When you are comparing your own code with for example Scikit-Learn's library, there are some technicalities 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 tend confuse us.
The intercept can be interpreted as the expected value of our target/output variables when all other predictors are set to zero. Thus, if we cannot assume that the expected outputs/targets are zero when all predictors are zero (the columns in the design matrix), it may be a bad idea to implement a model which penalizes the intercept. Furthermore, in for example Ridge and Lasso regression, the default solutions from the library Scikit-Learn (when not shrinking \( \beta_0 \)) for the unknown parameters \( \boldsymbol{\beta} \), are derived under the assumption that both \( \boldsymbol{y} \) and \( \boldsymbol{X} \) are zero centered, that is we subtract the mean values.
If our predictors represent different scales, then it is important to standardize the design matrix \( \boldsymbol{X} \) by subtracting the mean of each column from the corresponding column and dividing the column with its standard deviation. Most machine learning libraries do this as a default. This means that if you compare your code with the results from a given library, the results may differ.
The Standardscaler function in Scikit-Learn does this for us. For the data sets we have been studying in our various examples, the data are in many cases already scaled and there is no need to scale them. You as a user of different machine learning algorithms, should always perform a survey of your data, with a critical assessment of them in case you need to scale the data.
If you need to scale the data, not doing so will give an unfair penalization of the parameters since their magnitude depends on the scale of their corresponding predictor.
The Scikit-Learn site https://scikit-learn.org/stable/auto_examples/preprocessing/plot_all_scaling.html#plot-all-scaling-standard-scaler-section has a good discussion of different ways of preprocessing data.
Suppose as an example that you you have an input variable given by the heights of different persons. 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. This can clearly lead to problems in evaluating the cost/loss functions.
Keep in mind that when you transform your data set before training a model, the same transformation needs to be done on your eventual new data set before making a prediction. If we translate this into a Python code, it would could be implemented as
"""
#Model training, we compute the mean value of y and X
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
# The we fit our model with the training data
trained_model = some_model.fit(X_train,y_train)
#Model prediction, we need also to transform our data set used for the 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
"""
Let us try to understand what this may imply mathematically when we subtract the mean values, also known as zero centering. For simplicity, we will focus on ordinary regression, as done in the above example.
The cost/loss function for regression is
$$ C(\beta_0, \beta_1, ... , \beta_{p-1}) = \frac{1}{n}\sum_{i=0}^{n} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij}\beta_j\right)^2,. $$Recall also that we use the squared value. This expression can lead to an increased penalty for higher differences between predicted and output/target values.
What we have done is to single out the \( \beta_0 \) term in the definition of the mean squared error (MSE). The design matrix \( X \) does in this case not contain any intercept column. When we take the derivative with respect to \( \beta_0 \), we want the derivative to obey
$$ \frac{\partial C}{\partial \beta_j} = 0, $$for all \( j \). For \( \beta_0 \) we have
$$ \frac{\partial C}{\partial \beta_0} = -\frac{2}{n}\sum_{i=0}^{n-1} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij} \beta_j\right). $$Multiplying away the constant \( 2/n \), we obtain
$$ \sum_{i=0}^{n-1} \beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} \sum_{j=1}^{p-1} X_{ij} \beta_j. $$Let us specialize first to the case where we have only two parameters \( \beta_0 \) and \( \beta_1 \). Our result for \( \beta_0 \) simplifies then to
$$ n\beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} X_{i1} \beta_1. $$We obtain then
$$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \beta_1\frac{1}{n}\sum_{i=0}^{n-1} X_{i1}. $$If we define
$$ \mu_{\boldsymbol{x}_1}=\frac{1}{n}\sum_{i=0}^{n-1} X_{i1}, $$and the mean value of the outputs as
$$ \mu_y=\frac{1}{n}\sum_{i=0}^{n-1}y_i, $$we have
$$ \beta_0 = \mu_y - \beta_1\mu_{\boldsymbol{x}_1}. $$In the general case with more parameters than \( \beta_0 \) and \( \beta_1 \), we have
$$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \frac{1}{n}\sum_{i=0}^{n-1}\sum_{j=1}^{p-1} X_{ij}\beta_j. $$We can rewrite the latter equation as
$$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \sum_{j=1}^{p-1} \mu_{\boldsymbol{x}_j}\beta_j, $$where we have defined
$$ \mu_{\boldsymbol{x}_j}=\frac{1}{n}\sum_{i=0}^{n-1} X_{ij}, $$the mean value for all elements of the column vector \( \boldsymbol{x}_j \).
Replacing \( y_i \) with \( y_i - y_i - \overline{\boldsymbol{y}} \) and centering also our design matrix results in a cost function (in vector-matrix disguise)
$$ C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}). $$If we minimize with respect to \( \boldsymbol{\beta} \) we have then
$$ \hat{\boldsymbol{\beta}} = (\tilde{X}^T\tilde{X})^{-1}\tilde{X}^T\boldsymbol{\tilde{y}}, $$where \( \boldsymbol{\tilde{y}} = \boldsymbol{y} - \overline{\boldsymbol{y}} \) and \( \tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=0}^{n-1}X_{kj} \).
For Ridge regression we need to add \( \lambda \boldsymbol{\beta}^T\boldsymbol{\beta} \) to the cost function and get then
$$ \hat{\boldsymbol{\beta}} = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}}. $$What does this mean? And why do we insist on all this? Let us look at some examples.
This code shows a simple first-order fit to a data set using the above transformed data, where we consider the role of the intercept first, by either excluding it or including it (code example thanks to Øyvind Sigmundson Schøyen). Here our scaling of the data is done by subtracting the mean values only. Note also that we do not split the data into training and test.
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
np.random.seed(2021)
def MSE(y_data,y_model):
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
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
skl = LinearRegression(fit_intercept=False).fit(X, y)
print(f"True beta: {true_beta}")
print(f"Fitted beta: {beta}")
print(f"Sklearn fitted beta: {skl.coef_}")
ypredictOwn = X @ beta
ypredictSKL = skl.predict(X)
print(f"MSE with intercept column")
print(MSE(y,ypredictOwn))
print(f"MSE with intercept column from SKL")
print(MSE(y,ypredictSKL))
plt.figure()
plt.scatter(x, y, label="Data")
plt.plot(x, X @ beta, label="Fit")
plt.plot(x, skl.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
skl = 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 (without intercept): {beta}")
print(f"Sklearn intercept: {skl.intercept_}")
print(f"Sklearn fitted beta (without intercept): {skl.coef_}")
ypredictOwn = X @ beta
ypredictSKL = skl.predict(X)
print(f"MSE with Manual intercept")
print(MSE(y,ypredictOwn+intercept))
print(f"MSE with Sklearn intercept")
print(MSE(y,ypredictSKL))
plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
plt.plot(x, skl.predict(X), "--", label="Sklearn (fit_intercept=True)")
plt.grid()
plt.legend()
plt.show()
The intercept is the value of our output/target variable when all our features are zero and our function crosses the \( y \)-axis (for a one-dimensional case).
Printing the MSE, we see first that both methods give the same MSE, as they should. However, when we move to for example Ridge regression, the way we treat the intercept may give a larger or smaller MSE, meaning that the MSE can be penalized by the value of the intercept. Not including the intercept in the fit, means that the regularization term does not include \( \beta_0 \). For different values of \( \lambda \), this may lead to different MSE values.
To remind the reader, the regularization term, with the intercept in Ridge regression, is given by
$$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_2^2 = \lambda \sum_{j=0}^{p-1}\beta_j^2, $$but when we take out the intercept, this equation becomes
$$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_2^2 = \lambda \sum_{j=1}^{p-1}\beta_j^2. $$For Lasso regression we have
$$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_1 = \lambda \sum_{j=1}^{p-1}\vert\beta_j\vert. $$It means that, when scaling the design matrix and the outputs/targets, by subtracting the mean values, we have an optimization problem which is not penalized by the intercept. The MSE value can then be smaller since it focuses only on the remaining quantities. If we however bring back the intercept, we will get a MSE which then contains the intercept.
Armed with this wisdom, we attempt first to simply set the intercept equal to False in our implementation of Ridge regression for our well-known vanilla data set.
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 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)
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))
#We include explicitely the intercept column
for degree in range(Maxpolydegree):
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)
p = Maxpolydegree
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 6
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 2, 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
# Note: we include the intercept column and no scaling
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)
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
print("Beta values for own Ridge implementation")
print(OwnRidgeBeta)
print("Beta values for Scikit-Learn Ridge implementation")
print(RegRidge.coef_)
print("MSE values for own Ridge implementation")
print(MSEOwnRidgePredict[i])
print("MSE values for Scikit-Learn Ridge implementation")
print(MSERidgePredict[i])
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE own Ridge Test')
plt.plot(np.log10(lambdas), MSERidgePredict, 'g', label = 'MSE Ridge Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
The results here agree when we force Scikit-Learn's Ridge function to include the first column in our design matrix. We see that the results agree very well. Here we have thus explicitely included the intercept column in the design matrix. What happens if we do not include the intercept in our fit? Let us see how we can change this code by zero centering.
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 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 = 20
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)
#Center by removing mean from each feature
X_train_scaled = X_train - X_train_mean
X_test_scaled = X_test - X_train_mean
#The model intercept (called y_scaler) is given by the mean of the target variable (IF X is centered)
#Remove the intercept from the training data.
y_scaler = np.mean(y_train)
y_train_scaled = y_train - y_scaler
p = Maxpolydegree-1
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 6
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 2, 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
#Add intercept to prediction
ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler
RegRidge = linear_model.Ridge(lmb)
RegRidge.fit(X_train,y_train)
ypredictRidge = RegRidge.predict(X_test)
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_)
print("MSE values for own Ridge implementation")
print(MSEOwnRidgePredict[i])
print("MSE values for Scikit-Learn Ridge implementation")
print(MSERidgePredict[i])
# 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()
We see here, when compared to the code which includes explicitely the intercept column, that our MSE value is actually smaller. This is because the regularization term does not include the intercept value \( \beta_0 \) in the fitting. This applies to Lasso regularization as well. It means that our optimization is now done only with the centered matrix and/or vector that enter the fitting procedure.
Hitherto we have discussed Ridge and Lasso regression in terms of a linear analysis. This may to many of you feel rather technical and perhaps not that intuitive. The question is whether we can develop a more intuitive way of understanding what Ridge and Lasso express.
Before we proceed let us perform a Ridge, Lasso and OLS analysis of a polynomial fit.
We will play around with a study of the values for the optimal parameters \( \boldsymbol{\beta} \) using OLS, Ridge and Lasso regression. For OLS, you will notice as function of the noise and polynomial degree, that the parameters \( \beta \) will fluctuate from order to order in the polynomial fit and that for larger and larger polynomial degrees of freedom, the parameters will tend to increase in value for OLS.
For Ridge and Lasso regression, the higher order parameters will typically be reduced, providing thereby less fluctuations from one order to another one.
import numpy as np
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
# Make data set.
n = 10000
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
Maxpolydegree = 5
X = np.zeros((len(x),Maxpolydegree))
X[:,0] = 1.0
for polydegree in range(1,Maxpolydegree):
X[:,polydegree] = x**(polydegree)
# 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)
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 using Scikit-Learn
# Decide which values of lambda to use
nlambdas = 4
MSERidgePredict = np.zeros(nlambdas)
MSELassoPredict = np.zeros(nlambdas)
lambdas = np.logspace(-3, 1, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
# Make the fit using Ridge and Lasso
RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
RegRidge.fit(X_train,y_train)
RegLasso = linear_model.Lasso(lmb,fit_intercept=False)
RegLasso.fit(X_train,y_train)
# and then make the prediction
ypredictRidge = RegRidge.predict(X_test)
ypredictLasso = RegLasso.predict(X_test)
# Compute the MSE and print it
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
MSELassoPredict[i] = MSE(y_test,ypredictLasso)
print(lmb,RegRidge.coef_)
print(lmb,RegLasso.coef_)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSERidgePredict, 'b', label = 'MSE Ridge Test')
plt.plot(np.log10(lambdas), MSELassoPredict, 'r', label = 'MSE Lasso Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
How can we understand this?
We will now couple the discussions of ordinary least squares, Ridge and Lasso regression with a statistical interpretation, that is we move from a linear algebra analysis to a statistical analysis. In particular, we will focus on what the regularization terms can result in. We will amongst other things show that the regularization parameter can reduce considerably the variance of the parameters \( \beta \).
The advantage of doing linear regression is that we actually end up with analytical expressions for several statistical quantities. Standard least squares and Ridge regression allow us to derive quantities like the variance and other expectation values in a rather straightforward way.
It is assumed that \( \varepsilon_i \sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are independent, i.e.:
$$ \begin{align*} \mbox{Cov}(\varepsilon_{i_1}, \varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} & i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. \end{align*} $$The randomness of \( \varepsilon_i \) implies that \( \mathbf{y}_i \) is also a random variable. In particular, \( \mathbf{y}_i \) is normally distributed, because \( \varepsilon_i \sim \mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \boldsymbol{\beta} \) is a non-random scalar. To specify the parameters of the distribution of \( \mathbf{y}_i \) we need to calculate its first two moments.
Recall that \( \boldsymbol{X} \) is a matrix of dimensionality \( n\times p \). The notation above \( \mathbf{X}_{i,\ast} \) means that we are looking at the row number \( i \) and perform a sum over all values \( p \).
The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) that there exists a function \( f(\boldsymbol{x}) \) and a normal distributed error \( \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) \) which describe our data
$$ \boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon} $$We approximate this function with our model from the solution of the linear regression equations, that is our function \( f \) is approximated by \( \boldsymbol{\tilde{y}} \) where we want to minimize \( (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2 \), our MSE, with
$$ \boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}. $$We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \)
$$ \begin{align*} \mathbb{E}(y_i) & = \mathbb{E}(\mathbf{X}_{i, \ast} \, \boldsymbol{\beta}) + \mathbb{E}(\varepsilon_i) \, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, \end{align*} $$while its variance is
$$ \begin{align*} \mbox{Var}(y_i) & = \mathbb{E} \{ [y_i - \mathbb{E}(y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( y_i^2 ) - [\mathbb{E}(y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \varepsilon_i \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, \mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. \end{align*} $$Hence, \( y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2) \), that is \( \boldsymbol{y} \) follows a normal distribution with mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (not be confused with the singular values of the SVD).
With the OLS expressions for the optimal parameters \( \boldsymbol{\hat{\beta}} \) we can evaluate the expectation value
$$ \mathbb{E}(\boldsymbol{\hat{\beta}}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}. $$This means that the estimator of the regression parameters is unbiased.
We can also calculate the variance
The variance of the optimal value \( \boldsymbol{\hat{\beta}} \) is
$$ \begin{eqnarray*} \mbox{Var}(\boldsymbol{\hat{\beta}}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \} \\ & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \} \\ % & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} % \\ % & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} % \\ & = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \\ & = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} % \\ % & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} % \\ % & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T \\ & = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1}, \end{eqnarray*} $$where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the variance of the estimate of the \( j \)-th regression coefficient: \( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \). This may be used to construct a confidence interval for the estimates.
In a similar way, we can obtain analytical expressions for say the expectation values of the parameters \( \boldsymbol{\beta} \) and their variance when we employ Ridge regression, allowing us again to define a confidence interval.
It is rather straightforward to show that
$$ \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}. $$We see clearly that \( \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} \) for any \( \lambda > 0 \). We say then that the ridge estimator is biased.
We can also compute the variance as
$$ \mbox{Var}[\boldsymbol{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, $$and it is easy to see that if the parameter \( \lambda \) goes to infinity then the variance of Ridge parameters \( \boldsymbol{\beta} \) goes to zero.
With this, we can compute the difference
$$ \mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. $$The difference is non-negative definite since each component of the matrix product is non-negative definite. This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.
Our basic assumption when we derived the OLS equations was to assume that our output is determined by a given continuous function \( f(\boldsymbol{x}) \) and a random noise \( \boldsymbol{\epsilon} \) given by the normal distribution with zero mean value and an undetermined variance \( \sigma^2 \).
We found above that the outputs \( \boldsymbol{y} \) have a mean value given by \( \boldsymbol{X}\hat{\boldsymbol{\beta}} \) and variance \( \sigma^2 \). Since the entries to the design matrix are not stochastic variables, we can assume that the probability distribution of our targets is also a normal distribution but now with mean value \( \boldsymbol{X}\hat{\boldsymbol{\beta}} \). This means that a single output \( y_i \) is given by the Gaussian distribution
$$ y_i\sim \mathcal{N}(\boldsymbol{X}_{i,*}\boldsymbol{\beta}, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$We assume now that the various \( y_i \) values are stochastically distributed according to the above Gaussian distribution. We define this distribution as
$$ p(y_i, \boldsymbol{X}\vert\boldsymbol{\beta})=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}, $$which reads as finding the likelihood of an event \( y_i \) with the input variables \( \boldsymbol{X} \) given the parameters (to be determined) \( \boldsymbol{\beta} \).
Since these events are assumed to be independent and identicall distributed we can build the probability distribution function (PDF) for all possible event \( \boldsymbol{y} \) as the product of the single events, that is we have
$$ p(\boldsymbol{y},\boldsymbol{X}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}=\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta}). $$We will write this in a more compact form reserving \( \boldsymbol{D} \) for the domain of events, including the ouputs (targets) and the inputs. That is in case we have a simple one-dimensional input and output case
$$ \boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})]. $$In the more general case the various inputs should be replaced by the possible features represented by the input data set \( \boldsymbol{X} \). We can now rewrite the above probability as
$$ p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$It is a conditional probability (see below) and reads as the likelihood of a domain of events \( \boldsymbol{D} \) given a set of parameters \( \boldsymbol{\beta} \).
In statistics, maximum likelihood estimation (MLE) is a method of estimating the parameters of an assumed probability distribution, given some observed data. This is achieved by maximizing a likelihood function so that, under the assumed statistical model, the observed data is the most probable.
We will assume here that our events are given by the above Gaussian distribution and we will determine the optimal parameters \( \beta \) by maximizing the above PDF. However, computing the derivatives of a product function is cumbersome and can easily lead to overflow and/or underflowproblems, with potentials for loss of numerical precision.
In practice, it is more convenient to maximize the logarithm of the PDF because it is a monotonically increasing function of the argument. Alternatively, and this will be our option, we will minimize the negative of the logarithm since this is a monotonically decreasing function.
Note also that maximization/minimization of the logarithm of the PDF is equivalent to the maximization/minimization of the function itself.
We could now define a new cost function to minimize, namely the negative logarithm of the above PDF
$$ C(\boldsymbol{\beta}=-\log{\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}=-\sum_{i=0}^{n-1}\log{p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}, $$which becomes
$$ C(\boldsymbol{\beta}=\frac{n}{2}\log{2\pi\sigma^2}+\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}. $$Taking the derivative of the new cost function with respect to the parameters \( \beta \) we recognize our familiar OLS equation, namely
$$ \boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right) =0, $$which leads to the well-known OLS equation for the optimal paramters \( \beta \)
$$ \hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}! $$Before we make a similar analysis for Ridge and Lasso regression, we need a short reminder on statistics.
A central theorem in statistics is Bayes' theorem. This theorem plays a similar role as the good old Pythagoras' theorem in geometry. Bayes' theorem is extremely simple to derive. But to do so we need some basic axioms from statistics.
Assume we have two domains of events \( X=[x_0,x_1,\dots,x_{n-1}] \) and \( Y=[y_0,y_1,\dots,y_{n-1}] \).
We define also the likelihood for \( X \) and \( Y \) as \( p(X) \) and \( p(Y) \) respectively. The likelihood of a specific event \( x_i \) (or \( y_i \)) is then written as \( p(X=x_i) \) or just \( p(x_i)=p_i \).
$$ p(X \cup Y)= p(X)+p(Y)-p(X \cap Y). $$
$$ p(X \cup Y)= p(X,Y)= p(X\vert Y)p(Y)=p(Y\vert X)p(X), $$
where we read \( p(X\vert Y) \) as the likelihood of obtaining \( X \) given \( Y \).
If we have independent events then \( p(X,Y)=p(X)p(Y) \).
The marginal probability is defined in terms of only one of the set of variables \( X,Y \). For a discrete probability we have
$$ p(X)=\sum_{i=0}^{n-1}p(X,Y=y_i)=\sum_{i=0}^{n-1}p(X\vert Y=y_i)p(Y=y_i)=\sum_{i=0}^{n-1}p(X\vert y_i)p(y_i). $$
The conditional probability, if \( p(Y) > 0 \), is
$$ p(X\vert Y)= \frac{p(X,Y)}{p(Y)}=\frac{p(X,Y)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}. $$
If we combine the conditional probability with the marginal probability and the standard product rule, we have
$$ p(X\vert Y)= \frac{p(X,Y)}{p(Y)}, $$which we can rewrite as
$$ p(X\vert Y)= \frac{p(X,Y)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}=\frac{p(Y\vert X)p(X)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}, $$which is Bayes' theorem. It allows us to evaluate the uncertainty in in \( X \) after we have observed \( Y \). We can easily interchange \( X \) with \( Y \).
The quantity \( p(Y\vert X) \) on the right-hand side of the theorem is evaluated for the observed data \( Y \) and can be viewed as a function of the parameter space represented by \( X \). This function is not necesseraly normalized and is normally called the likelihood function.
The function \( p(X) \) on the right hand side is called the prior while the function on the left hand side is the called the posterior probability. The denominator on the right hand side serves as a normalization factor for the posterior distribution.
Let us try to illustrate Bayes' theorem through an example.
Let us suppose that you are undergoing a series of mammography scans in order to rule out possible breast cancer cases. We define the sensitivity for a positive event by the variable \( X \). It takes binary values with \( X=1 \) representing a positive event and \( X=0 \) being a negative event. We reserve \( Y \) as a classification parameter for either a negative or a positive breast cancer confirmation. (Short note on wordings: positive here means having breast cancer, although none of us would consider this being a positive thing).
We let \( Y=1 \) represent the the case of having breast cancer and \( Y=0 \) as not.
Let us assume that if you have breast cancer, the test will be positive with a probability of \( 0.8 \), that is we have
$$ p(X=1\vert Y=1) =0.8. $$This obviously sounds scary since many would conclude that if the test is positive, there is a likelihood of \( 80\% \) for having cancer. It is however not correct, as the following Bayesian analysis shows.
If we look at various national surveys on breast cancer, the general likelihood of developing breast cancer is a very small number. Let us assume that the prior probability in the population as a whole is
$$ p(Y=1) =0.004. $$We need also to account for the fact that the test may produce a false positive result (false alarm). Let us here assume that we have
$$ p(X=1\vert Y=0) =0.1. $$Using Bayes' theorem we can then find the posterior probability that the person has breast cancer in case of a positive test, that is we can compute
$$ p(Y=1\vert X=1)=\frac{p(X=1\vert Y=1)p(Y=1)}{p(X=1\vert Y=1)p(Y=1)+p(X=1\vert Y=0)p(Y=0)}=\frac{0.8\times 0.004}{0.8\times 0.004+0.1\times 0.996}=0.031. $$That is, in case of a positive test, there is only a \( 3\% \) chance of having breast cancer!
Using Bayes' theorem we can gain a better intuition about Ridge and Lasso regression.
For ordinary least squares we postulated that the maximum likelihood for the doamin of events \( \boldsymbol{D} \) (one-dimensional case)
$$ \boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})], $$is given by
$$ p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$In Bayes' theorem this function plays the role of the so-called likelihood. We could now ask the question what is the posterior probability of a parameter set \( \boldsymbol{\beta} \) given a domain of events \( \boldsymbol{D} \)? That is, how can we define the posterior probability
$$ p(\boldsymbol{\beta}\vert\boldsymbol{D}). $$Bayes' theorem comes to our rescue here since (omitting the normalization constant)
$$ p(\boldsymbol{\beta}\vert\boldsymbol{D})\propto p(\boldsymbol{D}\vert\boldsymbol{\beta})p(\boldsymbol{\beta}). $$We have a model for \( p(\boldsymbol{D}\vert\boldsymbol{\beta}) \) but need one for the prior \( p(\boldsymbol{\beta} \)!
With the posterior probability defined by a likelihood which we have already modeled and an unknown prior, we are now ready to make additional models for the prior.
We can, based on our discussions of the variance of \( \boldsymbol{\beta} \) and the mean value, assume that the prior for the values \( \boldsymbol{\beta} \) is given by a Gaussian with mean value zero and variance \( \tau^2 \), that is
$$ p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$Our posterior probability becomes then (omitting the normalization factor which is just a constant)
$$ p(\boldsymbol{\beta\vert\boldsymbol{D})}=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$We can now optimize this quantity with respect to \( \boldsymbol{\beta} \). As we did for OLS, this is most conveniently done by taking the negative logarithm of the posterior probability. Doing so and leaving out the constants terms that do not depend on \( \beta \), we have
$$ C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{2\tau^2}\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$and replacing \( 1/2\tau^2 \) with \( \lambda \) we have
$$ C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$which is our Ridge cost function! Nice, isn't it?
To derive the Lasso cost function, we simply replace the Gaussian prior with an exponential distribution (Laplace in this case) with zero mean value, that is
$$ p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. $$Our posterior probability becomes then (omitting the normalization factor which is just a constant)
$$ p(\boldsymbol{\beta}\vert\boldsymbol{D})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. $$Taking the negative logarithm of the posterior probability and leaving out the constants terms that do not depend on \( \beta \), we have
$$ C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{\tau}\vert\vert\boldsymbol{\beta}\vert\vert_1, $$and replacing \( 1/\tau \) with \( \lambda \) we have
$$ C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1, $$which is our Lasso cost function!
The material here contains a summary from last week 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.
If we now recall the definition of the covariance matrix (not using Bessel's correction) we have
$$ \boldsymbol{C}[\boldsymbol{X}]=\frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}, $$meaning that every squared non-singular value of \( \boldsymbol{X} \) divided by \( n \) ( the number of samples) are the eigenvalues of the covariance matrix. Every singular value of \( \boldsymbol{X} \) is thus a positive square root of an eigenvalue of \( \boldsymbol{X}^T\boldsymbol{X} \). If the matrix \( \boldsymbol{X} \) is self-adjoint, the singular values of \( \boldsymbol{X} \) are equal to the absolute value of the eigenvalues of \( \boldsymbol{X} \).
For \( \boldsymbol{X}\boldsymbol{X}^T \) we found
$$ \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T=\boldsymbol{U}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{U}^T. $$Since the matrices here have dimension \( n\times n \), we have
$$ \boldsymbol{\Sigma}\boldsymbol{\Sigma}^T = \begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \\ \boldsymbol{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \boldsymbol{0}\\ \end{bmatrix}=\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}, $$leading to
$$ \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}\boldsymbol{U}^T. $$Multiplying with \( \boldsymbol{U} \) from the right gives us the eigenvalue problem
$$ (\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U}=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}. $$It means that the eigenvalues of \( \boldsymbol{X}\boldsymbol{X}^T \) are again given by the non-zero singular values plus now a series of zeros. The column vectors of \( \boldsymbol{U} \) are the eigenvectors of \( \boldsymbol{X}\boldsymbol{X}^T \) and measure how much correlations are contained in the rows of \( \boldsymbol{X} \).
Since we will mainly be interested in the correlations among the features of our data (the columns of \( \boldsymbol{X} \), the quantity of interest for us are the non-zero singular values and the column vectors of \( \boldsymbol{V} \).
How do we use the SVD to invert a matrix \( \boldsymbol{X}^\boldsymbol{X} \) which is singular or near singular? The simple answer is to use the linear algebra function for the computation of the pseudoinverse of a given matrix \( \boldsymbol{X} \), that is
import numpy as np
X = np.array( [ [1,2,3],[2,4,5],[3,5,6]])
Xinv = np.linlag.pinv(X)
Let us first look at a matrix which does not causes problems and write our own function where we just use the SVD.
import numpy as np
# SVD inversion
def SVDinv(A):
''' Takes as input a numpy matrix A and returns inv(A) based on singular value decomposition (SVD).
SVD is numerically more stable than the inversion algorithms provided by
numpy and scipy.linalg at the cost of being slower.
'''
U, s, VT = np.linalg.svd(A)
print('test U')
print( (np.transpose(U) @ U - U @np.transpose(U)))
print('test VT')
print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))
D = np.zeros((len(U),len(VT)))
D = np.diag(s)
UT = np.transpose(U); V = np.transpose(VT); invD = np.linalg.inv(D)
return np.matmul(V,np.matmul(invD,UT))
#X = np.array([ [1.0, -1.0, 2.0], [1.0, 0.0, 1.0], [1.0, 2.0, -1.0], [1.0, 1.0, 0.0] ])
# Non-singular square matrix
X = np.array( [ [1,2,3],[2,4,5],[3,5,6]])
print(X)
A = np.transpose(X) @ X
# Brute force inversion
B = np.linalg.inv(A) # here we could use np.linalg.pinv(A)
C = SVDinv(A)
print(np.abs(B-C))
Although our matrix to invert \( \boldsymbol{X}^T\boldsymbol{X} \) is a square matrix, our matrix may be singular.
The pseudoinverse is the generalization of the matrix inverse for square matrices to rectangular matrices where the number of rows and columns are not equal.
It is also called the the Moore-Penrose Inverse after two independent discoverers of the method or the Generalized Inverse. It is used for the calculation of the inverse for singular or near singular matrices and for rectangular matrices.
Using the SVD we can obtain the pseudoinverse of a matrix \( \boldsymbol{A} \) (labeled here as \( \boldsymbol{A}_{\mathrm{PI}} \))
$$ \boldsymbol{A}_{\mathrm{PI}}= \boldsymbol{V}\boldsymbol{D}_{\mathrm{PI}}\boldsymbol{U}^T, $$where \( \boldsymbol{D}_{\mathrm{PI}} \) can be calculated by creating a diagonal matrix from \( \boldsymbol{\Sigma} \) where we only keep the singular values (the non-zero values). The following code computes the pseudoinvers of the matrix based on the SVD.
import numpy as np
# SVD inversion
def SVDinv(A):
U, s, VT = np.linalg.svd(A)
# reciprocals of singular values of s
d = 1.0 / s
# create m x n D matrix
D = np.zeros(A.shape)
# populate D with n x n diagonal matrix
D[:A.shape[1], :A.shape[1]] = np.diag(d)
UT = np.transpose(U)
V = np.transpose(VT)
return np.matmul(V,np.matmul(D.T,UT))
A = np.array([ [0.3, 0.4], [0.5, 0.6], [0.7, 0.8],[0.9, 1.0]])
print(A)
# Brute force inversion of super-collinear matrix
B = np.linalg.pinv(A)
print(B)
# Compare our own algorithm with pinv
C = SVDinv(A)
print(np.abs(C-B))
As you can see from this example, our own decomposition based on the SVD agrees with the pseudoinverse algorithm provided by Numpy.
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.
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()