The material here contains a summary of the lecture on Monday and discussion of SVD, Ridge and Lasso regression with examples
We used the SVD to analyse the matrix to invert in ordinary lineat regression
$$ \boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T=\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T. $$Since the matrices here have dimension \( p\times p \), with \( p \) corresponding to the singular values, we defined last week the matrix
$$ \boldsymbol{\Sigma}^T\boldsymbol{\Sigma} = \begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \\ \boldsymbol{0}\end{bmatrix}, $$where the tilde-matrix \( \tilde{\boldsymbol{\Sigma}} \) is a matrix of dimension \( p\times p \) containing only the singular values \( \sigma_i \), that is
$$ \tilde{\boldsymbol{\Sigma}}=\begin{bmatrix} \sigma_0 & 0 & 0 & \dots & 0 & 0 \\ 0 & \sigma_1 & 0 & \dots & 0 & 0 \\ 0 & 0 & \sigma_2 & \dots & 0 & 0 \\ 0 & 0 & 0 & \dots & \sigma_{p-2} & 0 \\ 0 & 0 & 0 & \dots & 0 & \sigma_{p-1} \\ \end{bmatrix}, $$meaning we can write
$$ \boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{V}\tilde{\boldsymbol{\Sigma}}^2\boldsymbol{V}^T. $$Multiplying from the right with \( \boldsymbol{V} \) (using the orthogonality of \( \boldsymbol{V} \)) we get
$$ \left(\boldsymbol{X}^T\boldsymbol{X}\right)\boldsymbol{V}=\boldsymbol{V}\tilde{\boldsymbol{\Sigma}}^2. $$This means the vectors \( \boldsymbol{v}_i \) of the orthogonal matrix \( \boldsymbol{V} \) are the eigenvectors of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) with eigenvalues given by the singular values squared, that is
$$ \left(\boldsymbol{X}^T\boldsymbol{X}\right)\boldsymbol{v}_i=\boldsymbol{v}_i\sigma_i^2. $$In other words, each non-zero singular value of \( \boldsymbol{X} \) is a positive square root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). It means also that the columns of \( \boldsymbol{V} \) are the eigenvectors of \( \boldsymbol{X}^T\boldsymbol{X} \). Since we have ordered the singular values of \( \boldsymbol{X} \) in a descending order, it means that the column vectors \( \boldsymbol{v}_i \) are hierarchically ordered by how much correlation they encode from the columns of \( \boldsymbol{X} \).
Note that these are also the eigenvectors and eigenvalues of the Hessian matrix.
Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is our optimization problem is
$$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. $$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. We do not include such a constraints in the discussions here.
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
$$ \vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. $$Using the matrix-vector expression for Ridge regression and dropping the parameter \( 1/n \) in front of the standard means squared error equation, we have
$$ C(\boldsymbol{X},\boldsymbol{\beta})=\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta}, $$and taking the derivatives with respect to \( \boldsymbol{\beta} \) we obtain then a slightly modified matrix inversion problem which for finite values of \( \lambda \) does not suffer from singularity problems. We obtain the optimal parameters
$$ \hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}, $$with \( \boldsymbol{I} \) being a \( p\times p \) identity matrix with the constraint that
$$ \sum_{i=0}^{p-1} \beta_i^2 \leq t, $$with \( t \) a finite positive number.
Note well that a library like Scikit-Learn does not include the \( 1/n \) factor in the expression for the mean-squared error. If you include it, the optimal parameter \( \beta \) becomes
$$ \hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+n\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$In our codes where we compare our own codes with Scikit-Learn, we do thus not include the \( 1/n \) factor in the cost function.
When we compare this with the ordinary least squares result we have
$$ \hat{\boldsymbol{\beta}}_{\mathrm{OLS}} = \left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}, $$which can lead to singular matrices. However, with the SVD, we can always compute the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \).
We see that Ridge regression is nothing but the standard OLS with a modified diagonal term added to \( \boldsymbol{X}^T\boldsymbol{X} \). The consequences, in particular for our discussion of the bias-variance tradeoff are rather interesting. We will see that for specific values of \( \lambda \), we may even reduce the variance of the optimal parameters \( \boldsymbol{\beta} \). These topics and other related ones, will be discussed after the more linear algebra oriented analysis here.
Using our insights about the SVD of the design matrix \( \boldsymbol{X} \) We have already analyzed the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as
$$ \tilde{\boldsymbol{y}}_{\mathrm{OLS}}=\boldsymbol{X}\boldsymbol{\beta} =\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}. $$For Ridge regression this becomes
$$ \tilde{\boldsymbol{y}}_{\mathrm{Ridge}}=\boldsymbol{X}\boldsymbol{\beta}_{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{\Sigma}^2\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y}, $$with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \) from the SVD of the matrix \( \boldsymbol{X} \).
Since \( \lambda \geq 0 \), it means that compared to OLS, we have
$$ \frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1. $$Ridge regression finds the coordinates of \( \boldsymbol{y} \) with respect to the orthonormal basis \( \boldsymbol{U} \), it then shrinks the coordinates by \( \frac{\sigma_j^2}{\sigma_j^2+\lambda} \). Recall that the SVD has eigenvalues ordered in a descending way, that is \( \sigma_i \geq \sigma_{i+1} \).
For small eigenvalues \( \sigma_i \) it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. More about this when we have covered the material on a statistical interpretation of various linear regression methods.
For the sake of simplicity, let us assume that the design matrix is orthonormal, that is
$$ \boldsymbol{X}^T\boldsymbol{X}=(\boldsymbol{X}^T\boldsymbol{X})^{-1} =\boldsymbol{I}. $$In this case the standard OLS results in
$$ \boldsymbol{\beta}^{\mathrm{OLS}} = \boldsymbol{X}^T\boldsymbol{y}=\sum_{i=0}^{n-1}\boldsymbol{u}_i\boldsymbol{u}_i^T\boldsymbol{y}, $$and
$$ \boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{I}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\left(1+\lambda\right)^{-1}\boldsymbol{\beta}^{\mathrm{OLS}}, $$that is the Ridge estimator scales the OLS estimator by the inverse of a factor \( 1+\lambda \), and the Ridge estimator converges to zero when the hyperparameter goes to infinity.
We will come back to more interpreations after we have gone through some of the statistical analysis part.
For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. Similarly, Mehta et al's article is also recommended.
Using the matrix-vector expression for Lasso regression, we have the following cost function
$$ C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1, $$Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicity)
$$ \frac{d \vert \beta\vert}{d \beta}=\mathrm{sgn}(\beta)=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right. $$we have that the derivative of the cost function is
$$ \frac{\partial C(\boldsymbol{X},\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=-\frac{2}{n}\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})+\lambda sgn(\boldsymbol{\beta})=0, $$and reordering we have
$$ \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}+\lambda sgn(\boldsymbol{\beta})=\boldsymbol{X}^T\boldsymbol{y}. $$This equation does not lead to a nice analytical equation as in Ridge regression or ordinary least squares. We have absorbed the factor \( 2/n \) in a redefinition of the parameter \( \lambda \). We will solve this type of problems using libraries like scikit-learn and using our own gradient descent code in project 1.
Let us assume that our design matrix is given by unit (identity) matrix, that is a square diagonal matrix with ones only along the diagonal. In this case we have an equal number of rows and columns \( n=p \).
Our model approximation is just \( \tilde{\boldsymbol{y}}=\boldsymbol{\beta} \) and the mean squared error and thereby the cost function for ordinary least sqquares (OLS) is then (we drop the term \( 1/n \))
$$ C(\boldsymbol{\beta})=\sum_{i=0}^{p-1}(y_i-\beta_i)^2, $$and minimizing we have that
$$ \hat{\beta}_i^{\mathrm{OLS}} = y_i. $$For Ridge regression our cost function is
$$ C(\boldsymbol{\beta})=\sum_{i=0}^{p-1}(y_i-\beta_i)^2+\lambda\sum_{i=0}^{p-1}\beta_i^2, $$and minimizing we have that
$$ \hat{\beta}_i^{\mathrm{Ridge}} = \frac{y_i}{1+\lambda}. $$For Lasso regression our cost function is
$$ C(\boldsymbol{\beta})=\sum_{i=0}^{p-1}(y_i-\beta_i)^2+\lambda\sum_{i=0}^{p-1}\vert\beta_i\vert=\sum_{i=0}^{p-1}(y_i-\beta_i)^2+\lambda\sum_{i=0}^{p-1}\sqrt{\beta_i^2}, $$and minimizing we have that
$$ -2\sum_{i=0}^{p-1}(y_i-\beta_i)+\lambda \sum_{i=0}^{p-1}\frac{(\beta_i)}{\vert\beta_i\vert}=0, $$which leads to
$$ \hat{\boldsymbol{\beta}}_i^{\mathrm{Lasso}} = \left\{\begin{array}{ccc}y_i-\frac{\lambda}{2} &\mathrm{if} & y_i> \frac{\lambda}{2}\\ y_i+\frac{\lambda}{2} &\mathrm{if} & y_i < -\frac{\lambda}{2}\\ 0 &\mathrm{if} & \vert y_i\vert\le \frac{\lambda}{2}\end{array}\right.\\. $$Plotting these results shows clearly that Lasso regression suppresses (sets to zero) values of \( \beta_i \) for specific values of \( \lambda \). Ridge regression reduces on the other hand the values of \( \beta_i \) as function of \( \lambda \).
Let us assume we have a data set with outputs/targets given by the vector
$$ \boldsymbol{y}=\begin{bmatrix}4 \\ 2 \\3\end{bmatrix}, $$and our inputs as a \( 3\times 2 \) design matrix
$$ \boldsymbol{X}=\begin{bmatrix}2 & 0\\ 0 & 1 \\ 0 & 0\end{bmatrix}, $$meaning that we have two features and two unknown parameters \( \beta_0 \) and \( \beta_1 \) to be determined either by ordinary least squares, Ridge or Lasso regression.
For ordinary least squares (OLS) we know that the optimal solution is
$$ \hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\left( \boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$Inserting the above values we obtain that
$$ \hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\begin{bmatrix}2 \\ 2\end{bmatrix}, $$The code which implements this simpler case is presented after the discussion of Ridge and Lasso.
For Ridge regression we have
$$ \hat{\boldsymbol{\beta}}^{\mathrm{Ridge}}=\left( \boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$Inserting the above values we obtain that
$$ \hat{\boldsymbol{\beta}}^{\mathrm{Ridge}}=\begin{bmatrix}\frac{8}{4+\lambda} \\ \frac{2}{1+\lambda}\end{bmatrix}, $$There is normally a constraint on the value of \( \vert\vert \boldsymbol{\beta}\vert\vert_2 \) via the parameter \( \lambda \). Let us for simplicity assume that \( \beta_0^2+\beta_1^2=1 \) as constraint. This will allow us to find an expression for the optimal values of \( \beta \) and \( \lambda \).
To see this, let us write the cost function for Ridge regression.
We define the MSE without the \( 1/n \) factor and have then, using that
$$ \boldsymbol{X}\boldsymbol{\beta}=\begin{bmatrix} 2\beta_0 \\ \beta_1 \\0 \end{bmatrix}, $$ $$ C(\boldsymbol{\beta})=(4-2\beta_0)^2+(2-\beta_1)^2+\lambda(\beta_0^2+\beta_1^2), $$and taking the derivative with respect to \( \beta_0 \) we get
$$ \beta_0=\frac{8}{4+\lambda}, $$and for \( \beta_1 \) we obtain
$$ \beta_1=\frac{2}{1+\lambda}, $$Using the constraint for \( \beta_0^2+\beta_1^2=1 \) we can constrain \( \lambda \) by solving
$$ \left(\frac{8}{4+\lambda}\right)^2+\left(\frac{2}{1+\lambda}\right)^2=1, $$which gives \( \lambda=4.571 \) and \( \beta_0=0.933 \) and \( \beta_1=0.359 \).
For Lasso we need now, keeping a constraint on \( \vert\beta_0\vert+\vert\beta_1\vert=1 \), to take the derivative of the absolute values of \( \beta_0 \) and \( \beta_1 \). This gives us the following derivatives of the cost function
$$ C(\boldsymbol{\beta})=(4-2\beta_0)^2+(2-\beta_1)^2+\lambda(\vert\beta_0\vert+\vert\beta_1\vert), $$ $$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_0}=-4(4-2\beta_0)+\lambda\mathrm{sgn}(\beta_0)=0, $$and
$$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_1}=-2(2-\beta_1)+\lambda\mathrm{sgn}(\beta_1)=0. $$We have now four cases to solve besides the trivial cases \( \beta_0 \) and/or \( \beta_1 \) are zero, namely
If we consider the first case, we have then
$$ -4(4-2\beta_0)+\lambda=0, $$and
$$ -2(2-\beta_1)+\lambda=0. $$which yields
$$ \beta_0=\frac{16+\lambda}{8}, $$and
$$ \beta_1=\frac{4+\lambda}{2}. $$Using the constraint on \( \beta_0 \) and \( \beta_1 \) we can then find the optimal value of \( \lambda \) for the different cases. We leave this as an exercise to you.
Here we set up the OLS, Ridge and Lasso functionality in order to study the above example. Note that here we have opted for a set of values of \( \lambda \), meaning that we need to perform a search in order to find the optimal values.
First we study and compare the OLS and Ridge results. The next code compares all three methods.
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
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.
X = np.array( [ [ 2, 0], [0, 1], [0,0]])
y = np.array( [4, 2, 3])
# matrix inversion to find beta
OLSbeta = np.linalg.inv(X.T @ X) @ X.T @ y
print(OLSbeta)
# and then make the prediction
ytildeOLS = X @ OLSbeta
print("Training MSE for OLS")
print(MSE(y,ytildeOLS))
ypredictOLS = X @ OLSbeta
# Repeat now for Ridge regression and various values of the regularization parameter
I = np.eye(2,2)
# Decide which values of lambda to use
nlambdas = 100
MSEPredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 4, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
Ridgebeta = np.linalg.inv(X.T @ X+lmb*I) @ X.T @ y
# print(Ridgebeta)
# and then make the prediction
ypredictRidge = X @ Ridgebeta
MSEPredict[i] = MSE(y,ypredictRidge)
# print(MSEPredict[i])
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSEPredict, 'r--', label = 'MSE Ridge Train')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
We see here that we reach a plateau. What is actually happening?
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
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
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
X = np.array( [ [ 2, 0], [0, 1], [0,0]])
y = np.array( [4, 2, 3])
# matrix inversion to find beta
OLSbeta = np.linalg.inv(X.T @ X) @ X.T @ y
print(OLSbeta)
# and then make the prediction
ytildeOLS = X @ OLSbeta
print("Training MSE for OLS")
print(MSE(y,ytildeOLS))
ypredictOLS = X @ OLSbeta
# Repeat now for Ridge regression and various values of the regularization parameter
I = np.eye(2,2)
# Decide which values of lambda to use
nlambdas = 100
MSERidgePredict = np.zeros(nlambdas)
MSELassoPredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 4, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
Ridgebeta = np.linalg.inv(X.T @ X+lmb*I) @ X.T @ y
print(Ridgebeta)
# and then make the prediction
ypredictRidge = X @ Ridgebeta
MSERidgePredict[i] = MSE(y,ypredictRidge)
RegLasso = linear_model.Lasso(lmb,fit_intercept=False)
RegLasso.fit(X,y)
ypredictLasso = RegLasso.predict(X)
print(RegLasso.coef_)
MSELassoPredict[i] = MSE(y,ypredictLasso)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSERidgePredict, 'r--', label = 'MSE Ridge Train')
plt.plot(np.log10(lambdas), MSELassoPredict, 'r--', label = 'MSE Lasso Train')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
import os
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
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)
x = np.random.rand(100)
y = 2.0+5*x*x+0.1*np.random.randn(100)
# number of features p (here degree of polynomial
p = 3
# The design matrix now as function of a given polynomial
X = np.zeros((len(x),p))
X[:,0] = 1.0
X[:,1] = x
X[:,2] = x*x
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
# matrix inversion to find beta
OLSbeta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(OLSbeta)
# and then make the prediction
ytildeOLS = X_train @ OLSbeta
print("Training MSE for OLS")
print(MSE(y_train,ytildeOLS))
ypredictOLS = X_test @ OLSbeta
print("Test MSE OLS")
print(MSE(y_test,ypredictOLS))
# Repeat now for Lasso and Ridge regression and various values of the regularization parameter
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 100
MSEPredict = np.zeros(nlambdas)
MSETrain = np.zeros(nlambdas)
MSELassoPredict = np.zeros(nlambdas)
MSELassoTrain = np.zeros(nlambdas)
lambdas = np.logspace(-4, 4, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
Ridgebeta = np.linalg.inv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
# include lasso using Scikit-Learn
RegLasso = linear_model.Lasso(lmb,fit_intercept=False)
RegLasso.fit(X_train,y_train)
# and then make the prediction
ytildeRidge = X_train @ Ridgebeta
ypredictRidge = X_test @ Ridgebeta
ytildeLasso = RegLasso.predict(X_train)
ypredictLasso = RegLasso.predict(X_test)
MSEPredict[i] = MSE(y_test,ypredictRidge)
MSETrain[i] = MSE(y_train,ytildeRidge)
MSELassoPredict[i] = MSE(y_test,ypredictLasso)
MSELassoTrain[i] = MSE(y_train,ytildeLasso)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSETrain, label = 'MSE Ridge train')
plt.plot(np.log10(lambdas), MSEPredict, 'r--', label = 'MSE Ridge Test')
plt.plot(np.log10(lambdas), MSELassoTrain, label = 'MSE Lasso train')
plt.plot(np.log10(lambdas), MSELassoPredict, 'r--', label = 'MSE Lasso Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()