsymbol{\tilde{y}}\ne \boldsymbol{y} \).
This means the vectors \( \boldsymbol{v}_i \) of the orthogonal matrix \( \boldsymbol{V} \) are the eigenvectors of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)
-with eigenvalues given by the singular values squared, that is
-
+
We define \( \boldsymbol{\Sigma}^T\boldsymbol{\Sigma}=\tilde{\boldsymbol{\Sigma}}^2 \) which is a diagonal matrix containing only the singular values squared. It has dimensionality \( p \times p \).
+
+
We can now insert the result for the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) into our equation for ordinary least squares where
This means the vectors \( \boldsymbol{u}_i \) of the orthogonal matrix \( \boldsymbol{U} \) are the eigenvectors of the matrix \( \boldsymbol{X}\boldsymbol{X}^T \)
-with eigenvalues given by the singular values squared, that is
-
Important note: we have defined our design matrix \( \boldsymbol{X} \) to be an
-\( n\times p \) matrix. In most supervised learning cases we have that \( n
-\ge p \), and quite often we have \( n >> p \). For linear algebra based methods like ordinary least squares or Ridge regression, this leads to a matrix \( \boldsymbol{X}^T\boldsymbol{X} \) which is small and thereby easier to handle from a computational point of view (in terms of number of floating point operations).
-
-
-
In our lectures, the number of columns will
-always refer to the number of features in our data set, while the
-number of rows represents the number of data inputs. Note that in
-other texts you may find the opposite notation. This has consequences
-for the definition of for example the covariance matrix and its relation to the SVD.
+
It means that the ordinary least square model (with the optimal
+parameters) \( \boldsymbol{\tilde{y}} \), corresponds to an orthogonal
+transformation of the output (or target) vector \( \boldsymbol{y} \) by the
+vectors of the matrix \( \boldsymbol{U} \). Note that the summation ends at \( p-1 \),
+that is \( \boldsymbol{\tilde{y}}\ne \boldsymbol{y} \).
@@ -464,7 +459,7 @@ for the definition of for example the covariance matrix and its relation to the
We have already noted that the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) in ordinary
-least squares is proportional to the second derivative of the cost
-function, that is we have
+
If we now multiply from the right with \( \boldsymbol{V} \) (using the orthogonality of \( \boldsymbol{V} \)) we get
This means the vectors \( \boldsymbol{v}_i \) of the orthogonal matrix \( \boldsymbol{V} \) are the eigenvectors of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)
+with eigenvalues given by the singular values squared, that is
+
This means the vectors \( \boldsymbol{u}_i \) of the orthogonal matrix \( \boldsymbol{U} \) are the eigenvectors of the matrix \( \boldsymbol{X}\boldsymbol{X}^T \)
+with eigenvalues given by the singular values squared, that is
+
Important note: we have defined our design matrix \( \boldsymbol{X} \) to be an
+\( n\times p \) matrix. In most supervised learning cases we have that \( n
+\ge p \), and quite often we have \( n >> p \). For linear algebra based methods like ordinary least squares or Ridge regression, this leads to a matrix \( \boldsymbol{X}^T\boldsymbol{X} \) which is small and thereby easier to handle from a computational point of view (in terms of number of floating point operations).
The Hessian matrix for ordinary least squares is also proportional to
-the covariance matrix. This means also that we can use the SVD to find
-the eigenvalues of the covariance matrix and the Hessian matrix in
-terms of the singular values. Let us develop these arguments, as they will play an important role in our machine learning studies.
+
In our lectures, the number of columns will
+always refer to the number of features in our data set, while the
+number of rows represents the number of data inputs. Note that in
+other texts you may find the opposite notation. This has consequences
+for the definition of for example the covariance matrix and its relation to the SVD.
@@ -443,7 +466,7 @@ terms of the singular values. Let us develop these arguments, as they will pla
Introducing the Covariance and Correlation functions
+
Meet the Covariance Matrix
-
Before we discuss the link between for example Ridge regression and the singular value decomposition, we need to remind ourselves about
-the definition of the covariance and the correlation function. These are quantities that play a central role in machine learning methods.
+
Before we move on to a discussion of Ridge and Lasso regression, we want to show an important example of the above.
+
+
We have already noted that the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) in ordinary
+least squares is proportional to the second derivative of the cost
+function, that is we have
-
Suppose we have defined two vectors
-\( \hat{x} \) and \( \hat{y} \) with \( n \) elements each. The covariance matrix \( \boldsymbol{C} \) is defined as
-
Note: we have used \( 1/n \) in the above definitions of the sample variance and covariance. We assume then that we can calculate the exact mean value.
-What you will find in essentially all statistics texts are equations
-with a factor \( 1/(n-1) \). This is called Bessel's correction. This
-method corrects the bias in the estimation of the population variance
-and covariance. It also partially corrects the bias in the estimation
-of the population standard deviation. If you use a library like
-Scikit-Learn or nunmpy's function calculate the covariance, this
-quantity will be computed with a factor \( 1/(n-1) \).
+
The Hessian matrix for ordinary least squares is also proportional to
+the covariance matrix. This means also that we can use the SVD to find
+the eigenvalues of the covariance matrix and the Hessian matrix in
+terms of the singular values. Let us develop these arguments, as they will play an important role in our machine learning studies.
@@ -458,7 +445,7 @@ quantity will be computed with a factor \( 1/(n-1) \).
Introducing the Covariance and Correlation functions
-
The covariance takes values between zero and infinity and may thus
-lead to problems with loss of numerical precision for particularly
-large values. It is common to scale the covariance matrix by
-introducing instead the correlation matrix defined via the so-called
-correlation function
+
Before we discuss the link between for example Ridge regression and the singular value decomposition, we need to remind ourselves about
+the definition of the covariance and the correlation function. These are quantities that play a central role in machine learning methods.
The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]
-\in [-1,1] \). This avoids eventual problems with too large values. We
-can then define the correlation matrix for the two vectors \( \boldsymbol{x} \)
-and \( \boldsymbol{y} \) as
+
Suppose we have defined two vectors
+\( \hat{x} \) and \( \hat{y} \) with \( n \) elements each. The covariance matrix \( \boldsymbol{C} \) is defined as
Note: we have used \( 1/n \) in the above definitions of the sample variance and covariance. We assume then that we can calculate the exact mean value.
+What you will find in essentially all statistics texts are equations
+with a factor \( 1/(n-1) \). This is called Bessel's correction. This
+method corrects the bias in the estimation of the population variance
+and covariance. It also partially corrects the bias in the estimation
+of the population standard deviation. If you use a library like
+Scikit-Learn or nunmpy's function calculate the covariance, this
+quantity will be computed with a factor \( 1/(n-1) \).
+
In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression
-we defined the design/feature matrix \( \boldsymbol{X} \) as
+
The covariance takes values between zero and infinity and may thus
+lead to problems with loss of numerical precision for particularly
+large values. It is common to scale the covariance matrix by
+introducing instead the correlation matrix defined via the so-called
+correlation function
with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors/features \( p \) refering to the column numbers and the
-entries \( n \) being the row elements.
-We can rewrite the design/feature matrix in terms of its column vectors as
-
With these definitions, we can now rewrite our \( 2\times 2 \)
-correlation/covariance matrix in terms of a moe general design/feature
-matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \)
-covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \)
+
The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]
+\in [-1,1] \). This avoids eventual problems with too large values. We
+can then define the correlation matrix for the two vectors \( \boldsymbol{x} \)
+and \( \boldsymbol{y} \) as
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 values. The following simple function uses the
-np.vstack function which takes each vector of dimension \( 1\times n \)
-and produces a \( 2\times n \) matrix \( \boldsymbol{W} \)
+
In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression
+we defined the design/feature matrix \( \boldsymbol{X} \) as
-
Note that this assumes you have the features as the rows, and the inputs as columns, that is
which in turn is converted into into the \( 2\times 2 \) covariance matrix
-\( \boldsymbol{C} \) 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.
+
with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors/features \( p \) refering to the column numbers and the
+entries \( n \) being the row elements.
+We can rewrite the design/feature matrix in terms of its column vectors as
+
With these definitions, we can now rewrite our \( 2\times 2 \)
+correlation/covariance matrix in terms of a moe general design/feature
+matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \)
+covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \)
The previous example can be converted into the correlation matrix by
-simply scaling the matrix elements with the variances. We should also
-subtract the mean values for each column. This leads to the following
-code which sets up the correlations matrix for the previous example in
-a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the \( 2\times 2 \) correlation matrix (since we have only two vectors).
+
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 values. The following simple function uses the
+np.vstack function which takes each vector of dimension \( 1\times n \)
+and produces a \( 2\times n \) matrix \( \boldsymbol{W} \)
+
+
+
Note that this assumes you have the features as the rows, and the inputs as columns, that is
which in turn is converted into into the \( 2\times 2 \) covariance matrix
+\( \boldsymbol{C} \) 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.
@@ -407,26 +423,15 @@ a more brute force way. Here we scale the mean values for each column of the des
-
importnumpyasnp
+
# Importing various packages
+importnumpyasnp
n =100
-# define two vectors
-x = np.random.random(size=n)
+x = np.random.normal(size=n)
+print(np.mean(x))
y =4+3*x+np.random.normal(size=n)
-#scaling the x and y vectors
-x = x - np.mean(x)
-y = y - np.mean(y)
-variance_x = np.sum(x@x)/n
-variance_y = np.sum(y@y)/n
-print(variance_x)
-print(variance_y)
-cov_xy = np.sum(x@y)/n
-cov_xx = np.sum(x@x)/n
-cov_yy = np.sum(y@y)/n
-C = np.zeros((2,2))
-C[0,0]= cov_xx/variance_x
-C[1,1]= cov_yy/variance_y
-C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
-C[1,0]= C[0,1]
+print(np.mean(y))
+W = np.vstack((x, y))
+C = np.cov(W)
print(C)
@@ -443,12 +448,6 @@ C[1,0]
-
We see that the matrix elements along the diagonal are one as they
-should be and that the matrix is symmetric. Furthermore, diagonalizing
-this matrix we easily see that it is a positive definite matrix.
-
-
-
The above procedure with numpy can be made more compact if we use pandas.
@@ -475,7 +474,7 @@ this matrix we easily see that it is a positive definite matrix.
The previous example can be converted into the correlation matrix by
+simply scaling the matrix elements with the variances. We should also
+subtract the mean values for each column. This leads to the following
+code which sets up the correlations matrix for the previous example in
+a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the \( 2\times 2 \) correlation matrix (since we have only two vectors).
+
-
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
@@ -402,19 +410,26 @@ MathJax.Hub.Config({
importnumpyasnp
-importpandasaspd
-n =10
-x = np.random.normal(size=n)
-x = x - np.mean(x)
+n =100
+# define two vectors
+x = np.random.random(size=n)
y =4+3*x+np.random.normal(size=n)
+#scaling the x and y vectors
+x = x - np.mean(x)
y = y - np.mean(y)
-# Note that we transpose the matrix in order to stay with our ordering n x p
-X = (np.vstack((x, y))).T
-print(X)
-Xpd = pd.DataFrame(X)
-print(Xpd)
-correlation_matrix = Xpd.corr()
-print(correlation_matrix)
+variance_x = np.sum(x@x)/n
+variance_y = np.sum(y@y)/n
+print(variance_x)
+print(variance_y)
+cov_xy = np.sum(x@y)/n
+cov_xx = np.sum(x@x)/n
+cov_yy = np.sum(y@y)/n
+C = np.zeros((2,2))
+C[0,0]= cov_xx/variance_x
+C[1,1]= cov_yy/variance_y
+C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
+C[1,0]= C[0,1]
+print(C)
Correlation Matrix with Pandas and the Franke function
+
Correlation Matrix with Pandas
+
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
@@ -400,49 +403,20 @@ MathJax.Hub.Config({
-
# Common imports
-importnumpyasnp
+
importnumpyasnpimportpandasaspd
-
-
-defFrankeFunction(x,y):
- term1 =0.75*np.exp(-(0.25*(9*x-2)**2) -0.25*((9*y-2)**2))
- term2 =0.75*np.exp(-((9*x+1)**2)/49.0-0.1*(9*y+1))
- term3 =0.5*np.exp(-(9*x-7)**2/4.0-0.25*((9*y-3)**2))
- term4 =-0.2*np.exp(-(9*x-4)**2- (9*y-7)**2)
- return term1 + term2 + term3 + term4
-
-
-defcreate_X(x, y, n ):
- iflen(x.shape) >1:
- x = np.ravel(x)
- y = np.ravel(y)
-
- N =len(x)
- l =int((n+1)*(n+2)/2) # Number of elements in beta
- X = np.ones((N,l))
-
- for i inrange(1,n+1):
- q =int((i)*(i+1)/2)
- for k inrange(i+1):
- X[:,q+k] = (x**(i-k))*(y**k)
-
- return X
-
-
-# Making meshgrid of datapoints and compute Franke's function
-n =4
-N =100
-x = np.sort(np.random.uniform(0, 1, N))
-y = np.sort(np.random.uniform(0, 1, N))
-z = FrankeFunction(x, y)
-X = create_X(x, y, n=n)
-
+n =10
+x = np.random.normal(size=n)
+x = x - np.mean(x)
+y =4+3*x+np.random.normal(size=n)
+y = y - np.mean(y)
+# Note that we transpose the matrix in order to stay with our ordering n x p
+X = (np.vstack((x, y))).T
+print(X)
Xpd = pd.DataFrame(X)
-# subtract the mean values and set up the covariance matrix
-Xpd = Xpd - Xpd.mean()
-covariance_matrix = Xpd.cov()
-print(covariance_matrix)
+print(Xpd)
+correlation_matrix = Xpd.corr()
+print(correlation_matrix)
where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this is the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
-
It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
+defcreate_X(x, y, n ):
+ iflen(x.shape) >1:
+ x = np.ravel(x)
+ y = np.ravel(y)
+
+ N =len(x)
+ l =int((n+1)*(n+2)/2) # Number of elements in beta
+ X = np.ones((N,l))
+
+ for i inrange(1,n+1):
+ q =int((i)*(i+1)/2)
+ for k inrange(i+1):
+ X[:,q+k] = (x**(i-k))*(y**k)
+
+ return X
+
+
+# Making meshgrid of datapoints and compute Franke's function
+n =4
+N =100
+x = np.sort(np.random.uniform(0, 1, N))
+y = np.sort(np.random.uniform(0, 1, N))
+z = FrankeFunction(x, y)
+X = create_X(x, y, n=n)
+
+Xpd = pd.DataFrame(X)
+# subtract the mean values and set up the covariance matrix
+Xpd = Xpd - Xpd.mean()
+covariance_matrix = Xpd.cov()
+print(covariance_matrix)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
We note here that the covariance is zero for the first rows and
+columns since all matrix elements in the design matrix were set to one
+(we are fitting the function in terms of a polynomial of degree \( n \)).
+
+
+
This means that the variance for these elements will be zero and will
+cause problems when we set up the correlation matrix. We can simply
+drop these elements and construct a correlation
+matrix without these elements.
+
where the tilde-matrix \( \tilde{\boldsymbol{\Sigma}} \) is a matrix of dimension \( p\times p \) containing only the singular values \( \sigma_i \), that is
-
+
If we then compute the expectation value (note the \( 1/n \) factor instead of \( 1/(n-1) \))
where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this is the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
+
It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
This means the vectors \( \boldsymbol{v}_i \) of the orthogonal matrix \( \boldsymbol{V} \)
-are the eigenvectors of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) with eigenvalues
-given by the singular values squared, that is
-
In other words, each non-zero singular value of \( \boldsymbol{X} \) is a positive
-square root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). It means also that
-the columns of \( \boldsymbol{V} \) are the eigenvectors of
-\( \boldsymbol{X}^T\boldsymbol{X} \). Since we have ordered the singular values of
-\( \boldsymbol{X} \) in a descending order, it means that the column vectors
-\( \boldsymbol{v}_i \) are hierarchically ordered by how much correlation they
-encode from the columns of \( \boldsymbol{X} \).
-
-
-
Note that these are also the eigenvectors and eigenvalues of the
-Hessian matrix.
-
-
-
If we now recall the definition of the covariance matrix (not using
-Bessel's correction) we have
-
-
+
Since the matrices here have dimension \( p\times p \), with \( p \) corresponding to the singular values, we defined earlier the matrix
where the tilde-matrix \( \tilde{\boldsymbol{\Sigma}} \) is a matrix of dimension \( p\times p \) containing only the singular values \( \sigma_i \), that is
meaning that every squared non-singular value of \( \boldsymbol{X} \) divided by \( n \) (
-the number of samples) are the eigenvalues of the covariance
-matrix. Every singular value of \( \boldsymbol{X} \) is thus a positive square
-root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). If the matrix \( \boldsymbol{X} \) is
-self-adjoint, the singular values of \( \boldsymbol{X} \) are equal to the
-absolute value of the eigenvalues of \( \boldsymbol{X} \).
-
@@ -453,6 +449,7 @@ absolute value of the eigenvalues of \( \boldsymbol{X} \).
It means that the eigenvalues of \( \boldsymbol{X}\boldsymbol{X}^T \) are again given by
-the non-zero singular values plus now a series of zeros. The column
-vectors of \( \boldsymbol{U} \) are the eigenvectors of \( \boldsymbol{X}\boldsymbol{X}^T \) and
-measure how much correlations are contained in the rows of \( \boldsymbol{X} \).
+
This means the vectors \( \boldsymbol{v}_i \) of the orthogonal matrix \( \boldsymbol{V} \)
+are the eigenvectors of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) with eigenvalues
+given by the singular values squared, that is
-
Since we will mainly be interested in the correlations among the features
-of our data (the columns of \( \boldsymbol{X} \), the quantity of interest for us are the non-zero singular
-values and the column vectors of \( \boldsymbol{V} \).
+$$
+\left(\boldsymbol{X}^T\boldsymbol{X}\right)\boldsymbol{v}_i=\boldsymbol{v}_i\sigma_i^2.
+$$
+
+
In other words, each non-zero singular value of \( \boldsymbol{X} \) is a positive
+square root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). It means also that
+the columns of \( \boldsymbol{V} \) are the eigenvectors of
+\( \boldsymbol{X}^T\boldsymbol{X} \). Since we have ordered the singular values of
+\( \boldsymbol{X} \) in a descending order, it means that the column vectors
+\( \boldsymbol{v}_i \) are hierarchically ordered by how much correlation they
+encode from the columns of \( \boldsymbol{X} \).
+
+
+
Note that these are also the eigenvectors and eigenvalues of the
+Hessian matrix.
+
+
+
If we now recall the definition of the covariance matrix (not using
+Bessel's correction) we have
+
meaning that every squared non-singular value of \( \boldsymbol{X} \) divided by \( n \) (
+the number of samples) are the eigenvalues of the covariance
+matrix. Every singular value of \( \boldsymbol{X} \) is thus a positive square
+root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). If the matrix \( \boldsymbol{X} \) is
+self-adjoint, the singular values of \( \boldsymbol{X} \) are equal to the
+absolute value of the eigenvalues of \( \boldsymbol{X} \).
@@ -446,6 +454,7 @@ values and the column vectors of \( \boldsymbol{V} \).
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
-
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
+
Multiplying with \( \boldsymbol{U} \) from the right gives us the eigenvalue problem
It means that the eigenvalues of \( \boldsymbol{X}\boldsymbol{X}^T \) are again given by
+the non-zero singular values plus now a series of zeros. The column
+vectors of \( \boldsymbol{U} \) are the eigenvectors of \( \boldsymbol{X}\boldsymbol{X}^T \) and
+measure how much correlations are contained in the rows of \( \boldsymbol{X} \).
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
+
Since we will mainly be interested in the correlations among the features
+of our data (the columns of \( \boldsymbol{X} \), the quantity of interest for us are the non-zero singular
+values and the column vectors of \( \boldsymbol{V} \).
Using the matrix-vector expression for Ridge regression and dropping the parameter \( 1/n \) in front of the standard means squared error equation, we have
and
-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
-the optimal parameters
+
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
In many textbooks the \( 1/n \) term is often omitted. Note that a library like Scikit-Learn does not include the \( 1/n \) factor in the setup of the cost function.
-
-
When we compare this with the ordinary least squares result we have
which can lead to singular matrices. However, with the SVD, we can always compute the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \).
-
-
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. We will see that for specific values of \( \lambda \), we may
-even reduce the variance of the optimal parameters \( \boldsymbol{\beta} \). These topics and other related ones, will be discussed after the more linear algebra oriented analysis here.
+
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
-
Using our insights about the SVD of the design matrix \( \boldsymbol{X} \)
-We have already analyzed the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as
+$$
+{\displaystyle \min_{\boldsymbol{\beta}\in
+{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2
+$$
+
+
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
Since \( \lambda \geq 0 \), it means that compared to OLS, we have
+
Using the matrix-vector expression for Ridge regression and dropping the parameter \( 1/n \) in front of the standard means squared error equation, we have
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
+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
+the optimal parameters
+
In many textbooks the \( 1/n \) term is often omitted. Note that a library like Scikit-Learn does not include the \( 1/n \) factor in the setup of the cost function.
+
+
When we compare this with the ordinary least squares result we have
which can lead to singular matrices. However, with the SVD, we can always compute the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \).
+
+
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. We will see that for specific values of \( \lambda \), we may
+even reduce the variance of the optimal parameters \( \boldsymbol{\beta} \). These topics and other related ones, will be discussed after the more linear algebra oriented analysis here.
-
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. More about this when we have covered the material on a statistical interpretation of various linear regression methods.
+
Using our insights about the SVD of the design matrix \( \boldsymbol{X} \)
+We have already analyzed the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as
+
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.
+
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 \( \sigma_i \geq
+\sigma_{i+1} \).
-
We will come back to more interpreations after we have gone through some of the statistical analysis part.
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. More about this when we have covered the material on a statistical interpretation of various linear regression methods.
Using the matrix-vector expression for Lasso regression and dropping the parameter \( 1/n \) in front of the standard means squared error equation, we have the following cost function
+
For the sake of simplicity, let us assume that the design matrix is orthonormal, that is
Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)
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.
+
-
This equation does not lead to a nice analytical equation as in either Ridge regression or ordinary least squares. This equation can however be solved by using standard convex optimization algorithms using for example the Python package CVXOPT. We will discuss this later.
+
We will come back to more interpreations after we have gone through some of the statistical analysis part.
The exercises here are meant to prepare you for work with project 1. The first exercise is a follow-up of exercise 2 from week 35 August 30-September 3).
-
-
-
Exercise 1: Setting up various Python environments
-
-
The first exercise here is of a mere technical art. We want you to have
-
-
git as a version control software and to establish a user account on a provider like GitHub. Other providers like GitLab etc are equally fine. You can also use the University of Oslo GitHub facilities.
-
Install various Python packages
-
-
We will make extensive use of Python as programming language and its
-myriad of available libraries. You will find
-IPython/Jupyter notebooks invaluable in your work. You can run R
-codes in the Jupyter/IPython notebooks, with the immediate benefit of
-visualizing your data. You can also use compiled languages like C++,
-Rust, Fortran etc if you prefer. The focus in these lectures will be
-on Python.
-
-
-
If you have Python installed (we recommend Python3) and you feel
-pretty familiar with installing different packages, we recommend that
-you install the following Python packages via pip as
-
For OSX users we recommend, after having installed Xcode, to
-install brew. Brew allows for a seamless installation of additional
-software via for example
-
-
-
-
brew install python3
-
-
For Linux users, with its variety of distributions like for example the widely popular Ubuntu distribution,
-you can use pip as well and simply install Python as
-
-
-
-
sudo apt-get install python3 (or python for Python2.7)
-
-
If you don't want to perform these operations separately and venture
-into the hassle of exploring how to set up dependencies and paths, we
-recommend two widely used distrubutions which set up all relevant
-dependencies for Python, namely
-
which is an open source
-distribution of the Python and R programming languages for large-scale
-data processing, predictive analytics, and scientific computing, that
-aims to simplify package management and deployment. Package versions
-are managed by the package management system conda.
-
is a Python
-distribution for scientific and analytic computing distribution and
-analysis environment, available for free and under a commercial
-license.
-
-
-
We recommend using Anaconda if you are not too familiar with setting paths in a terminal environment.
-
-
-
-
-
Exercise 2: making your own data and exploring scikit-learn
-
-
We will generate our own dataset for a function \( y(x) \) where \( x \in [0,1] \) and defined by random numbers computed with the uniform distribution. The function \( y \) is a quadratic polynomial in \( x \) with added stochastic noise according to the normal distribution \( \cal {N}(0,1) \).
-The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
-
-
-
-
-
-
-
-
-
x = np.random.rand(100,1)
-y =2.0+5*x*x+0.1*np.random.randn(100,1)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Write your own code (following the examples under the regression notes) for computing the parametrization of the data set fitting a second-order polynomial.
and the \( R^2 \) score function.
-If \( \tilde{\boldsymbol{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
-
You can use the functionality included in scikit-learn. If you feel for it, you can use your own program and define functions which compute the above two functions.
-Discuss the meaning of these results. Try also to vary the coefficient in front of the added stochastic noise term and discuss the quality of the fits.
-
The code here is an example of where we define our own design matrix and fit parameters \( \beta \).
-
-
-
-
-
-
-
-
importos
-importnumpyasnp
-importpandasaspd
-importmatplotlib.pyplotasplt
-fromsklearn.model_selectionimport train_test_split
-
-defsave_fig(fig_id):
- plt.savefig(image_path(fig_id) +".png", format='png')
-
-defR2(y_data, y_model):
- return1- np.sum((y_data - y_model) **2) / np.sum((y_data - np.mean(y_data)) **2)
-defMSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-x = np.random.rand(100)
-y =2.0+5*x*x+0.1*np.random.randn(100)
-
-
-# The design matrix now as function of a given polynomial
-X = np.zeros((len(x),3))
-X[:,0] =1.0
-X[:,1] = x
-X[:,2] = x**2
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-# matrix inversion to find beta
-beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
-print(beta)
-# 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))
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Exercise 3: Normalizing our data
-
-
A much used approach before starting to train the data is to preprocess our
-data. Normally the data may need a rescaling and/or may be sensitive
-to extreme values. Scaling the data renders our inputs much more
-suitable for the algorithms we want to employ.
-
-
-
Scikit-Learn has several functions which allow us to rescale the
-data, normally resulting in much better results in terms of various
-accuracy scores. The StandardScaler function in Scikit-Learn
-ensures that for each feature/predictor we study the mean value is
-zero and the variance is one (every column in the design/feature
-matrix). This scaling has the drawback that it does not ensure that
-we have a particular maximum or minimum in our data set. Another
-function included in Scikit-Learn is the MinMaxScaler which
-ensures that all features are exactly between \( 0 \) and \( 1 \). The
-
-
-
The Normalizer scales each data
-point such that the feature vector has a euclidean length of one. In other words, it
-projects a data point on the circle (or sphere in the case of higher dimensions) with a
-radius of 1. This means every data point is scaled by a different number (by the
-inverse of it’s length).
-This normalization is often used when only the direction (or angle) of the data matters,
-not the length of the feature vector.
-
-
-
The RobustScaler works similarly to the StandardScaler in that it
-ensures statistical properties for each feature that guarantee that
-they are on the same scale. However, the RobustScaler uses the median
-and quartiles, instead of mean and variance. This makes the
-RobustScaler ignore data points that are very different from the rest
-(like measurement errors). These odd data points are also called
-outliers, and might often lead to trouble for other scaling
-techniques.
-
-
-
It also common to split the data in a training set and a testing set. A typical split is to use \( 80\% \) of the data for training and the rest
-for testing. This can be done as follows with our design matrix \( \boldsymbol{X} \) and data \( \boldsymbol{y} \) (remember to import scikit-learn)
-
-
-
-
-
-
-
-
-
# split in training and test data
-X_train, X_test, y_train, y_test = train_test_split(X,y,test_size=0.2)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Then we can use the standard scaler to scale our data as
In this exercise we want you to to compute the MSE for the training
-data and the test data as function of the complexity of a polynomial,
-that is the degree of a given polynomial. We want you also to compute the \( R2 \) score as function of the complexity of the model for both training data and test data. You should also run the calculation with and without scaling.
-
-
-
One of
-the aims is to reproduce Figure 2.11 of Hastie et al.
-
-
-
Our data is defined by \( x\in [-3,3] \) with a total of for example \( 100 \) data points.
where \( y \) is the function we want to fit with a given polynomial.
-
-
-
-a)
-Write a first code which sets up a design matrix \( X \) defined by a fifth-order polynomial. Scale your data and split it in training and test data.
-
-
-
-
-
-
-b)
-Perform an ordinary least squares and compute the means squared error and the \( R2 \) factor for the training data and the test data, with and without scaling.
-
-
-
-
-
-
-c)
-Add now a model which allows you to make polynomials up to degree \( 15 \). Perform a standard OLS fitting of the training data and compute the MSE and \( R2 \) for the training and test data and plot both test and training data MSE and \( R2 \) as functions of the polynomial degree. Compare what you see with Figure 2.11 of Hastie et al. Comment your results. For which polynomial degree do you find an optimal MSE (smallest value)?
-
-
-
-
-
-
-
-
Exercise 4: Adding Ridge Regression
-
-
This exercise is a continuation of exercise 2. We will use the same function to
-generate our data set, still staying with a simple function \( y(x) \)
-which we want to fit using linear regression, but now extending the
-analysis to include the Ridge regression method.
-
-
-
We will thus again generate our own dataset for a function \( y(x) \) where
-\( x \in [0,1] \) and defined by random numbers computed with the uniform
-distribution. The function \( y \) is a quadratic polynomial in \( x \) with
-added stochastic noise according to the normal distribution \( \cal{N}(0,1) \).
-
-
-
The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
-
-
-
-
-
-
-
-
x = np.random.rand(100)
-y =2.0+5*x*x+0.1*np.random.randn(100)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 3. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
-
-
The code here allows you to perform your own Ridge calculation and
-perform calculations for various values of the regularization
-parameter \( \lambda \). This program can easily be extended upon.
-
-
-
-
-
-
-
-
-
-
importos
-importnumpyasnp
-importpandasaspd
-importmatplotlib.pyplotasplt
-fromsklearn.model_selectionimport train_test_split
-fromsklearn.preprocessingimport StandardScaler
-
-defR2(y_data, y_model):
- return1- np.sum((y_data - y_model) **2) / np.sum((y_data - np.mean(y_data)) **2)
-defMSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-
-# A seed just to ensure that the random numbers are the same for every run.
-# Useful for eventual debugging.
-np.random.seed(3155)
-
-x = np.random.rand(100)
-y =2.0+5*x*x+0.1*np.random.randn(100)
-
-# number of features p (here degree of polynomial
-p =3
-# The design matrix now as function of a given polynomial
-X = np.zeros((len(x),p))
-X[:,0] =1.0
-X[:,1] = x
-X[:,2] = x*x
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-
-# matrix inversion to find beta
-OLSbeta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
-print(OLSbeta)
-# and then make the prediction
-ytildeOLS = X_train @ OLSbeta
-print("Training R2 for OLS")
-print(R2(y_train,ytildeOLS))
-print("Training MSE for OLS")
-print(MSE(y_train,ytildeOLS))
-ypredictOLS = X_test @ OLSbeta
-print("Test R2 for OLS")
-print(R2(y_test,ypredictOLS))
-print("Test MSE OLS")
-print(MSE(y_test,ypredictOLS))
-
-# Repeat now for Ridge regression and various values of the regularization parameter
-I = np.eye(p,p)
-# Decide which values of lambda to use
-nlambdas =20
-MSEPredict = np.zeros(nlambdas)
-MSETrain = np.zeros(nlambdas)
-lambdas = np.logspace(-4, 1, nlambdas)
-for i inrange(nlambdas):
- lmb = lambdas[i]
- Ridgebeta = np.linalg.inv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
- # and then make the prediction
- ytildeRidge = X_train @ Ridgebeta
- ypredictRidge = X_test @ Ridgebeta
- MSEPredict[i] = MSE(y_test,ypredictRidge)
- MSETrain[i] = MSE(y_train,ytildeRidge)
-# Now plot the results
-plt.figure()
-plt.plot(np.log10(lambdas), MSETrain, label ='MSE Ridge train')
-plt.plot(np.log10(lambdas), MSEPredict, 'r--', label ='MSE Ridge Test')
-plt.xlabel('log10(lambda)')
-plt.ylabel('MSE')
-plt.legend()
-plt.show()
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Repeat the above but using the functionality of
-Scikit-Learn. Compare your code with the results from
-Scikit-Learn. Remember to run with the same random numbers for
-generating \( x \) and \( y \). Observe also that when you compare with Scikit-Learn, you need to pay attention to how the intercept is dealt with.
-
-
-
Finally, using Scikit-Learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
and the \( R^2 \) score function.
-If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
-
Discuss these quantities as functions of the variable \( \lambda \) in Ridge regression.
-
-
-
-
-
Exercise 5: Analytical exercises
-
-
In this exercise we derive the expressions for various derivatives of
-products of vectors and matrices. Such derivatives are central to the
-optimization of various cost functions. Although we will often use
-automatic differentiation in actual calculations, to be able to have
-analytical expressions is extremely helpful in case we have simpler
-derivatives as well as when we analyze various properties (like second
-derivatives) of the chosen cost functions. Vectors are always written
-as boldfaced lower case letters and matrices as upper case boldfaced
-letters.
-
and finally find the second derivative of this function with respect to the vector \( \boldsymbol{s} \).
-
-
Hint: In these exercises it is always useful to write out with summation indices the various quantities.
-As an example, consider the function
-
+
Using the matrix-vector expression for Lasso regression and dropping the parameter \( 1/n \) in front of the standard means squared error equation, we have the following cost function
which reads for a specific component \( f_i \) (we define the matrix \( \boldsymbol{A} \) to have dimension \( n\times n \) and the vector $\boldsymbol{x} to have length \( n \))
-
+
Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)
This equation does not lead to a nice analytical equation as in either Ridge regression or ordinary least squares. This equation can however be solved by using standard convex optimization algorithms using for example the Python package CVXOPT. We will discuss this later.
The exercises here are meant to prepare you for work with project 1. The first exercise is a follow-up of exercise 2 from week 35 August 30-September 3).
-
-The exercises here are meant to prepare you for work with project 1. The first exercise is a follow-up of exercise 2 from week 35 August 30-September 3).
-
-
+
Exercise 1: Setting up various Python environments
-
Exercise 1: Adding Ridge and Lasso Regression
+
The first exercise here is of a mere technical art. We want you to have
+
+
git as a version control software and to establish a user account on a provider like GitHub. Other providers like GitLab etc are equally fine. You can also use the University of Oslo GitHub facilities.
+
Install various Python packages
+
+
We will make extensive use of Python as programming language and its
+myriad of available libraries. You will find
+IPython/Jupyter notebooks invaluable in your work. You can run R
+codes in the Jupyter/IPython notebooks, with the immediate benefit of
+visualizing your data. You can also use compiled languages like C++,
+Rust, Fortran etc if you prefer. The focus in these lectures will be
+on Python.
+
+
+
If you have Python installed (we recommend Python3) and you feel
+pretty familiar with installing different packages, we recommend that
+you install the following Python packages via pip as
+
For OSX users we recommend, after having installed Xcode, to
+install brew. Brew allows for a seamless installation of additional
+software via for example
+
+
+
+
brew install python3
+
+
For Linux users, with its variety of distributions like for example the widely popular Ubuntu distribution,
+you can use pip as well and simply install Python as
+
+
+
+
sudo apt-get install python3 (or python for Python2.7)
+
+
If you don't want to perform these operations separately and venture
+into the hassle of exploring how to set up dependencies and paths, we
+recommend two widely used distrubutions which set up all relevant
+dependencies for Python, namely
+
which is an open source
+distribution of the Python and R programming languages for large-scale
+data processing, predictive analytics, and scientific computing, that
+aims to simplify package management and deployment. Package versions
+are managed by the package management system conda.
+
is a Python
+distribution for scientific and analytic computing distribution and
+analysis environment, available for free and under a commercial
+license.
+
+
+
We recommend using Anaconda if you are not too familiar with setting paths in a terminal environment.
+
+
+
+
+
Exercise 2: making your own data and exploring scikit-learn
+
+
We will generate our own dataset for a function \( y(x) \) where \( x \in [0,1] \) and defined by random numbers computed with the uniform distribution. The function \( y \) is a quadratic polynomial in \( x \) with added stochastic noise according to the normal distribution \( \cal {N}(0,1) \).
+The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
+
+
+
+
+
+
+
+
+
x = np.random.rand(100,1)
+y =2.0+5*x*x+0.1*np.random.randn(100,1)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Write your own code (following the examples under the regression notes) for computing the parametrization of the data set fitting a second-order polynomial.
and the \( R^2 \) score function.
+If \( \tilde{\boldsymbol{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
+
You can use the functionality included in scikit-learn. If you feel for it, you can use your own program and define functions which compute the above two functions.
+Discuss the meaning of these results. Try also to vary the coefficient in front of the added stochastic noise term and discuss the quality of the fits.
+
+
+
-This exercise is a continuation of exercise 2 from exercise set 1
-(week 35, August 30-September 3). We will use the same function to
+
The code here is an example of where we define our own design matrix and fit parameters \( \beta \).
+
+
+
+
+
+
+
+
importos
+importnumpyasnp
+importpandasaspd
+importmatplotlib.pyplotasplt
+fromsklearn.model_selectionimport train_test_split
+
+defsave_fig(fig_id):
+ plt.savefig(image_path(fig_id) +".png", format='png')
+
+defR2(y_data, y_model):
+ return1- np.sum((y_data - y_model) **2) / np.sum((y_data - np.mean(y_data)) **2)
+defMSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+x = np.random.rand(100)
+y =2.0+5*x*x+0.1*np.random.randn(100)
+
+
+# The design matrix now as function of a given polynomial
+X = np.zeros((len(x),3))
+X[:,0] =1.0
+X[:,1] = x
+X[:,2] = x**2
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
+print(beta)
+# 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))
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Exercise 3: Normalizing our data
+
+
A much used approach before starting to train the data is to preprocess our
+data. Normally the data may need a rescaling and/or may be sensitive
+to extreme values. Scaling the data renders our inputs much more
+suitable for the algorithms we want to employ.
+
+
+
Scikit-Learn has several functions which allow us to rescale the
+data, normally resulting in much better results in terms of various
+accuracy scores. The StandardScaler function in Scikit-Learn
+ensures that for each feature/predictor we study the mean value is
+zero and the variance is one (every column in the design/feature
+matrix). This scaling has the drawback that it does not ensure that
+we have a particular maximum or minimum in our data set. Another
+function included in Scikit-Learn is the MinMaxScaler which
+ensures that all features are exactly between \( 0 \) and \( 1 \). The
+
+
+
The Normalizer scales each data
+point such that the feature vector has a euclidean length of one. In other words, it
+projects a data point on the circle (or sphere in the case of higher dimensions) with a
+radius of 1. This means every data point is scaled by a different number (by the
+inverse of it’s length).
+This normalization is often used when only the direction (or angle) of the data matters,
+not the length of the feature vector.
+
+
+
The RobustScaler works similarly to the StandardScaler in that it
+ensures statistical properties for each feature that guarantee that
+they are on the same scale. However, the RobustScaler uses the median
+and quartiles, instead of mean and variance. This makes the
+RobustScaler ignore data points that are very different from the rest
+(like measurement errors). These odd data points are also called
+outliers, and might often lead to trouble for other scaling
+techniques.
+
+
+
It also common to split the data in a training set and a testing set. A typical split is to use \( 80\% \) of the data for training and the rest
+for testing. This can be done as follows with our design matrix \( \boldsymbol{X} \) and data \( \boldsymbol{y} \) (remember to import scikit-learn)
+
+
+
+
+
+
+
+
+
# split in training and test data
+X_train, X_test, y_train, y_test = train_test_split(X,y,test_size=0.2)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Then we can use the standard scaler to scale our data as
In this exercise we want you to to compute the MSE for the training
+data and the test data as function of the complexity of a polynomial,
+that is the degree of a given polynomial. We want you also to compute the \( R2 \) score as function of the complexity of the model for both training data and test data. You should also run the calculation with and without scaling.
+
+
+
One of
+the aims is to reproduce Figure 2.11 of Hastie et al.
+
+
+
Our data is defined by \( x\in [-3,3] \) with a total of for example \( 100 \) data points.
where \( y \) is the function we want to fit with a given polynomial.
+
+
+
+a)
+Write a first code which sets up a design matrix \( X \) defined by a fifth-order polynomial. Scale your data and split it in training and test data.
+
+
+
+
+
+
+b)
+Perform an ordinary least squares and compute the means squared error and the \( R2 \) factor for the training data and the test data, with and without scaling.
+
+
+
+
+
+
+c)
+Add now a model which allows you to make polynomials up to degree \( 15 \). Perform a standard OLS fitting of the training data and compute the MSE and \( R2 \) for the training and test data and plot both test and training data MSE and \( R2 \) as functions of the polynomial degree. Compare what you see with Figure 2.11 of Hastie et al. Comment your results. For which polynomial degree do you find an optimal MSE (smallest value)?
+
+
+
+
+
+
+
+
Exercise 4: Adding Ridge Regression
+
+
This exercise is a continuation of exercise 2. We will use the same function to
generate our data set, still staying with a simple function \( y(x) \)
which we want to fit using linear regression, but now extending the
-analysis to include the Ridge and the Lasso regression methods.
+analysis to include the Ridge regression method.
+
-
-We will thus again generate our own dataset for a function \( y(x) \) where
+
We will thus again generate our own dataset for a function \( y(x) \) where
\( x \in [0,1] \) and defined by random numbers computed with the uniform
distribution. The function \( y \) is a quadratic polynomial in \( x \) with
added stochastic noise according to the normal distribution \( \cal{N}(0,1) \).
+
-
-The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
-
+
The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
-
x = np.random.rand(100)
+
+
+
+
+
+
x = np.random.rand(100)
y =2.0+5*x*x+0.1*np.random.randn(100)
-
-
-Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 3. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
-The code here allows you to perform your own Ridge calculation and
+
Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 3. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
+
+
The code here allows you to perform your own Ridge calculation and
perform calculations for various values of the regularization
parameter \( \lambda \). This program can easily be extended upon.
+
-
-
importos
+
+
+
+
+
+
importosimportnumpyasnpimportpandasaspdimportmatplotlib.pyplotasplt
@@ -463,10 +859,6 @@ X[:,1] =
X[:,2] = x*x
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
# matrix inversion to find beta
OLSbeta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
@@ -506,204 +898,105 @@ plt.xlabel('
plt.ylabel('MSE')
plt.legend()
plt.show()
-
-
-Repeat the above but using the functionality of
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Repeat the above but using the functionality of
Scikit-Learn. Compare your code with the results from
Scikit-Learn. Remember to run with the same random numbers for
generating \( x \) and \( y \). Observe also that when you compare with Scikit-Learn, you need to pay attention to how the intercept is dealt with.
+
-
-Finally, using Scikit-Learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
+
Finally, using Scikit-Learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
and the \( R^2 \) score function.
If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
+
$$
R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
$$
-where we have defined the mean value of \( \hat{y} \) as
+
where we have defined the mean value of \( \hat{y} \) as
$$
\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
$$
-Discuss these quantities as functions of the variable \( \lambda \) in the Ridge and Lasso regression methods.
+
Discuss these quantities as functions of the variable \( \lambda \) in Ridge regression.
-
Exercise: Linear Regression for a two-dimensional function
-
-
-This is a longer exercise and the aim is to study in more detail various
-regression methods, including the Ordinary Least Squares (OLS) method,
-Ridge regression and finally Lasso regression.
-This exercise forms a part of project 1.
-
-
-We will study how to fit polynomials to a specific
-two-dimensional function called Franke's
-function. This
-is a function which has been widely used when testing various
-interpolation and fitting algorithms.
-
-
-The Franke function, which is a weighted sum of four exponentials reads as follows
-$$
-\begin{align*}
-f(x,y) &= \frac{3}{4}\exp{\left(-\frac{(9x-2)^2}{4} - \frac{(9y-2)^2}{4}\right)}+\frac{3}{4}\exp{\left(-\frac{(9x+1)^2}{49}- \frac{(9y+1)}{10}\right)} \\
-&+\frac{1}{2}\exp{\left(-\frac{(9x-7)^2}{4} - \frac{(9y-3)^2}{4}\right)} -\frac{1}{5}\exp{\left(-(9x-4)^2 - (9y-7)^2\right) }.
-\end{align*}
-$$
-
-
-The function will be defined for \( x,y\in [0,1] \). Our first step will
-be to perform an OLS regression analysis of this function, trying out
-a polynomial fit with an \( x \) and \( y \) dependence of the form \( [x, y,
-x^2, y^2, xy, \dots] \). We will fit a
-function (for example a polynomial) of \( x \) and \( y \). Thereafter we
-will repeat much of the same procedure using the Ridge and Lasso
-regression methods, introducing thus a dependence on the bias
-(penalty) \( \lambda \).
-
-
-The Python fucntion for the Franke function is included here (it performs also a three-dimensional plot of it)
-
-
-
-
frommpl_toolkits.mplot3dimport Axes3D
-importmatplotlib.pyplotasplt
-frommatplotlibimport cm
-frommatplotlib.tickerimport LinearLocator, FormatStrFormatter
-importnumpyasnp
-fromrandomimport random, seed
-
-fig = plt.figure()
-ax = fig.gca(projection='3d')
-
-# Make data.
-x = np.arange(0, 1, 0.05)
-y = np.arange(0, 1, 0.05)
-x, y = np.meshgrid(x,y)
-
-
-defFrankeFunction(x,y):
- term1 =0.75*np.exp(-(0.25*(9*x-2)**2) -0.25*((9*y-2)**2))
- term2 =0.75*np.exp(-((9*x+1)**2)/49.0-0.1*(9*y+1))
- term3 =0.5*np.exp(-(9*x-7)**2/4.0-0.25*((9*y-3)**2))
- term4 =-0.2*np.exp(-(9*x-4)**2- (9*y-7)**2)
- return term1 + term2 + term3 + term4
-
-
-z = FrankeFunction(x, y)
-
-# Plot the surface.
-surf = ax.plot_surface(x, y, z, cmap=cm.coolwarm,
- linewidth=0, antialiased=False)
-
-# Customize the z axis.
-ax.set_zlim(-0.10, 1.40)
-ax.zaxis.set_major_locator(LinearLocator(10))
-ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
-
-# Add a color bar which maps values to colors.
-fig.colorbar(surf, shrink=0.5, aspect=5)
-
-plt.show()
-
-
-We will generate our own dataset for a function
-\( \mathrm{FrankeFunction}(x,y) \) with \( x,y \in [0,1] \). The function
-\( f(x,y) \) is the Franke function. You should explore also the addition
-an added stochastic noise to this function using the normal
-distribution \( \cal{N}(0,1) \).
-
-
-Write your own code (using either a matrix inversion or a singular
-value decomposition from e.g., numpy ) or use your code and perform a standard least square regression
-analysis using polynomials in \( x \) and \( y \) up to fifth order. You can use scikit-learn as well.
-
-
-Evaluate the Mean Squared error (MSE)
-
-$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
-\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2,
-$$
-
-
-and the \( R^2 \) score function. If \( \tilde{\hat{y}}_i \) is the predicted
-value of the \( i-th \) sample and \( y_i \) is the corresponding true value,
-then the score \( R^2 \) is defined as
-
-$$
-R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
-$$
-
-
-where we have defined the mean value of \( \hat{y} \) as
-
-$$
-\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
-$$
-
-
-You should split your data in train and test and also consider scaling the data.
-
-
-To set up the design matrix, the following code can be used
-
-
-
-
defFrankeFunction(x,y):
- term1 =0.75*np.exp(-(0.25*(9*x-2)**2) -0.25*((9*y-2)**2))
- term2 =0.75*np.exp(-((9*x+1)**2)/49.0-0.1*(9*y+1))
- term3 =0.5*np.exp(-(9*x-7)**2/4.0-0.25*((9*y-3)**2))
- term4 =-0.2*np.exp(-(9*x-4)**2- (9*y-7)**2)
- return term1 + term2 + term3 + term4
-
-
-defcreate_X(x, y, n ):
- iflen(x.shape) >1:
- x = np.ravel(x)
- y = np.ravel(y)
-
- N =len(x)
- l =int((n+1)*(n+2)/2) # Number of elements in beta
- X = np.ones((N,l))
-
- for i inrange(1,n+1):
- q =int((i)*(i+1)/2)
- for k inrange(i+1):
- X[:,q+k] = (x**(i-k))*(y**k)
-
- return X
-
-
-# Making meshgrid of datapoints and compute Franke's function
-n =5
-N =1000
-x = np.sort(np.random.uniform(0, 1, N))
-y = np.sort(np.random.uniform(0, 1, N))
-z = FrankeFunction(x, y)
-X = create_X(x, y, n=n)
-
-
-Write then your own code for the Ridge method or use Scikit-Learn.
-Perform the same analysis as you did for ordinary Least Squares (for the same polynomials) but now for different values of \( \lambda \). Compare and
-analyze your results with those obtained with ordinary Least Squares. Study the
-dependence on \( \lambda \).
-
-
-This part is essentially a repeat of the previous ones, but now
-with Lasso regression. Write either your own code or
-use the functionalities of Scikit-Learn (recommended).
-Give a
-critical discussion of the three methods and a judgement of which
-model fits the data best.
-
-
+
+
Exercise 5: Analytical exercises
+
+
In this exercise we derive the expressions for various derivatives of
+products of vectors and matrices. Such derivatives are central to the
+optimization of various cost functions. Although we will often use
+automatic differentiation in actual calculations, to be able to have
+analytical expressions is extremely helpful in case we have simpler
+derivatives as well as when we analyze various properties (like second
+derivatives) of the chosen cost functions. Vectors are always written
+as boldfaced lower case letters and matrices as upper case boldfaced
+letters.
+
which reads for a specific component \( f_i \) (we define the matrix \( \boldsymbol{A} \) to have dimension \( n\times n \) and the vector $\boldsymbol{x} to have length \( n \))
diff --git a/doc/pub/week35/html/week35-reveal.html b/doc/pub/week35/html/week35-reveal.html
index 1da9a75fa..6050668ba 100644
--- a/doc/pub/week35/html/week35-reveal.html
+++ b/doc/pub/week35/html/week35-reveal.html
@@ -806,6 +806,12 @@ Jacobian is only a single-column vector, or an \( m\times 1 \) matrix. If
on the other hand \( \boldsymbol{y} \) is a scalar, the Jacobian becomes a
\( 1\times n \) matrix.
+
+
When this matrix is a square matrix \( m=n \), its determinant is often referred to as the Jacobian
+determinant. Both the matrix and (if \( m=n \)) the determinant are
+often referred to simply as the Jacobian. The Jacobian matrix represents the differential of \( \boldsymbol{y} \) at every point where the
+vector is differentiable.
+
We define a scalar (our cast functions are in general also scalars, think of the mean squared error) as the result of some matrix vector multiplications
with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors.
+In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first
+\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then
+
+$$
+
+\alpha = \boldsymbol{z}^T\boldsymbol{x}$,
+
+$$
+
+
which means that (using our previous example) we have
Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
See the jupyter-book (complete lecture notes) for the derivations of these relations.
diff --git a/doc/pub/week35/html/week35-solarized.html b/doc/pub/week35/html/week35-solarized.html
index ea38042b3..e3ca99cc1 100644
--- a/doc/pub/week35/html/week35-solarized.html
+++ b/doc/pub/week35/html/week35-solarized.html
@@ -123,6 +123,7 @@ div.toc p,a {
'some-useful-matrix-and-vector-expressions'),
('The Jacobian', 2, None, 'the-jacobian'),
('Derivatives, example 1', 2, None, 'derivatives-example-1'),
+ ('Example 2', 2, None, 'example-2'),
('Meet the Hessian Matrix', 2, None, 'meet-the-hessian-matrix'),
('Interpretations and optimizing our parameters',
2,
@@ -888,6 +889,12 @@ on the other hand \( \boldsymbol{y} \) is a scalar, the Jacobian becomes a
\( 1\times n \) matrix.
+
When this matrix is a square matrix \( m=n \), its determinant is often referred to as the Jacobian
+determinant. Both the matrix and (if \( m=n \)) the determinant are
+often referred to simply as the Jacobian. The Jacobian matrix represents the differential of \( \boldsymbol{y} \) at every point where the
+vector is differentiable.
+
+
Derivatives, example 1
@@ -911,6 +918,34 @@ $$
$$
+
+
Example 2
+
+
We define a scalar (our cast functions are in general also scalars, think of the mean squared error) as the result of some matrix vector multiplications
with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors.
+In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first
+\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then
+
Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
+$$
+\frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T..
+$$
+
+
$$
\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
$$
diff --git a/doc/pub/week35/html/week35.html b/doc/pub/week35/html/week35.html
index 4a8b811ab..0125443ab 100644
--- a/doc/pub/week35/html/week35.html
+++ b/doc/pub/week35/html/week35.html
@@ -200,6 +200,7 @@ div.toc p,a {
'some-useful-matrix-and-vector-expressions'),
('The Jacobian', 2, None, 'the-jacobian'),
('Derivatives, example 1', 2, None, 'derivatives-example-1'),
+ ('Example 2', 2, None, 'example-2'),
('Meet the Hessian Matrix', 2, None, 'meet-the-hessian-matrix'),
('Interpretations and optimizing our parameters',
2,
@@ -965,6 +966,12 @@ on the other hand \( \boldsymbol{y} \) is a scalar, the Jacobian becomes a
\( 1\times n \) matrix.
+
When this matrix is a square matrix \( m=n \), its determinant is often referred to as the Jacobian
+determinant. Both the matrix and (if \( m=n \)) the determinant are
+often referred to simply as the Jacobian. The Jacobian matrix represents the differential of \( \boldsymbol{y} \) at every point where the
+vector is differentiable.
+
+
Derivatives, example 1
@@ -988,6 +995,34 @@ $$
$$
+
+
Example 2
+
+
We define a scalar (our cast functions are in general also scalars, think of the mean squared error) as the result of some matrix vector multiplications
with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors.
+In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first
+\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then
+
Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that