-The expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is -our optimization problem is -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. -$$ - -or we can state it as -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2, -$$ - -where we have used the definition of a norm-2 vector, that is -$$ -\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. -$$ - -
-By 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 - -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2 -$$ - -
-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 - -$$ -C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1, -$$ - -
-we have a new optimization equation -$$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1 -$$ - -which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. - -
-Here we have defined the norm-1 as -$$ -\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. -$$ +
@@ -310,7 +415,7 @@ $$
- + -
-When the repetitive splitting of the data set is done randomly, -samples may accidently end up in a fast majority of the splits in -either training or test set. Such samples may have an unbalanced -influence on either model building or prediction evaluation. To avoid -this \( k \)-fold cross-validation structures the data splitting. The -samples are divided into \( k \) more or less equally sized exhaustive and -mutually exclusive subsets. In turn (at each split) one of these -subsets plays the role of the test set while the union of the -remaining subsets constitutes the training set. Such a splitting -warrants a balanced representation of each sample in both training and -test set over the splits. Still the division into the \( k \) subsets -involves a degree of randomness. This may be fully excluded when -choosing \( k=n \). This particular case is referred to as leave-one-out -cross-validation (LOOCV). +The expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is +our optimization problem is +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +$$ + +or we can state it as +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2, +$$ + +where we have used the definition of a norm-2 vector, that is +$$ +\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. +$$ + +
+By 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 + +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2 +$$ + +
+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 + +$$ +C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1, +$$ + +
+we have a new optimization equation +$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1 +$$ + +which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. + +
+Here we have defined the norm-1 as +$$ +\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. +$$
@@ -274,7 +469,7 @@ cross-validation (LOOCV).
+When the repetitive splitting of the data set is done randomly, +samples may accidently end up in a fast majority of the splits in +either training or test set. Such samples may have an unbalanced +influence on either model building or prediction evaluation. To avoid +this \( k \)-fold cross-validation structures the data splitting. The +samples are divided into \( k \) more or less equally sized exhaustive and +mutually exclusive subsets. In turn (at each split) one of these +subsets plays the role of the test set while the union of the +remaining subsets constitutes the training set. Such a splitting +warrants a balanced representation of each sample in both training and +test set over the splits. Still the division into the \( k \) subsets +involves a degree of randomness. This may be fully excluded when +choosing \( k=n \). This particular case is referred to as leave-one-out +cross-validation (LOOCV).
@@ -286,7 +433,7 @@ $$
- + -
-For the various values of \( k \) - -
diff --git a/doc/pub/week38/html/._week38-bs006.html b/doc/pub/week38/html/._week38-bs006.html index c03c94679..9e7552f09 100644 --- a/doc/pub/week38/html/._week38-bs006.html +++ b/doc/pub/week38/html/._week38-bs006.html @@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source @@ -185,43 +299,87 @@ MathJax.Hub.Config({ Contents @@ -237,104 +395,26 @@ MathJax.Hub.Config({ -
-The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial. -
+For the various values of \( k \) - -
import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.model_selection import KFold
-from sklearn.linear_model import Ridge
-from sklearn.model_selection import cross_val_score
-from sklearn.preprocessing import PolynomialFeatures
+
+- shuffle the dataset randomly.
+- Split the dataset into \( k \) groups.
+- For each unique group:
-# A seed just to ensure that the random numbers are the same for every run.
-# Useful for eventual debugging.
-np.random.seed(3155)
+
+- Decide which group to use as set for test data
+- Take the remaining groups as a training data set
+- Fit a model on the training set and evaluate it on the test set
+- Retain the evaluation score and discard the model
+
-# Generate the data.
-nsamples = 100
-x = np.random.randn(nsamples)
-y = 3*x**2 + np.random.randn(nsamples)
+
diff --git a/doc/pub/week38/html/._week38-bs007.html b/doc/pub/week38/html/._week38-bs007.html index dd37227aa..f8572aec5 100644 --- a/doc/pub/week38/html/._week38-bs007.html +++ b/doc/pub/week38/html/._week38-bs007.html @@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source @@ -185,43 +299,87 @@ MathJax.Hub.Config({ Contents @@ -237,416 +395,99 @@ MathJax.Hub.Config({ -
-When you are comparing your own code with for example Scikit-Learn's -library, there are some things to keep in mind. The examples -here demonstrate some of these aspects with potential pitfalls. - -
-The discussion here focuses on the role of the intercept, how we can -set up the design matrix, what scaling we should use and other topics -which may confuse us. - -
-Yes, it could be a bad idea to include the intercept column for the -exact reason you stated. If no transformation is applied to your data, -the intercept can be interpreted as the expected value of your target -variable when all your predictors are put to zero. Therefore, whenever -you cannot assume that the expected target variable is zero when all -your predictors are zero, it could be a bad idea to apply a model -which penalizes the intercept. Also, the analytical solution to the -ridge regression coefficients (when not shrinking $$\beta_0$$) is -derived under the assumption that both y and X are zero centered (mean -subtracted). What you are doing is correct, but you should also zero -center X (subtracting the mean of each column from the corresponding -column). - -
-If your predictors are of different scales, I would advice you to -standardize X by subtracting the mean of each column from the -corresponding column and dividing the column with its standard -deviation. If you dont do this, you will give an "unfair" penalization -of the parameters since their magnitude depends on the scale of their -corresponding predictor. Suppose that you have an input variable -"height". Human height might be measured in inches or meters or -kilometers. If measured in kilometers, a standard linear regression -model with this predictor would probably give a much bigger -coefficient term, than if measured in millimeters. You may see how -this could become a problem when considering the loss function for -ridge regression. - -
-Remember that when you do any transformation to your dataset before -training, the exact same transformation has to be applied to new data -before making a prediction. In your case, this means: - -
- - -
#Model training:
-y_train_mean = np.mean(y_train)
-X_train_mean = np.mean(X_train,axis=0)
-X_train = X_train - X_train_mean
-y_train = y_train - y_train_mean
-
-trained_model = some_model.fit(X_train,y_train)
-
-#Model prediction:
-X_test = X_test - X_train_mean #Use mean from training data
-y_pred = trained_model(X_test)
-y_pred = y_pred + y_train_mean
--Here is a mathematical explanation of the zero centering: - -
-The cost/loss function for Ridge regression is: - -$$ -C(\beta_0, \beta_1, ... , \beta_P) = \sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip}\beta_p)^2 + \lambda \sum_{p=1}^P \beta_p^2. -$$ - -
-Notice that the intercept is left out of the \( L_2 \) regularization term. The design matrix -\( X \) does in this case not contain any intercept column. We want - -$$ -\frac{\partial L}{\partial \beta_j} = 0, -$$ - -
-for all \( j \), so lets start with \( \beta_0 \). This means that we have - -$$ -\frac{\partial L}{\partial \beta_0} = -2\sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip} \beta_p). -$$ - -
-We want to solve -$$ --2\sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip} \beta_p) = 0, -$$ - -
-which gives -$$ -\sum_{i=1}^{n} \beta_0 = \sum_{i=1}^{n}y_i - \sum_{i=1}^{n} \sum_{p=1}^P X_{ip} \beta_p, -$$ - -
-or -$ n\beta_0 = \sum_{i=1}^{n} y_i - \sum_{p=1}^P\beta_p \sum_{i=1}^{n} X_{ip}$. - -
-If we assume that every column of \( X \) is centered, whic we can do by subtracting the mean, -
- - -
X = X - np.mean(X,axis=0)
--the sum $ \sum_{i=1}^{n} X_{ip} $ - -
-can be rewritten as -$$ -\sum_{i=1}^{n} (X_{ip} - \frac{1}{n}\sum_{i=1}^{n} X_{ip}) = \sum_{i=1}^{n} X_{ip} - \sum_{i=1}^{n} \frac{1}{n} \sum_{i=1}^{n}X_{ip}, -$$ - -resulting in -$$ -\sum_{i=1}^{n} X_{ip} - n \frac{1}{n} \sum_{i=1}^{n}X_{ip} = 0. -$$ - -
-Finally we have -$$ -n\beta_0 = \sum_{i=1}^{n} y_i - \sum_{p=1}^P\beta_p \sum_{i=1}^{n} X_{ip}, -$$ - -or -$$ -\beta_0 = \frac{1}{n}\sum_{i=1}^{n} y_i = y_{average}. -$$ - -
-Replacing \( y_i \) with \( y_i - \beta_0 = y_i - y_{average} \) in the loss function will give us (in vector-matrix disguise) -$$ -C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}, -$$ - -
-which has the solution - -
-\( \beta = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}} \). -where \( \boldsymbol{\tilde{y}} = \boldsymbol{y} - y_{average} \) -and \( \tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=1}^{n-1}X_{kj} \). - +The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial.
import numpy as np
-import pandas as pd
import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
-from sklearn import linear_model
-
-def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
-def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
+from sklearn.model_selection import KFold
+from sklearn.linear_model import Ridge
+from sklearn.model_selection import cross_val_score
+from sklearn.preprocessing import PolynomialFeatures
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)
-n = 100
-x = np.random.rand(n)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+# Generate the data.
+nsamples = 100
+x = np.random.randn(nsamples)
+y = 3*x**2 + np.random.randn(nsamples)
-Maxpolydegree = 20
-X = np.zeros((n,Maxpolydegree))
-X[:,0] = 1.0
+## Cross-validation on Ridge regression using KFold only
-for polydegree in range(1, Maxpolydegree):
- for degree in range(polydegree):
- X[:,degree] = x**degree
+# Decide degree on polynomial to fit
+poly = PolynomialFeatures(degree = 6)
-
-# 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.pinv(X_train.T @ X_train) @ X_train.T @ y_train
-print(OLSbeta)
-# and then make the prediction
-ytildeOLS = X_train @ OLSbeta
-print("Training MSE for OLS")
-print(MSE(y_train,ytildeOLS))
-ypredictOLS = X_test @ OLSbeta
-print("Test MSE OLS")
-print(MSE(y_test,ypredictOLS))
-
-p = len(OLSbeta)
-I = np.eye(p,p)
# Decide which values of lambda to use
-nlambdas = 4
-MSEOwnRidgePredict = np.zeros(nlambdas)
-MSEOwnRidgeTrain = np.zeros(nlambdas)
-MSERidgePredict = np.zeros(nlambdas)
-MSERidgeTrain = np.zeros(nlambdas)
+nlambdas = 500
+lambdas = np.logspace(-3, 5, nlambdas)
+
+# Initialize a KFold instance
+k = 5
+kfold = KFold(n_splits = k)
+
+# Perform the cross-validation to estimate MSE
+scores_KFold = np.zeros((nlambdas, k))
+
+i = 0
+for lmb in lambdas:
+ ridge = Ridge(alpha = lmb)
+ j = 0
+ for train_inds, test_inds in kfold.split(x):
+ xtrain = x[train_inds]
+ ytrain = y[train_inds]
+
+ xtest = x[test_inds]
+ ytest = y[test_inds]
+
+ Xtrain = poly.fit_transform(xtrain[:, np.newaxis])
+ ridge.fit(Xtrain, ytrain[:, np.newaxis])
+
+ Xtest = poly.fit_transform(xtest[:, np.newaxis])
+ ypred = ridge.predict(Xtest)
+
+ scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)
+
+ j += 1
+ i += 1
+
+
+estimated_mse_KFold = np.mean(scores_KFold, axis = 1)
+
+## Cross-validation using cross_val_score from sklearn along with KFold
+
+# kfold is an instance initialized above as:
+# kfold = KFold(n_splits = k)
+
+estimated_mse_sklearn = np.zeros(nlambdas)
+i = 0
+for lmb in lambdas:
+ ridge = Ridge(alpha = lmb)
+
+ X = poly.fit_transform(x[:, np.newaxis])
+ estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)
+
+ # cross_val_score return an array containing the estimated negative mse for every fold.
+ # we have to the the mean of every array in order to get an estimate of the mse of the model
+ estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
+
+ i += 1
+
+## Plot and compare the slightly different ways to perform cross-validation
-lambdas = np.logspace(-4, 4, nlambdas)
-for i in range(nlambdas):
- lmb = lambdas[i]
- OwnRidgeBeta = np.linalg.pinv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
- # include lasso using Scikit-Learn
- # Note: we include the intercept
- RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
- RegRidge.fit(X_train,y_train)
- # and then make the prediction
- ytildeOwnRidge = X_train @ OwnRidgeBeta
- ypredictOwnRidge = X_test @ OwnRidgeBeta
- ytildeRidge = RegRidge.predict(X_train)
- ypredictRidge = RegRidge.predict(X_test)
- MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
- MSEOwnRidgeTrain[i] = MSE(y_train,ytildeOwnRidge)
- MSERidgePredict[i] = MSE(y_test,ypredictRidge)
- MSERidgeTrain[i] = MSE(y_train,ytildeRidge)
- print("Beta values for own Ridge implementation")
- print(OwnRidgeBeta)
- print("Beta values for Scikit-Learn Ridge implementation")
- print(RegRidge.coef_)
-# Now plot the results
plt.figure()
-plt.plot(np.log10(lambdas), MSEOwnRidgeTrain, 'b', label = 'MSE Ridge train')
-plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE Ridge Test')
-plt.plot(np.log10(lambdas), MSERidgeTrain, 'y', label = 'MSE Ridge train')
-plt.plot(np.log10(lambdas), MSERidgePredict, 'g', label = 'MSE Ridge Test')
+
+plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
+plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')
plt.xlabel('log10(lambda)')
-plt.ylabel('MSE')
-plt.legend()
-plt.show()
-+plt.ylabel('mse') - -
import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
-from sklearn import linear_model
-from sklearn.preprocessing import StandardScaler
-
-def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
-def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-
-# A seed just to ensure that the random numbers are the same for every run.
-# Useful for eventual debugging.
-np.random.seed(315)
-
-n = 100
-x = np.random.rand(n)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
-
-Maxpolydegree = 5
-X = np.zeros((n,Maxpolydegree-1))
-
-for degree in range(1,Maxpolydegree): #No intercept column
- X[:,degree-1] = x**(degree)
-
-
-
-
-# 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)
-
-
-
-
-
-#For our own implementation, we will need to deal with the intercept by centering the design matrix and the target variable
-X_train_mean = np.mean(X_train,axis=0)
-X_train_scaled = X_train - X_train_mean #Center by removing mean from each feature
-X_test_scaled = X_test - X_train_mean
-
-y_scaler = np.mean(y_train) #The model intercept (called y_scaler) is given by the mean of target variable (IF X is centered)
-y_train_scaled = y_train - y_scaler #Remove the intercept from the training data.
-
-
-p = Maxpolydegree-1
-I = np.eye(p,p)
-# Decide which values of lambda to use
-nlambdas = 4
-MSEOwnRidgePredict = np.zeros(nlambdas)
-MSERidgePredict = np.zeros(nlambdas)
-
-lambdas = np.logspace(-4, 1, nlambdas)
-for i in range(nlambdas):
- lmb = lambdas[i]
- OwnRidgeBeta = np.linalg.pinv(X_train_scaled.T @ X_train_scaled+lmb*I) @ X_train_scaled.T @ (y_train_scaled)
- intercept_ = y_scaler - X_train_mean@OwnRidgeBeta #The intercept can be shifted so the model can predict on uncentered data
-
- ypredictOwnRidge = X_test @ OwnRidgeBeta + intercept_ #Add intercept to prediction
- #EQUIVALENT PREDICTION:
- ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler #Add intercept to prediction
- print("Values for own Ridge prediction")
- print(ypredictOwnRidge)
-
-
-
- RegRidge = linear_model.Ridge(lmb)
- RegRidge.fit(X_train,y_train)
- ypredictRidge = RegRidge.predict(X_test)
- print("Values for SL Ridge prediction")
- print(ypredictRidge)
-
-
- MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
- MSERidgePredict[i] = MSE(y_test,ypredictRidge)
-
- print("Beta values for own Ridge implementation")
- print(OwnRidgeBeta) #Intercept is given by mean of target variable
- print("Beta values for Scikit-Learn Ridge implementation")
- print(RegRidge.coef_)
- print('Intercept from own implementation:')
- print(intercept_)
- print('Intercept from Scikit-Learn Ridge implementation')
- print(RegRidge.intercept_)
-
-
-
-# Now plot the results
-
-plt.figure()
-plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'b--', label = 'MSE own Ridge Test')
-plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
-
-plt.xlabel('log10(lambda)')
-plt.ylabel('MSE')
-plt.legend()
-plt.show()
-- - -
import numpy as np
-import matplotlib.pyplot as plt
-
-from sklearn.linear_model import LinearRegression
-
-
-np.random.seed(2021)
-
-
-def fit_beta(X, y):
- return np.linalg.pinv(X.T @ X) @ X.T @ y
-
-
-true_beta = [2, 0.5, 3.7]
-
-x = np.linspace(0, 1, 11)
-y = np.sum(
- np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
-) + 0.1 * np.random.normal(size=len(x))
-
-degree = 3
-X = np.zeros((len(x), degree))
-
-# Include the intercept in the design matrix
-for p in range(degree):
- X[:, p] = x ** p
-
-beta = fit_beta(X, y)
-
-# Intercept is included in the design matrix
-clf = LinearRegression(fit_intercept=False).fit(X, y)
-
-print(f"True beta: {true_beta}")
-print(f"Fitted beta: {beta}")
-print(f"Sklearn fitted beta: {clf.coef_}")
-
-
-plt.figure()
-plt.scatter(x, y, label="Data")
-plt.plot(x, X @ beta, label="Fit")
-plt.plot(x, clf.predict(X), label="Sklearn (fit_intercept=False)")
-
-
-# Do not include the intercept in the design matrix
-X = np.zeros((len(x), degree - 1))
-
-for p in range(degree - 1):
- X[:, p] = x ** (p + 1)
-
-# Intercept is not included in the design matrix
-clf = LinearRegression(fit_intercept=True).fit(X, y)
-
-# Use centered values for X and y when computing coefficients
-y_offset = np.average(y, axis=0)
-X_offset = np.average(X, axis=0)
-
-beta = fit_beta(X - X_offset, y - y_offset)
-intercept = np.mean(y_offset - X_offset @ beta)
-
-print(f"Manual intercept: {intercept}")
-print(f"Fitted beta (sans intercept): {beta}")
-print(f"Sklearn intercept: {clf.intercept_}")
-print(f"Sklearn fitted beta (sans intercept): {clf.coef_}")
-
-plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
-plt.plot(x, clf.predict(X), "--", label="Sklearn (fit_intercept=True)")
-plt.grid()
plt.legend()
plt.show()
@@ -674,7 +515,7 @@ plt.show()
-The one-dimensional Ising model with nearest neighbor interaction, no -external field and a constant coupling constant \( J \) is given by +When you are comparing your own code with for example Scikit-Learn's +library, there are some things to keep in mind. The examples +here demonstrate some of these aspects with potential pitfalls. + +
+The discussion here focuses on the role of the intercept, how we can +set up the design matrix, what scaling we should use and other topics +which may confuse us. + +
+Yes, it could be a bad idea to include the intercept column for the +exact reason you stated. If no transformation is applied to your data, +the intercept can be interpreted as the expected value of your target +variable when all your predictors are put to zero. Therefore, whenever +you cannot assume that the expected target variable is zero when all +your predictors are zero, it could be a bad idea to apply a model +which penalizes the intercept. Also, the analytical solution to the +ridge regression coefficients (when not shrinking $$\beta_0$$) is +derived under the assumption that both y and X are zero centered (mean +subtracted). What you are doing is correct, but you should also zero +center X (subtracting the mean of each column from the corresponding +column). + +
+If your predictors are of different scales, I would advice you to +standardize X by subtracting the mean of each column from the +corresponding column and dividing the column with its standard +deviation. If you dont do this, you will give an "unfair" penalization +of the parameters since their magnitude depends on the scale of their +corresponding predictor. Suppose that you have an input variable +"height". Human height might be measured in inches or meters or +kilometers. If measured in kilometers, a standard linear regression +model with this predictor would probably give a much bigger +coefficient term, than if measured in millimeters. You may see how +this could become a problem when considering the loss function for +ridge regression. + +
+Remember that when you do any transformation to your dataset before +training, the exact same transformation has to be applied to new data +before making a prediction. In your case, this means: + +
+ + +
#Model training:
+y_train_mean = np.mean(y_train)
+X_train_mean = np.mean(X_train,axis=0)
+X_train = X_train - X_train_mean
+y_train = y_train - y_train_mean
+
+trained_model = some_model.fit(X_train,y_train)
+
+#Model prediction:
+X_test = X_test - X_train_mean #Use mean from training data
+y_pred = trained_model(X_test)
+y_pred = y_pred + y_train_mean
++Here is a mathematical explanation of the zero centering: + +
+The cost/loss function for Ridge regression is: $$ -\begin{align} - H = -J \sum_{k}^L s_k s_{k + 1}, -\tag{1} -\end{align} +C(\beta_0, \beta_1, ... , \beta_P) = \sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip}\beta_p)^2 + \lambda \sum_{p=1}^P \beta_p^2. $$
-where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins -in the system is determined by \( L \). For the one-dimensional system -there is no phase transition. +Notice that the intercept is left out of the \( L_2 \) regularization term. The design matrix +\( X \) does in this case not contain any intercept column. We want + +$$ +\frac{\partial L}{\partial \beta_j} = 0, +$$
-We will look at a system of \( L = 40 \) spins with a coupling constant of -\( J = 1 \). To get enough training data we will generate 10000 states -with their respective energies. +for all \( j \), so lets start with \( \beta_0 \). This means that we have + +$$ +\frac{\partial L}{\partial \beta_0} = -2\sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip} \beta_p). +$$ + +
+We want to solve +$$ +-2\sum_{i=1}^{n} (y_i - \beta_0 - \sum_{p=1}^P X_{ip} \beta_p) = 0, +$$ + +
+which gives +$$ +\sum_{i=1}^{n} \beta_0 = \sum_{i=1}^{n}y_i - \sum_{i=1}^{n} \sum_{p=1}^P X_{ip} \beta_p, +$$ + +
+or +$ n\beta_0 = \sum_{i=1}^{n} y_i - \sum_{p=1}^P\beta_p \sum_{i=1}^{n} X_{ip}$. + +
+If we assume that every column of \( X \) is centered, whic we can do by subtracting the mean, +
+ + +
X = X - np.mean(X,axis=0)
++the sum $ \sum_{i=1}^{n} X_{ip} $ + +
+can be rewritten as +$$ +\sum_{i=1}^{n} (X_{ip} - \frac{1}{n}\sum_{i=1}^{n} X_{ip}) = \sum_{i=1}^{n} X_{ip} - \sum_{i=1}^{n} \frac{1}{n} \sum_{i=1}^{n}X_{ip}, +$$ + +resulting in +$$ +\sum_{i=1}^{n} X_{ip} - n \frac{1}{n} \sum_{i=1}^{n}X_{ip} = 0. +$$ + +
+Finally we have +$$ +n\beta_0 = \sum_{i=1}^{n} y_i - \sum_{p=1}^P\beta_p \sum_{i=1}^{n} X_{ip}, +$$ + +or +$$ +\beta_0 = \frac{1}{n}\sum_{i=1}^{n} y_i = y_{average}. +$$ + +
+Replacing \( y_i \) with \( y_i - \beta_0 = y_i - y_{average} \) in the loss function will give us (in vector-matrix disguise) +$$ +C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}, +$$ + +
+which has the solution + +
+\( \beta = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}} \). +where \( \boldsymbol{\tilde{y}} = \boldsymbol{y} - y_{average} \) +and \( \tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=1}^{n-1}X_{kj} \).
import numpy as np
+import pandas as pd
import matplotlib.pyplot as plt
-from mpl_toolkits.axes_grid1 import make_axes_locatable
-import seaborn as sns
-import scipy.linalg as scl
from sklearn.model_selection import train_test_split
-import tqdm
-sns.set(color_codes=True)
-cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+from sklearn import linear_model
-L = 40
-n = int(1e4)
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
-spins = np.random.choice([-1, 1], size=(n, L))
-J = 1.0
-energies = np.zeros(n)
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(3155)
-for i in range(n):
- energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 20
+X = np.zeros((n,Maxpolydegree))
+X[:,0] = 1.0
+
+for polydegree in range(1, Maxpolydegree):
+ for degree in range(polydegree):
+ X[:,degree] = x**degree
+
+
+# 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.pinv(X_train.T @ X_train) @ X_train.T @ y_train
+print(OLSbeta)
+# and then make the prediction
+ytildeOLS = X_train @ OLSbeta
+print("Training MSE for OLS")
+print(MSE(y_train,ytildeOLS))
+ypredictOLS = X_test @ OLSbeta
+print("Test MSE OLS")
+print(MSE(y_test,ypredictOLS))
+
+p = len(OLSbeta)
+I = np.eye(p,p)
+# Decide which values of lambda to use
+nlambdas = 4
+MSEOwnRidgePredict = np.zeros(nlambdas)
+MSEOwnRidgeTrain = np.zeros(nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+MSERidgeTrain = np.zeros(nlambdas)
+
+lambdas = np.logspace(-4, 4, nlambdas)
+for i in range(nlambdas):
+ lmb = lambdas[i]
+ OwnRidgeBeta = np.linalg.pinv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
+ # include lasso using Scikit-Learn
+ # Note: we include the intercept
+ RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
+ RegRidge.fit(X_train,y_train)
+ # and then make the prediction
+ ytildeOwnRidge = X_train @ OwnRidgeBeta
+ ypredictOwnRidge = X_test @ OwnRidgeBeta
+ ytildeRidge = RegRidge.predict(X_train)
+ ypredictRidge = RegRidge.predict(X_test)
+ MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
+ MSEOwnRidgeTrain[i] = MSE(y_train,ytildeOwnRidge)
+ MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+ MSERidgeTrain[i] = MSE(y_train,ytildeRidge)
+ print("Beta values for own Ridge implementation")
+ print(OwnRidgeBeta)
+ print("Beta values for Scikit-Learn Ridge implementation")
+ print(RegRidge.coef_)
+# Now plot the results
+plt.figure()
+plt.plot(np.log10(lambdas), MSEOwnRidgeTrain, 'b', label = 'MSE Ridge train')
+plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE Ridge Test')
+plt.plot(np.log10(lambdas), MSERidgeTrain, 'y', label = 'MSE Ridge train')
+plt.plot(np.log10(lambdas), MSERidgePredict, 'g', label = 'MSE Ridge Test')
+
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
-Here we use ordinary least squares -regression to predict the energy for the nearest neighbor -one-dimensional Ising model on a ring, i.e., the endpoints wrap -around. We will use linear regression to fit a value for -the coupling constant to achieve this. + +
import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+from sklearn.preprocessing import StandardScaler
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(315)
+
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 5
+X = np.zeros((n,Maxpolydegree-1))
+
+for degree in range(1,Maxpolydegree): #No intercept column
+ X[:,degree-1] = x**(degree)
+
+
+
+
+# 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)
+
+
+
+
+
+#For our own implementation, we will need to deal with the intercept by centering the design matrix and the target variable
+X_train_mean = np.mean(X_train,axis=0)
+X_train_scaled = X_train - X_train_mean #Center by removing mean from each feature
+X_test_scaled = X_test - X_train_mean
+
+y_scaler = np.mean(y_train) #The model intercept (called y_scaler) is given by the mean of target variable (IF X is centered)
+y_train_scaled = y_train - y_scaler #Remove the intercept from the training data.
+
+
+p = Maxpolydegree-1
+I = np.eye(p,p)
+# Decide which values of lambda to use
+nlambdas = 4
+MSEOwnRidgePredict = np.zeros(nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+
+lambdas = np.logspace(-4, 1, nlambdas)
+for i in range(nlambdas):
+ lmb = lambdas[i]
+ OwnRidgeBeta = np.linalg.pinv(X_train_scaled.T @ X_train_scaled+lmb*I) @ X_train_scaled.T @ (y_train_scaled)
+ intercept_ = y_scaler - X_train_mean@OwnRidgeBeta #The intercept can be shifted so the model can predict on uncentered data
+
+ ypredictOwnRidge = X_test @ OwnRidgeBeta + intercept_ #Add intercept to prediction
+ #EQUIVALENT PREDICTION:
+ ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler #Add intercept to prediction
+ print("Values for own Ridge prediction")
+ print(ypredictOwnRidge)
+
+
+
+ RegRidge = linear_model.Ridge(lmb)
+ RegRidge.fit(X_train,y_train)
+ ypredictRidge = RegRidge.predict(X_test)
+ print("Values for SL Ridge prediction")
+ print(ypredictRidge)
+
+
+ MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
+ MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+
+ print("Beta values for own Ridge implementation")
+ print(OwnRidgeBeta) #Intercept is given by mean of target variable
+ print("Beta values for Scikit-Learn Ridge implementation")
+ print(RegRidge.coef_)
+ print('Intercept from own implementation:')
+ print(intercept_)
+ print('Intercept from Scikit-Learn Ridge implementation')
+ print(RegRidge.intercept_)
+
+
+
+# Now plot the results
+
+plt.figure()
+plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'b--', label = 'MSE own Ridge Test')
+plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
+
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
++ + +
import numpy as np
+import matplotlib.pyplot as plt
+
+from sklearn.linear_model import LinearRegression
+
+
+np.random.seed(2021)
+
+
+def fit_beta(X, y):
+ return np.linalg.pinv(X.T @ X) @ X.T @ y
+
+
+true_beta = [2, 0.5, 3.7]
+
+x = np.linspace(0, 1, 11)
+y = np.sum(
+ np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
+) + 0.1 * np.random.normal(size=len(x))
+
+degree = 3
+X = np.zeros((len(x), degree))
+
+# Include the intercept in the design matrix
+for p in range(degree):
+ X[:, p] = x ** p
+
+beta = fit_beta(X, y)
+
+# Intercept is included in the design matrix
+clf = LinearRegression(fit_intercept=False).fit(X, y)
+
+print(f"True beta: {true_beta}")
+print(f"Fitted beta: {beta}")
+print(f"Sklearn fitted beta: {clf.coef_}")
+
+
+plt.figure()
+plt.scatter(x, y, label="Data")
+plt.plot(x, X @ beta, label="Fit")
+plt.plot(x, clf.predict(X), label="Sklearn (fit_intercept=False)")
+
+
+# Do not include the intercept in the design matrix
+X = np.zeros((len(x), degree - 1))
+
+for p in range(degree - 1):
+ X[:, p] = x ** (p + 1)
+
+# Intercept is not included in the design matrix
+clf = LinearRegression(fit_intercept=True).fit(X, y)
+
+# Use centered values for X and y when computing coefficients
+y_offset = np.average(y, axis=0)
+X_offset = np.average(X, axis=0)
+
+beta = fit_beta(X - X_offset, y - y_offset)
+intercept = np.mean(y_offset - X_offset @ beta)
+
+print(f"Manual intercept: {intercept}")
+print(f"Fitted beta (sans intercept): {beta}")
+print(f"Sklearn intercept: {clf.intercept_}")
+print(f"Sklearn fitted beta (sans intercept): {clf.coef_}")
+
+plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
+plt.plot(x, clf.predict(X), "--", label="Sklearn (fit_intercept=True)")
+plt.grid()
+plt.legend()
+
+plt.show()
+
@@ -315,7 +833,7 @@ the coupling constant to achieve this.
-A more general form for the one-dimensional Ising model is +The one-dimensional Ising model with nearest neighbor interaction, no +external field and a constant coupling constant \( J \) is given by $$ \begin{align} - H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. -\tag{2} + H = -J \sum_{k}^L s_k s_{k + 1}, +\tag{1} \end{align} $$
-Here we allow for interactions beyond the nearest neighbors and a state dependent -coupling constant. This latter expression can be formulated as -a matrix-product -$$ -\begin{align} - \boldsymbol{H} = \boldsymbol{X} J, -\tag{3} -\end{align} -$$ +where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins +in the system is determined by \( L \). For the one-dimensional system +there is no phase transition.
-where \( X_{jk} = s_j s_k \) and \( J \) is a matrix which consists of the -elements \( -J_{jk} \). This form of writing the energy fits perfectly -with the form utilized in linear regression, that is - -$$ -\begin{align} - \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, -\tag{4} -\end{align} -$$ - -
-We split the data in training and test data as discussed in the previous example +We will look at a system of \( L = 40 \) spins with a coupling constant of +\( J = 1 \). To get enough training data we will generate 10000 states +with their respective energies.
-
X = np.zeros((n, L ** 2))
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.axes_grid1 import make_axes_locatable
+import seaborn as sns
+import scipy.linalg as scl
+from sklearn.model_selection import train_test_split
+import tqdm
+sns.set(color_codes=True)
+cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+
+L = 40
+n = int(1e4)
+
+spins = np.random.choice([-1, 1], size=(n, L))
+J = 1.0
+
+energies = np.zeros(n)
+
for i in range(n):
- X[i] = np.outer(spins[i], spins[i]).ravel()
-y = energies
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+ energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
+
+Here we use ordinary least squares
+regression to predict the energy for the nearest neighbor
+one-dimensional Ising model on a ring, i.e., the endpoints wrap
+around. We will use linear regression to fit a value for
+the coupling constant to achieve this.
+
@@ -309,7 +474,7 @@ X_train, X_test, y_train, y_test = train_tes
-In the ordinary least squares method we choose the cost function +A more general form for the one-dimensional Ising model is $$ \begin{align} - C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}. -\tag{5} + H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. +\tag{2} \end{align} $$
-We then find the extremal point of \( C \) by taking the derivative with respect to \( \boldsymbol{\beta} \) as discussed above. -This yields the expression for \( \boldsymbol{\beta} \) to be - +Here we allow for interactions beyond the nearest neighbors and a state dependent +coupling constant. This latter expression can be formulated as +a matrix-product $$ - \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}}, +\begin{align} + \boldsymbol{H} = \boldsymbol{X} J, +\tag{3} +\end{align} $$
-which immediately imposes some requirements on \( \boldsymbol{X} \) as there must exist -an inverse of \( \boldsymbol{X}^T \boldsymbol{X} \). If the expression we are modeling contains an -intercept, i.e., a constant term, we must make sure that the -first column of \( \boldsymbol{X} \) consists of \( 1 \). We do this here +where \( X_{jk} = s_j s_k \) and \( J \) is a matrix which consists of the +elements \( -J_{jk} \). This form of writing the energy fits perfectly +with the form utilized in linear regression, that is + +$$ +\begin{align} + \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, +\tag{4} +\end{align} +$$ + +
+We split the data in training and test data as discussed in the previous example
-
X_train_own = np.concatenate(
- (np.ones(len(X_train))[:, np.newaxis], X_train),
- axis=1
-)
-X_test_own = np.concatenate(
- (np.ones(len(X_test))[:, np.newaxis], X_test),
- axis=1
-)
-- - -
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
- return scl.inv(x.T @ x) @ (x.T @ y)
-beta = ols_inv(X_train_own, y_train)
+X = np.zeros((n, L ** 2))
+for i in range(n):
+ X[i] = np.outer(spins[i], spins[i]).ravel()
+y = energies
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
@@ -308,7 +468,7 @@ beta = ols_inv(X_train_own, y_train)
-Doing the inversion directly turns out to be a bad idea since the matrix -\( \boldsymbol{X}^T\boldsymbol{X} \) is singular. An alternative approach is to use the singular -value decomposition. Using the definition of the Moore-Penrose -pseudoinverse we can write the equation for \( \boldsymbol{\beta} \) as +In the ordinary least squares method we choose the cost function -$$ - \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y}, -$$ - -
-where the pseudoinverse of \( \boldsymbol{X} \) is given by - -$$ - \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}. -$$ - -
-Using singular value decomposition we can decompose the matrix \( \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T \), -where \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal(unitary) matrices and \( \boldsymbol{\Sigma} \) contains the singular values (more details below). -where \( X^{+} = V\Sigma^{+} U^T \). This reduces the equation for -\( \omega \) to $$ \begin{align} - \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}. -\tag{6} + C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}. +\tag{5} \end{align} $$
-Note that solving this equation by actually doing the pseudoinverse -(which is what we will do) is not a good idea as this operation scales -as \( \mathcal{O}(n^3) \), where \( n \) is the number of elements in a -general matrix. Instead, doing \( QR \)-factorization and solving the -linear system as an equation would reduce this down to -\( \mathcal{O}(n^2) \) operations. +We then find the extremal point of \( C \) by taking the derivative with respect to \( \boldsymbol{\beta} \) as discussed above. +This yields the expression for \( \boldsymbol{\beta} \) to be + +$$ + \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}}, +$$ + +
+which immediately imposes some requirements on \( \boldsymbol{X} \) as there must exist +an inverse of \( \boldsymbol{X}^T \boldsymbol{X} \). If the expression we are modeling contains an +intercept, i.e., a constant term, we must make sure that the +first column of \( \boldsymbol{X} \) consists of \( 1 \). We do this here
-
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
- u, s, v = scl.svd(x)
- return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
+X_train_own = np.concatenate(
+ (np.ones(len(X_train))[:, np.newaxis], X_train),
+ axis=1
+)
+X_test_own = np.concatenate(
+ (np.ones(len(X_test))[:, np.newaxis], X_test),
+ axis=1
+)
-
beta = ols_svd(X_train_own,y_train)
+def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
+ return scl.inv(x.T @ x) @ (x.T @ y)
+beta = ols_inv(X_train_own, y_train)
-
-When extracting the \( J \)-matrix we need to make sure that we remove the intercept, as is done here
-
-
-
-
-
J = beta[1:].reshape(L, L)
-
-
-A way of looking at the coefficients in \( J \) is to plot the matrices as images.
-
-
-
-
-
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J, **cmap_args)
-plt.title("OLS", fontsize=18)
-plt.xticks(fontsize=18)
-plt.yticks(fontsize=18)
-cb = fig.colorbar(im)
-cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
-plt.show()
-
-
-It is interesting to note that OLS
-considers both \( J_{j, j + 1} = -0.5 \) and \( J_{j, j - 1} = -0.5 \) as
-valid matrix elements for \( J \).
-In our discussion below on hyperparameters and Ridge and Lasso regression we will see that
-this problem can be removed, partly and only with Lasso regression.
-
-
-In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
-
@@ -347,7 +466,7 @@ In this case our matrix inversion was actually possible. The obvious question no
20
21
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs012.html b/doc/pub/week38/html/._week38-bs012.html
index eceb92ea3..6abfcd51a 100644
--- a/doc/pub/week38/html/._week38-bs012.html
+++ b/doc/pub/week38/html/._week38-bs012.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,128 +395,74 @@ MathJax.Hub.Config({
-The one-dimensional Ising model
+Singular Value decomposition
-Let us bring back the Ising model again, but now with an additional
-focus on Ridge and Lasso regression as well. We repeat some of the
-basic parts of the Ising model and the setup of the training and test
-data. The one-dimensional Ising model with nearest neighbor
-interaction, no external field and a constant coupling constant \( J \) is
-given by
+Doing the inversion directly turns out to be a bad idea since the matrix
+\( \boldsymbol{X}^T\boldsymbol{X} \) is singular. An alternative approach is to use the singular
+value decomposition. Using the definition of the Moore-Penrose
+pseudoinverse we can write the equation for \( \boldsymbol{\beta} \) as
+$$
+ \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y},
+$$
+
+
+where the pseudoinverse of \( \boldsymbol{X} \) is given by
+
+$$
+ \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}.
+$$
+
+
+Using singular value decomposition we can decompose the matrix \( \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T \),
+where \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal(unitary) matrices and \( \boldsymbol{\Sigma} \) contains the singular values (more details below).
+where \( X^{+} = V\Sigma^{+} U^T \). This reduces the equation for
+\( \omega \) to
$$
\begin{align}
- H = -J \sum_{k}^L s_k s_{k + 1},
-\tag{7}
-\end{align}
-$$
-
-where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
-
-
-We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
-
-
-
-
-
import numpy as np
-import matplotlib.pyplot as plt
-from mpl_toolkits.axes_grid1 import make_axes_locatable
-import seaborn as sns
-import scipy.linalg as scl
-from sklearn.model_selection import train_test_split
-import sklearn.linear_model as skl
-import tqdm
-sns.set(color_codes=True)
-cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
-
-L = 40
-n = int(1e4)
-
-spins = np.random.choice([-1, 1], size=(n, L))
-J = 1.0
-
-energies = np.zeros(n)
-
-for i in range(n):
- energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
-
-
-A more general form for the one-dimensional Ising model is
-
-$$
-\begin{align}
- H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
-\tag{8}
+ \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}.
+\tag{6}
\end{align}
$$
-Here we allow for interactions beyond the nearest neighbors and a more
-adaptive coupling matrix. This latter expression can be formulated as
-a matrix-product on the form
-$$
-\begin{align}
- H = X J,
-\tag{9}
-\end{align}
-$$
-
-
-where \( X_{jk} = s_j s_k \) and \( J \) is the matrix consisting of the
-elements \( -J_{jk} \). This form of writing the energy fits perfectly
-with the form utilized in linear regression, viz.
-$$
-\begin{align}
- \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}.
-\tag{10}
-\end{align}
-$$
-
-We organize the data as we did above
-
-
-
-
X = np.zeros((n, L ** 2))
-for i in range(n):
- X[i] = np.outer(spins[i], spins[i]).ravel()
-y = energies
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
-
-X_train_own = np.concatenate(
- (np.ones(len(X_train))[:, np.newaxis], X_train),
- axis=1
-)
-
-X_test_own = np.concatenate(
- (np.ones(len(X_test))[:, np.newaxis], X_test),
- axis=1
-)
-
-
-We will do all fitting with Scikit-Learn,
+Note that solving this equation by actually doing the pseudoinverse
+(which is what we will do) is not a good idea as this operation scales
+as \( \mathcal{O}(n^3) \), where \( n \) is the number of elements in a
+general matrix. Instead, doing \( QR \)-factorization and solving the
+linear system as an equation would reduce this down to
+\( \mathcal{O}(n^2) \) operations.
-
clf = skl.LinearRegression().fit(X_train, y_train)
+def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
+ u, s, v = scl.svd(x)
+ return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
-When extracting the \( J \)-matrix we make sure to remove the intercept
-
-
J_sk = clf.coef_.reshape(L, L)
+beta = ols_svd(X_train_own,y_train)
-And then we plot the results
+When extracting the \( J \)-matrix we need to make sure that we remove the intercept, as is done here
+
+
+
+
+
J = beta[1:].reshape(L, L)
+
+
+A way of looking at the coefficients in \( J \) is to plot the matrices as images.
+
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_sk, **cmap_args)
-plt.title("LinearRegression from Scikit-learn", fontsize=18)
+im = plt.imshow(J, **cmap_args)
+plt.title("OLS", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
@@ -366,7 +470,14 @@ cb.ax.se
plt.show()
-The results perfectly with our previous discussion where we used our own code.
+It is interesting to note that OLS
+considers both \( J_{j, j + 1} = -0.5 \) and \( J_{j, j - 1} = -0.5 \) as
+valid matrix elements for \( J \).
+In our discussion below on hyperparameters and Ridge and Lasso regression we will see that
+this problem can be removed, partly and only with Lasso regression.
+
+
+In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
@@ -394,7 +505,7 @@ The results perfectly with our previous discussion where we used our own code.
21
22
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs013.html b/doc/pub/week38/html/._week38-bs013.html
index 823d063a8..c3f414954 100644
--- a/doc/pub/week38/html/._week38-bs013.html
+++ b/doc/pub/week38/html/._week38-bs013.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,38 +395,137 @@ MathJax.Hub.Config({
-Ridge regression
+The one-dimensional Ising model
-Having explored the ordinary least squares we move on to ridge
-regression. In ridge regression we include a regularizer. This
-involves a new cost function which leads to a new estimate for the
-weights \( \boldsymbol{\beta} \). This results in a penalized regression problem. The
-cost function is given by
+Let us bring back the Ising model again, but now with an additional
+focus on Ridge and Lasso regression as well. We repeat some of the
+basic parts of the Ising model and the setup of the training and test
+data. The one-dimensional Ising model with nearest neighbor
+interaction, no external field and a constant coupling constant \( J \) is
+given by
$$
\begin{align}
- C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}.
-\tag{11}
+ H = -J \sum_{k}^L s_k s_{k + 1},
+\tag{7}
\end{align}
$$
+where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
+
+
+We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
+
-
_lambda = 0.1
-clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
-J_ridge_sk = clf_ridge.coef_.reshape(L, L)
-fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_ridge_sk, **cmap_args)
-plt.title("Ridge from Scikit-learn", fontsize=18)
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.axes_grid1 import make_axes_locatable
+import seaborn as sns
+import scipy.linalg as scl
+from sklearn.model_selection import train_test_split
+import sklearn.linear_model as skl
+import tqdm
+sns.set(color_codes=True)
+cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
+
+L = 40
+n = int(1e4)
+
+spins = np.random.choice([-1, 1], size=(n, L))
+J = 1.0
+
+energies = np.zeros(n)
+
+for i in range(n):
+ energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
+
+
+A more general form for the one-dimensional Ising model is
+
+$$
+\begin{align}
+ H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
+\tag{8}
+\end{align}
+$$
+
+
+Here we allow for interactions beyond the nearest neighbors and a more
+adaptive coupling matrix. This latter expression can be formulated as
+a matrix-product on the form
+$$
+\begin{align}
+ H = X J,
+\tag{9}
+\end{align}
+$$
+
+
+where \( X_{jk} = s_j s_k \) and \( J \) is the matrix consisting of the
+elements \( -J_{jk} \). This form of writing the energy fits perfectly
+with the form utilized in linear regression, viz.
+$$
+\begin{align}
+ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}.
+\tag{10}
+\end{align}
+$$
+
+We organize the data as we did above
+
+
+
+
X = np.zeros((n, L ** 2))
+for i in range(n):
+ X[i] = np.outer(spins[i], spins[i]).ravel()
+y = energies
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
+
+X_train_own = np.concatenate(
+ (np.ones(len(X_train))[:, np.newaxis], X_train),
+ axis=1
+)
+
+X_test_own = np.concatenate(
+ (np.ones(len(X_test))[:, np.newaxis], X_test),
+ axis=1
+)
+
+
+We will do all fitting with Scikit-Learn,
+
+
+
+
+
clf = skl.LinearRegression().fit(X_train, y_train)
+
+
+When extracting the \( J \)-matrix we make sure to remove the intercept
+
+
+
+
J_sk = clf.coef_.reshape(L, L)
+
+
+And then we plot the results
+
+
+
+
fig = plt.figure(figsize=(20, 14))
+im = plt.imshow(J_sk, **cmap_args)
+plt.title("LinearRegression from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
-
plt.show()
+
+The results perfectly with our previous discussion where we used our own code.
+
@@ -295,7 +552,7 @@ plt.show()
22
23
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs014.html b/doc/pub/week38/html/._week38-bs014.html
index 3aa6a42f3..54908b0cf 100644
--- a/doc/pub/week38/html/._week38-bs014.html
+++ b/doc/pub/week38/html/._week38-bs014.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,29 +395,31 @@ MathJax.Hub.Config({
-LASSO regression
+Ridge regression
-In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function.
+Having explored the ordinary least squares we move on to ridge
+regression. In ridge regression we include a regularizer. This
+involves a new cost function which leads to a new estimate for the
+weights \( \boldsymbol{\beta} \). This results in a penalized regression problem. The
+cost function is given by
$$
\begin{align}
- C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \sqrt{\boldsymbol{\beta}^T\boldsymbol{\beta}}.
-\tag{12}
+ C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}.
+\tag{11}
\end{align}
$$
-
-Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from Scikit-Learn.
-
-
clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
-J_lasso_sk = clf_lasso.coef_.reshape(L, L)
+_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
+J_ridge_sk = clf_ridge.coef_.reshape(L, L)
fig = plt.figure(figsize=(20, 14))
-im = plt.imshow(J_lasso_sk, **cmap_args)
-plt.title("Lasso from Scikit-learn", fontsize=18)
+im = plt.imshow(J_ridge_sk, **cmap_args)
+plt.title("Ridge from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
@@ -267,11 +427,6 @@ cb.ax.se
plt.show()
-
-It is quite striking how LASSO breaks the symmetry of the coupling
-constant as opposed to ridge and OLS. We get a sparse solution with
-\( J_{j, j + 1} = -1 \).
-
@@ -298,7 +453,7 @@ constant as opposed to ridge and OLS. We get a sparse solution with
23
24
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs015.html b/doc/pub/week38/html/._week38-bs015.html
index 50c489cf1..a97eb2415 100644
--- a/doc/pub/week38/html/._week38-bs015.html
+++ b/doc/pub/week38/html/._week38-bs015.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,56 +395,40 @@ MathJax.Hub.Config({
-Performance as function of the regularization parameter
+LASSO regression
-We see how the different models perform for a different set of values for \( \lambda \).
+In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function.
+
+$$
+\begin{align}
+ C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \sqrt{\boldsymbol{\beta}^T\boldsymbol{\beta}}.
+\tag{12}
+\end{align}
+$$
+
+
+Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from Scikit-Learn.
-
lambdas = np.logspace(-4, 5, 10)
-
-train_errors = {
- "ols_sk": np.zeros(lambdas.size),
- "ridge_sk": np.zeros(lambdas.size),
- "lasso_sk": np.zeros(lambdas.size)
-}
-
-test_errors = {
- "ols_sk": np.zeros(lambdas.size),
- "ridge_sk": np.zeros(lambdas.size),
- "lasso_sk": np.zeros(lambdas.size)
-}
-
-plot_counter = 1
-
-fig = plt.figure(figsize=(32, 54))
-
-for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
- for key, method in zip(
- ["ols_sk", "ridge_sk", "lasso_sk"],
- [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
- ):
- method = method.fit(X_train, y_train)
-
- train_errors[key][i] = method.score(X_train, y_train)
- test_errors[key][i] = method.score(X_test, y_test)
-
- omega = method.coef_.reshape(L, L)
-
- plt.subplot(10, 5, plot_counter)
- plt.imshow(omega, **cmap_args)
- plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
- plot_counter += 1
+clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
+J_lasso_sk = clf_lasso.coef_.reshape(L, L)
+fig = plt.figure(figsize=(20, 14))
+im = plt.imshow(J_lasso_sk, **cmap_args)
+plt.title("Lasso from Scikit-learn", fontsize=18)
+plt.xticks(fontsize=18)
+plt.yticks(fontsize=18)
+cb = fig.colorbar(im)
+cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
-We see that LASSO reaches a good solution for low
-values of \( \lambda \), but will "wither" when we increase \( \lambda \) too
-much. Ridge is more stable over a larger range of values for
-\( \lambda \), but eventually also fades away.
+It is quite striking how LASSO breaks the symmetry of the coupling
+constant as opposed to ridge and OLS. We get a sparse solution with
+\( J_{j, j + 1} = -1 \).
@@ -314,7 +456,7 @@ much. Ridge is more stable over a larger range of values for
24
25
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs016.html b/doc/pub/week38/html/._week38-bs016.html
index 068b9da2e..a6658ce81 100644
--- a/doc/pub/week38/html/._week38-bs016.html
+++ b/doc/pub/week38/html/._week38-bs016.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,54 +395,56 @@ MathJax.Hub.Config({
-Finding the optimal value of \( \lambda \)
+Performance as function of the regularization parameter
-To determine which value of \( \lambda \) is best we plot the accuracy of
-the models when predicting the training and the testing set. We expect
-the accuracy of the training set to be quite good, but if the accuracy
-of the testing set is much lower this tells us that we might be
-subject to an overfit model. The ideal scenario is an accuracy on the
-testing set that is close to the accuracy of the training set.
+We see how the different models perform for a different set of values for \( \lambda \).
-
fig = plt.figure(figsize=(20, 14))
+lambdas = np.logspace(-4, 5, 10)
-colors = {
- "ols_sk": "r",
- "ridge_sk": "y",
- "lasso_sk": "c"
+train_errors = {
+ "ols_sk": np.zeros(lambdas.size),
+ "ridge_sk": np.zeros(lambdas.size),
+ "lasso_sk": np.zeros(lambdas.size)
}
-for key in train_errors:
- plt.semilogx(
- lambdas,
- train_errors[key],
- colors[key],
- label="Train {0}".format(key),
- linewidth=4.0
- )
+test_errors = {
+ "ols_sk": np.zeros(lambdas.size),
+ "ridge_sk": np.zeros(lambdas.size),
+ "lasso_sk": np.zeros(lambdas.size)
+}
+
+plot_counter = 1
+
+fig = plt.figure(figsize=(32, 54))
+
+for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
+ for key, method in zip(
+ ["ols_sk", "ridge_sk", "lasso_sk"],
+ [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
+ ):
+ method = method.fit(X_train, y_train)
+
+ train_errors[key][i] = method.score(X_train, y_train)
+ test_errors[key][i] = method.score(X_test, y_test)
+
+ omega = method.coef_.reshape(L, L)
+
+ plt.subplot(10, 5, plot_counter)
+ plt.imshow(omega, **cmap_args)
+ plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
+ plot_counter += 1
-for key in test_errors:
- plt.semilogx(
- lambdas,
- test_errors[key],
- colors[key] + "--",
- label="Test {0}".format(key),
- linewidth=4.0
- )
-plt.legend(loc="best", fontsize=18)
-plt.xlabel(r"$\lambda$", fontsize=18)
-plt.ylabel(r"$R^2$", fontsize=18)
-plt.tick_params(labelsize=18)
plt.show()
-From the above figure we can see that LASSO with \( \lambda = 10^{-2} \)
-achieves a very good accuracy on the test set. This by far surpasses the
-other models for all values of \( \lambda \).
+We see that LASSO reaches a good solution for low
+values of \( \lambda \), but will "wither" when we increase \( \lambda \) too
+much. Ridge is more stable over a larger range of values for
+\( \lambda \), but eventually also fades away.
@@ -312,7 +472,7 @@ other models for all values of \( \lambda \).
25
26
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs017.html b/doc/pub/week38/html/._week38-bs017.html
index 6b46482d7..158c5d73f 100644
--- a/doc/pub/week38/html/._week38-bs017.html
+++ b/doc/pub/week38/html/._week38-bs017.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,22 +393,56 @@ MathJax.Hub.Config({
-
+
-Logistic Regression
+Finding the optimal value of \( \lambda \)
-In linear regression our main interest was centered on learning the
-coefficients of a functional fit (say a polynomial) in order to be
-able to predict the response of a continuous variable on some unseen
-data. The fit to the continuous variable \( y_i \) is based on some
-independent variables \( \hat{x}_i \). Linear regression resulted in
-analytical expressions for standard ordinary Least Squares or Ridge
-regression (in terms of matrices to invert) for several quantities,
-ranging from the variance and thereby the confidence intervals of the
-parameters \( \hat{\beta} \) to the mean squared error. If we can invert
-the product of the design matrices, linear regression gives then a
-simple recipe for fitting our data.
+To determine which value of \( \lambda \) is best we plot the accuracy of
+the models when predicting the training and the testing set. We expect
+the accuracy of the training set to be quite good, but if the accuracy
+of the testing set is much lower this tells us that we might be
+subject to an overfit model. The ideal scenario is an accuracy on the
+testing set that is close to the accuracy of the training set.
+
+
+
+
+
fig = plt.figure(figsize=(20, 14))
+
+colors = {
+ "ols_sk": "r",
+ "ridge_sk": "y",
+ "lasso_sk": "c"
+}
+
+for key in train_errors:
+ plt.semilogx(
+ lambdas,
+ train_errors[key],
+ colors[key],
+ label="Train {0}".format(key),
+ linewidth=4.0
+ )
+
+for key in test_errors:
+ plt.semilogx(
+ lambdas,
+ test_errors[key],
+ colors[key] + "--",
+ label="Test {0}".format(key),
+ linewidth=4.0
+ )
+plt.legend(loc="best", fontsize=18)
+plt.xlabel(r"$\lambda$", fontsize=18)
+plt.ylabel(r"$R^2$", fontsize=18)
+plt.tick_params(labelsize=18)
+plt.show()
+
+
+From the above figure we can see that LASSO with \( \lambda = 10^{-2} \)
+achieves a very good accuracy on the test set. This by far surpasses the
+other models for all values of \( \lambda \).
@@ -278,7 +470,7 @@ simple recipe for fitting our data.
26
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs018.html b/doc/pub/week38/html/._week38-bs018.html
index 4582d24f7..d21450c24 100644
--- a/doc/pub/week38/html/._week38-bs018.html
+++ b/doc/pub/week38/html/._week38-bs018.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,25 +395,20 @@ MathJax.Hub.Config({
-Classification problems
+Logistic Regression
-Classification problems, however, are concerned with outcomes taking
-the form of discrete variables (i.e. categories). We may for example,
-on the basis of DNA sequencing for a number of patients, like to find
-out which mutations are important for a certain disease; or based on
-scans of various patients' brains, figure out if there is a tumor or
-not; or given a specific physical system, we'd like to identify its
-state, say whether it is an ordered or disordered system (typical
-situation in solid state physics); or classify the status of a
-patient, whether she/he has a stroke or not and many other similar
-situations.
-
-
-The most common situation we encounter when we apply logistic
-regression is that of two possible outcomes, normally denoted as a
-binary outcome, true or false, positive or negative, success or
-failure etc.
+In linear regression our main interest was centered on learning the
+coefficients of a functional fit (say a polynomial) in order to be
+able to predict the response of a continuous variable on some unseen
+data. The fit to the continuous variable \( y_i \) is based on some
+independent variables \( \hat{x}_i \). Linear regression resulted in
+analytical expressions for standard ordinary Least Squares or Ridge
+regression (in terms of matrices to invert) for several quantities,
+ranging from the variance and thereby the confidence intervals of the
+parameters \( \hat{\beta} \) to the mean squared error. If we can invert
+the product of the design matrices, linear regression gives then a
+simple recipe for fitting our data.
@@ -283,7 +436,7 @@ failure etc.
27
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs019.html b/doc/pub/week38/html/._week38-bs019.html
index c17b37d05..1e92edabc 100644
--- a/doc/pub/week38/html/._week38-bs019.html
+++ b/doc/pub/week38/html/._week38-bs019.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,25 +393,27 @@ MathJax.Hub.Config({
-
+
-Optimization and Deep learning
+Classification problems
-Logistic regression will also serve as our stepping stone towards
-neural network algorithms and supervised deep learning. For logistic
-learning, the minimization of the cost function leads to a non-linear
-equation in the parameters \( \hat{\beta} \). The optimization of the
-problem calls therefore for minimization algorithms. This forms the
-bottle neck of all machine learning algorithms, namely how to find
-reliable minima of a multi-variable function. This leads us to the
-family of gradient descent methods. The latter are the working horses
-of basically all modern machine learning algorithms.
+Classification problems, however, are concerned with outcomes taking
+the form of discrete variables (i.e. categories). We may for example,
+on the basis of DNA sequencing for a number of patients, like to find
+out which mutations are important for a certain disease; or based on
+scans of various patients' brains, figure out if there is a tumor or
+not; or given a specific physical system, we'd like to identify its
+state, say whether it is an ordered or disordered system (typical
+situation in solid state physics); or classify the status of a
+patient, whether she/he has a stroke or not and many other similar
+situations.
-We note also that many of the topics discussed here on logistic
-regression are also commonly used in modern supervised Deep Learning
-models, as we will see later.
+The most common situation we encounter when we apply logistic
+regression is that of two possible outcomes, normally denoted as a
+binary outcome, true or false, positive or negative, success or
+failure etc.
@@ -281,7 +441,7 @@ models, as we will see later.
28
29
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs020.html b/doc/pub/week38/html/._week38-bs020.html
index cf7240d7a..801222e6e 100644
--- a/doc/pub/week38/html/._week38-bs020.html
+++ b/doc/pub/week38/html/._week38-bs020.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,31 +393,25 @@ MathJax.Hub.Config({
-
+
-Basics
+Optimization and Deep learning
-We consider the case where the dependent variables, also called the
-responses or the outcomes, \( y_i \) are discrete and only take values
-from \( k=0,\dots,K-1 \) (i.e. \( K \) classes).
+Logistic regression will also serve as our stepping stone towards
+neural network algorithms and supervised deep learning. For logistic
+learning, the minimization of the cost function leads to a non-linear
+equation in the parameters \( \hat{\beta} \). The optimization of the
+problem calls therefore for minimization algorithms. This forms the
+bottle neck of all machine learning algorithms, namely how to find
+reliable minima of a multi-variable function. This leads us to the
+family of gradient descent methods. The latter are the working horses
+of basically all modern machine learning algorithms.
-The goal is to predict the
-output classes from the design matrix \( \hat{X}\in\mathbb{R}^{n\times p} \)
-made of \( n \) samples, each of which carries \( p \) features or predictors. The
-primary goal is to identify the classes to which new unseen samples
-belong.
-
-
-Let us specialize to the case of two classes only, with outputs
-\( y_i=0 \) and \( y_i=1 \). Our outcomes could represent the status of a
-credit card user that could default or not on her/his credit card
-debt. That is
-
-$$
-y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}.
-$$
+We note also that many of the topics discussed here on logistic
+regression are also commonly used in modern supervised Deep Learning
+models, as we will see later.
@@ -287,7 +439,7 @@ $$
29
30
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- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs021.html b/doc/pub/week38/html/._week38-bs021.html
index f047e072b..102c5fb17 100644
--- a/doc/pub/week38/html/._week38-bs021.html
+++ b/doc/pub/week38/html/._week38-bs021.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,28 +393,31 @@ MathJax.Hub.Config({
-
+
-Linear classifier
+Basics
-Before moving to the logistic model, let us try to use our linear
-regression model to classify these two outcomes. We could for example
-fit a linear model to the default case if \( y_i > 0.5 \) and the no
-default case \( y_i \leq 0.5 \).
+We consider the case where the dependent variables, also called the
+responses or the outcomes, \( y_i \) are discrete and only take values
+from \( k=0,\dots,K-1 \) (i.e. \( K \) classes).
-We would then have our
-weighted linear combination, namely
-$$
-\begin{equation}
-\hat{y} = \hat{X}^T\hat{\beta} + \hat{\epsilon},
-\tag{13}
-\end{equation}
-$$
+The goal is to predict the
+output classes from the design matrix \( \hat{X}\in\mathbb{R}^{n\times p} \)
+made of \( n \) samples, each of which carries \( p \) features or predictors. The
+primary goal is to identify the classes to which new unseen samples
+belong.
-where \( \hat{y} \) is a vector representing the possible outcomes, \( \hat{X} \) is our
-\( n\times p \) design matrix and \( \hat{\beta} \) represents our estimators/predictors.
+
+Let us specialize to the case of two classes only, with outputs
+\( y_i=0 \) and \( y_i=1 \). Our outcomes could represent the status of a
+credit card user that could default or not on her/his credit card
+debt. That is
+
+$$
+y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}.
+$$
@@ -284,7 +445,7 @@ where \( \hat{y} \) is a vector representing the possible outcomes, \( \hat{X} \
30
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»
diff --git a/doc/pub/week38/html/._week38-bs022.html b/doc/pub/week38/html/._week38-bs022.html
index 4766e6b66..d3383d840 100644
--- a/doc/pub/week38/html/._week38-bs022.html
+++ b/doc/pub/week38/html/._week38-bs022.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,24 +395,26 @@ MathJax.Hub.Config({
-Some selected properties
+Linear classifier
-The main problem with our function is that it takes values on the
-entire real axis. In the case of logistic regression, however, the
-labels \( y_i \) are discrete variables. A typical example is the credit
-card data discussed below here, where we can set the state of
-defaulting the debt to \( y_i=1 \) and not to \( y_i=0 \) for one the persons
-in the data set (see the full example below).
+Before moving to the logistic model, let us try to use our linear
+regression model to classify these two outcomes. We could for example
+fit a linear model to the default case if \( y_i > 0.5 \) and the no
+default case \( y_i \leq 0.5 \).
-One simple way to get a discrete output is to have sign
-functions that map the output of a linear regressor to values \( \{0,1\} \),
-\( f(s_i)=sign(s_i)=1 \) if \( s_i\ge 0 \) and 0 if otherwise.
-We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning
-literature. This model is extremely simple. However, in many cases it is more
-favorable to use a ``soft" classifier that outputs
-the probability of a given category. This leads us to the logistic function.
+We would then have our
+weighted linear combination, namely
+$$
+\begin{equation}
+\hat{y} = \hat{X}^T\hat{\beta} + \hat{\epsilon},
+\tag{13}
+\end{equation}
+$$
+
+where \( \hat{y} \) is a vector representing the possible outcomes, \( \hat{X} \) is our
+\( n\times p \) design matrix and \( \hat{\beta} \) represents our estimators/predictors.
@@ -282,7 +442,7 @@ the probability of a given category. This leads us to the logistic function.
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»
diff --git a/doc/pub/week38/html/._week38-bs023.html b/doc/pub/week38/html/._week38-bs023.html
index 844883928..90809c067 100644
--- a/doc/pub/week38/html/._week38-bs023.html
+++ b/doc/pub/week38/html/._week38-bs023.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,69 +395,25 @@ MathJax.Hub.Config({
-Simple example
+Some selected properties
-The following example on data for coronary heart disease (CHD) as function of age may serve as an illustration. In the code here we read and plot whether a person has had CHD (output = 1) or not (output = 0). This ouput is plotted the person's against age. Clearly, the figure shows that attempting to make a standard linear regression fit may not be very meaningful.
+The main problem with our function is that it takes values on the
+entire real axis. In the case of logistic regression, however, the
+labels \( y_i \) are discrete variables. A typical example is the credit
+card data discussed below here, where we can set the state of
+defaulting the debt to \( y_i=1 \) and not to \( y_i=0 \) for one the persons
+in the data set (see the full example below).
+One simple way to get a discrete output is to have sign
+functions that map the output of a linear regressor to values \( \{0,1\} \),
+\( f(s_i)=sign(s_i)=1 \) if \( s_i\ge 0 \) and 0 if otherwise.
+We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning
+literature. This model is extremely simple. However, in many cases it is more
+favorable to use a ``soft" classifier that outputs
+the probability of a given category. This leads us to the logistic function.
-
-
# Common imports
-import os
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression, Ridge, Lasso
-from sklearn.model_selection import train_test_split
-from sklearn.utils import resample
-from sklearn.metrics import mean_squared_error
-from IPython.display import display
-from pylab import plt, mpl
-plt.style.use('seaborn')
-mpl.rcParams['font.family'] = 'serif'
-
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
-
-if not os.path.exists(PROJECT_ROOT_DIR):
- os.mkdir(PROJECT_ROOT_DIR)
-
-if not os.path.exists(FIGURE_ID):
- os.makedirs(FIGURE_ID)
-
-if not os.path.exists(DATA_ID):
- os.makedirs(DATA_ID)
-
-def image_path(fig_id):
- return os.path.join(FIGURE_ID, fig_id)
-
-def data_path(dat_id):
- return os.path.join(DATA_ID, dat_id)
-
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-infile = open(data_path("chddata.csv"),'r')
-
-# Read the chd data as csv file and organize the data into arrays with age group, age, and chd
-chd = pd.read_csv(infile, names=('ID', 'Age', 'Agegroup', 'CHD'))
-chd.columns = ['ID', 'Age', 'Agegroup', 'CHD']
-output = chd['CHD']
-age = chd['Age']
-agegroup = chd['Agegroup']
-numberID = chd['ID']
-display(chd)
-
-plt.scatter(age, output, marker='o')
-plt.axis([18,70.0,-0.1, 1.2])
-plt.xlabel(r'Age')
-plt.ylabel(r'CHD')
-plt.title(r'Age distribution and Coronary heart disease')
-plt.show()
-
@@ -326,7 +440,7 @@ plt.show()
32
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»
diff --git a/doc/pub/week38/html/._week38-bs024.html b/doc/pub/week38/html/._week38-bs024.html
index cd61d3c47..b99e8a876 100644
--- a/doc/pub/week38/html/._week38-bs024.html
+++ b/doc/pub/week38/html/._week38-bs024.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,42 +395,69 @@ MathJax.Hub.Config({
-Plotting the mean value for each group
+Simple example
-What we could attempt however is to plot the mean value for each group.
+The following example on data for coronary heart disease (CHD) as function of age may serve as an illustration. In the code here we read and plot whether a person has had CHD (output = 1) or not (output = 0). This ouput is plotted the person's against age. Clearly, the figure shows that attempting to make a standard linear regression fit may not be very meaningful.
-
agegroupmean = np.array([0.1, 0.133, 0.250, 0.333, 0.462, 0.625, 0.765, 0.800])
-group = np.array([1, 2, 3, 4, 5, 6, 7, 8])
-plt.plot(group, agegroupmean, "r-")
-plt.axis([0,9,0, 1.0])
-plt.xlabel(r'Age group')
-plt.ylabel(r'CHD mean values')
-plt.title(r'Mean values for each age group')
+# Common imports
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression, Ridge, Lasso
+from sklearn.model_selection import train_test_split
+from sklearn.utils import resample
+from sklearn.metrics import mean_squared_error
+from IPython.display import display
+from pylab import plt, mpl
+plt.style.use('seaborn')
+mpl.rcParams['font.family'] = 'serif'
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("chddata.csv"),'r')
+
+# Read the chd data as csv file and organize the data into arrays with age group, age, and chd
+chd = pd.read_csv(infile, names=('ID', 'Age', 'Agegroup', 'CHD'))
+chd.columns = ['ID', 'Age', 'Agegroup', 'CHD']
+output = chd['CHD']
+age = chd['Age']
+agegroup = chd['Agegroup']
+numberID = chd['ID']
+display(chd)
+
+plt.scatter(age, output, marker='o')
+plt.axis([18,70.0,-0.1, 1.2])
+plt.xlabel(r'Age')
+plt.ylabel(r'CHD')
+plt.title(r'Age distribution and Coronary heart disease')
plt.show()
-
-We are now trying to find a function \( f(y\vert x) \), that is a function which gives us an expected value for the output \( y \) with a given input \( x \).
-In standard linear regression with a linear dependence on \( x \), we would write this in terms of our model
-$$
-f(y_i\vert x_i)=\beta_0+\beta_1 x_i.
-$$
-
-
-This expression implies however that \( f(y_i\vert x_i) \) could take any
-value from minus infinity to plus infinity. If we however let
-\( f(y\vert y) \) be represented by the mean value, the above example
-shows us that we can constrain the function to take values between
-zero and one, that is we have \( 0 \le f(y_i\vert x_i) \le 1 \). Looking
-at our last curve we see also that it has an S-shaped form. This leads
-us to a very popular model for the function \( f \), namely the so-called
-Sigmoid function or logistic model. We will consider this function as
-representing the probability for finding a value of \( y_i \) with a given
-\( x_i \).
-
@@ -299,7 +484,7 @@ representing the probability for finding a value of \( y_i \) with a given
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diff --git a/doc/pub/week38/html/._week38-bs025.html b/doc/pub/week38/html/._week38-bs025.html
index f9daf0b2a..85edea53b 100644
--- a/doc/pub/week38/html/._week38-bs025.html
+++ b/doc/pub/week38/html/._week38-bs025.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,25 +395,41 @@ MathJax.Hub.Config({
-The logistic function
+Plotting the mean value for each group
-Another widely studied model, is the so-called
-perceptron model, which is an example of a "hard classification" model. We
-will encounter this model when we discuss neural networks as
-well. Each datapoint is deterministically assigned to a category (i.e
-\( y_i=0 \) or \( y_i=1 \)). In many cases, and the coronary heart disease data forms one of many such examples, it is favorable to have a "soft"
-classifier that outputs the probability of a given category rather
-than a single value. For example, given \( x_i \), the classifier
-outputs the probability of being in a category \( k \). Logistic regression
-is the most common example of a so-called soft classifier. In logistic
-regression, the probability that a data point \( x_i \)
-belongs to a category \( y_i=\{0,1\} \) is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event,
+What we could attempt however is to plot the mean value for each group.
+
+
+
+
+
agegroupmean = np.array([0.1, 0.133, 0.250, 0.333, 0.462, 0.625, 0.765, 0.800])
+group = np.array([1, 2, 3, 4, 5, 6, 7, 8])
+plt.plot(group, agegroupmean, "r-")
+plt.axis([0,9,0, 1.0])
+plt.xlabel(r'Age group')
+plt.ylabel(r'CHD mean values')
+plt.title(r'Mean values for each age group')
+plt.show()
+
+
+We are now trying to find a function \( f(y\vert x) \), that is a function which gives us an expected value for the output \( y \) with a given input \( x \).
+In standard linear regression with a linear dependence on \( x \), we would write this in terms of our model
$$
-p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}.
+f(y_i\vert x_i)=\beta_0+\beta_1 x_i.
$$
-Note that \( 1-p(t)= p(-t) \).
+
+This expression implies however that \( f(y_i\vert x_i) \) could take any
+value from minus infinity to plus infinity. If we however let
+\( f(y\vert y) \) be represented by the mean value, the above example
+shows us that we can constrain the function to take values between
+zero and one, that is we have \( 0 \le f(y_i\vert x_i) \le 1 \). Looking
+at our last curve we see also that it has an S-shaped form. This leads
+us to a very popular model for the function \( f \), namely the so-called
+Sigmoid function or logistic model. We will consider this function as
+representing the probability for finding a value of \( y_i \) with a given
+\( x_i \).
@@ -283,7 +457,7 @@ Note that \( 1-p(t)= p(-t) \).
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diff --git a/doc/pub/week38/html/._week38-bs026.html b/doc/pub/week38/html/._week38-bs026.html
index ce0681e0b..dfd66cd8a 100644
--- a/doc/pub/week38/html/._week38-bs026.html
+++ b/doc/pub/week38/html/._week38-bs026.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,69 +395,26 @@ MathJax.Hub.Config({
-Examples of likelihood functions used in logistic regression and nueral networks
+The logistic function
-The following code plots the logistic function, the step function and other functions we will encounter from here and on.
+Another widely studied model, is the so-called
+perceptron model, which is an example of a "hard classification" model. We
+will encounter this model when we discuss neural networks as
+well. Each datapoint is deterministically assigned to a category (i.e
+\( y_i=0 \) or \( y_i=1 \)). In many cases, and the coronary heart disease data forms one of many such examples, it is favorable to have a "soft"
+classifier that outputs the probability of a given category rather
+than a single value. For example, given \( x_i \), the classifier
+outputs the probability of being in a category \( k \). Logistic regression
+is the most common example of a so-called soft classifier. In logistic
+regression, the probability that a data point \( x_i \)
+belongs to a category \( y_i=\{0,1\} \) is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event,
+$$
+p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}.
+$$
-
+Note that \( 1-p(t)= p(-t) \).
-
-
"""The sigmoid function (or the logistic curve) is a
-function that takes any real number, z, and outputs a number (0,1).
-It is useful in neural networks for assigning weights on a relative scale.
-The value z is the weighted sum of parameters involved in the learning algorithm."""
-
-import numpy
-import matplotlib.pyplot as plt
-import math as mt
-
-z = numpy.arange(-5, 5, .1)
-sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))
-sigma = sigma_fn(z)
-
-fig = plt.figure()
-ax = fig.add_subplot(111)
-ax.plot(z, sigma)
-ax.set_ylim([-0.1, 1.1])
-ax.set_xlim([-5,5])
-ax.grid(True)
-ax.set_xlabel('z')
-ax.set_title('sigmoid function')
-
-plt.show()
-
-"""Step Function"""
-z = numpy.arange(-5, 5, .02)
-step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)
-step = step_fn(z)
-
-fig = plt.figure()
-ax = fig.add_subplot(111)
-ax.plot(z, step)
-ax.set_ylim([-0.5, 1.5])
-ax.set_xlim([-5,5])
-ax.grid(True)
-ax.set_xlabel('z')
-ax.set_title('step function')
-
-plt.show()
-
-"""tanh Function"""
-z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)
-t = numpy.tanh(z)
-
-fig = plt.figure()
-ax = fig.add_subplot(111)
-ax.plot(z, t)
-ax.set_ylim([-1.0, 1.0])
-ax.set_xlim([-2*mt.pi,2*mt.pi])
-ax.grid(True)
-ax.set_xlabel('z')
-ax.set_title('tanh function')
-
-plt.show()
-
@@ -326,7 +441,7 @@ plt.show()
35
36
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs027.html b/doc/pub/week38/html/._week38-bs027.html
index 750a7fa05..090e8a876 100644
--- a/doc/pub/week38/html/._week38-bs027.html
+++ b/doc/pub/week38/html/._week38-bs027.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,25 +395,69 @@ MathJax.Hub.Config({
-Two parameters
+Examples of likelihood functions used in logistic regression and nueral networks
-We assume now that we have two classes with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assume also that we have only two parameters \( \beta \) in our fitting of the Sigmoid function, that is we define probabilities
-$$
-\begin{align*}
-p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
-p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}),
-\end{align*}
-$$
-
-where \( \hat{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
+The following code plots the logistic function, the step function and other functions we will encounter from here and on.
-Note that we used
-$$
-p(y_i=0\vert x_i, \hat{\beta}) = 1-p(y_i=1\vert x_i, \hat{\beta}).
-$$
+
+
"""The sigmoid function (or the logistic curve) is a
+function that takes any real number, z, and outputs a number (0,1).
+It is useful in neural networks for assigning weights on a relative scale.
+The value z is the weighted sum of parameters involved in the learning algorithm."""
+
+import numpy
+import matplotlib.pyplot as plt
+import math as mt
+
+z = numpy.arange(-5, 5, .1)
+sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))
+sigma = sigma_fn(z)
+
+fig = plt.figure()
+ax = fig.add_subplot(111)
+ax.plot(z, sigma)
+ax.set_ylim([-0.1, 1.1])
+ax.set_xlim([-5,5])
+ax.grid(True)
+ax.set_xlabel('z')
+ax.set_title('sigmoid function')
+
+plt.show()
+
+"""Step Function"""
+z = numpy.arange(-5, 5, .02)
+step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)
+step = step_fn(z)
+
+fig = plt.figure()
+ax = fig.add_subplot(111)
+ax.plot(z, step)
+ax.set_ylim([-0.5, 1.5])
+ax.set_xlim([-5,5])
+ax.grid(True)
+ax.set_xlabel('z')
+ax.set_title('step function')
+
+plt.show()
+
+"""tanh Function"""
+z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)
+t = numpy.tanh(z)
+
+fig = plt.figure()
+ax = fig.add_subplot(111)
+ax.plot(z, t)
+ax.set_ylim([-1.0, 1.0])
+ax.set_xlim([-2*mt.pi,2*mt.pi])
+ax.grid(True)
+ax.set_xlabel('z')
+ax.set_title('tanh function')
+
+plt.show()
+
@@ -282,7 +484,7 @@ $$
36
37
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs028.html b/doc/pub/week38/html/._week38-bs028.html
index cddda738c..7587a1a31 100644
--- a/doc/pub/week38/html/._week38-bs028.html
+++ b/doc/pub/week38/html/._week38-bs028.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,26 +393,25 @@ MathJax.Hub.Config({
-
+
-Maximum likelihood
+Two parameters
-In order to define the total likelihood for all possible outcomes from a
-dataset \( \mathcal{D}=\{(y_i,x_i)\} \), with the binary labels
-\( y_i\in\{0,1\} \) and where the data points are drawn independently, we use the so-called Maximum Likelihood Estimation (MLE) principle.
-We aim thus at maximizing
-the probability of seeing the observed data. We can then approximate the
-likelihood in terms of the product of the individual probabilities of a specific outcome \( y_i \), that is
+We assume now that we have two classes with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assume also that we have only two parameters \( \beta \) in our fitting of the Sigmoid function, that is we define probabilities
$$
\begin{align*}
-P(\mathcal{D}|\hat{\beta})& = \prod_{i=1}^n \left[p(y_i=1|x_i,\hat{\beta})\right]^{y_i}\left[1-p(y_i=1|x_i,\hat{\beta}))\right]^{1-y_i}\nonumber \\
+p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}),
\end{align*}
$$
-from which we obtain the log-likelihood and our cost/loss function
+where \( \hat{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
+
+
+Note that we used
$$
-\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\hat{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\hat{\beta}))\right]\right).
+p(y_i=0\vert x_i, \hat{\beta}) = 1-p(y_i=1\vert x_i, \hat{\beta}).
$$
@@ -283,7 +440,7 @@ $$
37
38
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs029.html b/doc/pub/week38/html/._week38-bs029.html
index bad6f5b87..6766bb751 100644
--- a/doc/pub/week38/html/._week38-bs029.html
+++ b/doc/pub/week38/html/._week38-bs029.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -235,26 +393,28 @@ MathJax.Hub.Config({
-
+
-The cost function rewritten
+Maximum likelihood
-Reordering the logarithms, we can rewrite the cost/loss function as
+In order to define the total likelihood for all possible outcomes from a
+dataset \( \mathcal{D}=\{(y_i,x_i)\} \), with the binary labels
+\( y_i\in\{0,1\} \) and where the data points are drawn independently, we use the so-called Maximum Likelihood Estimation (MLE) principle.
+We aim thus at maximizing
+the probability of seeing the observed data. We can then approximate the
+likelihood in terms of the product of the individual probabilities of a specific outcome \( y_i \), that is
$$
-\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
+\begin{align*}
+P(\mathcal{D}|\hat{\beta})& = \prod_{i=1}^n \left[p(y_i=1|x_i,\hat{\beta})\right]^{y_i}\left[1-p(y_i=1|x_i,\hat{\beta}))\right]^{1-y_i}\nonumber \\
+\end{align*}
$$
-
-The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to \( \beta \).
-Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that
+from which we obtain the log-likelihood and our cost/loss function
$$
-\mathcal{C}(\hat{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
+\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\hat{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\hat{\beta}))\right]\right).
$$
-This equation is known in statistics as the cross entropy. Finally, we note that just as in linear regression,
-in practice we often supplement the cross-entropy with additional regularization terms, usually \( L_1 \) and \( L_2 \) regularization as we did for Ridge and Lasso regression.
-
@@ -280,6 +440,8 @@ in practice we often supplement the cross-entropy with additional regularization
37
38
39
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs030.html b/doc/pub/week38/html/._week38-bs030.html
index 332ee8a13..956d56e8b 100644
--- a/doc/pub/week38/html/._week38-bs030.html
+++ b/doc/pub/week38/html/._week38-bs030.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,24 +395,23 @@ MathJax.Hub.Config({
-Minimizing the cross entropy
+The cost function rewritten
-The cross entropy is a convex function of the weights \( \hat{\beta} \) and,
-therefore, any local minimizer is a global minimizer.
+Reordering the logarithms, we can rewrite the cost/loss function as
+$$
+\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
+$$
-Minimizing this
-cost function with respect to the two parameters \( \beta_0 \) and \( \beta_1 \) we obtain
-
+The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to \( \beta \).
+Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that
$$
-\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right),
+\mathcal{C}(\hat{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
$$
-and
-$$
-\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right).
-$$
+This equation is known in statistics as the cross entropy. Finally, we note that just as in linear regression,
+in practice we often supplement the cross-entropy with additional regularization terms, usually \( L_1 \) and \( L_2 \) regularization as we did for Ridge and Lasso regression.
@@ -280,6 +437,9 @@ $$
37
38
39
+ 40
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs031.html b/doc/pub/week38/html/._week38-bs031.html
index 607fb9db8..cc52c0edc 100644
--- a/doc/pub/week38/html/._week38-bs031.html
+++ b/doc/pub/week38/html/._week38-bs031.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,24 +395,23 @@ MathJax.Hub.Config({
-A more compact expression
+Minimizing the cross entropy
-Let us now define a vector \( \hat{y} \) with \( n \) elements \( y_i \), an
-\( n\times p \) matrix \( \hat{X} \) which contains the \( x_i \) values and a
-vector \( \hat{p} \) of fitted probabilities \( p(y_i\vert x_i,\hat{\beta}) \). We can rewrite in a more compact form the first
-derivative of cost function as
-
-$$
-\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right).
-$$
+The cross entropy is a convex function of the weights \( \hat{\beta} \) and,
+therefore, any local minimizer is a global minimizer.
-If we in addition define a diagonal matrix \( \hat{W} \) with elements
-\( p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta}) \), we can obtain a compact expression of the second derivative as
+Minimizing this
+cost function with respect to the two parameters \( \beta_0 \) and \( \beta_1 \) we obtain
$$
-\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}.
+\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right),
+$$
+
+and
+$$
+\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right).
$$
@@ -280,6 +437,10 @@ $$
37
38
39
+ 40
+ 41
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs032.html b/doc/pub/week38/html/._week38-bs032.html
index 9580b3a82..d28f8f740 100644
--- a/doc/pub/week38/html/._week38-bs032.html
+++ b/doc/pub/week38/html/._week38-bs032.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,17 +395,24 @@ MathJax.Hub.Config({
-Extending to more predictors
+A more compact expression
-Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with \( p \) predictors
+Let us now define a vector \( \hat{y} \) with \( n \) elements \( y_i \), an
+\( n\times p \) matrix \( \hat{X} \) which contains the \( x_i \) values and a
+vector \( \hat{p} \) of fitted probabilities \( p(y_i\vert x_i,\hat{\beta}) \). We can rewrite in a more compact form the first
+derivative of cost function as
+
$$
-\log{ \frac{p(\hat{\beta}\hat{x})}{1-p(\hat{\beta}\hat{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p.
+\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right).
$$
-Here we defined \( \hat{x}=[1,x_1,x_2,\dots,x_p] \) and \( \hat{\beta}=[\beta_0, \beta_1, \dots, \beta_p] \) leading to
+
+If we in addition define a diagonal matrix \( \hat{W} \) with elements
+\( p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta}) \), we can obtain a compact expression of the second derivative as
+
$$
-p(\hat{\beta}\hat{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}.
+\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}.
$$
@@ -272,6 +437,11 @@ $$
37
38
39
+ 40
+ 41
+ 42
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs033.html b/doc/pub/week38/html/._week38-bs033.html
index ebf1825a0..144e22e47 100644
--- a/doc/pub/week38/html/._week38-bs033.html
+++ b/doc/pub/week38/html/._week38-bs033.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,31 +395,19 @@ MathJax.Hub.Config({
-Including more classes
+Extending to more predictors
-Till now we have mainly focused on two classes, the so-called binary
-system. Suppose we wish to extend to \( K \) classes. Let us for the sake
-of simplicity assume we have only two predictors. We have then following model
-
+Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with \( p \) predictors
$$
-\log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1,
+\log{ \frac{p(\hat{\beta}\hat{x})}{1-p(\hat{\beta}\hat{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p.
$$
-and
+Here we defined \( \hat{x}=[1,x_1,x_2,\dots,x_p] \) and \( \hat{\beta}=[\beta_0, \beta_1, \dots, \beta_p] \) leading to
$$
-\log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1,
+p(\hat{\beta}\hat{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}.
$$
-and so on till the class \( C=K-1 \) class
-$$
-\log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1,
-$$
-
-
-and the model is specified in term of \( K-1 \) so-called log-odds or
-logit transformations.
-
@@ -283,6 +429,12 @@ and the model is specified in term of \( K-1 \) so-called log-odds or
37
38
39
+ 40
+ 41
+ 42
+ 43
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs034.html b/doc/pub/week38/html/._week38-bs034.html
index fa3d8cd68..da329c4ce 100644
--- a/doc/pub/week38/html/._week38-bs034.html
+++ b/doc/pub/week38/html/._week38-bs034.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,45 +395,30 @@ MathJax.Hub.Config({
-More classes
+Including more classes
-In our discussion of neural networks we will encounter the above again
-in terms of a slightly modified function, the so-called Softmax function.
-
-
-The softmax function is used in various multiclass classification
-methods, such as multinomial logistic regression (also known as
-softmax regression), multiclass linear discriminant analysis, naive
-Bayes classifiers, and artificial neural networks. Specifically, in
-multinomial logistic regression and linear discriminant analysis, the
-input to the function is the result of \( K \) distinct linear functions,
-and the predicted probability for the \( k \)-th class given a sample
-vector \( \hat{x} \) and a weighting vector \( \hat{\beta} \) is (with two
-predictors):
+Till now we have mainly focused on two classes, the so-called binary
+system. Suppose we wish to extend to \( K \) classes. Let us for the sake
+of simplicity assume we have only two predictors. We have then following model
$$
-p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}.
+\log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1,
$$
-It is easy to extend to more predictors. The final class is
+and
$$
-p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}},
+\log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1,
+$$
+
+and so on till the class \( C=K-1 \) class
+$$
+\log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1,
$$
-and they sum to one. Our earlier discussions were all specialized to
-the case with two classes only. It is easy to see from the above that
-what we derived earlier is compatible with these equations.
-
-
-To find the optimal parameters we would typically use a gradient
-descent method. Newton's method and gradient descent methods are
-discussed in the material on optimization
-methods.
-
-
-This will be discussed next week. Before we develop our own codes for logistic regression, we end this lecture by studying the functionality that Scikit-learn offers.
+and the model is specified in term of \( K-1 \) so-called log-odds or
+logit transformations.
@@ -297,6 +440,13 @@ This will be discussed next week. Before we develop our own codes for logistic r
37
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+ 40
+ 41
+ 42
+ 43
+ 44
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs035.html b/doc/pub/week38/html/._week38-bs035.html
index b9c0e467f..175324d55 100644
--- a/doc/pub/week38/html/._week38-bs035.html
+++ b/doc/pub/week38/html/._week38-bs035.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,42 +395,43 @@ MathJax.Hub.Config({
-Wisconsin Cancer Data
+More classes
-We show here how we can use a simple regression case on the breast
-cancer data using Logistic regression as our algorithm for
-classification.
+In our discussion of neural networks we will encounter the above again
+in terms of a slightly modified function, the so-called Softmax function.
+The softmax function is used in various multiclass classification
+methods, such as multinomial logistic regression (also known as
+softmax regression), multiclass linear discriminant analysis, naive
+Bayes classifiers, and artificial neural networks. Specifically, in
+multinomial logistic regression and linear discriminant analysis, the
+input to the function is the result of \( K \) distinct linear functions,
+and the predicted probability for the \( k \)-th class given a sample
+vector \( \hat{x} \) and a weighting vector \( \hat{\beta} \) is (with two
+predictors):
-
-
import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import train_test_split
-from sklearn.datasets import load_breast_cancer
-from sklearn.linear_model import LogisticRegression
+$$
+p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}.
+$$
-# Load the data
-cancer = load_breast_cancer()
+It is easy to extend to more predictors. The final class is
+$$
+p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}},
+$$
+
+
+and they sum to one. Our earlier discussions were all specialized to
+the case with two classes only. It is easy to see from the above that
+what we derived earlier is compatible with these equations.
+
+
+To find the optimal parameters we would typically use a gradient
+descent method. Newton's method and gradient descent methods are
+discussed in the material on optimization
+methods.
-X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
-print(X_train.shape)
-print(X_test.shape)
-# Logistic Regression
-logreg = LogisticRegression(solver='lbfgs')
-logreg.fit(X_train, y_train)
-print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
-#now scale the data
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-# Logistic Regression
-logreg.fit(X_train_scaled, y_train)
-print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-
@@ -292,6 +451,14 @@ logreg.fit(X_train_scaled, y_train)
37
38
39
+ 40
+ 41
+ 42
+ 43
+ 44
+ 45
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs036.html b/doc/pub/week38/html/._week38-bs036.html
index 183febb49..a907143fd 100644
--- a/doc/pub/week38/html/._week38-bs036.html
+++ b/doc/pub/week38/html/._week38-bs036.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,49 +395,8 @@ MathJax.Hub.Config({
-Using the correlation matrix
+Friday September 24
-
-In addition to the above scores, we could also study the covariance (and the correlation matrix).
-We use Pandas to compute the correlation matrix.
-
-
-
-
import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import train_test_split
-from sklearn.datasets import load_breast_cancer
-from sklearn.linear_model import LogisticRegression
-cancer = load_breast_cancer()
-import pandas as pd
-# Making a data frame
-cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-
-fig, axes = plt.subplots(15,2,figsize=(10,20))
-malignant = cancer.data[cancer.target == 0]
-benign = cancer.data[cancer.target == 1]
-ax = axes.ravel()
-
-for i in range(30):
- _, bins = np.histogram(cancer.data[:,i], bins =50)
- ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
- ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
- ax[i].set_title(cancer.feature_names[i])
- ax[i].set_yticks(())
-ax[0].set_xlabel("Feature magnitude")
-ax[0].set_ylabel("Frequency")
-ax[0].legend(["Malignant", "Benign"], loc ="best")
-fig.tight_layout()
-plt.show()
-
-import seaborn as sns
-correlation_matrix = cancerpd.corr().round(1)
-# use the heatmap function from seaborn to plot the correlation matrix
-# annot = True to print the values inside the square
-plt.figure(figsize=(15,8))
-sns.heatmap(data=correlation_matrix, annot=True)
-plt.show()
-
@@ -298,6 +415,15 @@ plt.show()
37
38
39
+ 40
+ 41
+ 42
+ 43
+ 44
+ 45
+ 46
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs037.html b/doc/pub/week38/html/._week38-bs037.html
index 8df23b1bb..b3c416ee3 100644
--- a/doc/pub/week38/html/._week38-bs037.html
+++ b/doc/pub/week38/html/._week38-bs037.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,42 +395,42 @@ MathJax.Hub.Config({
-Discussing the correlation data
+Wisconsin Cancer Data
-In the above example we note two things. In the first plot we display
-the overlap of benign and malignant tumors as functions of the various
-features in the Wisconsing breast cancer data set. We see that for
-some of the features we can distinguish clearly the benign and
-malignant cases while for other features we cannot. This can point to
-us which features may be of greater interest when we wish to classify
-a benign or not benign tumour.
+We show here how we can use a simple regression case on the breast
+cancer data using Logistic regression as our algorithm for
+classification.
-
-In the second figure we have computed the so-called correlation
-matrix, which in our case with thirty features becomes a \( 30\times 30 \)
-matrix.
-
-
-We constructed this matrix using pandas via the statements
-
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-
-
-and then
-
+
import matplotlib.pyplot as plt
+import numpy as np
+from sklearn.model_selection import train_test_split
+from sklearn.datasets import load_breast_cancer
+from sklearn.linear_model import LogisticRegression
-
-correlation_matrix = cancerpd.corr().round(1)
-
-
-Diagonalizing this matrix we can in turn say something about which
-features are of relevance and which are not. This leads us to
-the classical Principal Component Analysis (PCA) theorem with
-applications. This will be discussed later this semester (week 43).
+# Load the data
+cancer = load_breast_cancer()
+X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
+print(X_train.shape)
+print(X_test.shape)
+# Logistic Regression
+logreg = LogisticRegression(solver='lbfgs')
+logreg.fit(X_train, y_train)
+print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
+#now scale the data
+from sklearn.preprocessing import StandardScaler
+scaler = StandardScaler()
+scaler.fit(X_train)
+X_train_scaled = scaler.transform(X_train)
+X_test_scaled = scaler.transform(X_test)
+# Logistic Regression
+logreg.fit(X_train_scaled, y_train)
+print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+
@@ -290,6 +448,16 @@ applications. This will be discussed later this semester (37
38
39
+ 40
+ 41
+ 42
+ 43
+ 44
+ 45
+ 46
+ 47
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs038.html b/doc/pub/week38/html/._week38-bs038.html
index c2df66b5e..3ea361de6 100644
--- a/doc/pub/week38/html/._week38-bs038.html
+++ b/doc/pub/week38/html/._week38-bs038.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -237,7 +395,11 @@ MathJax.Hub.Config({
-Other measures in classification studies: Cancer Data again
+Using the correlation matrix
+
+
+In addition to the above scores, we could also study the covariance (and the correlation matrix).
+We use Pandas to compute the correlation matrix.
@@ -246,48 +408,37 @@ MathJax.Hub.Config({
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
-
-# Load the data
cancer = load_breast_cancer()
+import pandas as pd
+# Making a data frame
+cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
-print(X_train.shape)
-print(X_test.shape)
-# Logistic Regression
-logreg = LogisticRegression(solver='lbfgs')
-logreg.fit(X_train, y_train)
-print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
-#now scale the data
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-# Logistic Regression
-logreg.fit(X_train_scaled, y_train)
-print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+fig, axes = plt.subplots(15,2,figsize=(10,20))
+malignant = cancer.data[cancer.target == 0]
+benign = cancer.data[cancer.target == 1]
+ax = axes.ravel()
-
-from sklearn.preprocessing import LabelEncoder
-from sklearn.model_selection import cross_validate
-#Cross validation
-accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']
-print(accuracy)
-print("Test set accuracy with Logistic Regression and scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-
-
-import scikitplot as skplt
-y_pred = logreg.predict(X_test_scaled)
-skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
+for i in range(30):
+ _, bins = np.histogram(cancer.data[:,i], bins =50)
+ ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
+ ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
+ ax[i].set_title(cancer.feature_names[i])
+ ax[i].set_yticks(())
+ax[0].set_xlabel("Feature magnitude")
+ax[0].set_ylabel("Frequency")
+ax[0].legend(["Malignant", "Benign"], loc ="best")
+fig.tight_layout()
plt.show()
-y_probas = logreg.predict_proba(X_test_scaled)
-skplt.metrics.plot_roc(y_test, y_probas)
-plt.show()
-skplt.metrics.plot_cumulative_gain(y_test, y_probas)
+
+import seaborn as sns
+correlation_matrix = cancerpd.corr().round(1)
+# use the heatmap function from seaborn to plot the correlation matrix
+# annot = True to print the values inside the square
+plt.figure(figsize=(15,8))
+sns.heatmap(data=correlation_matrix, annot=True)
plt.show()
-
@@ -303,6 +454,18 @@ plt.show()
- 37
- 38
- 39
+ - 40
+ - 41
+ - 42
+ - 43
+ - 44
+ - 45
+ - 46
+ - 47
+ - 48
+ - ...
+ - 83
+ - »
diff --git a/doc/pub/week38/html/._week38-bs039.html b/doc/pub/week38/html/._week38-bs039.html
index 3fa44fc00..b532e7ba0 100644
--- a/doc/pub/week38/html/._week38-bs039.html
+++ b/doc/pub/week38/html/._week38-bs039.html
@@ -42,7 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -195,47 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -251,42 +395,42 @@ MathJax.Hub.Config({
-Wisconsin Cancer Data
+Discussing the correlation data
-We show here how we can use a simple regression case on the breast
-cancer data using Logistic regression as our algorithm for
-classification.
+In the above example we note two things. In the first plot we display
+the overlap of benign and malignant tumors as functions of the various
+features in the Wisconsing breast cancer data set. We see that for
+some of the features we can distinguish clearly the benign and
+malignant cases while for other features we cannot. This can point to
+us which features may be of greater interest when we wish to classify
+a benign or not benign tumour.
+
+In the second figure we have computed the so-called correlation
+matrix, which in our case with thirty features becomes a \( 30\times 30 \)
+matrix.
+
+
+We constructed this matrix using pandas via the statements
-
import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import train_test_split
-from sklearn.datasets import load_breast_cancer
-from sklearn.linear_model import LogisticRegression
-
-# Load the data
-cancer = load_breast_cancer()
-
-X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
-print(X_train.shape)
-print(X_test.shape)
-# Logistic Regression
-logreg = LogisticRegression(solver='lbfgs')
-logreg.fit(X_train, y_train)
-print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
-#now scale the data
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-# Logistic Regression
-logreg.fit(X_train_scaled, y_train)
-print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
+
+and then
+
+
+
+
correlation_matrix = cancerpd.corr().round(1)
+
+
+Diagonalizing this matrix we can in turn say something about which
+features are of relevance and which are not. This leads us to
+the classical Principal Component Analysis (PCA) theorem with
+applications. This will be discussed later this semester (week 43).
+
@@ -306,6 +450,14 @@ logreg.fit(X_train_scaled, y_train)
41
42
43
+ 44
+ 45
+ 46
+ 47
+ 48
+ 49
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs040.html b/doc/pub/week38/html/._week38-bs040.html
index 4ea1dbb67..47b176e16 100644
--- a/doc/pub/week38/html/._week38-bs040.html
+++ b/doc/pub/week38/html/._week38-bs040.html
@@ -42,7 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -195,47 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -251,11 +395,7 @@ MathJax.Hub.Config({
-Using the correlation matrix
-
-
-In addition to the above scores, we could also study the covariance (and the correlation matrix).
-We use Pandas to compute the correlation matrix.
+
Other measures in classification studies: Cancer Data again
@@ -264,34 +404,44 @@ We use Pandas to compute the correlation matrix.
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
+
+# Load the data
cancer = load_breast_cancer()
-import pandas as pd
-# Making a data frame
-cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-fig, axes = plt.subplots(15,2,figsize=(10,20))
-malignant = cancer.data[cancer.target == 0]
-benign = cancer.data[cancer.target == 1]
-ax = axes.ravel()
+X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
+print(X_train.shape)
+print(X_test.shape)
+# Logistic Regression
+logreg = LogisticRegression(solver='lbfgs')
+logreg.fit(X_train, y_train)
+print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
+#now scale the data
+from sklearn.preprocessing import StandardScaler
+scaler = StandardScaler()
+scaler.fit(X_train)
+X_train_scaled = scaler.transform(X_train)
+X_test_scaled = scaler.transform(X_test)
+# Logistic Regression
+logreg.fit(X_train_scaled, y_train)
+print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-for i in range(30):
- _, bins = np.histogram(cancer.data[:,i], bins =50)
- ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
- ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
- ax[i].set_title(cancer.feature_names[i])
- ax[i].set_yticks(())
-ax[0].set_xlabel("Feature magnitude")
-ax[0].set_ylabel("Frequency")
-ax[0].legend(["Malignant", "Benign"], loc ="best")
-fig.tight_layout()
+
+from sklearn.preprocessing import LabelEncoder
+from sklearn.model_selection import cross_validate
+#Cross validation
+accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']
+print(accuracy)
+print("Test set accuracy with Logistic Regression and scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+
+
+import scikitplot as skplt
+y_pred = logreg.predict(X_test_scaled)
+skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
plt.show()
-
-import seaborn as sns
-correlation_matrix = cancerpd.corr().round(1)
-# use the heatmap function from seaborn to plot the correlation matrix
-# annot = True to print the values inside the square
-plt.figure(figsize=(15,8))
-sns.heatmap(data=correlation_matrix, annot=True)
+y_probas = logreg.predict_proba(X_test_scaled)
+skplt.metrics.plot_roc(y_test, y_probas)
+plt.show()
+skplt.metrics.plot_cumulative_gain(y_test, y_probas)
plt.show()
@@ -312,6 +462,15 @@ plt.show()
41
42
43
+ 44
+ 45
+ 46
+ 47
+ 48
+ 49
+ 50
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs041.html b/doc/pub/week38/html/._week38-bs041.html
index 402bd29b0..b563384d6 100644
--- a/doc/pub/week38/html/._week38-bs041.html
+++ b/doc/pub/week38/html/._week38-bs041.html
@@ -42,7 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -195,47 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -251,41 +395,20 @@ MathJax.Hub.Config({
-Discussing the correlation data
+Optimization, the central part of any Machine Learning algortithm
-In the above example we note two things. In the first plot we display
-the overlap of benign and malignant tumors as functions of the various
-features in the Wisconsing breast cancer data set. We see that for
-some of the features we can distinguish clearly the benign and
-malignant cases while for other features we cannot. This can point to
-us which features may be of greater interest when we wish to classify
-a benign or not benign tumour.
+Overview Video, why do we care about gradient methods?
-In the second figure we have computed the so-called correlation
-matrix, which in our case with thirty features becomes a \( 30\times 30 \)
-matrix.
-
-
-We constructed this matrix using pandas via the statements
-
-
-
-
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-
-
-and then
-
-
-
-
correlation_matrix = cancerpd.corr().round(1)
-
-
-Diagonalizing this matrix we can in turn say something about which
-features are of relevance and which are not. This leads us to
-the classical Principal Component Analysis (PCA) theorem with
-applications. This will be discussed later this semester (week 43).
+Almost every problem in machine learning and data science starts with
+a dataset \( X \), a model \( g(\beta) \), which is a function of the
+parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows
+us to judge how well the model \( g(\beta) \) explains the observations
+\( X \). The model is fit by finding the values of \( \beta \) that minimize
+the cost function. Ideally we would be able to solve for \( \beta \)
+analytically, however this is not possible in general and we must use
+some approximative/numerical method to compute the minimum.
@@ -304,6 +427,16 @@ applications. This will be discussed later this semester (41
42
43
+ 44
+ 45
+ 46
+ 47
+ 48
+ 49
+ 50
+ 51
+ ...
+ 83
»
diff --git a/doc/pub/week38/html/._week38-bs042.html b/doc/pub/week38/html/._week38-bs042.html
index 6b41df4bd..09c9ebe85 100644
--- a/doc/pub/week38/html/._week38-bs042.html
+++ b/doc/pub/week38/html/._week38-bs042.html
@@ -42,7 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -195,47 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -251,57 +395,26 @@ MathJax.Hub.Config({
-Other measures in classification studies: Cancer Data again
+Revisiting our Logistic Regression case
+
+In our discussion on Logistic Regression we studied the
+case of
+two classes, with \( y_i \) either
+\( 0 \) or \( 1 \). Furthermore we assumed also that we have only two
+parameters \( \beta \) in our fitting, that is we
+defined probabilities
-
-
import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import train_test_split
-from sklearn.datasets import load_breast_cancer
-from sklearn.linear_model import LogisticRegression
+$$
+\begin{align*}
+p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
+\end{align*}
+$$
-# Load the data
-cancer = load_breast_cancer()
+where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
-X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
-print(X_train.shape)
-print(X_test.shape)
-# Logistic Regression
-logreg = LogisticRegression(solver='lbfgs')
-logreg.fit(X_train, y_train)
-print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
-#now scale the data
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-# Logistic Regression
-logreg.fit(X_train_scaled, y_train)
-print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-
-
-from sklearn.preprocessing import LabelEncoder
-from sklearn.model_selection import cross_validate
-#Cross validation
-accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']
-print(accuracy)
-print("Test set accuracy with Logistic Regression and scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-
-
-import scikitplot as skplt
-y_pred = logreg.predict(X_test_scaled)
-skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
-plt.show()
-y_probas = logreg.predict_proba(X_test_scaled)
-skplt.metrics.plot_roc(y_test, y_probas)
-plt.show()
-skplt.metrics.plot_cumulative_gain(y_test, y_probas)
-plt.show()
-
-
@@ -317,6 +430,18 @@ plt.show()
- 41
- 42
- 43
+ - 44
+ - 45
+ - 46
+ - 47
+ - 48
+ - 49
+ - 50
+ - 51
+ - 52
+ - ...
+ - 83
+ - »
diff --git a/doc/pub/week38/html/week38-bs.html b/doc/pub/week38/html/week38-bs.html
index 9ae29ebbb..e179160ae 100644
--- a/doc/pub/week38/html/week38-bs.html
+++ b/doc/pub/week38/html/week38-bs.html
@@ -42,6 +42,7 @@ Automatically generated HTML file from DocOnce source
@@ -185,43 +299,87 @@ MathJax.Hub.Config({
Contents
@@ -280,7 +438,7 @@ MathJax.Hub.Config({
9
10
...
- 39
+ 83
»
diff --git a/doc/pub/week38/html/week38-reveal.html b/doc/pub/week38/html/week38-reveal.html
index 55ac915a9..9b83e2c64 100644
--- a/doc/pub/week38/html/week38-reveal.html
+++ b/doc/pub/week38/html/week38-reveal.html
@@ -168,6 +168,11 @@ MathJax.Hub.Config({
+
+Thursday September 23
+
+
+
Ridge and LASSO Regression, reminder
@@ -2011,9 +2016,11 @@ To find the optimal parameters we would typically use a gradient
descent method. Newton's method and gradient descent methods are
discussed in the material on optimization
methods.
+
-
-This will be discussed next week. Before we develop our own codes for logistic regression, we end this lecture by studying the functionality that Scikit-learn offers.
+
+
+Friday September 24
@@ -2196,6 +2203,1293 @@ plt.show()
+
+Optimization, the central part of any Machine Learning algortithm
+
+
+Overview Video, why do we care about gradient methods?
+
+
+Almost every problem in machine learning and data science starts with
+a dataset \( X \), a model \( g(\beta) \), which is a function of the
+parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows
+us to judge how well the model \( g(\beta) \) explains the observations
+\( X \). The model is fit by finding the values of \( \beta \) that minimize
+the cost function. Ideally we would be able to solve for \( \beta \)
+analytically, however this is not possible in general and we must use
+some approximative/numerical method to compute the minimum.
+
+
+
+
+Revisiting our Logistic Regression case
+
+
+In our discussion on Logistic Regression we studied the
+case of
+two classes, with \( y_i \) either
+\( 0 \) or \( 1 \). Furthermore we assumed also that we have only two
+parameters \( \beta \) in our fitting, that is we
+defined probabilities
+
+
+$$
+\begin{align*}
+p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
+\end{align*}
+$$
+
+
+where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
+
+
+
+
+The equations to solve
+
+
+Our compact equations used a definition of a vector \( \boldsymbol{y} \) with \( n \)
+elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the
+\( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities
+\( p(y_i\vert x_i,\boldsymbol{\beta}) \). We rewrote in a more compact form
+the first derivative of the cost function as
+
+
+$$
+\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
+$$
+
+
+
+If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements
+\( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as
+
+
+$$
+\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
+$$
+
+
+This defines what is called the Hessian matrix.
+
+
+
+
+Solving using Newton-Raphson's method
+
+
+If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
+
+
+Our iterative scheme is then given by
+
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}},
+$$
+
+
+or in matrix form as
+
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}.
+$$
+
+
+The right-hand side is computed with the old values of \( \beta \).
+
+
+If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
+
+
+
+
+Brief reminder on Newton-Raphson's method
+
+
+Let us quickly remind ourselves how we derive the above method.
+
+
+Perhaps the most celebrated of all one-dimensional root-finding
+routines is Newton's method, also called the Newton-Raphson
+method. This method requires the evaluation of both the
+function \( f \) and its derivative \( f' \) at arbitrary points.
+If you can only calculate the derivative
+numerically and/or your function is not of the smooth type, we
+normally discourage the use of this method.
+
+
+
+
+The equations
+
+
+The Newton-Raphson formula consists geometrically of extending the
+tangent line at a current point until it crosses zero, then setting
+the next guess to the abscissa of that zero-crossing. The mathematics
+behind this method is rather simple. Employing a Taylor expansion for
+\( x \) sufficiently close to the solution \( s \), we have
+
+
+$$
+ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
+ \tag{14}
+$$
+
+
+
+For small enough values of the function and for well-behaved
+functions, the terms beyond linear are unimportant, hence we obtain
+
+
+$$
+ f(x)+(s-x)f'(x)\approx 0,
+$$
+
+
+yielding
+
+$$
+ s\approx x-\frac{f(x)}{f'(x)}.
+$$
+
+
+
+Having in mind an iterative procedure, it is natural to start iterating with
+
+$$
+ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
+$$
+
+
+
+
+
+Simple geometric interpretation
+
+
+The above is Newton-Raphson's method. It has a simple geometric
+interpretation, namely \( x_{n+1} \) is the point where the tangent from
+\( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution,
+Newton-Raphson converges fast to the desired result. However, if we
+are far from a root, where the higher-order terms in the series are
+important, the Newton-Raphson formula can give grossly inaccurate
+results. For instance, the initial guess for the root might be so far
+from the true root as to let the search interval include a local
+maximum or minimum of the function. If an iteration places a trial
+guess near such a local extremum, so that the first derivative nearly
+vanishes, then Newton-Raphson may fail totally
+
+
+
+
+Extending to more than one variable
+
+
+Newton's method can be generalized to systems of several non-linear equations
+and variables. Consider the case with two equations
+
+$$
+ \begin{array}{cc} f_1(x_1,x_2) &=0\\
+ f_2(x_1,x_2) &=0,\end{array}
+$$
+
+
+which we Taylor expand to obtain
+
+
+$$
+ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
+ \partial f_1/\partial x_1+h_2
+ \partial f_1/\partial x_2+\dots\\
+ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
+ \partial f_2/\partial x_1+h_2
+ \partial f_2/\partial x_2+\dots
+ \end{array}.
+$$
+
+
+Defining the Jacobian matrix \( {\bf \boldsymbol{J}} \) we have
+
+$$
+ {\bf \boldsymbol{J}}=\left( \begin{array}{cc}
+ \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
+ \partial f_2/\partial x_1 &\partial f_2/\partial x_2
+ \end{array} \right),
+$$
+
+
+we can rephrase Newton's method as
+
+$$
+\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
+\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
+\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
+$$
+
+
+where we have defined
+
+$$
+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
+ -{\bf \boldsymbol{J}}^{-1}
+ \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
+$$
+
+
+We need thus to compute the inverse of the Jacobian matrix and it
+is to understand that difficulties may
+arise in case \( {\bf \boldsymbol{J}} \) is nearly singular.
+
+
+It is rather straightforward to extend the above scheme to systems of
+more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
+
+
+
+
+Steepest descent
+
+
+The basic idea of gradient descent is
+that a function \( F(\mathbf{x}) \),
+\( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the
+direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
+
+
+It can be shown that if
+
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
+$$
+
+
+with \( \gamma_k > 0 \).
+
+
+For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \)
+we are always moving towards smaller function values, i.e a minimum.
+
+
+
+
+More on Steepest descent
+
+
+The previous observation is the basis of the method of steepest
+descent, which is also referred to as just gradient descent (GD). One
+starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and
+computes new approximations according to
+
+
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
+$$
+
+
+
+The parameter \( \gamma_k \) is often referred to as the step length or
+the learning rate within the context of Machine Learning.
+
+
+
+
+The ideal
+
+
+Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global
+minimum of the function \( F \). In general we do not know if we are in a
+global or local minimum. In the special case when \( F \) is a convex
+function, all local minima are also global minima, so in this case
+gradient descent can converge to the global solution. The advantage of
+this scheme is that it is conceptually simple and straightforward to
+implement. However the method in this form has some severe
+limitations:
+
+
+In machine learing we are often faced with non-convex high dimensional
+cost functions with many local minima. Since GD is deterministic we
+will get stuck in a local minimum, if the method converges, unless we
+have a very good intial guess. This also implies that the scheme is
+sensitive to the chosen initial condition.
+
+
+Note that the gradient is a function of \( \mathbf{x} =
+(x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
+
+
+
+
+The sensitiveness of the gradient descent
+
+
+The gradient descent method
+is sensitive to the choice of learning rate \( \gamma_k \). This is due
+to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to
+determine an optimal learning rate. If the learning rate is chosen too
+small the method will take a long time to converge and if it is too
+large we can experience erratic behavior.
+
+
+Many of these shortcomings can be alleviated by introducing
+randomness. One such method is that of Stochastic Gradient Descent
+(SGD), see below.
+
+
+
+
+Convex functions
+
+
+Ideally we want our cost/loss function to be convex(concave).
+
+
+First we give the definition of a convex set: A set \( C \) in
+\( \mathbb{R}^n \) is said to be convex if, for all \( x \) and \( y \) in \( C \) and
+all \( t \in (0,1) \) , the point \( (1 − t)x + ty \) also belongs to
+C. Geometrically this means that every point on the line segment
+connecting \( x \) and \( y \) is in \( C \) as discussed below.
+
+
+The convex subsets of \( \mathbb{R} \) are the intervals of
+\( \mathbb{R} \). Examples of convex sets of \( \mathbb{R}^2 \) are the
+regular polygons (triangles, rectangles, pentagons, etc...).
+
+
+
+
+Convex function
+
+
+Convex function: Let \( X \subset \mathbb{R}^n \) be a convex set. Assume that the function \( f: X \rightarrow \mathbb{R} \) is continuous, then \( f \) is said to be convex if
+$$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$
+
for all \( x_1, x_2 \in X \) and for all \( t \in [0,1] \). If \( \leq \) is replaced with a strict inequaltiy in the definition, we demand \( x_1 \neq x_2 \) and \( t\in(0,1) \) then \( f \) is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting \( f(x_1) \) and \( f(x_2) \), the value of the function on the interval \( [x_1,x_2] \) is always below the line as illustrated below.
+
+
+
+
+Conditions on convex functions
+
+
+In the following we state first and second-order conditions which
+ensures convexity of a function \( f \). We write \( D_f \) to denote the
+domain of \( f \), i.e the subset of \( R^n \) where \( f \) is defined. For more
+details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.
+
+
+
+First order condition
+
+Suppose \( f \) is differentiable (i.e \( \nabla f(x) \) is well defined for
+all \( x \) in the domain of \( f \)). Then \( f \) is convex if and only if \( D_f \)
+is a convex set and
+$$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$
+
holds
+for all \( x,y \in D_f \). This condition means that for a convex function
+the first order Taylor expansion (right hand side above) at any point
+a global under estimator of the function. To convince yourself you can
+make a drawing of \( f(x) = x^2+1 \) and draw the tangent line to \( f(x) \) and
+note that it is always below the graph.
+
+
+
+
+Second order condition
+
+Assume that \( f \) is twice
+differentiable, i.e the Hessian matrix exists at each point in
+\( D_f \). Then \( f \) is convex if and only if \( D_f \) is a convex set and its
+Hessian is positive semi-definite for all \( x\in D_f \). For a
+single-variable function this reduces to \( f''(x) \geq 0 \). Geometrically this means that \( f \) has nonnegative curvature
+everywhere.
+
+
+
+This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
+
+
+
+
+More on convex functions
+
+
+The next result is of great importance to us and the reason why we are
+going on about convex functions. In machine learning we frequently
+have to minimize a loss/cost function in order to find the best
+parameters for the model we are considering.
+
+
+Ideally we want the
+global minimum (for high-dimensional models it is hard to know
+if we have local or global minimum). However, if the cost/loss function
+is convex the following result provides invaluable information:
+
+
+
+Any minimum is global for convex functions
+
+Consider the problem of finding \( x \in \mathbb{R}^n \) such that \( f(x) \)
+is minimal, where \( f \) is convex and differentiable. Then, any point
+\( x^* \) that satisfies \( \nabla f(x^*) = 0 \) is a global minimum.
+
+
+
+This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
+
+
+
+
+Some simple problems
+
+
+- Show that \( f(x)=x^2 \) is convex for \( x \in \mathbb{R} \) using the definition of convexity. Hint: If you re-write the definition, \( f \) is convex if the following holds for all \( x,y \in D_f \) and any \( \lambda \in [0,1] \) $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$.
+- Using the second order condition show that the following functions are convex on the specified domain.
+
+
+ - \( f(x) = e^x \) is convex for \( x \in \mathbb{R} \).
+ - \( g(x) = -\ln(x) \) is convex for \( x \in (0,\infty) \).
+
+- Let \( f(x) = x^2 \) and \( g(x) = e^x \). Show that \( f(g(x)) \) and \( g(f(x)) \) is convex for \( x \in \mathbb{R} \). Also show that if \( f(x) \) is any convex function than \( h(x) = e^{f(x)} \) is convex.
+- A norm is any function that satisfy the following properties
+
+
+ - \( f(\alpha x) = |\alpha| f(x) \) for all \( \alpha \in \mathbb{R} \).
+ - \( f(x+y) \leq f(x) + f(y) \)
+ - \( f(x) \leq 0 \) for all \( x \in \mathbb{R}^n \) with equality if and only if \( x = 0 \)
+
+
+
+
+
+Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
+
+
+
+
+Friday September 25
+
+
+
+
+
+Standard steepest descent
+
+
+Before we proceed, we would like to discuss the approach called the
+standard Steepest descent (different from the above steepest descent discussion), which again leads to us having to be able
+to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
+
+
+The success of the CG method
+for finding solutions of non-linear problems is based on the theory
+of conjugate gradients for linear systems of equations. It belongs to
+the class of iterative methods for solving problems from linear
+algebra of the type
+
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}.
+\end{equation*}
+$$
+
+
+
+In the iterative process we end up with a problem like
+
+
+$$
+\begin{equation*}
+ \boldsymbol{r}= \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x},
+\end{equation*}
+$$
+
+
+where \( \boldsymbol{r} \) is the so-called residual or error in the iterative process.
+
+
+When we have found the exact solution, \( \boldsymbol{r}=0 \).
+
+
+
+
+Gradient method
+
+
+The residual is zero when we reach the minimum of the quadratic equation
+
+$$
+\begin{equation*}
+ P(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T\boldsymbol{b},
+\end{equation*}
+$$
+
+
+
+with the constraint that the matrix \( \boldsymbol{A} \) is positive definite and
+symmetric. This defines also the Hessian and we want it to be positive definite.
+
+
+
+
+Steepest descent method
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+
+or consider the system
+
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+
+instead.
+
+
+
+
+Steepest descent method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+
+This suggests taking the first basis vector \( \boldsymbol{r}_1 \) (see below for definition)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+
+
+
+
+
+
+
+Final expressions
+
+
+
+We can compute the residual iteratively as
+
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+
+which equals
+
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{r}_k),
+ \end{equation*}
+$$
+
+
+or
+
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{r}_k,
+ \end{equation*}
+$$
+
+
+which gives
+
+
+$$
+\alpha_k = \frac{\boldsymbol{r}_k^T\boldsymbol{r}_k}{\boldsymbol{r}_k^T\boldsymbol{A}\boldsymbol{r}_k}
+$$
+
+
+leading to the iterative scheme
+
+$$
+\begin{equation*}
+\boldsymbol{x}_{k+1}=\boldsymbol{x}_k-\alpha_k\boldsymbol{r}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+Steepest descent example
+
+
+
+
+
import numpy as np
+import numpy.linalg as la
+
+import scipy.optimize as sopt
+
+import matplotlib.pyplot as pt
+from mpl_toolkits.mplot3d import axes3d
+
+def f(x):
+ return 0.5*x[0]**2 + 2.5*x[1]**2
+
+def df(x):
+ return np.array([x[0], 5*x[1]])
+
+fig = pt.figure()
+ax = fig.gca(projection="3d")
+
+xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
+fmesh = f(np.array([xmesh, ymesh]))
+ax.plot_surface(xmesh, ymesh, fmesh)
+
+
+And then as countor plot
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh)
+guesses = [np.array([2, 2./5])]
+
+
+Find guesses
+
+
+
+
x = guesses[-1]
+s = -df(x)
+
+
+Run it!
+
+
+
+
def f1d(alpha):
+ return f(x + alpha*s)
+
+alpha_opt = sopt.golden(f1d)
+next_guess = x + alpha_opt * s
+guesses.append(next_guess)
+print(next_guess)
+
+
+What happened?
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh, 50)
+it_array = np.array(guesses)
+pt.plot(it_array.T[0], it_array.T[1], "x-")
+
+
+
+
+
+Conjugate gradient method
+
+
+
+In the CG method we define so-called conjugate directions and two vectors
+\( \boldsymbol{s} \) and \( \boldsymbol{t} \)
+are said to be
+conjugate if
+
+$$
+\begin{equation*}
+\boldsymbol{s}^T\boldsymbol{A}\boldsymbol{t}= 0.
+\end{equation*}
+$$
+
+
+The philosophy of the CG method is to perform searches in various conjugate directions
+of our vectors \( \boldsymbol{x}_i \) obeying the above criterion, namely
+
+$$
+\begin{equation*}
+\boldsymbol{x}_i^T\boldsymbol{A}\boldsymbol{x}_j= 0.
+\end{equation*}
+$$
+
+
+Two vectors are conjugate if they are orthogonal with respect to
+this inner product. Being conjugate is a symmetric relation: if \( \boldsymbol{s} \) is conjugate to \( \boldsymbol{t} \), then \( \boldsymbol{t} \) is conjugate to \( \boldsymbol{s} \).
+
+
+
+
+
+Conjugate gradient method
+
+
+
+An example is given by the eigenvectors of the matrix
+
+$$
+\begin{equation*}
+\boldsymbol{v}_i^T\boldsymbol{A}\boldsymbol{v}_j= \lambda\boldsymbol{v}_i^T\boldsymbol{v}_j,
+\end{equation*}
+$$
+
+
+which is zero unless \( i=j \).
+
+
+
+
+
+Conjugate gradient method
+
+
+
+Assume now that we have a symmetric positive-definite matrix \( \boldsymbol{A} \) of size
+\( n\times n \). At each iteration \( i+1 \) we obtain the conjugate direction of a vector
+
+$$
+\begin{equation*}
+\boldsymbol{x}_{i+1}=\boldsymbol{x}_{i}+\alpha_i\boldsymbol{p}_{i}.
+\end{equation*}
+$$
+
+
+We assume that \( \boldsymbol{p}_{i} \) is a sequence of \( n \) mutually conjugate directions.
+Then the \( \boldsymbol{p}_{i} \) form a basis of \( R^n \) and we can expand the solution
+$ \boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}$ in this basis, namely
+
+
+$$
+\begin{equation*}
+ \boldsymbol{x} = \sum^{n}_{i=1} \alpha_i \boldsymbol{p}_i.
+\end{equation*}
+$$
+
+
+
+
+
+
+Conjugate gradient method
+
+
+
+The coefficients are given by
+
+$$
+\begin{equation*}
+ \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
+\end{equation*}
+$$
+
+
+Multiplying with \( \boldsymbol{p}_k^T \) from the left gives
+
+
+$$
+\begin{equation*}
+ \boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{p}_i= \boldsymbol{p}_k^T \boldsymbol{b},
+\end{equation*}
+$$
+
+
+and we can define the coefficients \( \alpha_k \) as
+
+
+$$
+\begin{equation*}
+ \alpha_k = \frac{\boldsymbol{p}_k^T \boldsymbol{b}}{\boldsymbol{p}_k^T \boldsymbol{A} \boldsymbol{p}_k}
+\end{equation*}
+$$
+
+
+
+
+
+
+Conjugate gradient method and iterations
+
+
+
+If we choose the conjugate vectors \( \boldsymbol{p}_k \) carefully,
+then we may not need all of them to obtain a good approximation to the solution
+\( \boldsymbol{x} \).
+We want to regard the conjugate gradient method as an iterative method.
+This will us to solve systems where \( n \) is so large that the direct
+method would take too much time.
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+
+or consider the system
+
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+
+instead.
+
+
+
+
+
+Conjugate gradient method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+
+This suggests taking the first basis vector \( \boldsymbol{p}_1 \)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+The other vectors in the basis will be conjugate to the gradient,
+hence the name conjugate gradient method.
+
+
+
+
+
+Conjugate gradient method
+
+
+
+Let \( \boldsymbol{r}_k \) be the residual at the \( k \)-th step:
+
+$$
+\begin{equation*}
+\boldsymbol{r}_k=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k.
+\end{equation*}
+$$
+
+
+Note that \( \boldsymbol{r}_k \) is the negative gradient of \( f \) at
+\( \boldsymbol{x}=\boldsymbol{x}_k \),
+so the gradient descent method would be to move in the direction \( \boldsymbol{r}_k \).
+Here, we insist that the directions \( \boldsymbol{p}_k \) are conjugate to each other,
+so we take the direction closest to the gradient \( \boldsymbol{r}_k \)
+under the conjugacy constraint.
+This gives the following expression
+
+$$
+\begin{equation*}
+\boldsymbol{p}_{k+1}=\boldsymbol{r}_k-\frac{\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{r}_k}{\boldsymbol{p}_k^T\boldsymbol{A}\boldsymbol{p}_k} \boldsymbol{p}_k.
+\end{equation*}
+$$
+
+
+
+
+
+
+Conjugate gradient method
+
+
+
+We can also compute the residual iteratively as
+
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+
+which equals
+
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{p}_k),
+ \end{equation*}
+$$
+
+
+or
+
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{p}_k,
+ \end{equation*}
+$$
+
+
+which gives
+
+
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{r}_k-\boldsymbol{A}\boldsymbol{p}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+Revisiting some of our first Linear Regression Encounters
+
+
+We will use linear regression as a case study for the gradient descent
+methods. Linear regression is a great test case for the gradient
+descent methods discussed in the lectures since it has several
+desirable properties such as:
+
+
+- An analytical solution (recall homework set 1).
+- The gradient can be computed analytically.
+- The cost function is convex which guarantees that gradient descent converges for small enough learning rates
+
+
+
+We revisit an example similar to what we had in the first homework set. We had a function of the type
+
+
+
+
+
x = 2*np.random.rand(m,1)
+y = 4+3*x+np.random.randn(m,1)
+
+
+with \( x_i \in [0,1] \) is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution \( \cal {N}(0,1) \).
+The linear regression model is given by
+
+$$
+h_\beta(x) = \boldsymbol{y} = \beta_0 + \beta_1 x,
+$$
+
+
+such that
+
+$$
+\boldsymbol{y}_i = \beta_0 + \beta_1 x_i.
+$$
+
+
+
+
+
+Gradient descent example
+
+
+Let \( \mathbf{y} = (y_1,\cdots,y_n)^T \), \( \mathbf{\boldsymbol{y}} = (\boldsymbol{y}_1,\cdots,\boldsymbol{y}_n)^T \) and \( \beta = (\beta_0, \beta_1)^T \)
+
+
+It is convenient to write \( \mathbf{\boldsymbol{y}} = X\beta \) where \( X \in \mathbb{R}^{100 \times 2} \) is the design matrix given by (we keep the intercept here)
+
+$$
+X \equiv \begin{bmatrix}
+1 & x_1 \\
+\vdots & \vdots \\
+1 & x_{100} & \\
+\end{bmatrix}.
+$$
+
+
+The cost/loss/risk function is given by (
+
+$$
+C(\beta) = \frac{1}{n}||X\beta-\mathbf{y}||_{2}^{2} = \frac{1}{n}\sum_{i=1}^{100}\left[ (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2\right]
+$$
+
+
+and we want to find \( \beta \) such that \( C(\beta) \) is minimized.
+
+
+
+
+The derivative of the cost/loss function
+
+
+Computing \( \partial C(\beta) / \partial \beta_0 \) and \( \partial C(\beta) / \partial \beta_1 \) we can show that the gradient can be written as
+
+$$
+\nabla_{\beta} C(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} = \frac{2}{n}X^T(X\beta - \mathbf{y}),
+$$
+
+
+where \( X \) is the design matrix defined above.
+
+
+
+
+The Hessian matrix
+The Hessian matrix of \( C(\beta) \) is given by
+
+$$
+\boldsymbol{H} \equiv \begin{bmatrix}
+\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
+\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
+\end{bmatrix} = \frac{2}{n}X^T X.
+$$
+
+
+This result implies that \( C(\beta) \) is a convex function since the matrix \( X^T X \) always is positive semi-definite.
+
+
+
+
+Simple program
+
+
+We can now write a program that minimizes \( C(\beta) \) using the gradient descent method with a constant learning rate \( \gamma \) according to
+
+$$
+\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots
+$$
+
+
+
+We can use the expression we computed for the gradient and let use a
+\( \beta_0 \) be chosen randomly and let \( \gamma = 0.001 \). Stop iterating
+when \( ||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8} \). Note that the code below does not include the latter stop criterion.
+
+
+And finally we can compare our solution for \( \beta \) with the analytic result given by
+\( \beta= (X^TX)^{-1} X^T \mathbf{y} \).
+
+
+
+
+Gradient Descent Example
+
+
+Here our simple example
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+# Hessian matrix
+H = (2.0/n)* X.T @ X
+# Get the eigenvalues
+EigValues, EigVectors = np.linalg.eig(H)
+print(EigValues)
+
+beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y
+print(beta_linreg)
+beta = np.random.randn(2,1)
+
+eta = 1.0/np.max(EigValues)
+Niterations = 1000
+
+for iter in range(Niterations):
+ gradient = (2.0/n)*X.T @ (X @ beta-y)
+ beta -= eta*gradient
+
+print(beta)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(beta)
+ypredict2 = xbnew.dot(beta_linreg)
+plt.plot(xnew, ypredict, "r-")
+plt.plot(xnew, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example')
+plt.show()
+
+
+
+
+
+And a corresponding example using scikit-learn
+
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import SGDRegressor
+
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)
+print(beta_linreg)
+sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
+sgdreg.fit(x,y.ravel())
+print(sgdreg.intercept_, sgdreg.coef_)
+
+
+
+
+
+Gradient descent and Ridge
+
+
+We have also discussed Ridge regression where the loss function contains a regularized term given by the \( L_2 \) norm of \( \beta \),
+
+$$
+C_{\text{ridge}}(\beta) = \frac{1}{n}||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0.
+$$
+
+
+
+In order to minimize \( C_{\text{ridge}}(\beta) \) using GD we only have adjust the gradient as follows
+
+$$
+\nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta).
+$$
+
+
+
+We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by
+
+$$
+\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}.
+$$
+
+
+
+
+
+Program example for gradient descent with Ridge Regression
+
+
+
+
from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+XT_X = X.T @ X
+
+#Ridge parameter lambda
+lmbda = 0.001
+Id = lmbda* np.eye(XT_X.shape[0])
+
+beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y
+print(beta_linreg)
+# Start plain gradient descent
+beta = np.random.randn(2,1)
+
+eta = 0.1
+Niterations = 100
+
+for iter in range(Niterations):
+ gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta
+ beta -= eta*gradients
+
+print(beta)
+ypredict = X @ beta
+ypredict2 = X @ beta_linreg
+plt.plot(x, ypredict, "r-")
+plt.plot(x, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example for Ridge')
+plt.show()
+
+
+
+
+
+Using gradient descent methods, limitations
+
+
+- Gradient descent (GD) finds local minima of our function. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our cost/loss/risk function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance.
+- GD is sensitive to initial conditions. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.
+- Gradients are computationally expensive to calculate for large datasets. In many cases in statistics and ML, the cost/loss/risk function is a sum of terms, with one term for each data point. For example, in linear regression, \( E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2 \); for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over all \( n \) data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called "mini batches". This has the added benefit of introducing stochasticity into our algorithm.
+- GD is very sensitive to choices of learning rates. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would adaptively choose the learning rates to match the landscape.
+- GD treats all directions in parameter space uniformly. Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive.
+- GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. This leads us to our next topic, Stochastic Gradient Methods.
+
+
+
+
diff --git a/doc/pub/week38/html/week38-solarized.html b/doc/pub/week38/html/week38-solarized.html
index fb4f3fd78..f7d219ce9 100644
--- a/doc/pub/week38/html/week38-solarized.html
+++ b/doc/pub/week38/html/week38-solarized.html
@@ -26,6 +26,32 @@ pre {
border: 0pt solid #93a1a1;
box-shadow: none;
}
+.alert-text-small { font-size: 80%; }
+.alert-text-large { font-size: 130%; }
+.alert-text-normal { font-size: 90%; }
+.alert {
+ padding:8px 35px 8px 14px; margin-bottom:18px;
+ text-shadow:0 1px 0 rgba(255,255,255,0.5);
+ border:1px solid #93a1a1;
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+ -webkit-border-radius: 4px;
+ -moz-border-radius: 4px;
+ color: #555;
+ background-color: #eee8d5;
+ background-position: 10px 5px;
+ background-repeat: no-repeat;
+ background-size: 38px;
+ padding-left: 55px;
+ width: 75%;
+ }
+.alert-block {padding-top:14px; padding-bottom:14px}
+.alert-block > p, .alert-block > ul {margin-bottom:1em}
+.alert li {margin-top: 1em}
+.alert-block p+p {margin-top:5px}
+.alert-notice { background-image: url(https://cdn.rawgit.com/doconce/doconce/master/bundled/html_images/small_yellow_notice.png); }
+.alert-summary { background-image:url(https://cdn.rawgit.com/doconce/doconce/master/bundled/html_images/small_yellow_summary.png); }
+.alert-warning { background-image: url(https://cdn.rawgit.com/doconce/doconce/master/bundled/html_images/small_yellow_warning.png); }
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div { text-align: justify; text-justify: inter-word; }
@@ -36,6 +62,7 @@ div { text-align: justify; text-justify: inter-word; }
@@ -196,6 +336,11 @@ MathJax.Hub.Config({
+Thursday September 23
+
+
+
+
Ridge and LASSO Regression, reminder
@@ -1923,7 +2068,9 @@ discussed in the material on Friday September 24
@@ -2102,6 +2249,1186 @@ skplt.metrics.plot_cumulative_gain(y_test, y_probas)
plt.show()
+
+
+
Optimization, the central part of any Machine Learning algortithm
+
+
+Overview Video, why do we care about gradient methods?
+
+
+Almost every problem in machine learning and data science starts with
+a dataset \( X \), a model \( g(\beta) \), which is a function of the
+parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows
+us to judge how well the model \( g(\beta) \) explains the observations
+\( X \). The model is fit by finding the values of \( \beta \) that minimize
+the cost function. Ideally we would be able to solve for \( \beta \)
+analytically, however this is not possible in general and we must use
+some approximative/numerical method to compute the minimum.
+
+
+
+
+
Revisiting our Logistic Regression case
+
+
+In our discussion on Logistic Regression we studied the
+case of
+two classes, with \( y_i \) either
+\( 0 \) or \( 1 \). Furthermore we assumed also that we have only two
+parameters \( \beta \) in our fitting, that is we
+defined probabilities
+
+$$
+\begin{align*}
+p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
+\end{align*}
+$$
+
+where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
+
+
+
+
+
The equations to solve
+
+
+Our compact equations used a definition of a vector \( \boldsymbol{y} \) with \( n \)
+elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the
+\( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities
+\( p(y_i\vert x_i,\boldsymbol{\beta}) \). We rewrote in a more compact form
+the first derivative of the cost function as
+
+$$
+\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
+$$
+
+
+If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements
+\( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as
+
+$$
+\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
+$$
+
+This defines what is called the Hessian matrix.
+
+
+
+
+
Solving using Newton-Raphson's method
+
+
+If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
+
+
+Our iterative scheme is then given by
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}},
+$$
+
+or in matrix form as
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}.
+$$
+
+The right-hand side is computed with the old values of \( \beta \).
+
+
+If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
+
+
+
+
+
Brief reminder on Newton-Raphson's method
+
+
+Let us quickly remind ourselves how we derive the above method.
+
+
+Perhaps the most celebrated of all one-dimensional root-finding
+routines is Newton's method, also called the Newton-Raphson
+method. This method requires the evaluation of both the
+function \( f \) and its derivative \( f' \) at arbitrary points.
+If you can only calculate the derivative
+numerically and/or your function is not of the smooth type, we
+normally discourage the use of this method.
+
+
+
+
+
The equations
+
+
+The Newton-Raphson formula consists geometrically of extending the
+tangent line at a current point until it crosses zero, then setting
+the next guess to the abscissa of that zero-crossing. The mathematics
+behind this method is rather simple. Employing a Taylor expansion for
+\( x \) sufficiently close to the solution \( s \), we have
+
+$$
+ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
+ \label{eq:taylornr}
+$$
+
+
+For small enough values of the function and for well-behaved
+functions, the terms beyond linear are unimportant, hence we obtain
+
+$$
+ f(x)+(s-x)f'(x)\approx 0,
+$$
+
+yielding
+$$
+ s\approx x-\frac{f(x)}{f'(x)}.
+$$
+
+
+Having in mind an iterative procedure, it is natural to start iterating with
+$$
+ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
+$$
+
+
+
+
+
Simple geometric interpretation
+
+
+The above is Newton-Raphson's method. It has a simple geometric
+interpretation, namely \( x_{n+1} \) is the point where the tangent from
+\( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution,
+Newton-Raphson converges fast to the desired result. However, if we
+are far from a root, where the higher-order terms in the series are
+important, the Newton-Raphson formula can give grossly inaccurate
+results. For instance, the initial guess for the root might be so far
+from the true root as to let the search interval include a local
+maximum or minimum of the function. If an iteration places a trial
+guess near such a local extremum, so that the first derivative nearly
+vanishes, then Newton-Raphson may fail totally
+
+
+
+
+
Extending to more than one variable
+
+
+Newton's method can be generalized to systems of several non-linear equations
+and variables. Consider the case with two equations
+$$
+ \begin{array}{cc} f_1(x_1,x_2) &=0\\
+ f_2(x_1,x_2) &=0,\end{array}
+$$
+
+which we Taylor expand to obtain
+
+$$
+ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
+ \partial f_1/\partial x_1+h_2
+ \partial f_1/\partial x_2+\dots\\
+ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
+ \partial f_2/\partial x_1+h_2
+ \partial f_2/\partial x_2+\dots
+ \end{array}.
+$$
+
+Defining the Jacobian matrix \( {\bf \boldsymbol{J}} \) we have
+$$
+ {\bf \boldsymbol{J}}=\left( \begin{array}{cc}
+ \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
+ \partial f_2/\partial x_1 &\partial f_2/\partial x_2
+ \end{array} \right),
+$$
+
+we can rephrase Newton's method as
+$$
+\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
+\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
+\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
+$$
+
+where we have defined
+$$
+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
+ -{\bf \boldsymbol{J}}^{-1}
+ \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
+$$
+
+We need thus to compute the inverse of the Jacobian matrix and it
+is to understand that difficulties may
+arise in case \( {\bf \boldsymbol{J}} \) is nearly singular.
+
+
+It is rather straightforward to extend the above scheme to systems of
+more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
+
+
+
+
+
Steepest descent
+
+
+The basic idea of gradient descent is
+that a function \( F(\mathbf{x}) \),
+\( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the
+direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
+
+
+It can be shown that if
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
+$$
+
+with \( \gamma_k > 0 \).
+
+
+For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \)
+we are always moving towards smaller function values, i.e a minimum.
+
+
+
+
+
More on Steepest descent
+
+
+The previous observation is the basis of the method of steepest
+descent, which is also referred to as just gradient descent (GD). One
+starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and
+computes new approximations according to
+
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
+$$
+
+
+The parameter \( \gamma_k \) is often referred to as the step length or
+the learning rate within the context of Machine Learning.
+
+
+
+
+
The ideal
+
+
+Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global
+minimum of the function \( F \). In general we do not know if we are in a
+global or local minimum. In the special case when \( F \) is a convex
+function, all local minima are also global minima, so in this case
+gradient descent can converge to the global solution. The advantage of
+this scheme is that it is conceptually simple and straightforward to
+implement. However the method in this form has some severe
+limitations:
+
+
+In machine learing we are often faced with non-convex high dimensional
+cost functions with many local minima. Since GD is deterministic we
+will get stuck in a local minimum, if the method converges, unless we
+have a very good intial guess. This also implies that the scheme is
+sensitive to the chosen initial condition.
+
+
+Note that the gradient is a function of \( \mathbf{x} =
+(x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
+
+
+
+
+
The sensitiveness of the gradient descent
+
+
+The gradient descent method
+is sensitive to the choice of learning rate \( \gamma_k \). This is due
+to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to
+determine an optimal learning rate. If the learning rate is chosen too
+small the method will take a long time to converge and if it is too
+large we can experience erratic behavior.
+
+
+Many of these shortcomings can be alleviated by introducing
+randomness. One such method is that of Stochastic Gradient Descent
+(SGD), see below.
+
+
+
+
+
Convex functions
+
+
+Ideally we want our cost/loss function to be convex(concave).
+
+
+First we give the definition of a convex set: A set \( C \) in
+\( \mathbb{R}^n \) is said to be convex if, for all \( x \) and \( y \) in \( C \) and
+all \( t \in (0,1) \) , the point \( (1 − t)x + ty \) also belongs to
+C. Geometrically this means that every point on the line segment
+connecting \( x \) and \( y \) is in \( C \) as discussed below.
+
+
+The convex subsets of \( \mathbb{R} \) are the intervals of
+\( \mathbb{R} \). Examples of convex sets of \( \mathbb{R}^2 \) are the
+regular polygons (triangles, rectangles, pentagons, etc...).
+
+
+
+
+
Convex function
+
+
+Convex function: Let \( X \subset \mathbb{R}^n \) be a convex set. Assume that the function \( f: X \rightarrow \mathbb{R} \) is continuous, then \( f \) is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all \( x_1, x_2 \in X \) and for all \( t \in [0,1] \). If \( \leq \) is replaced with a strict inequaltiy in the definition, we demand \( x_1 \neq x_2 \) and \( t\in(0,1) \) then \( f \) is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting \( f(x_1) \) and \( f(x_2) \), the value of the function on the interval \( [x_1,x_2] \) is always below the line as illustrated below.
+
+
+
+
+
Conditions on convex functions
+
+
+In the following we state first and second-order conditions which
+ensures convexity of a function \( f \). We write \( D_f \) to denote the
+domain of \( f \), i.e the subset of \( R^n \) where \( f \) is defined. For more
+details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.
+
+
+
+First order condition
+
+Suppose \( f \) is differentiable (i.e \( \nabla f(x) \) is well defined for
+all \( x \) in the domain of \( f \)). Then \( f \) is convex if and only if \( D_f \)
+is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds
+for all \( x,y \in D_f \). This condition means that for a convex function
+the first order Taylor expansion (right hand side above) at any point
+a global under estimator of the function. To convince yourself you can
+make a drawing of \( f(x) = x^2+1 \) and draw the tangent line to \( f(x) \) and
+note that it is always below the graph.
+
+
+
+
+
+Second order condition
+
+Assume that \( f \) is twice
+differentiable, i.e the Hessian matrix exists at each point in
+\( D_f \). Then \( f \) is convex if and only if \( D_f \) is a convex set and its
+Hessian is positive semi-definite for all \( x\in D_f \). For a
+single-variable function this reduces to \( f''(x) \geq 0 \). Geometrically this means that \( f \) has nonnegative curvature
+everywhere.
+
+
+
+
+This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
+
+
+
+
+
More on convex functions
+
+
+The next result is of great importance to us and the reason why we are
+going on about convex functions. In machine learning we frequently
+have to minimize a loss/cost function in order to find the best
+parameters for the model we are considering.
+
+
+Ideally we want the
+global minimum (for high-dimensional models it is hard to know
+if we have local or global minimum). However, if the cost/loss function
+is convex the following result provides invaluable information:
+
+
+
+Any minimum is global for convex functions
+
+Consider the problem of finding \( x \in \mathbb{R}^n \) such that \( f(x) \)
+is minimal, where \( f \) is convex and differentiable. Then, any point
+\( x^* \) that satisfies \( \nabla f(x^*) = 0 \) is a global minimum.
+
+
+
+
+This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
+
+
+
+
+
Some simple problems
+
+
+- Show that \( f(x)=x^2 \) is convex for \( x \in \mathbb{R} \) using the definition of convexity. Hint: If you re-write the definition, \( f \) is convex if the following holds for all \( x,y \in D_f \) and any \( \lambda \in [0,1] \) $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$.
+- Using the second order condition show that the following functions are convex on the specified domain.
+
+
+ - \( f(x) = e^x \) is convex for \( x \in \mathbb{R} \).
+ - \( g(x) = -\ln(x) \) is convex for \( x \in (0,\infty) \).
+
+
+- Let \( f(x) = x^2 \) and \( g(x) = e^x \). Show that \( f(g(x)) \) and \( g(f(x)) \) is convex for \( x \in \mathbb{R} \). Also show that if \( f(x) \) is any convex function than \( h(x) = e^{f(x)} \) is convex.
+- A norm is any function that satisfy the following properties
+
+
+ - \( f(\alpha x) = |\alpha| f(x) \) for all \( \alpha \in \mathbb{R} \).
+ - \( f(x+y) \leq f(x) + f(y) \)
+ - \( f(x) \leq 0 \) for all \( x \in \mathbb{R}^n \) with equality if and only if \( x = 0 \)
+
+
+
+
+Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
+
+
+
+
+
Friday September 25
+
+
+Video of Lecture and link to handwritten notes.
+
+
+
+
+
Standard steepest descent
+
+
+Before we proceed, we would like to discuss the approach called the
+standard Steepest descent (different from the above steepest descent discussion), which again leads to us having to be able
+to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
+
+
+The success of the CG method
+for finding solutions of non-linear problems is based on the theory
+of conjugate gradients for linear systems of equations. It belongs to
+the class of iterative methods for solving problems from linear
+algebra of the type
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}.
+\end{equation*}
+$$
+
+
+In the iterative process we end up with a problem like
+
+$$
+\begin{equation*}
+ \boldsymbol{r}= \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x},
+\end{equation*}
+$$
+
+where \( \boldsymbol{r} \) is the so-called residual or error in the iterative process.
+
+
+When we have found the exact solution, \( \boldsymbol{r}=0 \).
+
+
+
+
+
Gradient method
+
+
+The residual is zero when we reach the minimum of the quadratic equation
+$$
+\begin{equation*}
+ P(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T\boldsymbol{b},
+\end{equation*}
+$$
+
+
+with the constraint that the matrix \( \boldsymbol{A} \) is positive definite and
+symmetric. This defines also the Hessian and we want it to be positive definite.
+
+
+
+
+
Steepest descent method
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+or consider the system
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+instead.
+
+
+
+
+
Steepest descent method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+This suggests taking the first basis vector \( \boldsymbol{r}_1 \) (see below for definition)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+
+
+
+
+
+
+
+
+
Final expressions
+
+
+
+We can compute the residual iteratively as
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+which equals
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{r}_k),
+ \end{equation*}
+$$
+
+or
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{r}_k,
+ \end{equation*}
+$$
+
+which gives
+
+$$
+\alpha_k = \frac{\boldsymbol{r}_k^T\boldsymbol{r}_k}{\boldsymbol{r}_k^T\boldsymbol{A}\boldsymbol{r}_k}
+$$
+
+leading to the iterative scheme
+$$
+\begin{equation*}
+\boldsymbol{x}_{k+1}=\boldsymbol{x}_k-\alpha_k\boldsymbol{r}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+
Steepest descent example
+
+
+
+
+
import numpy as np
+import numpy.linalg as la
+
+import scipy.optimize as sopt
+
+import matplotlib.pyplot as pt
+from mpl_toolkits.mplot3d import axes3d
+
+def f(x):
+ return 0.5*x[0]**2 + 2.5*x[1]**2
+
+def df(x):
+ return np.array([x[0], 5*x[1]])
+
+fig = pt.figure()
+ax = fig.gca(projection="3d")
+
+xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
+fmesh = f(np.array([xmesh, ymesh]))
+ax.plot_surface(xmesh, ymesh, fmesh)
+
+
+And then as countor plot
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh)
+guesses = [np.array([2, 2./5])]
+
+
+Find guesses
+
+
+
+
x = guesses[-1]
+s = -df(x)
+
+
+Run it!
+
+
+
+
def f1d(alpha):
+ return f(x + alpha*s)
+
+alpha_opt = sopt.golden(f1d)
+next_guess = x + alpha_opt * s
+guesses.append(next_guess)
+print(next_guess)
+
+
+What happened?
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh, 50)
+it_array = np.array(guesses)
+pt.plot(it_array.T[0], it_array.T[1], "x-")
+
+
+
+
+
Conjugate gradient method
+
+
+
+In the CG method we define so-called conjugate directions and two vectors
+\( \boldsymbol{s} \) and \( \boldsymbol{t} \)
+are said to be
+conjugate if
+$$
+\begin{equation*}
+\boldsymbol{s}^T\boldsymbol{A}\boldsymbol{t}= 0.
+\end{equation*}
+$$
+
+The philosophy of the CG method is to perform searches in various conjugate directions
+of our vectors \( \boldsymbol{x}_i \) obeying the above criterion, namely
+$$
+\begin{equation*}
+\boldsymbol{x}_i^T\boldsymbol{A}\boldsymbol{x}_j= 0.
+\end{equation*}
+$$
+
+Two vectors are conjugate if they are orthogonal with respect to
+this inner product. Being conjugate is a symmetric relation: if \( \boldsymbol{s} \) is conjugate to \( \boldsymbol{t} \), then \( \boldsymbol{t} \) is conjugate to \( \boldsymbol{s} \).
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+An example is given by the eigenvectors of the matrix
+$$
+\begin{equation*}
+\boldsymbol{v}_i^T\boldsymbol{A}\boldsymbol{v}_j= \lambda\boldsymbol{v}_i^T\boldsymbol{v}_j,
+\end{equation*}
+$$
+
+which is zero unless \( i=j \).
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+Assume now that we have a symmetric positive-definite matrix \( \boldsymbol{A} \) of size
+\( n\times n \). At each iteration \( i+1 \) we obtain the conjugate direction of a vector
+$$
+\begin{equation*}
+\boldsymbol{x}_{i+1}=\boldsymbol{x}_{i}+\alpha_i\boldsymbol{p}_{i}.
+\end{equation*}
+$$
+
+We assume that \( \boldsymbol{p}_{i} \) is a sequence of \( n \) mutually conjugate directions.
+Then the \( \boldsymbol{p}_{i} \) form a basis of \( R^n \) and we can expand the solution
+$ \boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}$ in this basis, namely
+
+$$
+\begin{equation*}
+ \boldsymbol{x} = \sum^{n}_{i=1} \alpha_i \boldsymbol{p}_i.
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+The coefficients are given by
+$$
+\begin{equation*}
+ \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
+\end{equation*}
+$$
+
+Multiplying with \( \boldsymbol{p}_k^T \) from the left gives
+
+$$
+\begin{equation*}
+ \boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{p}_i= \boldsymbol{p}_k^T \boldsymbol{b},
+\end{equation*}
+$$
+
+and we can define the coefficients \( \alpha_k \) as
+
+$$
+\begin{equation*}
+ \alpha_k = \frac{\boldsymbol{p}_k^T \boldsymbol{b}}{\boldsymbol{p}_k^T \boldsymbol{A} \boldsymbol{p}_k}
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method and iterations
+
+
+
+
+
+If we choose the conjugate vectors \( \boldsymbol{p}_k \) carefully,
+then we may not need all of them to obtain a good approximation to the solution
+\( \boldsymbol{x} \).
+We want to regard the conjugate gradient method as an iterative method.
+This will us to solve systems where \( n \) is so large that the direct
+method would take too much time.
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+or consider the system
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+instead.
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+This suggests taking the first basis vector \( \boldsymbol{p}_1 \)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+The other vectors in the basis will be conjugate to the gradient,
+hence the name conjugate gradient method.
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+Let \( \boldsymbol{r}_k \) be the residual at the \( k \)-th step:
+$$
+\begin{equation*}
+\boldsymbol{r}_k=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k.
+\end{equation*}
+$$
+
+Note that \( \boldsymbol{r}_k \) is the negative gradient of \( f \) at
+\( \boldsymbol{x}=\boldsymbol{x}_k \),
+so the gradient descent method would be to move in the direction \( \boldsymbol{r}_k \).
+Here, we insist that the directions \( \boldsymbol{p}_k \) are conjugate to each other,
+so we take the direction closest to the gradient \( \boldsymbol{r}_k \)
+under the conjugacy constraint.
+This gives the following expression
+$$
+\begin{equation*}
+\boldsymbol{p}_{k+1}=\boldsymbol{r}_k-\frac{\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{r}_k}{\boldsymbol{p}_k^T\boldsymbol{A}\boldsymbol{p}_k} \boldsymbol{p}_k.
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+We can also compute the residual iteratively as
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+which equals
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{p}_k),
+ \end{equation*}
+$$
+
+or
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{p}_k,
+ \end{equation*}
+$$
+
+which gives
+
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{r}_k-\boldsymbol{A}\boldsymbol{p}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+
Revisiting some of our first Linear Regression Encounters
+
+
+We will use linear regression as a case study for the gradient descent
+methods. Linear regression is a great test case for the gradient
+descent methods discussed in the lectures since it has several
+desirable properties such as:
+
+
+- An analytical solution (recall homework set 1).
+- The gradient can be computed analytically.
+- The cost function is convex which guarantees that gradient descent converges for small enough learning rates
+
+
+We revisit an example similar to what we had in the first homework set. We had a function of the type
+
+
+
+
+
x = 2*np.random.rand(m,1)
+y = 4+3*x+np.random.randn(m,1)
+
+
+with \( x_i \in [0,1] \) is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution \( \cal {N}(0,1) \).
+The linear regression model is given by
+$$
+h_\beta(x) = \boldsymbol{y} = \beta_0 + \beta_1 x,
+$$
+
+such that
+$$
+\boldsymbol{y}_i = \beta_0 + \beta_1 x_i.
+$$
+
+
+
+
+
Gradient descent example
+
+
+Let \( \mathbf{y} = (y_1,\cdots,y_n)^T \), \( \mathbf{\boldsymbol{y}} = (\boldsymbol{y}_1,\cdots,\boldsymbol{y}_n)^T \) and \( \beta = (\beta_0, \beta_1)^T \)
+
+
+It is convenient to write \( \mathbf{\boldsymbol{y}} = X\beta \) where \( X \in \mathbb{R}^{100 \times 2} \) is the design matrix given by (we keep the intercept here)
+$$
+X \equiv \begin{bmatrix}
+1 & x_1 \\
+\vdots & \vdots \\
+1 & x_{100} & \\
+\end{bmatrix}.
+$$
+
+The cost/loss/risk function is given by (
+$$
+C(\beta) = \frac{1}{n}||X\beta-\mathbf{y}||_{2}^{2} = \frac{1}{n}\sum_{i=1}^{100}\left[ (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2\right]
+$$
+
+and we want to find \( \beta \) such that \( C(\beta) \) is minimized.
+
+
+
+
+
The derivative of the cost/loss function
+
+
+Computing \( \partial C(\beta) / \partial \beta_0 \) and \( \partial C(\beta) / \partial \beta_1 \) we can show that the gradient can be written as
+$$
+\nabla_{\beta} C(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} = \frac{2}{n}X^T(X\beta - \mathbf{y}),
+$$
+
+where \( X \) is the design matrix defined above.
+
+
+
+
+
The Hessian matrix
+The Hessian matrix of \( C(\beta) \) is given by
+$$
+\boldsymbol{H} \equiv \begin{bmatrix}
+\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
+\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
+\end{bmatrix} = \frac{2}{n}X^T X.
+$$
+
+This result implies that \( C(\beta) \) is a convex function since the matrix \( X^T X \) always is positive semi-definite.
+
+
+
+
+
Simple program
+
+
+We can now write a program that minimizes \( C(\beta) \) using the gradient descent method with a constant learning rate \( \gamma \) according to
+$$
+\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots
+$$
+
+
+We can use the expression we computed for the gradient and let use a
+\( \beta_0 \) be chosen randomly and let \( \gamma = 0.001 \). Stop iterating
+when \( ||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8} \). Note that the code below does not include the latter stop criterion.
+
+
+And finally we can compare our solution for \( \beta \) with the analytic result given by
+\( \beta= (X^TX)^{-1} X^T \mathbf{y} \).
+
+
+
+
+
Gradient Descent Example
+
+
+Here our simple example
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+# Hessian matrix
+H = (2.0/n)* X.T @ X
+# Get the eigenvalues
+EigValues, EigVectors = np.linalg.eig(H)
+print(EigValues)
+
+beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y
+print(beta_linreg)
+beta = np.random.randn(2,1)
+
+eta = 1.0/np.max(EigValues)
+Niterations = 1000
+
+for iter in range(Niterations):
+ gradient = (2.0/n)*X.T @ (X @ beta-y)
+ beta -= eta*gradient
+
+print(beta)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(beta)
+ypredict2 = xbnew.dot(beta_linreg)
+plt.plot(xnew, ypredict, "r-")
+plt.plot(xnew, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example')
+plt.show()
+
+
+
+
+
And a corresponding example using scikit-learn
+
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import SGDRegressor
+
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)
+print(beta_linreg)
+sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
+sgdreg.fit(x,y.ravel())
+print(sgdreg.intercept_, sgdreg.coef_)
+
+
+
+
+
Gradient descent and Ridge
+
+
+We have also discussed Ridge regression where the loss function contains a regularized term given by the \( L_2 \) norm of \( \beta \),
+$$
+C_{\text{ridge}}(\beta) = \frac{1}{n}||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0.
+$$
+
+
+In order to minimize \( C_{\text{ridge}}(\beta) \) using GD we only have adjust the gradient as follows
+$$
+\nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta).
+$$
+
+
+We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by
+$$
+\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}.
+$$
+
+
+
+
+
Program example for gradient descent with Ridge Regression
+
+
+
+
from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+XT_X = X.T @ X
+
+#Ridge parameter lambda
+lmbda = 0.001
+Id = lmbda* np.eye(XT_X.shape[0])
+
+beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y
+print(beta_linreg)
+# Start plain gradient descent
+beta = np.random.randn(2,1)
+
+eta = 0.1
+Niterations = 100
+
+for iter in range(Niterations):
+ gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta
+ beta -= eta*gradients
+
+print(beta)
+ypredict = X @ beta
+ypredict2 = X @ beta_linreg
+plt.plot(x, ypredict, "r-")
+plt.plot(x, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example for Ridge')
+plt.show()
+
+
+
+
+
Using gradient descent methods, limitations
+
+
+- Gradient descent (GD) finds local minima of our function. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our cost/loss/risk function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance.
+- GD is sensitive to initial conditions. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.
+- Gradients are computationally expensive to calculate for large datasets. In many cases in statistics and ML, the cost/loss/risk function is a sum of terms, with one term for each data point. For example, in linear regression, \( E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2 \); for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over all \( n \) data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called "mini batches". This has the added benefit of introducing stochasticity into our algorithm.
+- GD is very sensitive to choices of learning rates. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would adaptively choose the learning rates to match the landscape.
+- GD treats all directions in parameter space uniformly. Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive.
+- GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. This leads us to our next topic, Stochastic Gradient Methods.
+
+
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+Thursday September 23
+
+
+
+
Ridge and LASSO Regression, reminder
@@ -1928,7 +2073,9 @@ discussed in the material on Friday September 24
@@ -2107,6 +2254,1186 @@ skplt.metrics.<
plt.show()
+
+
+
Optimization, the central part of any Machine Learning algortithm
+
+
+Overview Video, why do we care about gradient methods?
+
+
+Almost every problem in machine learning and data science starts with
+a dataset \( X \), a model \( g(\beta) \), which is a function of the
+parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows
+us to judge how well the model \( g(\beta) \) explains the observations
+\( X \). The model is fit by finding the values of \( \beta \) that minimize
+the cost function. Ideally we would be able to solve for \( \beta \)
+analytically, however this is not possible in general and we must use
+some approximative/numerical method to compute the minimum.
+
+
+
+
+
Revisiting our Logistic Regression case
+
+
+In our discussion on Logistic Regression we studied the
+case of
+two classes, with \( y_i \) either
+\( 0 \) or \( 1 \). Furthermore we assumed also that we have only two
+parameters \( \beta \) in our fitting, that is we
+defined probabilities
+
+$$
+\begin{align*}
+p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
+\end{align*}
+$$
+
+where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
+
+
+
+
+
The equations to solve
+
+
+Our compact equations used a definition of a vector \( \boldsymbol{y} \) with \( n \)
+elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the
+\( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities
+\( p(y_i\vert x_i,\boldsymbol{\beta}) \). We rewrote in a more compact form
+the first derivative of the cost function as
+
+$$
+\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
+$$
+
+
+If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements
+\( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as
+
+$$
+\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
+$$
+
+This defines what is called the Hessian matrix.
+
+
+
+
+
Solving using Newton-Raphson's method
+
+
+If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
+
+
+Our iterative scheme is then given by
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}},
+$$
+
+or in matrix form as
+
+$$
+\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}.
+$$
+
+The right-hand side is computed with the old values of \( \beta \).
+
+
+If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
+
+
+
+
+
Brief reminder on Newton-Raphson's method
+
+
+Let us quickly remind ourselves how we derive the above method.
+
+
+Perhaps the most celebrated of all one-dimensional root-finding
+routines is Newton's method, also called the Newton-Raphson
+method. This method requires the evaluation of both the
+function \( f \) and its derivative \( f' \) at arbitrary points.
+If you can only calculate the derivative
+numerically and/or your function is not of the smooth type, we
+normally discourage the use of this method.
+
+
+
+
+
The equations
+
+
+The Newton-Raphson formula consists geometrically of extending the
+tangent line at a current point until it crosses zero, then setting
+the next guess to the abscissa of that zero-crossing. The mathematics
+behind this method is rather simple. Employing a Taylor expansion for
+\( x \) sufficiently close to the solution \( s \), we have
+
+$$
+ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
+ \label{eq:taylornr}
+$$
+
+
+For small enough values of the function and for well-behaved
+functions, the terms beyond linear are unimportant, hence we obtain
+
+$$
+ f(x)+(s-x)f'(x)\approx 0,
+$$
+
+yielding
+$$
+ s\approx x-\frac{f(x)}{f'(x)}.
+$$
+
+
+Having in mind an iterative procedure, it is natural to start iterating with
+$$
+ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
+$$
+
+
+
+
+
Simple geometric interpretation
+
+
+The above is Newton-Raphson's method. It has a simple geometric
+interpretation, namely \( x_{n+1} \) is the point where the tangent from
+\( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution,
+Newton-Raphson converges fast to the desired result. However, if we
+are far from a root, where the higher-order terms in the series are
+important, the Newton-Raphson formula can give grossly inaccurate
+results. For instance, the initial guess for the root might be so far
+from the true root as to let the search interval include a local
+maximum or minimum of the function. If an iteration places a trial
+guess near such a local extremum, so that the first derivative nearly
+vanishes, then Newton-Raphson may fail totally
+
+
+
+
+
Extending to more than one variable
+
+
+Newton's method can be generalized to systems of several non-linear equations
+and variables. Consider the case with two equations
+$$
+ \begin{array}{cc} f_1(x_1,x_2) &=0\\
+ f_2(x_1,x_2) &=0,\end{array}
+$$
+
+which we Taylor expand to obtain
+
+$$
+ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
+ \partial f_1/\partial x_1+h_2
+ \partial f_1/\partial x_2+\dots\\
+ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
+ \partial f_2/\partial x_1+h_2
+ \partial f_2/\partial x_2+\dots
+ \end{array}.
+$$
+
+Defining the Jacobian matrix \( {\bf \boldsymbol{J}} \) we have
+$$
+ {\bf \boldsymbol{J}}=\left( \begin{array}{cc}
+ \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
+ \partial f_2/\partial x_1 &\partial f_2/\partial x_2
+ \end{array} \right),
+$$
+
+we can rephrase Newton's method as
+$$
+\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
+\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
+\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
+$$
+
+where we have defined
+$$
+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
+ -{\bf \boldsymbol{J}}^{-1}
+ \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
+$$
+
+We need thus to compute the inverse of the Jacobian matrix and it
+is to understand that difficulties may
+arise in case \( {\bf \boldsymbol{J}} \) is nearly singular.
+
+
+It is rather straightforward to extend the above scheme to systems of
+more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
+
+
+
+
+
Steepest descent
+
+
+The basic idea of gradient descent is
+that a function \( F(\mathbf{x}) \),
+\( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the
+direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
+
+
+It can be shown that if
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
+$$
+
+with \( \gamma_k > 0 \).
+
+
+For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \)
+we are always moving towards smaller function values, i.e a minimum.
+
+
+
+
+
More on Steepest descent
+
+
+The previous observation is the basis of the method of steepest
+descent, which is also referred to as just gradient descent (GD). One
+starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and
+computes new approximations according to
+
+$$
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
+$$
+
+
+The parameter \( \gamma_k \) is often referred to as the step length or
+the learning rate within the context of Machine Learning.
+
+
+
+
+
The ideal
+
+
+Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global
+minimum of the function \( F \). In general we do not know if we are in a
+global or local minimum. In the special case when \( F \) is a convex
+function, all local minima are also global minima, so in this case
+gradient descent can converge to the global solution. The advantage of
+this scheme is that it is conceptually simple and straightforward to
+implement. However the method in this form has some severe
+limitations:
+
+
+In machine learing we are often faced with non-convex high dimensional
+cost functions with many local minima. Since GD is deterministic we
+will get stuck in a local minimum, if the method converges, unless we
+have a very good intial guess. This also implies that the scheme is
+sensitive to the chosen initial condition.
+
+
+Note that the gradient is a function of \( \mathbf{x} =
+(x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
+
+
+
+
+
The sensitiveness of the gradient descent
+
+
+The gradient descent method
+is sensitive to the choice of learning rate \( \gamma_k \). This is due
+to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to
+determine an optimal learning rate. If the learning rate is chosen too
+small the method will take a long time to converge and if it is too
+large we can experience erratic behavior.
+
+
+Many of these shortcomings can be alleviated by introducing
+randomness. One such method is that of Stochastic Gradient Descent
+(SGD), see below.
+
+
+
+
+
Convex functions
+
+
+Ideally we want our cost/loss function to be convex(concave).
+
+
+First we give the definition of a convex set: A set \( C \) in
+\( \mathbb{R}^n \) is said to be convex if, for all \( x \) and \( y \) in \( C \) and
+all \( t \in (0,1) \) , the point \( (1 − t)x + ty \) also belongs to
+C. Geometrically this means that every point on the line segment
+connecting \( x \) and \( y \) is in \( C \) as discussed below.
+
+
+The convex subsets of \( \mathbb{R} \) are the intervals of
+\( \mathbb{R} \). Examples of convex sets of \( \mathbb{R}^2 \) are the
+regular polygons (triangles, rectangles, pentagons, etc...).
+
+
+
+
+
Convex function
+
+
+Convex function: Let \( X \subset \mathbb{R}^n \) be a convex set. Assume that the function \( f: X \rightarrow \mathbb{R} \) is continuous, then \( f \) is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all \( x_1, x_2 \in X \) and for all \( t \in [0,1] \). If \( \leq \) is replaced with a strict inequaltiy in the definition, we demand \( x_1 \neq x_2 \) and \( t\in(0,1) \) then \( f \) is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting \( f(x_1) \) and \( f(x_2) \), the value of the function on the interval \( [x_1,x_2] \) is always below the line as illustrated below.
+
+
+
+
+
Conditions on convex functions
+
+
+In the following we state first and second-order conditions which
+ensures convexity of a function \( f \). We write \( D_f \) to denote the
+domain of \( f \), i.e the subset of \( R^n \) where \( f \) is defined. For more
+details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.
+
+
+
+First order condition
+
+Suppose \( f \) is differentiable (i.e \( \nabla f(x) \) is well defined for
+all \( x \) in the domain of \( f \)). Then \( f \) is convex if and only if \( D_f \)
+is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds
+for all \( x,y \in D_f \). This condition means that for a convex function
+the first order Taylor expansion (right hand side above) at any point
+a global under estimator of the function. To convince yourself you can
+make a drawing of \( f(x) = x^2+1 \) and draw the tangent line to \( f(x) \) and
+note that it is always below the graph.
+
+
+
+
+
+Second order condition
+
+Assume that \( f \) is twice
+differentiable, i.e the Hessian matrix exists at each point in
+\( D_f \). Then \( f \) is convex if and only if \( D_f \) is a convex set and its
+Hessian is positive semi-definite for all \( x\in D_f \). For a
+single-variable function this reduces to \( f''(x) \geq 0 \). Geometrically this means that \( f \) has nonnegative curvature
+everywhere.
+
+
+
+
+This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
+
+
+
+
+
More on convex functions
+
+
+The next result is of great importance to us and the reason why we are
+going on about convex functions. In machine learning we frequently
+have to minimize a loss/cost function in order to find the best
+parameters for the model we are considering.
+
+
+Ideally we want the
+global minimum (for high-dimensional models it is hard to know
+if we have local or global minimum). However, if the cost/loss function
+is convex the following result provides invaluable information:
+
+
+
+Any minimum is global for convex functions
+
+Consider the problem of finding \( x \in \mathbb{R}^n \) such that \( f(x) \)
+is minimal, where \( f \) is convex and differentiable. Then, any point
+\( x^* \) that satisfies \( \nabla f(x^*) = 0 \) is a global minimum.
+
+
+
+
+This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
+
+
+
+
+
Some simple problems
+
+
+- Show that \( f(x)=x^2 \) is convex for \( x \in \mathbb{R} \) using the definition of convexity. Hint: If you re-write the definition, \( f \) is convex if the following holds for all \( x,y \in D_f \) and any \( \lambda \in [0,1] \) $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$.
+- Using the second order condition show that the following functions are convex on the specified domain.
+
+
+ - \( f(x) = e^x \) is convex for \( x \in \mathbb{R} \).
+ - \( g(x) = -\ln(x) \) is convex for \( x \in (0,\infty) \).
+
+
+- Let \( f(x) = x^2 \) and \( g(x) = e^x \). Show that \( f(g(x)) \) and \( g(f(x)) \) is convex for \( x \in \mathbb{R} \). Also show that if \( f(x) \) is any convex function than \( h(x) = e^{f(x)} \) is convex.
+- A norm is any function that satisfy the following properties
+
+
+ - \( f(\alpha x) = |\alpha| f(x) \) for all \( \alpha \in \mathbb{R} \).
+ - \( f(x+y) \leq f(x) + f(y) \)
+ - \( f(x) \leq 0 \) for all \( x \in \mathbb{R}^n \) with equality if and only if \( x = 0 \)
+
+
+
+
+Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
+
+
+
+
+
Friday September 25
+
+
+Video of Lecture and link to handwritten notes.
+
+
+
+
+
Standard steepest descent
+
+
+Before we proceed, we would like to discuss the approach called the
+standard Steepest descent (different from the above steepest descent discussion), which again leads to us having to be able
+to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
+
+
+The success of the CG method
+for finding solutions of non-linear problems is based on the theory
+of conjugate gradients for linear systems of equations. It belongs to
+the class of iterative methods for solving problems from linear
+algebra of the type
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}.
+\end{equation*}
+$$
+
+
+In the iterative process we end up with a problem like
+
+$$
+\begin{equation*}
+ \boldsymbol{r}= \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x},
+\end{equation*}
+$$
+
+where \( \boldsymbol{r} \) is the so-called residual or error in the iterative process.
+
+
+When we have found the exact solution, \( \boldsymbol{r}=0 \).
+
+
+
+
+
Gradient method
+
+
+The residual is zero when we reach the minimum of the quadratic equation
+$$
+\begin{equation*}
+ P(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T\boldsymbol{b},
+\end{equation*}
+$$
+
+
+with the constraint that the matrix \( \boldsymbol{A} \) is positive definite and
+symmetric. This defines also the Hessian and we want it to be positive definite.
+
+
+
+
+
Steepest descent method
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+or consider the system
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+instead.
+
+
+
+
+
Steepest descent method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+This suggests taking the first basis vector \( \boldsymbol{r}_1 \) (see below for definition)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+
+
+
+
+
+
+
+
+
Final expressions
+
+
+
+We can compute the residual iteratively as
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+which equals
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{r}_k),
+ \end{equation*}
+$$
+
+or
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{r}_k,
+ \end{equation*}
+$$
+
+which gives
+
+$$
+\alpha_k = \frac{\boldsymbol{r}_k^T\boldsymbol{r}_k}{\boldsymbol{r}_k^T\boldsymbol{A}\boldsymbol{r}_k}
+$$
+
+leading to the iterative scheme
+$$
+\begin{equation*}
+\boldsymbol{x}_{k+1}=\boldsymbol{x}_k-\alpha_k\boldsymbol{r}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+
Steepest descent example
+
+
+
+
+
import numpy as np
+import numpy.linalg as la
+
+import scipy.optimize as sopt
+
+import matplotlib.pyplot as pt
+from mpl_toolkits.mplot3d import axes3d
+
+def f(x):
+ return 0.5*x[0]**2 + 2.5*x[1]**2
+
+def df(x):
+ return np.array([x[0], 5*x[1]])
+
+fig = pt.figure()
+ax = fig.gca(projection="3d")
+
+xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
+fmesh = f(np.array([xmesh, ymesh]))
+ax.plot_surface(xmesh, ymesh, fmesh)
+
+
+And then as countor plot
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh)
+guesses = [np.array([2, 2./5])]
+
+
+Find guesses
+
+
+
+
x = guesses[-1]
+s = -df(x)
+
+
+Run it!
+
+
+
+
def f1d(alpha):
+ return f(x + alpha*s)
+
+alpha_opt = sopt.golden(f1d)
+next_guess = x + alpha_opt * s
+guesses.append(next_guess)
+print(next_guess)
+
+
+What happened?
+
+
+
+
pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh, 50)
+it_array = np.array(guesses)
+pt.plot(it_array.T[0], it_array.T[1], "x-")
+
+
+
+
+
Conjugate gradient method
+
+
+
+In the CG method we define so-called conjugate directions and two vectors
+\( \boldsymbol{s} \) and \( \boldsymbol{t} \)
+are said to be
+conjugate if
+$$
+\begin{equation*}
+\boldsymbol{s}^T\boldsymbol{A}\boldsymbol{t}= 0.
+\end{equation*}
+$$
+
+The philosophy of the CG method is to perform searches in various conjugate directions
+of our vectors \( \boldsymbol{x}_i \) obeying the above criterion, namely
+$$
+\begin{equation*}
+\boldsymbol{x}_i^T\boldsymbol{A}\boldsymbol{x}_j= 0.
+\end{equation*}
+$$
+
+Two vectors are conjugate if they are orthogonal with respect to
+this inner product. Being conjugate is a symmetric relation: if \( \boldsymbol{s} \) is conjugate to \( \boldsymbol{t} \), then \( \boldsymbol{t} \) is conjugate to \( \boldsymbol{s} \).
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+An example is given by the eigenvectors of the matrix
+$$
+\begin{equation*}
+\boldsymbol{v}_i^T\boldsymbol{A}\boldsymbol{v}_j= \lambda\boldsymbol{v}_i^T\boldsymbol{v}_j,
+\end{equation*}
+$$
+
+which is zero unless \( i=j \).
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+Assume now that we have a symmetric positive-definite matrix \( \boldsymbol{A} \) of size
+\( n\times n \). At each iteration \( i+1 \) we obtain the conjugate direction of a vector
+$$
+\begin{equation*}
+\boldsymbol{x}_{i+1}=\boldsymbol{x}_{i}+\alpha_i\boldsymbol{p}_{i}.
+\end{equation*}
+$$
+
+We assume that \( \boldsymbol{p}_{i} \) is a sequence of \( n \) mutually conjugate directions.
+Then the \( \boldsymbol{p}_{i} \) form a basis of \( R^n \) and we can expand the solution
+$ \boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}$ in this basis, namely
+
+$$
+\begin{equation*}
+ \boldsymbol{x} = \sum^{n}_{i=1} \alpha_i \boldsymbol{p}_i.
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+The coefficients are given by
+$$
+\begin{equation*}
+ \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
+\end{equation*}
+$$
+
+Multiplying with \( \boldsymbol{p}_k^T \) from the left gives
+
+$$
+\begin{equation*}
+ \boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{p}_i= \boldsymbol{p}_k^T \boldsymbol{b},
+\end{equation*}
+$$
+
+and we can define the coefficients \( \alpha_k \) as
+
+$$
+\begin{equation*}
+ \alpha_k = \frac{\boldsymbol{p}_k^T \boldsymbol{b}}{\boldsymbol{p}_k^T \boldsymbol{A} \boldsymbol{p}_k}
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method and iterations
+
+
+
+
+
+If we choose the conjugate vectors \( \boldsymbol{p}_k \) carefully,
+then we may not need all of them to obtain a good approximation to the solution
+\( \boldsymbol{x} \).
+We want to regard the conjugate gradient method as an iterative method.
+This will us to solve systems where \( n \) is so large that the direct
+method would take too much time.
+
+
+We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \).
+We can assume without loss of generality that
+$$
+\begin{equation*}
+\boldsymbol{x}_0=0,
+\end{equation*}
+$$
+
+or consider the system
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
+\end{equation*}
+$$
+
+instead.
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form
+$$
+\begin{equation*}
+ f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
+\end{equation*}
+$$
+
+This suggests taking the first basis vector \( \boldsymbol{p}_1 \)
+to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \),
+which equals
+$$
+\begin{equation*}
+\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
+\end{equation*}
+$$
+
+and
+\( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
+The other vectors in the basis will be conjugate to the gradient,
+hence the name conjugate gradient method.
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+Let \( \boldsymbol{r}_k \) be the residual at the \( k \)-th step:
+$$
+\begin{equation*}
+\boldsymbol{r}_k=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k.
+\end{equation*}
+$$
+
+Note that \( \boldsymbol{r}_k \) is the negative gradient of \( f \) at
+\( \boldsymbol{x}=\boldsymbol{x}_k \),
+so the gradient descent method would be to move in the direction \( \boldsymbol{r}_k \).
+Here, we insist that the directions \( \boldsymbol{p}_k \) are conjugate to each other,
+so we take the direction closest to the gradient \( \boldsymbol{r}_k \)
+under the conjugacy constraint.
+This gives the following expression
+$$
+\begin{equation*}
+\boldsymbol{p}_{k+1}=\boldsymbol{r}_k-\frac{\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{r}_k}{\boldsymbol{p}_k^T\boldsymbol{A}\boldsymbol{p}_k} \boldsymbol{p}_k.
+\end{equation*}
+$$
+
+
+
+
+
+
+
Conjugate gradient method
+
+
+
+We can also compute the residual iteratively as
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
+ \end{equation*}
+$$
+
+which equals
+$$
+\begin{equation*}
+\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{p}_k),
+ \end{equation*}
+$$
+
+or
+$$
+\begin{equation*}
+(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{p}_k,
+ \end{equation*}
+$$
+
+which gives
+
+$$
+\begin{equation*}
+\boldsymbol{r}_{k+1}=\boldsymbol{r}_k-\boldsymbol{A}\boldsymbol{p}_{k},
+ \end{equation*}
+$$
+
+
+
+
+
+
+
Revisiting some of our first Linear Regression Encounters
+
+
+We will use linear regression as a case study for the gradient descent
+methods. Linear regression is a great test case for the gradient
+descent methods discussed in the lectures since it has several
+desirable properties such as:
+
+
+- An analytical solution (recall homework set 1).
+- The gradient can be computed analytically.
+- The cost function is convex which guarantees that gradient descent converges for small enough learning rates
+
+
+We revisit an example similar to what we had in the first homework set. We had a function of the type
+
+
+
+
+
x = 2*np.random.rand(m,1)
+y = 4+3*x+np.random.randn(m,1)
+
+
+with \( x_i \in [0,1] \) is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution \( \cal {N}(0,1) \).
+The linear regression model is given by
+$$
+h_\beta(x) = \boldsymbol{y} = \beta_0 + \beta_1 x,
+$$
+
+such that
+$$
+\boldsymbol{y}_i = \beta_0 + \beta_1 x_i.
+$$
+
+
+
+
+
Gradient descent example
+
+
+Let \( \mathbf{y} = (y_1,\cdots,y_n)^T \), \( \mathbf{\boldsymbol{y}} = (\boldsymbol{y}_1,\cdots,\boldsymbol{y}_n)^T \) and \( \beta = (\beta_0, \beta_1)^T \)
+
+
+It is convenient to write \( \mathbf{\boldsymbol{y}} = X\beta \) where \( X \in \mathbb{R}^{100 \times 2} \) is the design matrix given by (we keep the intercept here)
+$$
+X \equiv \begin{bmatrix}
+1 & x_1 \\
+\vdots & \vdots \\
+1 & x_{100} & \\
+\end{bmatrix}.
+$$
+
+The cost/loss/risk function is given by (
+$$
+C(\beta) = \frac{1}{n}||X\beta-\mathbf{y}||_{2}^{2} = \frac{1}{n}\sum_{i=1}^{100}\left[ (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2\right]
+$$
+
+and we want to find \( \beta \) such that \( C(\beta) \) is minimized.
+
+
+
+
+
The derivative of the cost/loss function
+
+
+Computing \( \partial C(\beta) / \partial \beta_0 \) and \( \partial C(\beta) / \partial \beta_1 \) we can show that the gradient can be written as
+$$
+\nabla_{\beta} C(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} = \frac{2}{n}X^T(X\beta - \mathbf{y}),
+$$
+
+where \( X \) is the design matrix defined above.
+
+
+
+
+
The Hessian matrix
+The Hessian matrix of \( C(\beta) \) is given by
+$$
+\boldsymbol{H} \equiv \begin{bmatrix}
+\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
+\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
+\end{bmatrix} = \frac{2}{n}X^T X.
+$$
+
+This result implies that \( C(\beta) \) is a convex function since the matrix \( X^T X \) always is positive semi-definite.
+
+
+
+
+
Simple program
+
+
+We can now write a program that minimizes \( C(\beta) \) using the gradient descent method with a constant learning rate \( \gamma \) according to
+$$
+\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots
+$$
+
+
+We can use the expression we computed for the gradient and let use a
+\( \beta_0 \) be chosen randomly and let \( \gamma = 0.001 \). Stop iterating
+when \( ||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8} \). Note that the code below does not include the latter stop criterion.
+
+
+And finally we can compare our solution for \( \beta \) with the analytic result given by
+\( \beta= (X^TX)^{-1} X^T \mathbf{y} \).
+
+
+
+
+
Gradient Descent Example
+
+
+Here our simple example
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+# Hessian matrix
+H = (2.0/n)* X.T @ X
+# Get the eigenvalues
+EigValues, EigVectors = np.linalg.eig(H)
+print(EigValues)
+
+beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y
+print(beta_linreg)
+beta = np.random.randn(2,1)
+
+eta = 1.0/np.max(EigValues)
+Niterations = 1000
+
+for iter in range(Niterations):
+ gradient = (2.0/n)*X.T @ (X @ beta-y)
+ beta -= eta*gradient
+
+print(beta)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(beta)
+ypredict2 = xbnew.dot(beta_linreg)
+plt.plot(xnew, ypredict, "r-")
+plt.plot(xnew, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example')
+plt.show()
+
+
+
+
+
And a corresponding example using scikit-learn
+
+
+
+
+
# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import SGDRegressor
+
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)
+print(beta_linreg)
+sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
+sgdreg.fit(x,y.ravel())
+print(sgdreg.intercept_, sgdreg.coef_)
+
+
+
+
+
Gradient descent and Ridge
+
+
+We have also discussed Ridge regression where the loss function contains a regularized term given by the \( L_2 \) norm of \( \beta \),
+$$
+C_{\text{ridge}}(\beta) = \frac{1}{n}||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0.
+$$
+
+
+In order to minimize \( C_{\text{ridge}}(\beta) \) using GD we only have adjust the gradient as follows
+$$
+\nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta).
+$$
+
+
+We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by
+$$
+\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}.
+$$
+
+
+
+
+
Program example for gradient descent with Ridge Regression
+
+
+
+
from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+XT_X = X.T @ X
+
+#Ridge parameter lambda
+lmbda = 0.001
+Id = lmbda* np.eye(XT_X.shape[0])
+
+beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y
+print(beta_linreg)
+# Start plain gradient descent
+beta = np.random.randn(2,1)
+
+eta = 0.1
+Niterations = 100
+
+for iter in range(Niterations):
+ gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta
+ beta -= eta*gradients
+
+print(beta)
+ypredict = X @ beta
+ypredict2 = X @ beta_linreg
+plt.plot(x, ypredict, "r-")
+plt.plot(x, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example for Ridge')
+plt.show()
+
+
+
+
+
Using gradient descent methods, limitations
+
+
+- Gradient descent (GD) finds local minima of our function. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our cost/loss/risk function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance.
+- GD is sensitive to initial conditions. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.
+- Gradients are computationally expensive to calculate for large datasets. In many cases in statistics and ML, the cost/loss/risk function is a sum of terms, with one term for each data point. For example, in linear regression, \( E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2 \); for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over all \( n \) data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called "mini batches". This has the added benefit of introducing stochasticity into our algorithm.
+- GD is very sensitive to choices of learning rates. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would adaptively choose the learning rates to match the landscape.
+- GD treats all directions in parameter space uniformly. Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive.
+- GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. This leads us to our next topic, Stochastic Gradient Methods.
+
+
diff --git a/doc/pub/week38/ipynb/ipynb-week38-src.tar.gz b/doc/pub/week38/ipynb/ipynb-week38-src.tar.gz
index 41fc69dc0..19e55c41b 100644
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diff --git a/doc/pub/week38/ipynb/week38.ipynb b/doc/pub/week38/ipynb/week38.ipynb
index c54bc5049..dc2d08352 100644
--- a/doc/pub/week38/ipynb/week38.ipynb
+++ b/doc/pub/week38/ipynb/week38.ipynb
@@ -26,6 +26,9 @@
"\n",
"* Friday: Logistic Regression and Optimization methods\n",
"\n",
+ "## Thursday September 23\n",
+ "\n",
+ "\n",
"## Ridge and LASSO Regression, reminder\n",
"\n",
"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 \n",
@@ -2518,8 +2521,8 @@
"discussed in the material on [optimization\n",
"methods](https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html).\n",
"\n",
- "This will be discussed next week. Before we develop our own codes for logistic regression, we end this lecture by studying the functionality that **Scikit-learn** offers. \n",
"\n",
+ "## Friday September 24\n",
"\n",
"\n",
"\n",
@@ -2742,6 +2745,1648 @@
"skplt.metrics.plot_cumulative_gain(y_test, y_probas)\n",
"plt.show()"
]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Optimization, the central part of any Machine Learning algortithm\n",
+ "\n",
+ "[Overview Video, why do we care about gradient methods?](https://www.uio.no/studier/emner/matnat/fys/FYS-STK3155/h20/forelesningsvideoer/OverarchingAimsWeek39.mp4?vrtx=view-as-webpage)\n",
+ "\n",
+ "\n",
+ "\n",
+ "Almost every problem in machine learning and data science starts with\n",
+ "a dataset $X$, a model $g(\\beta)$, which is a function of the\n",
+ "parameters $\\beta$ and a cost function $C(X, g(\\beta))$ that allows\n",
+ "us to judge how well the model $g(\\beta)$ explains the observations\n",
+ "$X$. The model is fit by finding the values of $\\beta$ that minimize\n",
+ "the cost function. Ideally we would be able to solve for $\\beta$\n",
+ "analytically, however this is not possible in general and we must use\n",
+ "some approximative/numerical method to compute the minimum.\n",
+ "\n",
+ "\n",
+ "## Revisiting our Logistic Regression case\n",
+ "\n",
+ "In our discussion on Logistic Regression we studied the \n",
+ "case of\n",
+ "two classes, with $y_i$ either\n",
+ "$0$ or $1$. Furthermore we assumed also that we have only two\n",
+ "parameters $\\beta$ in our fitting, that is we\n",
+ "defined probabilities"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "p(y_i=1|x_i,\\boldsymbol{\\beta}) &= \\frac{\\exp{(\\beta_0+\\beta_1x_i)}}{1+\\exp{(\\beta_0+\\beta_1x_i)}},\\nonumber\\\\\n",
+ "p(y_i=0|x_i,\\boldsymbol{\\beta}) &= 1 - p(y_i=1|x_i,\\boldsymbol{\\beta}),\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\boldsymbol{\\beta}$ are the weights we wish to extract from data, in our case $\\beta_0$ and $\\beta_1$. \n",
+ "\n",
+ "## The equations to solve\n",
+ "\n",
+ "Our compact equations used a definition of a vector $\\boldsymbol{y}$ with $n$\n",
+ "elements $y_i$, an $n\\times p$ matrix $\\boldsymbol{X}$ which contains the\n",
+ "$x_i$ values and a vector $\\boldsymbol{p}$ of fitted probabilities\n",
+ "$p(y_i\\vert x_i,\\boldsymbol{\\beta})$. We rewrote in a more compact form\n",
+ "the first derivative of the cost function as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\mathcal{C}(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}} = -\\boldsymbol{X}^T\\left(\\boldsymbol{y}-\\boldsymbol{p}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "If we in addition define a diagonal matrix $\\boldsymbol{W}$ with elements \n",
+ "$p(y_i\\vert x_i,\\boldsymbol{\\beta})(1-p(y_i\\vert x_i,\\boldsymbol{\\beta})$, we can obtain a compact expression of the second derivative as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial^2 \\mathcal{C}(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}\\partial \\boldsymbol{\\beta}^T} = \\boldsymbol{X}^T\\boldsymbol{W}\\boldsymbol{X}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This defines what is called the Hessian matrix.\n",
+ "\n",
+ "## Solving using Newton-Raphson's method\n",
+ "\n",
+ "If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives. \n",
+ "\n",
+ "Our iterative scheme is then given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta}^{\\mathrm{new}} = \\boldsymbol{\\beta}^{\\mathrm{old}}-\\left(\\frac{\\partial^2 \\mathcal{C}(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}\\partial \\boldsymbol{\\beta}^T}\\right)^{-1}_{\\boldsymbol{\\beta}^{\\mathrm{old}}}\\times \\left(\\frac{\\partial \\mathcal{C}(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}}\\right)_{\\boldsymbol{\\beta}^{\\mathrm{old}}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or in matrix form as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta}^{\\mathrm{new}} = \\boldsymbol{\\beta}^{\\mathrm{old}}-\\left(\\boldsymbol{X}^T\\boldsymbol{W}\\boldsymbol{X} \\right)^{-1}\\times \\left(-\\boldsymbol{X}^T(\\boldsymbol{y}-\\boldsymbol{p}) \\right)_{\\boldsymbol{\\beta}^{\\mathrm{old}}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The right-hand side is computed with the old values of $\\beta$. \n",
+ "\n",
+ "If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement. \n",
+ "\n",
+ "\n",
+ "## Brief reminder on Newton-Raphson's method\n",
+ "\n",
+ "Let us quickly remind ourselves how we derive the above method.\n",
+ "\n",
+ "Perhaps the most celebrated of all one-dimensional root-finding\n",
+ "routines is Newton's method, also called the Newton-Raphson\n",
+ "method. This method requires the evaluation of both the\n",
+ "function $f$ and its derivative $f'$ at arbitrary points. \n",
+ "If you can only calculate the derivative\n",
+ "numerically and/or your function is not of the smooth type, we\n",
+ "normally discourage the use of this method.\n",
+ "\n",
+ "## The equations\n",
+ "\n",
+ "The Newton-Raphson formula consists geometrically of extending the\n",
+ "tangent line at a current point until it crosses zero, then setting\n",
+ "the next guess to the abscissa of that zero-crossing. The mathematics\n",
+ "behind this method is rather simple. Employing a Taylor expansion for\n",
+ "$x$ sufficiently close to the solution $s$, we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "f(s)=0=f(x)+(s-x)f'(x)+\\frac{(s-x)^2}{2}f''(x) +\\dots.\n",
+ " \\label{eq:taylornr} \\tag{14}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "For small enough values of the function and for well-behaved\n",
+ "functions, the terms beyond linear are unimportant, hence we obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(x)+(s-x)f'(x)\\approx 0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "yielding"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "s\\approx x-\\frac{f(x)}{f'(x)}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Having in mind an iterative procedure, it is natural to start iterating with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "x_{n+1}=x_n-\\frac{f(x_n)}{f'(x_n)}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Simple geometric interpretation\n",
+ "\n",
+ "The above is Newton-Raphson's method. It has a simple geometric\n",
+ "interpretation, namely $x_{n+1}$ is the point where the tangent from\n",
+ "$(x_n,f(x_n))$ crosses the $x$-axis. Close to the solution,\n",
+ "Newton-Raphson converges fast to the desired result. However, if we\n",
+ "are far from a root, where the higher-order terms in the series are\n",
+ "important, the Newton-Raphson formula can give grossly inaccurate\n",
+ "results. For instance, the initial guess for the root might be so far\n",
+ "from the true root as to let the search interval include a local\n",
+ "maximum or minimum of the function. If an iteration places a trial\n",
+ "guess near such a local extremum, so that the first derivative nearly\n",
+ "vanishes, then Newton-Raphson may fail totally\n",
+ "\n",
+ "\n",
+ "## Extending to more than one variable\n",
+ "\n",
+ "Newton's method can be generalized to systems of several non-linear equations\n",
+ "and variables. Consider the case with two equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{array}{cc} f_1(x_1,x_2) &=0\\\\\n",
+ " f_2(x_1,x_2) &=0,\\end{array}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which we Taylor expand to obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1\n",
+ " \\partial f_1/\\partial x_1+h_2\n",
+ " \\partial f_1/\\partial x_2+\\dots\\\\\n",
+ " 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1\n",
+ " \\partial f_2/\\partial x_1+h_2\n",
+ " \\partial f_2/\\partial x_2+\\dots\n",
+ " \\end{array}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Defining the Jacobian matrix ${\\bf \\boldsymbol{J}}$ we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\bf \\boldsymbol{J}}=\\left( \\begin{array}{cc}\n",
+ " \\partial f_1/\\partial x_1 & \\partial f_1/\\partial x_2 \\\\\n",
+ " \\partial f_2/\\partial x_1 &\\partial f_2/\\partial x_2\n",
+ " \\end{array} \\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we can rephrase Newton's method as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\left(\\begin{array}{c} x_1^{n+1} \\\\ x_2^{n+1} \\end{array} \\right)=\n",
+ "\\left(\\begin{array}{c} x_1^{n} \\\\ x_2^{n} \\end{array} \\right)+\n",
+ "\\left(\\begin{array}{c} h_1^{n} \\\\ h_2^{n} \\end{array} \\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have defined"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\left(\\begin{array}{c} h_1^{n} \\\\ h_2^{n} \\end{array} \\right)=\n",
+ " -{\\bf \\boldsymbol{J}}^{-1}\n",
+ " \\left(\\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\\\ f_2(x_1^{n},x_2^{n}) \\end{array} \\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We need thus to compute the inverse of the Jacobian matrix and it\n",
+ "is to understand that difficulties may\n",
+ "arise in case ${\\bf \\boldsymbol{J}}$ is nearly singular.\n",
+ "\n",
+ "It is rather straightforward to extend the above scheme to systems of\n",
+ "more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## Steepest descent\n",
+ "\n",
+ "The basic idea of gradient descent is\n",
+ "that a function $F(\\mathbf{x})$, \n",
+ "$\\mathbf{x} \\equiv (x_1,\\cdots,x_n)$, decreases fastest if one goes from $\\bf {x}$ in the\n",
+ "direction of the negative gradient $-\\nabla F(\\mathbf{x})$.\n",
+ "\n",
+ "It can be shown that if"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbf{x}_{k+1} = \\mathbf{x}_k - \\gamma_k \\nabla F(\\mathbf{x}_k),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\gamma_k > 0$.\n",
+ "\n",
+ "For $\\gamma_k$ small enough, then $F(\\mathbf{x}_{k+1}) \\leq\n",
+ "F(\\mathbf{x}_k)$. This means that for a sufficiently small $\\gamma_k$\n",
+ "we are always moving towards smaller function values, i.e a minimum.\n",
+ "\n",
+ "\n",
+ "## More on Steepest descent\n",
+ "\n",
+ "The previous observation is the basis of the method of steepest\n",
+ "descent, which is also referred to as just gradient descent (GD). One\n",
+ "starts with an initial guess $\\mathbf{x}_0$ for a minimum of $F$ and\n",
+ "computes new approximations according to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbf{x}_{k+1} = \\mathbf{x}_k - \\gamma_k \\nabla F(\\mathbf{x}_k), \\ \\ k \\geq 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The parameter $\\gamma_k$ is often referred to as the step length or\n",
+ "the learning rate within the context of Machine Learning.\n",
+ "\n",
+ "\n",
+ "## The ideal\n",
+ "\n",
+ "Ideally the sequence $\\{\\mathbf{x}_k \\}_{k=0}$ converges to a global\n",
+ "minimum of the function $F$. In general we do not know if we are in a\n",
+ "global or local minimum. In the special case when $F$ is a convex\n",
+ "function, all local minima are also global minima, so in this case\n",
+ "gradient descent can converge to the global solution. The advantage of\n",
+ "this scheme is that it is conceptually simple and straightforward to\n",
+ "implement. However the method in this form has some severe\n",
+ "limitations:\n",
+ "\n",
+ "In machine learing we are often faced with non-convex high dimensional\n",
+ "cost functions with many local minima. Since GD is deterministic we\n",
+ "will get stuck in a local minimum, if the method converges, unless we\n",
+ "have a very good intial guess. This also implies that the scheme is\n",
+ "sensitive to the chosen initial condition.\n",
+ "\n",
+ "Note that the gradient is a function of $\\mathbf{x} =\n",
+ "(x_1,\\cdots,x_n)$ which makes it expensive to compute numerically.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The sensitiveness of the gradient descent\n",
+ "\n",
+ "The gradient descent method \n",
+ "is sensitive to the choice of learning rate $\\gamma_k$. This is due\n",
+ "to the fact that we are only guaranteed that $F(\\mathbf{x}_{k+1}) \\leq\n",
+ "F(\\mathbf{x}_k)$ for sufficiently small $\\gamma_k$. The problem is to\n",
+ "determine an optimal learning rate. If the learning rate is chosen too\n",
+ "small the method will take a long time to converge and if it is too\n",
+ "large we can experience erratic behavior.\n",
+ "\n",
+ "Many of these shortcomings can be alleviated by introducing\n",
+ "randomness. One such method is that of Stochastic Gradient Descent\n",
+ "(SGD), see below.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Convex functions\n",
+ "\n",
+ "Ideally we want our cost/loss function to be convex(concave).\n",
+ "\n",
+ "First we give the definition of a convex set: A set $C$ in\n",
+ "$\\mathbb{R}^n$ is said to be convex if, for all $x$ and $y$ in $C$ and\n",
+ "all $t \\in (0,1)$ , the point $(1 − t)x + ty$ also belongs to\n",
+ "C. Geometrically this means that every point on the line segment\n",
+ "connecting $x$ and $y$ is in $C$ as discussed below.\n",
+ "\n",
+ "The convex subsets of $\\mathbb{R}$ are the intervals of\n",
+ "$\\mathbb{R}$. Examples of convex sets of $\\mathbb{R}^2$ are the\n",
+ "regular polygons (triangles, rectangles, pentagons, etc...).\n",
+ "\n",
+ "## Convex function\n",
+ "\n",
+ "**Convex function**: Let $X \\subset \\mathbb{R}^n$ be a convex set. Assume that the function $f: X \\rightarrow \\mathbb{R}$ is continuous, then $f$ is said to be convex if $$f(tx_1 + (1-t)x_2) \\leq tf(x_1) + (1-t)f(x_2) $$ for all $x_1, x_2 \\in X$ and for all $t \\in [0,1]$. If $\\leq$ is replaced with a strict inequaltiy in the definition, we demand $x_1 \\neq x_2$ and $t\\in(0,1)$ then $f$ is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting $f(x_1)$ and $f(x_2)$, the value of the function on the interval $[x_1,x_2]$ is always below the line as illustrated below.\n",
+ "\n",
+ "## Conditions on convex functions\n",
+ "\n",
+ "In the following we state first and second-order conditions which\n",
+ "ensures convexity of a function $f$. We write $D_f$ to denote the\n",
+ "domain of $f$, i.e the subset of $R^n$ where $f$ is defined. For more\n",
+ "details and proofs we refer to: [S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press](http://stanford.edu/boyd/cvxbook/, 2004).\n",
+ "\n",
+ "**First order condition.**\n",
+ "\n",
+ "Suppose $f$ is differentiable (i.e $\\nabla f(x)$ is well defined for\n",
+ "all $x$ in the domain of $f$). Then $f$ is convex if and only if $D_f$\n",
+ "is a convex set and $$f(y) \\geq f(x) + \\nabla f(x)^T (y-x) $$ holds\n",
+ "for all $x,y \\in D_f$. This condition means that for a convex function\n",
+ "the first order Taylor expansion (right hand side above) at any point\n",
+ "a global under estimator of the function. To convince yourself you can\n",
+ "make a drawing of $f(x) = x^2+1$ and draw the tangent line to $f(x)$ and\n",
+ "note that it is always below the graph.\n",
+ "\n",
+ "\n",
+ "\n",
+ "**Second order condition.**\n",
+ "\n",
+ "Assume that $f$ is twice\n",
+ "differentiable, i.e the Hessian matrix exists at each point in\n",
+ "$D_f$. Then $f$ is convex if and only if $D_f$ is a convex set and its\n",
+ "Hessian is positive semi-definite for all $x\\in D_f$. For a\n",
+ "single-variable function this reduces to $f''(x) \\geq 0$. Geometrically this means that $f$ has nonnegative curvature\n",
+ "everywhere.\n",
+ "\n",
+ "\n",
+ "\n",
+ "This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.\n",
+ "\n",
+ "## More on convex functions\n",
+ "\n",
+ "The next result is of great importance to us and the reason why we are\n",
+ "going on about convex functions. In machine learning we frequently\n",
+ "have to minimize a loss/cost function in order to find the best\n",
+ "parameters for the model we are considering. \n",
+ "\n",
+ "Ideally we want the\n",
+ "global minimum (for high-dimensional models it is hard to know\n",
+ "if we have local or global minimum). However, if the cost/loss function\n",
+ "is convex the following result provides invaluable information:\n",
+ "\n",
+ "**Any minimum is global for convex functions.**\n",
+ "\n",
+ "Consider the problem of finding $x \\in \\mathbb{R}^n$ such that $f(x)$\n",
+ "is minimal, where $f$ is convex and differentiable. Then, any point\n",
+ "$x^*$ that satisfies $\\nabla f(x^*) = 0$ is a global minimum.\n",
+ "\n",
+ "\n",
+ "\n",
+ "This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.\n",
+ "\n",
+ "## Some simple problems\n",
+ "\n",
+ "1. Show that $f(x)=x^2$ is convex for $x \\in \\mathbb{R}$ using the definition of convexity. Hint: If you re-write the definition, $f$ is convex if the following holds for all $x,y \\in D_f$ and any $\\lambda \\in [0,1]$ $\\lambda f(x)+(1-\\lambda)f(y)-f(\\lambda x + (1-\\lambda) y ) \\geq 0$.\n",
+ "\n",
+ "2. Using the second order condition show that the following functions are convex on the specified domain.\n",
+ "\n",
+ " * $f(x) = e^x$ is convex for $x \\in \\mathbb{R}$.\n",
+ "\n",
+ " * $g(x) = -\\ln(x)$ is convex for $x \\in (0,\\infty)$.\n",
+ "\n",
+ "\n",
+ "3. Let $f(x) = x^2$ and $g(x) = e^x$. Show that $f(g(x))$ and $g(f(x))$ is convex for $x \\in \\mathbb{R}$. Also show that if $f(x)$ is any convex function than $h(x) = e^{f(x)}$ is convex.\n",
+ "\n",
+ "4. A norm is any function that satisfy the following properties\n",
+ "\n",
+ " * $f(\\alpha x) = |\\alpha| f(x)$ for all $\\alpha \\in \\mathbb{R}$.\n",
+ "\n",
+ " * $f(x+y) \\leq f(x) + f(y)$\n",
+ "\n",
+ " * $f(x) \\leq 0$ for all $x \\in \\mathbb{R}^n$ with equality if and only if $x = 0$\n",
+ "\n",
+ "\n",
+ "Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).\n",
+ "\n",
+ "\n",
+ "## Friday September 25\n",
+ "\n",
+ "[Video of Lecture](https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureSeptember25.mp4?vrtx=view-as-webpage) and [link to handwritten notes](https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember25.pdf).\n",
+ "\n",
+ "\n",
+ "## Standard steepest descent\n",
+ "\n",
+ "\n",
+ "Before we proceed, we would like to discuss the approach called the\n",
+ "**standard Steepest descent** (different from the above steepest descent discussion), which again leads to us having to be able\n",
+ "to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).\n",
+ "\n",
+ "[The success of the CG method](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)\n",
+ "for finding solutions of non-linear problems is based on the theory\n",
+ "of conjugate gradients for linear systems of equations. It belongs to\n",
+ "the class of iterative methods for solving problems from linear\n",
+ "algebra of the type"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}\\boldsymbol{x} = \\boldsymbol{b}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In the iterative process we end up with a problem like"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{r}= \\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\boldsymbol{r}$ is the so-called residual or error in the iterative process.\n",
+ "\n",
+ "When we have found the exact solution, $\\boldsymbol{r}=0$.\n",
+ "\n",
+ "## Gradient method\n",
+ "\n",
+ "The residual is zero when we reach the minimum of the quadratic equation"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "P(\\boldsymbol{x})=\\frac{1}{2}\\boldsymbol{x}^T\\boldsymbol{A}\\boldsymbol{x} - \\boldsymbol{x}^T\\boldsymbol{b},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with the constraint that the matrix $\\boldsymbol{A}$ is positive definite and\n",
+ "symmetric. This defines also the Hessian and we want it to be positive definite. \n",
+ "\n",
+ "\n",
+ "## Steepest descent method\n",
+ "\n",
+ "We denote the initial guess for $\\boldsymbol{x}$ as $\\boldsymbol{x}_0$. \n",
+ "We can assume without loss of generality that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_0=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or consider the system"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}\\boldsymbol{z} = \\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "instead.\n",
+ "\n",
+ "\n",
+ "## Steepest descent method\n",
+ "One can show that the solution $\\boldsymbol{x}$ is also the unique minimizer of the quadratic form"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(\\boldsymbol{x}) = \\frac{1}{2}\\boldsymbol{x}^T\\boldsymbol{A}\\boldsymbol{x} - \\boldsymbol{x}^T \\boldsymbol{x} , \\quad \\boldsymbol{x}\\in\\mathbf{R}^n.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This suggests taking the first basis vector $\\boldsymbol{r}_1$ (see below for definition) \n",
+ "to be the gradient of $f$ at $\\boldsymbol{x}=\\boldsymbol{x}_0$, \n",
+ "which equals"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}\\boldsymbol{x}_0-\\boldsymbol{b},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and \n",
+ "$\\boldsymbol{x}_0=0$ it is equal $-\\boldsymbol{b}$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Final expressions\n",
+ "We can compute the residual iteratively as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{r}_{k+1}=\\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_{k+1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which equals"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{b}-\\boldsymbol{A}(\\boldsymbol{x}_k+\\alpha_k\\boldsymbol{r}_k),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_k)-\\alpha_k\\boldsymbol{A}\\boldsymbol{r}_k,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which gives"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\alpha_k = \\frac{\\boldsymbol{r}_k^T\\boldsymbol{r}_k}{\\boldsymbol{r}_k^T\\boldsymbol{A}\\boldsymbol{r}_k}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "leading to the iterative scheme"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_{k+1}=\\boldsymbol{x}_k-\\alpha_k\\boldsymbol{r}_{k},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Steepest descent example"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import numpy.linalg as la\n",
+ "\n",
+ "import scipy.optimize as sopt\n",
+ "\n",
+ "import matplotlib.pyplot as pt\n",
+ "from mpl_toolkits.mplot3d import axes3d\n",
+ "\n",
+ "def f(x):\n",
+ " return 0.5*x[0]**2 + 2.5*x[1]**2\n",
+ "\n",
+ "def df(x):\n",
+ " return np.array([x[0], 5*x[1]])\n",
+ "\n",
+ "fig = pt.figure()\n",
+ "ax = fig.gca(projection=\"3d\")\n",
+ "\n",
+ "xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]\n",
+ "fmesh = f(np.array([xmesh, ymesh]))\n",
+ "ax.plot_surface(xmesh, ymesh, fmesh)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "And then as countor plot"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "pt.axis(\"equal\")\n",
+ "pt.contour(xmesh, ymesh, fmesh)\n",
+ "guesses = [np.array([2, 2./5])]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Find guesses"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "x = guesses[-1]\n",
+ "s = -df(x)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Run it!"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "def f1d(alpha):\n",
+ " return f(x + alpha*s)\n",
+ "\n",
+ "alpha_opt = sopt.golden(f1d)\n",
+ "next_guess = x + alpha_opt * s\n",
+ "guesses.append(next_guess)\n",
+ "print(next_guess)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "What happened?"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "pt.axis(\"equal\")\n",
+ "pt.contour(xmesh, ymesh, fmesh, 50)\n",
+ "it_array = np.array(guesses)\n",
+ "pt.plot(it_array.T[0], it_array.T[1], \"x-\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Conjugate gradient method\n",
+ "In the CG method we define so-called conjugate directions and two vectors \n",
+ "$\\boldsymbol{s}$ and $\\boldsymbol{t}$\n",
+ "are said to be\n",
+ "conjugate if"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{s}^T\\boldsymbol{A}\\boldsymbol{t}= 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The philosophy of the CG method is to perform searches in various conjugate directions\n",
+ "of our vectors $\\boldsymbol{x}_i$ obeying the above criterion, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_i^T\\boldsymbol{A}\\boldsymbol{x}_j= 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Two vectors are conjugate if they are orthogonal with respect to \n",
+ "this inner product. Being conjugate is a symmetric relation: if $\\boldsymbol{s}$ is conjugate to $\\boldsymbol{t}$, then $\\boldsymbol{t}$ is conjugate to $\\boldsymbol{s}$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Conjugate gradient method\n",
+ "An example is given by the eigenvectors of the matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{v}_i^T\\boldsymbol{A}\\boldsymbol{v}_j= \\lambda\\boldsymbol{v}_i^T\\boldsymbol{v}_j,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which is zero unless $i=j$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Conjugate gradient method\n",
+ "Assume now that we have a symmetric positive-definite matrix $\\boldsymbol{A}$ of size\n",
+ "$n\\times n$. At each iteration $i+1$ we obtain the conjugate direction of a vector"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_{i+1}=\\boldsymbol{x}_{i}+\\alpha_i\\boldsymbol{p}_{i}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We assume that $\\boldsymbol{p}_{i}$ is a sequence of $n$ mutually conjugate directions. \n",
+ "Then the $\\boldsymbol{p}_{i}$ form a basis of $R^n$ and we can expand the solution \n",
+ "$ \\boldsymbol{A}\\boldsymbol{x} = \\boldsymbol{b}$ in this basis, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x} = \\sum^{n}_{i=1} \\alpha_i \\boldsymbol{p}_i.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Conjugate gradient method\n",
+ "The coefficients are given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbf{A}\\mathbf{x} = \\sum^{n}_{i=1} \\alpha_i \\mathbf{A} \\mathbf{p}_i = \\mathbf{b}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Multiplying with $\\boldsymbol{p}_k^T$ from the left gives"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{p}_k^T \\boldsymbol{A}\\boldsymbol{x} = \\sum^{n}_{i=1} \\alpha_i\\boldsymbol{p}_k^T \\boldsymbol{A}\\boldsymbol{p}_i= \\boldsymbol{p}_k^T \\boldsymbol{b},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and we can define the coefficients $\\alpha_k$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\alpha_k = \\frac{\\boldsymbol{p}_k^T \\boldsymbol{b}}{\\boldsymbol{p}_k^T \\boldsymbol{A} \\boldsymbol{p}_k}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Conjugate gradient method and iterations\n",
+ "\n",
+ "If we choose the conjugate vectors $\\boldsymbol{p}_k$ carefully, \n",
+ "then we may not need all of them to obtain a good approximation to the solution \n",
+ "$\\boldsymbol{x}$. \n",
+ "We want to regard the conjugate gradient method as an iterative method. \n",
+ "This will us to solve systems where $n$ is so large that the direct \n",
+ "method would take too much time.\n",
+ "\n",
+ "We denote the initial guess for $\\boldsymbol{x}$ as $\\boldsymbol{x}_0$. \n",
+ "We can assume without loss of generality that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_0=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or consider the system"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}\\boldsymbol{z} = \\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "instead.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Conjugate gradient method\n",
+ "One can show that the solution $\\boldsymbol{x}$ is also the unique minimizer of the quadratic form"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(\\boldsymbol{x}) = \\frac{1}{2}\\boldsymbol{x}^T\\boldsymbol{A}\\boldsymbol{x} - \\boldsymbol{x}^T \\boldsymbol{x} , \\quad \\boldsymbol{x}\\in\\mathbf{R}^n.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This suggests taking the first basis vector $\\boldsymbol{p}_1$ \n",
+ "to be the gradient of $f$ at $\\boldsymbol{x}=\\boldsymbol{x}_0$, \n",
+ "which equals"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}\\boldsymbol{x}_0-\\boldsymbol{b},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and \n",
+ "$\\boldsymbol{x}_0=0$ it is equal $-\\boldsymbol{b}$.\n",
+ "The other vectors in the basis will be conjugate to the gradient, \n",
+ "hence the name conjugate gradient method.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Conjugate gradient method\n",
+ "Let $\\boldsymbol{r}_k$ be the residual at the $k$-th step:"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{r}_k=\\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_k.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Note that $\\boldsymbol{r}_k$ is the negative gradient of $f$ at \n",
+ "$\\boldsymbol{x}=\\boldsymbol{x}_k$, \n",
+ "so the gradient descent method would be to move in the direction $\\boldsymbol{r}_k$. \n",
+ "Here, we insist that the directions $\\boldsymbol{p}_k$ are conjugate to each other, \n",
+ "so we take the direction closest to the gradient $\\boldsymbol{r}_k$ \n",
+ "under the conjugacy constraint. \n",
+ "This gives the following expression"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{p}_{k+1}=\\boldsymbol{r}_k-\\frac{\\boldsymbol{p}_k^T \\boldsymbol{A}\\boldsymbol{r}_k}{\\boldsymbol{p}_k^T\\boldsymbol{A}\\boldsymbol{p}_k} \\boldsymbol{p}_k.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Conjugate gradient method\n",
+ "We can also compute the residual iteratively as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{r}_{k+1}=\\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_{k+1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which equals"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{b}-\\boldsymbol{A}(\\boldsymbol{x}_k+\\alpha_k\\boldsymbol{p}_k),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{x}_k)-\\alpha_k\\boldsymbol{A}\\boldsymbol{p}_k,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which gives"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{r}_{k+1}=\\boldsymbol{r}_k-\\boldsymbol{A}\\boldsymbol{p}_{k},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## Revisiting some of our first Linear Regression Encounters\n",
+ "\n",
+ "We will use linear regression as a case study for the gradient descent\n",
+ "methods. Linear regression is a great test case for the gradient\n",
+ "descent methods discussed in the lectures since it has several\n",
+ "desirable properties such as:\n",
+ "\n",
+ "1. An analytical solution (recall homework set 1).\n",
+ "\n",
+ "2. The gradient can be computed analytically.\n",
+ "\n",
+ "3. The cost function is convex which guarantees that gradient descent converges for small enough learning rates\n",
+ "\n",
+ "We revisit an example similar to what we had in the first homework set. We had a function of the type"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "x = 2*np.random.rand(m,1)\n",
+ "y = 4+3*x+np.random.randn(m,1)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $x_i \\in [0,1] $ is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution $\\cal {N}(0,1)$. \n",
+ "The linear regression model is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "h_\\beta(x) = \\boldsymbol{y} = \\beta_0 + \\beta_1 x,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "such that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y}_i = \\beta_0 + \\beta_1 x_i.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## Gradient descent example\n",
+ "\n",
+ "Let $\\mathbf{y} = (y_1,\\cdots,y_n)^T$, $\\mathbf{\\boldsymbol{y}} = (\\boldsymbol{y}_1,\\cdots,\\boldsymbol{y}_n)^T$ and $\\beta = (\\beta_0, \\beta_1)^T$\n",
+ "\n",
+ "It is convenient to write $\\mathbf{\\boldsymbol{y}} = X\\beta$ where $X \\in \\mathbb{R}^{100 \\times 2} $ is the design matrix given by (we keep the intercept here)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "X \\equiv \\begin{bmatrix}\n",
+ "1 & x_1 \\\\\n",
+ "\\vdots & \\vdots \\\\\n",
+ "1 & x_{100} & \\\\\n",
+ "\\end{bmatrix}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The cost/loss/risk function is given by ("
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\beta) = \\frac{1}{n}||X\\beta-\\mathbf{y}||_{2}^{2} = \\frac{1}{n}\\sum_{i=1}^{100}\\left[ (\\beta_0 + \\beta_1 x_i)^2 - 2 y_i (\\beta_0 + \\beta_1 x_i) + y_i^2\\right]\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and we want to find $\\beta$ such that $C(\\beta)$ is minimized.\n",
+ "\n",
+ "## The derivative of the cost/loss function\n",
+ "\n",
+ "Computing $\\partial C(\\beta) / \\partial \\beta_0$ and $\\partial C(\\beta) / \\partial \\beta_1$ we can show that the gradient can be written as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\nabla_{\\beta} C(\\beta) = \\frac{2}{n}\\begin{bmatrix} \\sum_{i=1}^{100} \\left(\\beta_0+\\beta_1x_i-y_i\\right) \\\\\n",
+ "\\sum_{i=1}^{100}\\left( x_i (\\beta_0+\\beta_1x_i)-y_ix_i\\right) \\\\\n",
+ "\\end{bmatrix} = \\frac{2}{n}X^T(X\\beta - \\mathbf{y}),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $X$ is the design matrix defined above.\n",
+ "\n",
+ "## The Hessian matrix\n",
+ "The Hessian matrix of $C(\\beta)$ is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{H} \\equiv \\begin{bmatrix}\n",
+ "\\frac{\\partial^2 C(\\beta)}{\\partial \\beta_0^2} & \\frac{\\partial^2 C(\\beta)}{\\partial \\beta_0 \\partial \\beta_1} \\\\\n",
+ "\\frac{\\partial^2 C(\\beta)}{\\partial \\beta_0 \\partial \\beta_1} & \\frac{\\partial^2 C(\\beta)}{\\partial \\beta_1^2} & \\\\\n",
+ "\\end{bmatrix} = \\frac{2}{n}X^T X.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This result implies that $C(\\beta)$ is a convex function since the matrix $X^T X$ always is positive semi-definite.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Simple program\n",
+ "\n",
+ "We can now write a program that minimizes $C(\\beta)$ using the gradient descent method with a constant learning rate $\\gamma$ according to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\beta_{k+1} = \\beta_k - \\gamma \\nabla_\\beta C(\\beta_k), \\ k=0,1,\\cdots\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can use the expression we computed for the gradient and let use a\n",
+ "$\\beta_0$ be chosen randomly and let $\\gamma = 0.001$. Stop iterating\n",
+ "when $||\\nabla_\\beta C(\\beta_k) || \\leq \\epsilon = 10^{-8}$. **Note that the code below does not include the latter stop criterion**.\n",
+ "\n",
+ "And finally we can compare our solution for $\\beta$ with the analytic result given by \n",
+ "$\\beta= (X^TX)^{-1} X^T \\mathbf{y}$.\n",
+ "\n",
+ "## Gradient Descent Example\n",
+ "\n",
+ "Here our simple example"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "\n",
+ "# Importing various packages\n",
+ "from random import random, seed\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from mpl_toolkits.mplot3d import Axes3D\n",
+ "from matplotlib import cm\n",
+ "from matplotlib.ticker import LinearLocator, FormatStrFormatter\n",
+ "import sys\n",
+ "\n",
+ "# the number of datapoints\n",
+ "n = 100\n",
+ "x = 2*np.random.rand(n,1)\n",
+ "y = 4+3*x+np.random.randn(n,1)\n",
+ "\n",
+ "X = np.c_[np.ones((n,1)), x]\n",
+ "# Hessian matrix\n",
+ "H = (2.0/n)* X.T @ X\n",
+ "# Get the eigenvalues\n",
+ "EigValues, EigVectors = np.linalg.eig(H)\n",
+ "print(EigValues)\n",
+ "\n",
+ "beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y\n",
+ "print(beta_linreg)\n",
+ "beta = np.random.randn(2,1)\n",
+ "\n",
+ "eta = 1.0/np.max(EigValues)\n",
+ "Niterations = 1000\n",
+ "\n",
+ "for iter in range(Niterations):\n",
+ " gradient = (2.0/n)*X.T @ (X @ beta-y)\n",
+ " beta -= eta*gradient\n",
+ "\n",
+ "print(beta)\n",
+ "xnew = np.array([[0],[2]])\n",
+ "xbnew = np.c_[np.ones((2,1)), xnew]\n",
+ "ypredict = xbnew.dot(beta)\n",
+ "ypredict2 = xbnew.dot(beta_linreg)\n",
+ "plt.plot(xnew, ypredict, \"r-\")\n",
+ "plt.plot(xnew, ypredict2, \"b-\")\n",
+ "plt.plot(x, y ,'ro')\n",
+ "plt.axis([0,2.0,0, 15.0])\n",
+ "plt.xlabel(r'$x$')\n",
+ "plt.ylabel(r'$y$')\n",
+ "plt.title(r'Gradient descent example')\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## And a corresponding example using **scikit-learn**"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "# Importing various packages\n",
+ "from random import random, seed\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.linear_model import SGDRegressor\n",
+ "\n",
+ "n = 100\n",
+ "x = 2*np.random.rand(n,1)\n",
+ "y = 4+3*x+np.random.randn(n,1)\n",
+ "\n",
+ "X = np.c_[np.ones((n,1)), x]\n",
+ "beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)\n",
+ "print(beta_linreg)\n",
+ "sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)\n",
+ "sgdreg.fit(x,y.ravel())\n",
+ "print(sgdreg.intercept_, sgdreg.coef_)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## Gradient descent and Ridge\n",
+ "\n",
+ "We have also discussed Ridge regression where the loss function contains a regularized term given by the $L_2$ norm of $\\beta$,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C_{\\text{ridge}}(\\beta) = \\frac{1}{n}||X\\beta -\\mathbf{y}||^2 + \\lambda ||\\beta||^2, \\ \\lambda \\geq 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In order to minimize $C_{\\text{ridge}}(\\beta)$ using GD we only have adjust the gradient as follows"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\nabla_\\beta C_{\\text{ridge}}(\\beta) = \\frac{2}{n}\\begin{bmatrix} \\sum_{i=1}^{100} \\left(\\beta_0+\\beta_1x_i-y_i\\right) \\\\\n",
+ "\\sum_{i=1}^{100}\\left( x_i (\\beta_0+\\beta_1x_i)-y_ix_i\\right) \\\\\n",
+ "\\end{bmatrix} + 2\\lambda\\begin{bmatrix} \\beta_0 \\\\ \\beta_1\\end{bmatrix} = 2 (X^T(X\\beta - \\mathbf{y})+\\lambda \\beta).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can easily extend our program to minimize $C_{\\text{ridge}}(\\beta)$ using gradient descent and compare with the analytical solution given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\beta_{\\text{ridge}} = \\left(X^T X + \\lambda I_{2 \\times 2} \\right)^{-1} X^T \\mathbf{y}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Program example for gradient descent with Ridge Regression"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": false,
+ "editable": true
+ },
+ "outputs": [],
+ "source": [
+ "from random import random, seed\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from mpl_toolkits.mplot3d import Axes3D\n",
+ "from matplotlib import cm\n",
+ "from matplotlib.ticker import LinearLocator, FormatStrFormatter\n",
+ "import sys\n",
+ "\n",
+ "# the number of datapoints\n",
+ "n = 100\n",
+ "x = 2*np.random.rand(n,1)\n",
+ "y = 4+3*x+np.random.randn(n,1)\n",
+ "\n",
+ "X = np.c_[np.ones((n,1)), x]\n",
+ "XT_X = X.T @ X\n",
+ "\n",
+ "#Ridge parameter lambda\n",
+ "lmbda = 0.001\n",
+ "Id = lmbda* np.eye(XT_X.shape[0])\n",
+ "\n",
+ "beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y\n",
+ "print(beta_linreg)\n",
+ "# Start plain gradient descent\n",
+ "beta = np.random.randn(2,1)\n",
+ "\n",
+ "eta = 0.1\n",
+ "Niterations = 100\n",
+ "\n",
+ "for iter in range(Niterations):\n",
+ " gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta\n",
+ " beta -= eta*gradients\n",
+ "\n",
+ "print(beta)\n",
+ "ypredict = X @ beta\n",
+ "ypredict2 = X @ beta_linreg\n",
+ "plt.plot(x, ypredict, \"r-\")\n",
+ "plt.plot(x, ypredict2, \"b-\")\n",
+ "plt.plot(x, y ,'ro')\n",
+ "plt.axis([0,2.0,0, 15.0])\n",
+ "plt.xlabel(r'$x$')\n",
+ "plt.ylabel(r'$y$')\n",
+ "plt.title(r'Gradient descent example for Ridge')\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Using gradient descent methods, limitations\n",
+ "\n",
+ "* **Gradient descent (GD) finds local minima of our function**. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our cost/loss/risk function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance.\n",
+ "\n",
+ "* **GD is sensitive to initial conditions**. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.\n",
+ "\n",
+ "* **Gradients are computationally expensive to calculate for large datasets**. In many cases in statistics and ML, the cost/loss/risk function is a sum of terms, with one term for each data point. For example, in linear regression, $E \\propto \\sum_{i=1}^n (y_i - \\mathbf{w}^T\\cdot\\mathbf{x}_i)^2$; for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over *all* $n$ data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called \"mini batches\". This has the added benefit of introducing stochasticity into our algorithm.\n",
+ "\n",
+ "* **GD is very sensitive to choices of learning rates**. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would *adaptively* choose the learning rates to match the landscape.\n",
+ "\n",
+ "* **GD treats all directions in parameter space uniformly.** Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive. \n",
+ "\n",
+ "* GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. This leads us to our next topic, Stochastic Gradient Methods."
+ ]
}
],
"metadata": {},
diff --git a/doc/src/week38/week38.do.txt b/doc/src/week38/week38.do.txt
index d21cdc7a1..965ada8e8 100644
--- a/doc/src/week38/week38.do.txt
+++ b/doc/src/week38/week38.do.txt
@@ -13,6 +13,10 @@ DATE: today
* Friday: Logistic Regression and Optimization methods
+!split
+===== Thursday September 23 =====
+
+
!split
===== Ridge and LASSO Regression, reminder =====
@@ -1654,8 +1658,9 @@ descent method. Newton's method and gradient descent methods are
discussed in the material on "optimization
methods":"https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html".
-This will be discussed next week. Before we develop our own codes for logistic regression, we end this lecture by studying the functionality that _Scikit-learn_ offers.
+!split
+===== Friday September 24 =====
@@ -1826,8 +1831,1024 @@ plt.show()
+!split
+===== Optimization, the central part of any Machine Learning algortithm =====
+
+"Overview Video, why do we care about gradient methods?":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK3155/h20/forelesningsvideoer/OverarchingAimsWeek39.mp4?vrtx=view-as-webpage"
+
+
+
+Almost every problem in machine learning and data science starts with
+a dataset $X$, a model $g(\beta)$, which is a function of the
+parameters $\beta$ and a cost function $C(X, g(\beta))$ that allows
+us to judge how well the model $g(\beta)$ explains the observations
+$X$. The model is fit by finding the values of $\beta$ that minimize
+the cost function. Ideally we would be able to solve for $\beta$
+analytically, however this is not possible in general and we must use
+some approximative/numerical method to compute the minimum.
+
+
+!split
+===== Revisiting our Logistic Regression case =====
+
+In our discussion on Logistic Regression we studied the
+case of
+two classes, with $y_i$ either
+$0$ or $1$. Furthermore we assumed also that we have only two
+parameters $\beta$ in our fitting, that is we
+defined probabilities
+
+!bt
+\begin{align*}
+p(y_i=1|x_i,\bm{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
+p(y_i=0|x_i,\bm{\beta}) &= 1 - p(y_i=1|x_i,\bm{\beta}),
+\end{align*}
+!et
+where $\bm{\beta}$ are the weights we wish to extract from data, in our case $\beta_0$ and $\beta_1$.
+
+!split
+===== The equations to solve =====
+
+Our compact equations used a definition of a vector $\bm{y}$ with $n$
+elements $y_i$, an $n\times p$ matrix $\bm{X}$ which contains the
+$x_i$ values and a vector $\bm{p}$ of fitted probabilities
+$p(y_i\vert x_i,\bm{\beta})$. We rewrote in a more compact form
+the first derivative of the cost function as
+
+!bt
+\[
+\frac{\partial \mathcal{C}(\bm{\beta})}{\partial \bm{\beta}} = -\bm{X}^T\left(\bm{y}-\bm{p}\right).
+\]
+!et
+
+If we in addition define a diagonal matrix $\bm{W}$ with elements
+$p(y_i\vert x_i,\bm{\beta})(1-p(y_i\vert x_i,\bm{\beta})$, we can obtain a compact expression of the second derivative as
+
+!bt
+\[
+\frac{\partial^2 \mathcal{C}(\bm{\beta})}{\partial \bm{\beta}\partial \bm{\beta}^T} = \bm{X}^T\bm{W}\bm{X}.
+\]
+!et
+This defines what is called the Hessian matrix.
+
+!split
+===== Solving using Newton-Raphson's method =====
+
+If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
+
+Our iterative scheme is then given by
+
+!bt
+\[
+\bm{\beta}^{\mathrm{new}} = \bm{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\bm{\beta})}{\partial \bm{\beta}\partial \bm{\beta}^T}\right)^{-1}_{\bm{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\bm{\beta})}{\partial \bm{\beta}}\right)_{\bm{\beta}^{\mathrm{old}}},
+\]
+!et
+or in matrix form as
+
+!bt
+\[
+\bm{\beta}^{\mathrm{new}} = \bm{\beta}^{\mathrm{old}}-\left(\bm{X}^T\bm{W}\bm{X} \right)^{-1}\times \left(-\bm{X}^T(\bm{y}-\bm{p}) \right)_{\bm{\beta}^{\mathrm{old}}}.
+\]
+!et
+The right-hand side is computed with the old values of $\beta$.
+
+If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
+
+
+!split
+===== Brief reminder on Newton-Raphson's method =====
+
+Let us quickly remind ourselves how we derive the above method.
+
+Perhaps the most celebrated of all one-dimensional root-finding
+routines is Newton's method, also called the Newton-Raphson
+method. This method requires the evaluation of both the
+function $f$ and its derivative $f'$ at arbitrary points.
+If you can only calculate the derivative
+numerically and/or your function is not of the smooth type, we
+normally discourage the use of this method.
+
+!split
+===== The equations =====
+
+The Newton-Raphson formula consists geometrically of extending the
+tangent line at a current point until it crosses zero, then setting
+the next guess to the abscissa of that zero-crossing. The mathematics
+behind this method is rather simple. Employing a Taylor expansion for
+$x$ sufficiently close to the solution $s$, we have
+
+
+!bt
+\[
+ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
+ \label{eq:taylornr}
+\]
+!et
+
+For small enough values of the function and for well-behaved
+functions, the terms beyond linear are unimportant, hence we obtain
+
+
+!bt
+\[
+ f(x)+(s-x)f'(x)\approx 0,
+\]
+!et
+yielding
+!bt
+\[
+ s\approx x-\frac{f(x)}{f'(x)}.
+\]
+!et
+
+Having in mind an iterative procedure, it is natural to start iterating with
+!bt
+\[
+ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
+\]
+!et
+
+!split
+===== Simple geometric interpretation =====
+
+The above is Newton-Raphson's method. It has a simple geometric
+interpretation, namely $x_{n+1}$ is the point where the tangent from
+$(x_n,f(x_n))$ crosses the $x$-axis. Close to the solution,
+Newton-Raphson converges fast to the desired result. However, if we
+are far from a root, where the higher-order terms in the series are
+important, the Newton-Raphson formula can give grossly inaccurate
+results. For instance, the initial guess for the root might be so far
+from the true root as to let the search interval include a local
+maximum or minimum of the function. If an iteration places a trial
+guess near such a local extremum, so that the first derivative nearly
+vanishes, then Newton-Raphson may fail totally
+
+
+!split
+===== Extending to more than one variable =====
+
+Newton's method can be generalized to systems of several non-linear equations
+and variables. Consider the case with two equations
+!bt
+\[
+ \begin{array}{cc} f_1(x_1,x_2) &=0\\
+ f_2(x_1,x_2) &=0,\end{array}
+\]
+!et
+which we Taylor expand to obtain
+
+!bt
+\[
+ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
+ \partial f_1/\partial x_1+h_2
+ \partial f_1/\partial x_2+\dots\\
+ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
+ \partial f_2/\partial x_1+h_2
+ \partial f_2/\partial x_2+\dots
+ \end{array}.
+\]
+!et
+Defining the Jacobian matrix ${\bf \bm{J}}$ we have
+!bt
+\[
+ {\bf \bm{J}}=\left( \begin{array}{cc}
+ \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
+ \partial f_2/\partial x_1 &\partial f_2/\partial x_2
+ \end{array} \right),
+\]
+!et
+we can rephrase Newton's method as
+!bt
+\[
+\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
+\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
+\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
+\]
+!et
+where we have defined
+!bt
+\[
+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
+ -{\bf \bm{J}}^{-1}
+ \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
+\]
+!et
+We need thus to compute the inverse of the Jacobian matrix and it
+is to understand that difficulties may
+arise in case ${\bf \bm{J}}$ is nearly singular.
+
+It is rather straightforward to extend the above scheme to systems of
+more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
+
+
+
+!split
+===== Steepest descent =====
+
+The basic idea of gradient descent is
+that a function $F(\mathbf{x})$,
+$\mathbf{x} \equiv (x_1,\cdots,x_n)$, decreases fastest if one goes from $\bf {x}$ in the
+direction of the negative gradient $-\nabla F(\mathbf{x})$.
+
+It can be shown that if
+!bt
+\[
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
+\]
+!et
+with $\gamma_k > 0$.
+
+For $\gamma_k$ small enough, then $F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k)$. This means that for a sufficiently small $\gamma_k$
+we are always moving towards smaller function values, i.e a minimum.
+
+!split
+===== More on Steepest descent =====
+
+The previous observation is the basis of the method of steepest
+descent, which is also referred to as just gradient descent (GD). One
+starts with an initial guess $\mathbf{x}_0$ for a minimum of $F$ and
+computes new approximations according to
+
+!bt
+\[
+\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
+\]
+!et
+
+The parameter $\gamma_k$ is often referred to as the step length or
+the learning rate within the context of Machine Learning.
+
+!split
+===== The ideal =====
+
+Ideally the sequence $\{\mathbf{x}_k \}_{k=0}$ converges to a global
+minimum of the function $F$. In general we do not know if we are in a
+global or local minimum. In the special case when $F$ is a convex
+function, all local minima are also global minima, so in this case
+gradient descent can converge to the global solution. The advantage of
+this scheme is that it is conceptually simple and straightforward to
+implement. However the method in this form has some severe
+limitations:
+
+In machine learing we are often faced with non-convex high dimensional
+cost functions with many local minima. Since GD is deterministic we
+will get stuck in a local minimum, if the method converges, unless we
+have a very good intial guess. This also implies that the scheme is
+sensitive to the chosen initial condition.
+
+Note that the gradient is a function of $\mathbf{x} =
+(x_1,\cdots,x_n)$ which makes it expensive to compute numerically.
+
+
+!split
+===== The sensitiveness of the gradient descent =====
+
+The gradient descent method
+is sensitive to the choice of learning rate $\gamma_k$. This is due
+to the fact that we are only guaranteed that $F(\mathbf{x}_{k+1}) \leq
+F(\mathbf{x}_k)$ for sufficiently small $\gamma_k$. The problem is to
+determine an optimal learning rate. If the learning rate is chosen too
+small the method will take a long time to converge and if it is too
+large we can experience erratic behavior.
+
+Many of these shortcomings can be alleviated by introducing
+randomness. One such method is that of Stochastic Gradient Descent
+(SGD), see below.
+
+
+!split
+===== Convex functions =====
+
+Ideally we want our cost/loss function to be convex(concave).
+
+First we give the definition of a convex set: A set $C$ in
+$\mathbb{R}^n$ is said to be convex if, for all $x$ and $y$ in $C$ and
+all $t \in (0,1)$ , the point $(1 − t)x + ty$ also belongs to
+C. Geometrically this means that every point on the line segment
+connecting $x$ and $y$ is in $C$ as discussed below.
+
+The convex subsets of $\mathbb{R}$ are the intervals of
+$\mathbb{R}$. Examples of convex sets of $\mathbb{R}^2$ are the
+regular polygons (triangles, rectangles, pentagons, etc...).
+
+!split
+===== Convex function =====
+
+_Convex function_: Let $X \subset \mathbb{R}^n$ be a convex set. Assume that the function $f: X \rightarrow \mathbb{R}$ is continuous, then $f$ is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all $x_1, x_2 \in X$ and for all $t \in [0,1]$. If $\leq$ is replaced with a strict inequaltiy in the definition, we demand $x_1 \neq x_2$ and $t\in(0,1)$ then $f$ is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting $f(x_1)$ and $f(x_2)$, the value of the function on the interval $[x_1,x_2]$ is always below the line as illustrated below.
+
+!split
+===== Conditions on convex functions =====
+
+In the following we state first and second-order conditions which
+ensures convexity of a function $f$. We write $D_f$ to denote the
+domain of $f$, i.e the subset of $R^n$ where $f$ is defined. For more
+details and proofs we refer to: "S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press":"http://stanford.edu/boyd/cvxbook/, 2004".
+
+!bblock First order condition
+Suppose $f$ is differentiable (i.e $\nabla f(x)$ is well defined for
+all $x$ in the domain of $f$). Then $f$ is convex if and only if $D_f$
+is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds
+for all $x,y \in D_f$. This condition means that for a convex function
+the first order Taylor expansion (right hand side above) at any point
+a global under estimator of the function. To convince yourself you can
+make a drawing of $f(x) = x^2+1$ and draw the tangent line to $f(x)$ and
+note that it is always below the graph.
+!eblock
+
+!bblock Second order condition
+Assume that $f$ is twice
+differentiable, i.e the Hessian matrix exists at each point in
+$D_f$. Then $f$ is convex if and only if $D_f$ is a convex set and its
+Hessian is positive semi-definite for all $x\in D_f$. For a
+single-variable function this reduces to $f''(x) \geq 0$. Geometrically this means that $f$ has nonnegative curvature
+everywhere.
+!eblock
+
+This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
+
+!split
+===== More on convex functions =====
+
+The next result is of great importance to us and the reason why we are
+going on about convex functions. In machine learning we frequently
+have to minimize a loss/cost function in order to find the best
+parameters for the model we are considering.
+
+Ideally we want the
+global minimum (for high-dimensional models it is hard to know
+if we have local or global minimum). However, if the cost/loss function
+is convex the following result provides invaluable information:
+
+!bblock Any minimum is global for convex functions
+Consider the problem of finding $x \in \mathbb{R}^n$ such that $f(x)$
+is minimal, where $f$ is convex and differentiable. Then, any point
+$x^*$ that satisfies $\nabla f(x^*) = 0$ is a global minimum.
+!eblock
+
+This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
+
+!split
+===== Some simple problems =====
+
+o Show that $f(x)=x^2$ is convex for $x \in \mathbb{R}$ using the definition of convexity. Hint: If you re-write the definition, $f$ is convex if the following holds for all $x,y \in D_f$ and any $\lambda \in [0,1]$ $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$.
+
+o Using the second order condition show that the following functions are convex on the specified domain.
+ * $f(x) = e^x$ is convex for $x \in \mathbb{R}$.
+ * $g(x) = -\ln(x)$ is convex for $x \in (0,\infty)$.
+o Let $f(x) = x^2$ and $g(x) = e^x$. Show that $f(g(x))$ and $g(f(x))$ is convex for $x \in \mathbb{R}$. Also show that if $f(x)$ is any convex function than $h(x) = e^{f(x)}$ is convex.
+
+o A norm is any function that satisfy the following properties
+ * $f(\alpha x) = |\alpha| f(x)$ for all $\alpha \in \mathbb{R}$.
+ * $f(x+y) \leq f(x) + f(y)$
+ * $f(x) \leq 0$ for all $x \in \mathbb{R}^n$ with equality if and only if $x = 0$
+
+Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
+
+
+!split
+===== Friday September 25 =====
+
+"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureSeptember25.mp4?vrtx=view-as-webpage" and "link to handwritten notes":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember25.pdf".
+
+
+!split
+===== Standard steepest descent =====
+
+
+Before we proceed, we would like to discuss the approach called the
+_standard Steepest descent_ (different from the above steepest descent discussion), which again leads to us having to be able
+to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
+
+"The success of the CG method":"https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf"
+for finding solutions of non-linear problems is based on the theory
+of conjugate gradients for linear systems of equations. It belongs to
+the class of iterative methods for solving problems from linear
+algebra of the type
+!bt
+\begin{equation*}
+\bm{A}\bm{x} = \bm{b}.
+\end{equation*}
+!et
+
+In the iterative process we end up with a problem like
+
+!bt
+\begin{equation*}
+ \bm{r}= \bm{b}-\bm{A}\bm{x},
+\end{equation*}
+!et
+where $\bm{r}$ is the so-called residual or error in the iterative process.
+
+When we have found the exact solution, $\bm{r}=0$.
+
+!split
+===== Gradient method =====
+
+The residual is zero when we reach the minimum of the quadratic equation
+!bt
+\begin{equation*}
+ P(\bm{x})=\frac{1}{2}\bm{x}^T\bm{A}\bm{x} - \bm{x}^T\bm{b},
+\end{equation*}
+!et
+
+with the constraint that the matrix $\bm{A}$ is positive definite and
+symmetric. This defines also the Hessian and we want it to be positive definite.
+
+
+!split
+===== Steepest descent method =====
+
+We denote the initial guess for $\bm{x}$ as $\bm{x}_0$.
+We can assume without loss of generality that
+!bt
+\begin{equation*}
+\bm{x}_0=0,
+\end{equation*}
+!et
+or consider the system
+!bt
+\begin{equation*}
+\bm{A}\bm{z} = \bm{b}-\bm{A}\bm{x}_0,
+\end{equation*}
+!et
+instead.
+
+
+!split
+===== Steepest descent method =====
+!bblock
+One can show that the solution $\bm{x}$ is also the unique minimizer of the quadratic form
+!bt
+\begin{equation*}
+ f(\bm{x}) = \frac{1}{2}\bm{x}^T\bm{A}\bm{x} - \bm{x}^T \bm{x} , \quad \bm{x}\in\mathbf{R}^n.
+\end{equation*}
+!et
+This suggests taking the first basis vector $\bm{r}_1$ (see below for definition)
+to be the gradient of $f$ at $\bm{x}=\bm{x}_0$,
+which equals
+!bt
+\begin{equation*}
+\bm{A}\bm{x}_0-\bm{b},
+\end{equation*}
+!et
+and
+$\bm{x}_0=0$ it is equal $-\bm{b}$.
+
+!eblock
+
+!split
+===== Final expressions =====
+!bblock
+We can compute the residual iteratively as
+!bt
+\begin{equation*}
+\bm{r}_{k+1}=\bm{b}-\bm{A}\bm{x}_{k+1},
+ \end{equation*}
+!et
+which equals
+!bt
+\begin{equation*}
+\bm{b}-\bm{A}(\bm{x}_k+\alpha_k\bm{r}_k),
+ \end{equation*}
+!et
+or
+!bt
+\begin{equation*}
+(\bm{b}-\bm{A}\bm{x}_k)-\alpha_k\bm{A}\bm{r}_k,
+ \end{equation*}
+!et
+which gives
+
+!bt
+\[
+\alpha_k = \frac{\bm{r}_k^T\bm{r}_k}{\bm{r}_k^T\bm{A}\bm{r}_k}
+\]
+!et
+leading to the iterative scheme
+!bt
+\begin{equation*}
+\bm{x}_{k+1}=\bm{x}_k-\alpha_k\bm{r}_{k},
+ \end{equation*}
+!et
+!eblock
+
+
+
+!split
+===== Steepest descent example =====
+
+!bc pycod
+import numpy as np
+import numpy.linalg as la
+
+import scipy.optimize as sopt
+
+import matplotlib.pyplot as pt
+from mpl_toolkits.mplot3d import axes3d
+
+def f(x):
+ return 0.5*x[0]**2 + 2.5*x[1]**2
+
+def df(x):
+ return np.array([x[0], 5*x[1]])
+
+fig = pt.figure()
+ax = fig.gca(projection="3d")
+
+xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
+fmesh = f(np.array([xmesh, ymesh]))
+ax.plot_surface(xmesh, ymesh, fmesh)
+!ec
+And then as countor plot
+!bc pycod
+pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh)
+guesses = [np.array([2, 2./5])]
+!ec
+Find guesses
+!bc pycod
+x = guesses[-1]
+s = -df(x)
+!ec
+Run it!
+!bc pycod
+def f1d(alpha):
+ return f(x + alpha*s)
+
+alpha_opt = sopt.golden(f1d)
+next_guess = x + alpha_opt * s
+guesses.append(next_guess)
+print(next_guess)
+!ec
+What happened?
+!bc pycod
+pt.axis("equal")
+pt.contour(xmesh, ymesh, fmesh, 50)
+it_array = np.array(guesses)
+pt.plot(it_array.T[0], it_array.T[1], "x-")
+!ec
+
+!split
+===== Conjugate gradient method =====
+!bblock
+In the CG method we define so-called conjugate directions and two vectors
+$\bm{s}$ and $\bm{t}$
+are said to be
+conjugate if
+!bt
+\begin{equation*}
+\bm{s}^T\bm{A}\bm{t}= 0.
+\end{equation*}
+!et
+The philosophy of the CG method is to perform searches in various conjugate directions
+of our vectors $\bm{x}_i$ obeying the above criterion, namely
+!bt
+\begin{equation*}
+\bm{x}_i^T\bm{A}\bm{x}_j= 0.
+\end{equation*}
+!et
+Two vectors are conjugate if they are orthogonal with respect to
+this inner product. Being conjugate is a symmetric relation: if $\bm{s}$ is conjugate to $\bm{t}$, then $\bm{t}$ is conjugate to $\bm{s}$.
+!eblock
+
+!split
+===== Conjugate gradient method =====
+!bblock
+An example is given by the eigenvectors of the matrix
+!bt
+\begin{equation*}
+\bm{v}_i^T\bm{A}\bm{v}_j= \lambda\bm{v}_i^T\bm{v}_j,
+\end{equation*}
+!et
+which is zero unless $i=j$.
+!eblock
+
+
+!split
+===== Conjugate gradient method =====
+!bblock
+Assume now that we have a symmetric positive-definite matrix $\bm{A}$ of size
+$n\times n$. At each iteration $i+1$ we obtain the conjugate direction of a vector
+!bt
+\begin{equation*}
+\bm{x}_{i+1}=\bm{x}_{i}+\alpha_i\bm{p}_{i}.
+\end{equation*}
+!et
+We assume that $\bm{p}_{i}$ is a sequence of $n$ mutually conjugate directions.
+Then the $\bm{p}_{i}$ form a basis of $R^n$ and we can expand the solution
+$ \bm{A}\bm{x} = \bm{b}$ in this basis, namely
+
+!bt
+\begin{equation*}
+ \bm{x} = \sum^{n}_{i=1} \alpha_i \bm{p}_i.
+\end{equation*}
+!et
+!eblock
+
+!split
+===== Conjugate gradient method =====
+!bblock
+The coefficients are given by
+!bt
+\begin{equation*}
+ \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
+\end{equation*}
+!et
+Multiplying with $\bm{p}_k^T$ from the left gives
+
+!bt
+\begin{equation*}
+ \bm{p}_k^T \bm{A}\bm{x} = \sum^{n}_{i=1} \alpha_i\bm{p}_k^T \bm{A}\bm{p}_i= \bm{p}_k^T \bm{b},
+\end{equation*}
+!et
+and we can define the coefficients $\alpha_k$ as
+
+!bt
+\begin{equation*}
+ \alpha_k = \frac{\bm{p}_k^T \bm{b}}{\bm{p}_k^T \bm{A} \bm{p}_k}
+\end{equation*}
+!et
+!eblock
+
+!split
+===== Conjugate gradient method and iterations =====
+!bblock
+
+If we choose the conjugate vectors $\bm{p}_k$ carefully,
+then we may not need all of them to obtain a good approximation to the solution
+$\bm{x}$.
+We want to regard the conjugate gradient method as an iterative method.
+This will us to solve systems where $n$ is so large that the direct
+method would take too much time.
+
+We denote the initial guess for $\bm{x}$ as $\bm{x}_0$.
+We can assume without loss of generality that
+!bt
+\begin{equation*}
+\bm{x}_0=0,
+\end{equation*}
+!et
+or consider the system
+!bt
+\begin{equation*}
+\bm{A}\bm{z} = \bm{b}-\bm{A}\bm{x}_0,
+\end{equation*}
+!et
+instead.
+!eblock
+
+
+!split
+===== Conjugate gradient method =====
+!bblock
+One can show that the solution $\bm{x}$ is also the unique minimizer of the quadratic form
+!bt
+\begin{equation*}
+ f(\bm{x}) = \frac{1}{2}\bm{x}^T\bm{A}\bm{x} - \bm{x}^T \bm{x} , \quad \bm{x}\in\mathbf{R}^n.
+\end{equation*}
+!et
+This suggests taking the first basis vector $\bm{p}_1$
+to be the gradient of $f$ at $\bm{x}=\bm{x}_0$,
+which equals
+!bt
+\begin{equation*}
+\bm{A}\bm{x}_0-\bm{b},
+\end{equation*}
+!et
+and
+$\bm{x}_0=0$ it is equal $-\bm{b}$.
+The other vectors in the basis will be conjugate to the gradient,
+hence the name conjugate gradient method.
+!eblock
+
+
+!split
+===== Conjugate gradient method =====
+!bblock
+Let $\bm{r}_k$ be the residual at the $k$-th step:
+!bt
+\begin{equation*}
+\bm{r}_k=\bm{b}-\bm{A}\bm{x}_k.
+\end{equation*}
+!et
+Note that $\bm{r}_k$ is the negative gradient of $f$ at
+$\bm{x}=\bm{x}_k$,
+so the gradient descent method would be to move in the direction $\bm{r}_k$.
+Here, we insist that the directions $\bm{p}_k$ are conjugate to each other,
+so we take the direction closest to the gradient $\bm{r}_k$
+under the conjugacy constraint.
+This gives the following expression
+!bt
+\begin{equation*}
+\bm{p}_{k+1}=\bm{r}_k-\frac{\bm{p}_k^T \bm{A}\bm{r}_k}{\bm{p}_k^T\bm{A}\bm{p}_k} \bm{p}_k.
+\end{equation*}
+!et
+!eblock
+
+!split
+===== Conjugate gradient method =====
+!bblock
+We can also compute the residual iteratively as
+!bt
+\begin{equation*}
+\bm{r}_{k+1}=\bm{b}-\bm{A}\bm{x}_{k+1},
+ \end{equation*}
+!et
+which equals
+!bt
+\begin{equation*}
+\bm{b}-\bm{A}(\bm{x}_k+\alpha_k\bm{p}_k),
+ \end{equation*}
+!et
+or
+!bt
+\begin{equation*}
+(\bm{b}-\bm{A}\bm{x}_k)-\alpha_k\bm{A}\bm{p}_k,
+ \end{equation*}
+!et
+which gives
+
+!bt
+\begin{equation*}
+\bm{r}_{k+1}=\bm{r}_k-\bm{A}\bm{p}_{k},
+ \end{equation*}
+!et
+!eblock
+!split
+===== Revisiting some of our first Linear Regression Encounters =====
+
+We will use linear regression as a case study for the gradient descent
+methods. Linear regression is a great test case for the gradient
+descent methods discussed in the lectures since it has several
+desirable properties such as:
+
+o An analytical solution (recall homework set 1).
+o The gradient can be computed analytically.
+o The cost function is convex which guarantees that gradient descent converges for small enough learning rates
+
+We revisit an example similar to what we had in the first homework set. We had a function of the type
+
+!bc pycod
+x = 2*np.random.rand(m,1)
+y = 4+3*x+np.random.randn(m,1)
+!ec
+with $x_i \in [0,1] $ is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution $\cal {N}(0,1)$.
+The linear regression model is given by
+!bt
+\[
+h_\beta(x) = \bm{y} = \beta_0 + \beta_1 x,
+\]
+!et
+such that
+!bt
+\[
+\bm{y}_i = \beta_0 + \beta_1 x_i.
+\]
+!et
+
+!split
+===== Gradient descent example =====
+
+Let $\mathbf{y} = (y_1,\cdots,y_n)^T$, $\mathbf{\bm{y}} = (\bm{y}_1,\cdots,\bm{y}_n)^T$ and $\beta = (\beta_0, \beta_1)^T$
+
+It is convenient to write $\mathbf{\bm{y}} = X\beta$ where $X \in \mathbb{R}^{100 \times 2} $ is the design matrix given by (we keep the intercept here)
+!bt
+\[
+X \equiv \begin{bmatrix}
+1 & x_1 \\
+\vdots & \vdots \\
+1 & x_{100} & \\
+\end{bmatrix}.
+\]
+!et
+The cost/loss/risk function is given by (
+!bt
+\[
+C(\beta) = \frac{1}{n}||X\beta-\mathbf{y}||_{2}^{2} = \frac{1}{n}\sum_{i=1}^{100}\left[ (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2\right]
+\]
+!et
+and we want to find $\beta$ such that $C(\beta)$ is minimized.
+
+!split
+===== The derivative of the cost/loss function =====
+
+Computing $\partial C(\beta) / \partial \beta_0$ and $\partial C(\beta) / \partial \beta_1$ we can show that the gradient can be written as
+!bt
+\[
+\nabla_{\beta} C(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} = \frac{2}{n}X^T(X\beta - \mathbf{y}),
+\]
+!et
+where $X$ is the design matrix defined above.
+
+!split
+===== The Hessian matrix =====
+The Hessian matrix of $C(\beta)$ is given by
+!bt
+\[
+\bm{H} \equiv \begin{bmatrix}
+\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
+\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
+\end{bmatrix} = \frac{2}{n}X^T X.
+\]
+!et
+This result implies that $C(\beta)$ is a convex function since the matrix $X^T X$ always is positive semi-definite.
+
+
+
+
+!split
+===== Simple program =====
+
+We can now write a program that minimizes $C(\beta)$ using the gradient descent method with a constant learning rate $\gamma$ according to
+!bt
+\[
+\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots
+\]
+!et
+
+We can use the expression we computed for the gradient and let use a
+$\beta_0$ be chosen randomly and let $\gamma = 0.001$. Stop iterating
+when $||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8}$. _Note that the code below does not include the latter stop criterion_.
+
+And finally we can compare our solution for $\beta$ with the analytic result given by
+$\beta= (X^TX)^{-1} X^T \mathbf{y}$.
+
+!split
+===== Gradient Descent Example =====
+
+Here our simple example
+!bc pycod
+
+# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+# Hessian matrix
+H = (2.0/n)* X.T @ X
+# Get the eigenvalues
+EigValues, EigVectors = np.linalg.eig(H)
+print(EigValues)
+
+beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y
+print(beta_linreg)
+beta = np.random.randn(2,1)
+
+eta = 1.0/np.max(EigValues)
+Niterations = 1000
+
+for iter in range(Niterations):
+ gradient = (2.0/n)*X.T @ (X @ beta-y)
+ beta -= eta*gradient
+
+print(beta)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(beta)
+ypredict2 = xbnew.dot(beta_linreg)
+plt.plot(xnew, ypredict, "r-")
+plt.plot(xnew, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example')
+plt.show()
+
+!ec
+
+!split
+===== And a corresponding example using _scikit-learn_ =====
+
+!bc pycod
+# Importing various packages
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import SGDRegressor
+
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)
+print(beta_linreg)
+sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
+sgdreg.fit(x,y.ravel())
+print(sgdreg.intercept_, sgdreg.coef_)
+
+!ec
+
+
+
+!split
+===== Gradient descent and Ridge =====
+
+We have also discussed Ridge regression where the loss function contains a regularized term given by the $L_2$ norm of $\beta$,
+!bt
+\[
+C_{\text{ridge}}(\beta) = \frac{1}{n}||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0.
+\]
+!et
+
+In order to minimize $C_{\text{ridge}}(\beta)$ using GD we only have adjust the gradient as follows
+!bt
+\[
+\nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
+\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
+\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta).
+\]
+!et
+
+We can easily extend our program to minimize $C_{\text{ridge}}(\beta)$ using gradient descent and compare with the analytical solution given by
+!bt
+\[
+\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}.
+\]
+!et
+
+
+!split
+===== Program example for gradient descent with Ridge Regression =====
+!bc pycod
+from random import random, seed
+import numpy as np
+import matplotlib.pyplot as plt
+from mpl_toolkits.mplot3d import Axes3D
+from matplotlib import cm
+from matplotlib.ticker import LinearLocator, FormatStrFormatter
+import sys
+
+# the number of datapoints
+n = 100
+x = 2*np.random.rand(n,1)
+y = 4+3*x+np.random.randn(n,1)
+
+X = np.c_[np.ones((n,1)), x]
+XT_X = X.T @ X
+
+#Ridge parameter lambda
+lmbda = 0.001
+Id = lmbda* np.eye(XT_X.shape[0])
+
+beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y
+print(beta_linreg)
+# Start plain gradient descent
+beta = np.random.randn(2,1)
+
+eta = 0.1
+Niterations = 100
+
+for iter in range(Niterations):
+ gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta
+ beta -= eta*gradients
+
+print(beta)
+ypredict = X @ beta
+ypredict2 = X @ beta_linreg
+plt.plot(x, ypredict, "r-")
+plt.plot(x, ypredict2, "b-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Gradient descent example for Ridge')
+plt.show()
+
+
+!ec
+
+!split
+===== Using gradient descent methods, limitations =====
+
+* _Gradient descent (GD) finds local minima of our function_. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our cost/loss/risk function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance.
+
+* _GD is sensitive to initial conditions_. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.
+
+* _Gradients are computationally expensive to calculate for large datasets_. In many cases in statistics and ML, the cost/loss/risk function is a sum of terms, with one term for each data point. For example, in linear regression, $E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2$; for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over *all* $n$ data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called ``mini batches''. This has the added benefit of introducing stochasticity into our algorithm.
+
+* _GD is very sensitive to choices of learning rates_. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would *adaptively* choose the learning rates to match the landscape.
+
+* _GD treats all directions in parameter space uniformly._ Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive.
+
+* GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. This leads us to our next topic, Stochastic Gradient Methods.
+