The main topics on Thursday are:
We need first a reminder from last week about linear regression.
Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
Consider an experiment in which \( p \) characteristics of \( n \) samples are measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix \( \mathbf{X} \).
The matrix \( \mathbf{X} \) is called the design matrix. Additional information of the samples is available in the form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is generally referred to as the response variable. The aim of regression analysis is to explain \( \boldsymbol{y} \) in terms of \( \boldsymbol{X} \) through a functional relationship like \( y_i = f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of \( f(\cdot) \) is available, it is common to assume a linear relationship between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots, \beta_{p-1}]^{T} \) are the regression parameters.
Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \), consider the model we discussed for describing nuclear binding energies.
There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. Assuming $$ BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, $$ we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms. This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a \( p\times n \) matrix \( \boldsymbol{X} \).
Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression.
Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is $$ y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, $$ where \( \epsilon_i \) is the error in our approximation.
For every set of values \( y_i,x_i \) we have thus the corresponding set of equations $$ \begin{align*} y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ \dots & \dots \\ y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ \end{align*} $$
Defining the vectors $$ \boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$ and $$ \boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, $$ and $$ \boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, $$ and the design matrix $$ \boldsymbol{X}= \begin{bmatrix} 1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ 1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ 1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ 1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ \end{bmatrix} $$ we can rewrite our equations as $$ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. $$ The above design matrix is called a Vandermonde matrix.
We are obviously not limited to the above polynomial expansions. We could replace the various powers of \( x \) with elements of Fourier series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. For every set of values \( y_i,x_i \) we can then generalize the equations to $$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$
Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
We redefine in turn the matrix \( \boldsymbol{X} \) as $$ \boldsymbol{X}= \begin{bmatrix} x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ \end{bmatrix} $$ and without loss of generality we rewrite again our equations as $$ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. $$ The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
We have defined the matrix \( \boldsymbol{X} \) via the equations $$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$
As we noted above, we stayed with a system with the design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements.
In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
We restate the parts of the code we are most interested in.
# Common imports
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from IPython.display import display
import os
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
os.makedirs(DATA_ID)
def image_path(fig_id):
return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("MassEval2016.dat"),'r')
# Read the experimental data with Pandas
Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
names=('N', 'Z', 'A', 'Element', 'Ebinding'),
widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
header=39,
index_col=False)
# Extrapolated values are indicated by '#' in place of the decimal place, so
# the Ebinding column won't be numeric. Coerce to float and drop these entries.
Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
Masses = Masses.dropna()
# Convert from keV to MeV.
Masses['Ebinding'] /= 1000
# Group the DataFrame by nucleon number, A.
Masses = Masses.groupby('A')
# Find the rows of the grouped DataFrame with the maximum binding energy.
Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
A = Masses['A']
Z = Masses['Z']
N = Masses['N']
Element = Masses['Element']
Energies = Masses['Ebinding']
# Now we set up the design matrix X
X = np.zeros((len(A),5))
X[:,0] = 1
X[:,1] = A
X[:,2] = A**(2.0/3.0)
X[:,3] = A**(-1.0/3.0)
X[:,4] = A**(-1.0)
# Then nice printout using pandas
DesignMatrix = pd.DataFrame(X)
DesignMatrix.index = A
DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
display(DesignMatrix)
With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as $$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$ throughout these lectures.
With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as $$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$ and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely $$ C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, $$ or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as $$ C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. $$ This function is one possible way to define the so-called cost function.
It is also common to define the function \( C \) as $$ C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, $$ since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
The function $$ C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, $$ can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value. When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value $$ y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, $$
where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable.
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem $$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. $$ In practical terms it means we will require $$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, $$ which results in $$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, $$ or in a matrix-vector form as $$ \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). $$
We can rewrite $$ \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), $$ as $$ \boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, $$ and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution $$ \boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$
We note also that since our design matrix is defined as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in {\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \), in our case \( p=5 \) meaning that we end up with inverting a small \( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional matrices to invert. The methods discussed here and for many other supervised learning algorithms like classification with logistic regression or support vector machines, exhibit dimensionalities which allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \).
Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and matrices as upper case boldfaced letters. $$ \frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b}, $$ $$ \frac{\partial (\boldsymbol{a}^T\boldsymbol{A}\boldsymbol{a})}{\partial \boldsymbol{a}} = (\boldsymbol{A}+\boldsymbol{A}^T)\boldsymbol{a}, $$ $$ \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T, $$ $$ \frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T. $$
A very important matrix we will meet again and again in Machine Learning is the Hessian. It is given by the second derivative of the cost function with respect to the parameter \( \beta \). Using the above expression for derivatives of vectors and matrices, we find that the second derivative of the cost function is, $$ \frac{\partial}{\partial \boldsymbol{\beta}^T}\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} =\frac{\partial}{\partial \boldsymbol{\beta}^T}\left[-\frac{2}{n}\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right]=\frac{2}{n}\boldsymbol{X}^T\boldsymbol{X}. $$
The Hessian matrix plays an important role and is defined here as $$ \boldsymbol{H}=\boldsymbol{X}^T\boldsymbol{X}. $$
For ordinary least squares, it is inversely proportional (derivation next week) with the variance of the optimal parameters \( \hat{\boldsymbol{\beta}} \). Furthermore, we will see later this week that is (beside \( 1/n \)) equal to the covariance matrix. It plays also a very important role in optmization algorithms and Principal Component Analysis as a way to reduce the dimensionality of a machine learning problem.
Linear algebra question: Can we use the Hessian matrix to say something about properties of the cost function (our optmization problem)? (hint: think about convex or concave problems and how to relate these to a matrix!).
The residuals \( \boldsymbol{\epsilon} \) are in turn given by $$ \boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}, $$ and with $$ \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, $$ we have $$ \boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, $$ meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to write
# matrix inversion to find beta
beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
# and then make the prediction
ytilde = X @ beta
Alternatively, you can use the least squares functionality in Numpy as
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
ytildenp = np.dot(fit,X.T)
And finally we plot our fit with and compare with data
Masses['Eapprox'] = ytilde
# Generate a plot comparing the experimental with the fitted values values.
fig, ax = plt.subplots()
ax.set_xlabel(r'$A = N + Z$')
ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
label='Ame2016')
ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
label='Fit')
ax.legend()
save_fig("Masses2016OLS")
plt.show()
We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of Scikit-Learn in the introductory slides. Since we are not using Scikit-Learn here we can define our own \( R2 \) function as
def R2(y_data, y_model):
return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
and we would be using it as
print(R2(Energies,ytilde))
We can easily add our MSE score as
def MSE(y_data,y_model):
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
print(MSE(Energies,ytilde))
and finally the relative error as
def RelativeError(y_data,y_model):
return abs((y_data-y_model)/y_data)
print(RelativeError(Energies, ytilde))
It is normal in essentially all Machine Learning studies to split the data in a training set and a test set (sometimes also an additional validation set). Scikit-Learn has an own function for this. There is no explicit recipe for how much data should be included as training data and say test data. An accepted rule of thumb is to use approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will postpone a discussion of this splitting to the end of these notes and our discussion of the so-called bias-variance tradeoff. Here we limit ourselves to repeat the above equation of state fitting example but now splitting the data into a training set and a test set.
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
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
x = np.random.rand(100)
y = 2.0+5*x*x+0.1*np.random.randn(100)
# The design matrix now as function of a given polynomial
X = np.zeros((len(x),3))
X[:,0] = 1.0
X[:,1] = x
X[:,2] = x**2
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
# matrix inversion to find beta
beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(beta)
# and then make the prediction
ytilde = X_train @ beta
print("Training R2")
print(R2(y_train,ytilde))
print("Training MSE")
print(MSE(y_train,ytilde))
ypredict = X_test @ beta
print("Test R2")
print(R2(y_test,ypredict))
print("Test MSE")
print(MSE(y_test,ypredict))
# equivalently in numpy
def train_test_split_numpy(inputs, labels, train_size, test_size):
n_inputs = len(inputs)
inputs_shuffled = inputs.copy()
labels_shuffled = labels.copy()
np.random.shuffle(inputs_shuffled)
np.random.shuffle(labels_shuffled)
train_end = int(n_inputs*train_size)
X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]
Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]
return X_train, X_test, Y_train, Y_test
But since scikit-learn has its own function for doing this and since it interfaces easily with tensorflow and other libraries, we normally recommend using the latter functionality.
The Boston housing data set was originally a part of UCI Machine Learning Repository and has been removed now. The data set is now included in Scikit-Learn's library. There are 506 samples and 13 feature (predictor) variables in this data set. The objective is to predict the value of prices of the house using the features (predictors) listed here.
The features/predictors are
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
and load the Boston Housing DataSet from Scikit-Learn
from sklearn.datasets import load_boston
boston_dataset = load_boston()
# boston_dataset is a dictionary
# let's check what it contains
boston_dataset.keys()
Then we invoke Pandas
boston = pd.DataFrame(boston_dataset.data, columns=boston_dataset.feature_names)
boston.head()
boston['MEDV'] = boston_dataset.target
and preprocess the data
# check for missing values in all the columns
boston.isnull().sum()
We can then visualize the data
# set the size of the figure
sns.set(rc={'figure.figsize':(11.7,8.27)})
# plot a histogram showing the distribution of the target values
sns.distplot(boston['MEDV'], bins=30)
plt.show()
It is now useful to look at the correlation matrix
# compute the pair wise correlation for all columns
correlation_matrix = boston.corr().round(2)
# use the heatmap function from seaborn to plot the correlation matrix
# annot = True to print the values inside the square
sns.heatmap(data=correlation_matrix, annot=True)
From the above coorelation plot we can see that MEDV is strongly correlated to LSTAT and RM. We see also that RAD and TAX are stronly correlated, but we don't include this in our features together to avoid multi-colinearity
plt.figure(figsize=(20, 5))
features = ['LSTAT', 'RM']
target = boston['MEDV']
for i, col in enumerate(features):
plt.subplot(1, len(features) , i+1)
x = boston[col]
y = target
plt.scatter(x, y, marker='o')
plt.title(col)
plt.xlabel(col)
plt.ylabel('MEDV')
Now we start training our model
X = pd.DataFrame(np.c_[boston['LSTAT'], boston['RM']], columns = ['LSTAT','RM'])
Y = boston['MEDV']
We split the data into training and test sets
from sklearn.model_selection import train_test_split
# splits the training and test data set in 80% : 20%
# assign random_state to any value.This ensures consistency.
X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state=5)
print(X_train.shape)
print(X_test.shape)
print(Y_train.shape)
print(Y_test.shape)
Then we use the linear regression functionality from Scikit-Learn
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error, r2_score
lin_model = LinearRegression()
lin_model.fit(X_train, Y_train)
# model evaluation for training set
y_train_predict = lin_model.predict(X_train)
rmse = (np.sqrt(mean_squared_error(Y_train, y_train_predict)))
r2 = r2_score(Y_train, y_train_predict)
print("The model performance for training set")
print("--------------------------------------")
print('RMSE is {}'.format(rmse))
print('R2 score is {}'.format(r2))
print("\n")
# model evaluation for testing set
y_test_predict = lin_model.predict(X_test)
# root mean square error of the model
rmse = (np.sqrt(mean_squared_error(Y_test, y_test_predict)))
# r-squared score of the model
r2 = r2_score(Y_test, y_test_predict)
print("The model performance for testing set")
print("--------------------------------------")
print('RMSE is {}'.format(rmse))
print('R2 score is {}'.format(r2))
# plotting the y_test vs y_pred
# ideally should have been a straight line
plt.scatter(Y_test, y_test_predict)
plt.show()
Many Machine Learning problems involve thousands or even millions of features for each training instance. Not only does this make training extremely slow, it can also make it much harder to find a good solution, as we will see. This problem is often referred to as the curse of dimensionality. Fortunately, in real-world problems, it is often possible to reduce the number of features considerably, turning an intractable problem into a tractable one.
Later we will discuss some of the most popular dimensionality reduction techniques: the principal component analysis (PCA), Kernel PCA, and Locally Linear Embedding (LLE).
Principal component analysis and its various variants deal with the problem of fitting a low-dimensional affine subspace to a set of of data points in a high-dimensional space. With its family of methods it is one of the most used tools in data modeling, compression and visualization.
Before we proceed however, we will discuss how to preprocess our data. Till now and in connection with our previous examples we have not met so many cases where we are too sensitive to the scaling of our data. Normally the data may need a rescaling and/or may be sensitive to extreme values. Scaling the data renders our inputs much more suitable for the algorithms we want to employ.
For data sets gathered for real world applications, it is rather normal that different features have very different units and numerical scales. For example, a data set detailing health habits may include features such as age in the range \( 0-80 \), and caloric intake of order \( 2000 \). Many machine learning methods sensitive to the scales of the features and may perform poorly if they are very different scales. Therefore, it is typical to scale the features in a way to avoid such outlier values.
Scikit-Learn has several functions which allow us to rescale the data, normally resulting in much better results in terms of various accuracy scores. The StandardScaler function in Scikit-Learn ensures that for each feature/predictor we study the mean value is zero and the variance is one (every column in the design/feature matrix). This scaling has the drawback that it does not ensure that we have a particular maximum or minimum in our data set. Another function included in Scikit-Learn is the MinMaxScaler which ensures that all features are exactly between \( 0 \) and \( 1 \). The
The Normalizer scales each data point such that the feature vector has a euclidean length of one. In other words, it projects a data point on the circle (or sphere in the case of higher dimensions) with a radius of 1. This means every data point is scaled by a different number (by the inverse of it’s length). This normalization is often used when only the direction (or angle) of the data matters, not the length of the feature vector.
The RobustScaler works similarly to the StandardScaler in that it ensures statistical properties for each feature that guarantee that they are on the same scale. However, the RobustScaler uses the median and quartiles, instead of mean and variance. This makes the RobustScaler ignore data points that are very different from the rest (like measurement errors). These odd data points are also called outliers, and might often lead to trouble for other scaling techniques.
Many features are often scaled using standardization to improve performance. In Scikit-Learn this is given by the StandardScaler function as discussed above. It is easy however to write your own. Mathematically, this involves subtracting the mean and divide by the standard deviation over the data set, for each feature: $$ x_j^{(i)} \rightarrow \frac{x_j^{(i)} - \overline{x}_j}{\sigma(x_j)}, $$ where \( \overline{x}_j \) and \( \sigma(x_j) \) are the mean and standard deviation, respectively, of the feature \( x_j \). This ensures that each feature has zero mean and unit standard deviation. For data sets where we do not have the standard deviation or don't wish to calculate it, it is then common to simply set it to one.
Let us consider the following vanilla example where we use both Scikit-Learn and write our own function as well. We produce a simple test design matrix with random numbers. Each column could then represent a specific feature whose mean value is subracted.
import sklearn.linear_model as skl
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import MinMaxScaler, StandardScaler, Normalizer
import numpy as np
import pandas as pd
from IPython.display import display
np.random.seed(100)
# setting up a 10 x 5 matrix
rows = 10
cols = 5
X = np.random.randn(rows,cols)
XPandas = pd.DataFrame(X)
display(XPandas)
print(XPandas.mean())
print(XPandas.std())
XPandas = (XPandas -XPandas.mean())
display(XPandas)
scaler = StandardScaler(with_std=False)
scaler.fit(X)
Xscaled = scaler.transform(X)
display(XPandas-Xscaled)
Small exercise: perform the standard scaling by including the standard deviation.
Another commonly used scaling method is min-max scaling. This is very useful for when we want the features to lie in a certain interval. To scale the feature \( x_j \) to the interval \( [a, b] \), we can apply the transformation $$ x_j^{(i)} \rightarrow (b-a)\frac{x_j^{(i)} - \min(x_j)}{\max(x_j) - \min(x_j)} - a $$ where \( \min(x_j) \) and \( \max(x_j) \) return the minimum and maximum value of \( x_j \) over the data set, respectively.
Our data is defined by \( x\in [-3,3] \) with a total of for example \( 100 \) data points.
np.random.seed()
n = 100
maxdegree = 14
# Make data set.
x = np.linspace(-3, 3, n).reshape(-1, 1)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
where \( y \) is the function we want to fit with a given polynomial.
Write a first code which sets up a design matrix \( X \) defined by a fifth-order polynomial. Scale your data and split it in training and test data.
import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.preprocessing import PolynomialFeatures
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline
np.random.seed(2018)
n = 50
maxdegree = 5
# Make data set.
x = np.linspace(-3, 3, n).reshape(-1, 1)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
TestError = np.zeros(maxdegree)
TrainError = np.zeros(maxdegree)
polydegree = np.zeros(maxdegree)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
scaler = StandardScaler()
scaler.fit(x_train)
x_train_scaled = scaler.transform(x_train)
x_test_scaled = scaler.transform(x_test)
for degree in range(maxdegree):
model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
clf = model.fit(x_train_scaled,y_train)
y_fit = clf.predict(x_train_scaled)
y_pred = clf.predict(x_test_scaled)
polydegree[degree] = degree
TestError[degree] = np.mean( np.mean((y_test - y_pred)**2) )
TrainError[degree] = np.mean( np.mean((y_train - y_fit)**2) )
plt.plot(polydegree, TestError, label='Test Error')
plt.plot(polydegree, TrainError, label='Train Error')
plt.legend()
plt.show()
# Common imports
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import sklearn.linear_model as skl
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import MinMaxScaler, StandardScaler, Normalizer
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
os.makedirs(DATA_ID)
def image_path(fig_id):
return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
plt.savefig(image_path(fig_id) + ".png", format='png')
def FrankeFunction(x,y):
term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
return term1 + term2 + term3 + term4
def create_X(x, y, n ):
if len(x.shape) > 1:
x = np.ravel(x)
y = np.ravel(y)
N = len(x)
l = int((n+1)*(n+2)/2) # Number of elements in beta
X = np.ones((N,l))
for i in range(1,n+1):
q = int((i)*(i+1)/2)
for k in range(i+1):
X[:,q+k] = (x**(i-k))*(y**k)
return X
# Making meshgrid of datapoints and compute Franke's function
n = 5
N = 1000
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)
# split in training and test data
X_train, X_test, y_train, y_test = train_test_split(X,z,test_size=0.2)
clf = skl.LinearRegression().fit(X_train, y_train)
# The mean squared error and R2 score
print("MSE before scaling: {:.2f}".format(mean_squared_error(clf.predict(X_test), y_test)))
print("R2 score before scaling {:.2f}".format(clf.score(X_test,y_test)))
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
print("Feature min values before scaling:\n {}".format(X_train.min(axis=0)))
print("Feature max values before scaling:\n {}".format(X_train.max(axis=0)))
print("Feature min values after scaling:\n {}".format(X_train_scaled.min(axis=0)))
print("Feature max values after scaling:\n {}".format(X_train_scaled.max(axis=0)))
clf = skl.LinearRegression().fit(X_train_scaled, y_train)
print("MSE after scaling: {:.2f}".format(mean_squared_error(clf.predict(X_test_scaled), y_test)))
print("R2 score for scaled data: {:.2f}".format(clf.score(X_test_scaled,y_test)))
What is presented here is a mathematical analysis of various regression algorithms (ordinary least squares, Ridge and Lasso Regression). The analysis is based on an important algorithm in linear algebra, the so-called Singular Value Decomposition (SVD).
We have shown that in ordinary least squares that the optimal parameter \( \beta \) are given by $$ \hat{\boldsymbol{\beta}} = \left(\boldsymbol{X}^T\boldsymbol{X}\right^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$
This means that our best model is defined as $$ \tilde{\boldsymbol{y}}=\boldsymbol{X}\hat{\boldsymbol{\beta}} = \boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$
We define now a matrix $$ \boldsymbol{A}=\boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right^{-1}\boldsymbol{X}^T. $$
It means that we can rewrite $$ \tilde{\boldsymbol{y}}=\boldsymbol{X}\hat{\boldsymbol{\beta}} = \boldsymbol{A}\boldsymbol{y}. $$
The matrix \( \boldsymbol{A} \) has the important property that \( \boldsymbol{A}^2=\boldsymbol{A} \). This is the definition of a projection matrix. We can then interpret that our optimal model \( \tilde{\boldsymbol{y}} \) ir represented by an orthogonal (it has to be a square matrix like ours, if not we have an oblique projection matrix ) projection of \( \boldsymbol{y} \) onto a space defined by the column vectors of \( \boldsymbol{X} \).
We have defined the residual error as $$ \boldsymbol{\epsilon}=\boldsymbol{y}-\tilde{\boldsymbol{y}}=(\boldsymbol{I}-\boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right^{-1}\boldsymbol{X}^T)\boldsymbol{y}. $$
The residual errors are then the projections of \( \boldsymbol{y} \) onto the orthogonal components of the space defined by the column vectors of \( \boldsymbol{X} \).
If the matrix \( \boldsymbol{X} \) is an orthogonal (or unitary in case of complex values) matrix, we have $$ \boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{X}\boldsymbol{X}^T = \boldsymbol{I}. $$
In this case the matrix \( \boldsymbol{A} \) becomes $$ \boldsymbol{A}=\boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right^{-1}\boldsymbol{X}^T)=\boldsymbol{I}, $$ and we have the obvious case $$ \boldsymbol{\epsilon}=\boldsymbol{y}-\tilde{\boldsymbol{y}}=0. $$
The examples we have looked at so far are cases where we normally can invert the matrix \( \boldsymbol{X}^T\boldsymbol{X} \). Using a polynomial expansion as we did both for the masses and the fitting of various functions leads to row vectors of the design matrix which are essentially orthogonal due to the polynomial character of our model. Obtaining the inverse of the design matrix is then often done via a so-called LU, QR or Cholesky decomposition.
As we will also see in the first project, this may however not the be case in general and a standard matrix inversion algorithm based on say LU, QR or Cholesky decomposition may lead to singularities. We will see examples of this below.
There is however a way to partially circumvent this problem and also gain some insights about the ordinary least squares approach, and later shrinkage methods like Ridge and Lasso regressions.
This is given by the Singular Value Decomposition algorithm, perhaps the most powerful linear algebra algorithm. Let us look at a different example where we may have problems with the standard matrix inversion algorithm. Thereafter we dive into the math of the SVD.
One of the typical problems we encounter with linear regression, in particular when the matrix \( \boldsymbol{X} \) (our so-called design matrix) is high-dimensional, are problems with near singular or singular matrices. The column vectors of \( \boldsymbol{X} \) may be linearly dependent, normally referred to as super-collinearity. This means that the matrix may be rank deficient and it is basically impossible to to model the data using linear regression. As an example, consider the matrix $$ \begin{align*} \mathbf{X} & = \left[ \begin{array}{rrr} 1 & -1 & 2 \\ 1 & 0 & 1 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \end{array} \right] \end{align*} $$
The columns of \( \boldsymbol{X} \) are linearly dependent. We see this easily since the the first column is the row-wise sum of the other two columns. The rank (more correct, the column rank) of a matrix is the dimension of the space spanned by the column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number of linearly independent columns. In this particular case the matrix has rank 2.
Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies that the inverse of the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) (the matrix we need to invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this $$ \begin{align*} \boldsymbol{X} & = \left[ \begin{array}{rr} 1 & -1 \\ 1 & -1 \end{array} \right]. \end{align*} $$ We see easily that \( \mbox{det}(\boldsymbol{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. This is equivalent to saying that the matrix \( \boldsymbol{X} \) has at least an eigenvalue which is zero.
If our design matrix \( \boldsymbol{X} \) which enters the linear regression problem $$ \begin{align} \boldsymbol{\beta} & = (\boldsymbol{X}^{T} \boldsymbol{X})^{-1} \boldsymbol{X}^{T} \boldsymbol{y}, \label{_auto1} \end{align} $$ has linearly dependent column vectors, we will not be able to compute the inverse of \( \boldsymbol{X}^T\boldsymbol{X} \) and we cannot find the parameters (estimators) \( \beta_i \). The estimators are only well-defined if \( (\boldsymbol{X}^{T}\boldsymbol{X})^{-1} \) exits. This is more likely to happen when the matrix \( \boldsymbol{X} \) is high-dimensional. In this case it is likely to encounter a situation where the regression parameters \( \beta_i \) cannot be estimated.
A cheap ad hoc approach is simply to add a small diagonal component to the matrix to invert, that is we change $$ \boldsymbol{X}^{T} \boldsymbol{X} \rightarrow \boldsymbol{X}^{T} \boldsymbol{X}+\lambda \boldsymbol{I}, $$ where \( \boldsymbol{I} \) is the identity matrix. When we discuss Ridge regression this is actually what we end up evaluating. The parameter \( \lambda \) is called a hyperparameter. More about this later.
From standard linear algebra we know that a square matrix \( \boldsymbol{X} \) can be diagonalized if and only it is a so-called normal matrix, that is if \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \) we have \( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) or if \( \boldsymbol{X}\in {\mathbb{C}}^{n\times n} \) we have \( \boldsymbol{X}\boldsymbol{X}^{\dagger}=\boldsymbol{X}^{\dagger}\boldsymbol{X} \). The matrix has then a set of eigenpairs $$ (\lambda_1,\boldsymbol{u}_1),\dots, (\lambda_n,\boldsymbol{u}_n), $$ and the eigenvalues are given by the diagonal matrix $$ \boldsymbol{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n). $$ The matrix \( \boldsymbol{X} \) can be written in terms of an orthogonal/unitary transformation \( \boldsymbol{U} \) $$ \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, $$ with \( \boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) or \( \boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I} \).
Not all square matrices are diagonalizable. A matrix like the one discussed above $$ \boldsymbol{X} = \begin{bmatrix} 1& -1 \\ 1& -1\\ \end{bmatrix} $$ is not diagonalizable, it is a so-called defective matrix. It is easy to see that the condition \( \boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X} \) is not fulfilled.
However, and this is the strength of the SVD algorithm, any general matrix \( \boldsymbol{X} \) can be decomposed in terms of a diagonal matrix and two orthogonal/unitary matrices. The Singular Value Decompostion (SVD) theorem states that a general \( m\times n \) matrix \( \boldsymbol{X} \) can be written in terms of a diagonal matrix \( \boldsymbol{\Sigma} \) of dimensionality \( m\times n \) and two orthognal matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \), where the first has dimensionality \( m \times m \) and the last dimensionality \( n\times n \). We have then $$ \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T $$
As an example, the above defective matrix can be decomposed as $$ \boldsymbol{X} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1& 1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 0\\ \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -1 \\ 1& 1\\ \end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, $$
with eigenvalues \( \sigma_1=2 \) and \( \sigma_2=0 \). The SVD exits always!
The SVD decomposition (singular values) gives eigenvalues \( \sigma_i\geq\sigma_{i+1} \) for all \( i \) and for dimensions larger than \( i=p \), the eigenvalues (singular values) are zero.
In the general case, where our design matrix \( \boldsymbol{X} \) has dimension \( n\times p \), the matrix is thus decomposed into an \( n\times n \) orthogonal matrix \( \boldsymbol{U} \), a \( p\times p \) orthogonal matrix \( \boldsymbol{V} \) and a diagonal matrix \( \boldsymbol{\Sigma} \) with \( r=\mathrm{min}(n,p) \) singular values \( \sigma_i\geq 0 \) on the main diagonal and zeros filling the rest of the matrix. There are at most \( p \) singular values assuming that \( n > p \). In our regression examples for the nuclear masses and the equation of state this is indeed the case, while for the Ising model we have \( p > n \). These are often cases that lead to near singular or singular matrices.
The columns of \( \boldsymbol{U} \) are called the left singular vectors while the columns of \( \boldsymbol{V} \) are the right singular vectors.
If we assume that \( n > p \), then our matrix \( \boldsymbol{U} \) has dimension \( n \times n \). The last \( n-p \) columns of \( \boldsymbol{U} \) become however irrelevant in our calculations since they are multiplied with the zeros in \( \boldsymbol{\Sigma} \).
The economy-size decomposition removes extra rows or columns of zeros from the diagonal matrix of singular values, \( \boldsymbol{\Sigma} \), along with the columns in either \( \boldsymbol{U} \) or \( \boldsymbol{V} \) that multiply those zeros in the expression. Removing these zeros and columns can improve execution time and reduce storage requirements without compromising the accuracy of the decomposition.
If \( n > p \), we keep only the first \( p \) columns of \( \boldsymbol{U} \) and \( \boldsymbol{\Sigma} \) has dimension \( p\times p \). If \( p > n \), then only the first \( n \) columns of \( \boldsymbol{V} \) are computed and \( \boldsymbol{\Sigma} \) has dimension \( n\times n \). The \( n=p \) case is obvious, we retain the full SVD. In general the economy-size SVD leads to less FLOPS and still conserving the desired accuracy.
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,full_matrices=True)
print('test U')
print( (np.transpose(U) @ U - U @np.transpose(U)))
print('test VT')
print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))
print(U)
print(S)
print(VT)
D = np.zeros((len(U),len(VT)))
for i in range(0,len(VT)):
D[i,i]=S[i]
return U @ D @ VT
X = np.array([ [1.0,-1.0], [1.0,-1.0]])
#X = np.array([[1, 2], [3, 4], [5, 6]])
print(X)
C = SVDinv(X)
# Print the difference between the original matrix and the SVD one
print(C-X)
The matrix \( \boldsymbol{X} \) has columns that are linearly dependent. The first column is the row-wise sum of the other two columns. The rank of a matrix (the column rank) is the dimension of space spanned by the column vectors. The rank of the matrix is the number of linearly independent columns, in this case just \( 2 \). We see this from the singular values when running the above code. Running the standard inversion algorithm for matrix inversion with \( \boldsymbol{X}^T\boldsymbol{X} \) results in the program terminating due to a singular matrix.
The \( U \), \( S \), and \( V \) matrices returned from the svd() function cannot be multiplied directly.
As you can see from the code, the \( S \) vector must be converted into a diagonal matrix. This may cause a as the size of the matrices do not fit the rules of matrix multiplication, where the number of columns in a matrix must match the number of rows in the subsequent matrix.
If you wish to include the zero singular values, you will need to resize the matrices. More about this later.
There are several interesting mathematical properties which will be relevant when we are going to discuss the differences between say ordinary least squares (OLS) and Ridge regression.
We have from OLS that the parameters of the linear approximation are given by $$ \boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$
The matrix to invert can be rewritten in terms of our SVD decomposition as $$ \boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. $$ Using the orthogonality properties of \( \boldsymbol{U} \) we have $$ \boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T = \boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T, $$ with \( \boldsymbol{D} \) being a diagonal matrix with values along the diagonal given by the singular values squared.
This means that $$ (\boldsymbol{X}^T\boldsymbol{X})\boldsymbol{V} = \boldsymbol{V}\boldsymbol{D}, $$ that is the eigenvectors of \( (\boldsymbol{X}^T\boldsymbol{X}) \) are given by the columns of the right singular matrix of \( \boldsymbol{X} \) and the eigenvalues are the squared singular values. It is easy to show (show this) that $$ (\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}, $$ that is, the eigenvectors of \( (\boldsymbol{X}\boldsymbol{X})^T \) are the columns of the left singular matrix and the eigenvalues are the same.
Going back to our OLS equation we have $$ \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}. $$ We will come back to this expression when we discuss Ridge regression. $$ \tilde{y}^{OLS}={\bf X}\hat{\beta}^{OLS}=\sum_{j=1}^p {\bf u}_j{\bf u}_j^T{\bf y}$$ and for Ridge we have $$ \tilde{y}^{Ridge}={\bf X}\hat{\beta}^{Ridge}=\sum_{j=1}^p {\bf u}_j\frac{\sigma_j^2}{\sigma_j^2+\lambda}{\bf u}_j^T{\bf y}$$ .
It is indeed the economy-sized SVD, note the summation runs up tp $$p$$ only and not $$n$$.
Here we have that $${\bf X} = {\bf U}{\bf \Sigma}{\bf V}^T$$, with $$\Sigma$$ being an $$ n\times p$$ matrix and $${\bf V}$$ being a $$ p\times p$$ matrix. We also have assumed here that $$ n > p$$.
Video of Lecture from 2020 and handwritten notes
More material will be added here, see handwritten notes also.
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. 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, $$ C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta}, $$
by taking the derivatives with respect to \( \boldsymbol{\beta} \) we obtain then a slightly modified matrix inversion problem which for finite values of \( \lambda \) does not suffer from singularity problems. We obtain $$ \boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}, $$
with \( \boldsymbol{I} \) being a \( p\times p \) identity matrix with the constraint that $$ \sum_{i=0}^{p-1} \beta_i^2 \leq t, $$
with \( t \) a finite positive number.
We see that Ridge regression is nothing but the standard OLS with a modified diagonal term added to \( \boldsymbol{X}^T\boldsymbol{X} \). The consequences, in particular for our discussion of the bias-variance tradeoff are rather interesting.
Furthermore, if we use the result above in terms of the SVD decomposition (our analysis was done for the OLS method), we had $$ (\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}. $$
We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \( \boldsymbol{U} \) as $$ \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y} $$
For Ridge regression this becomes $$ \boldsymbol{X}\boldsymbol{\beta}^{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y}, $$
with the vectors \( \boldsymbol{u}_j \) being the columns of \( \boldsymbol{U} \).
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. Actually, calculating the variance of \( \boldsymbol{X}\boldsymbol{v}_j \) shows that this quantity is equal to \( \sigma_j^2/n \). With a parameter \( \lambda \) we can thus shrink the role of specific parameters.
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 $$ \boldsymbol{\beta}^{\mathrm{OLS}} = \boldsymbol{X}^T\boldsymbol{y}=\sum_{i=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\boldsymbol{y}, $$
and $$ \boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{I}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\left(1+\lambda\right)^{-1}\boldsymbol{\beta}^{\mathrm{OLS}}, $$
that is the Ridge estimator scales the OLS estimator by the inverse of a factor \( 1+\lambda \), and the Ridge estimator converges to zero when the hyperparameter goes to infinity.
We will come back to more interpreations after we have gone through some of the statistical analysis part.
For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. Similarly, Mehta et al's article is also recommended.
The parameter \( \lambda \) that we have introduced in the Ridge (and Lasso as well) regression is often called a regularization parameter or shrinkage parameter. It is common to call it a hyperparameter. What does it mean mathemtically?
Here we will first look at how to analyze the difference between the standard OLS equations and the Ridge expressions in terms of a linear algebra analysis using the SVD algorithm. Thereafter, we will link (see the material on the bias-variance tradeoff below) these observation to the statisical analysis of the results. In particular we consider how the variance of the parameters \( \boldsymbol{\beta} \) is affected by changing the parameter \( \lambda \).
We have our design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). With the SVD we decompose it as $$ \boldsymbol{X} = \boldsymbol{U\Sigma V^T}, $$
with \( \boldsymbol{U}\in {\mathbb{R}}^{n\times n} \), \( \boldsymbol{\Sigma}\in {\mathbb{R}}^{n\times p} \) and \( \boldsymbol{V}\in {\mathbb{R}}^{p\times p} \).
The matrices \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are unitary/orthonormal matrices, that is in case the matrices are real we have \( \boldsymbol{U}^T\boldsymbol{U}=\boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I} \) and \( \boldsymbol{V}^T\boldsymbol{V}=\boldsymbol{V}\boldsymbol{V}^T=\boldsymbol{I} \).
Before we discuss the link between for example Ridge regression and the singular value decomposition, we need to remind ourselves about the definition of the covariance and the correlation function. These are quantities
Suppose we have defined two vectors \( \hat{x} \) and \( \hat{y} \) with \( n \) elements each. The covariance matrix \( \boldsymbol{C} \) is defined as $$ \boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{cov}[\boldsymbol{x},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{cov}[\boldsymbol{y},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{y},\boldsymbol{y}] \\ \end{bmatrix}, $$ where for example $$ \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). $$ With this definition and recalling that the variance is defined as $$ \mathrm{var}[\boldsymbol{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, $$ we can rewrite the covariance matrix as $$ \boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] & \mathrm{var}[\boldsymbol{y}] \\ \end{bmatrix}. $$
The covariance takes values between zero and infinity and may thus lead to problems with loss of numerical precision for particularly large values. It is common to scale the covariance matrix by introducing instead the correlation matrix defined via the so-called correlation function $$ \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]=\frac{\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}]}{\sqrt{\mathrm{var}[\boldsymbol{x}] \mathrm{var}[\boldsymbol{y}]}}. $$
The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \in [-1,1] \). This avoids eventual problems with too large values. We can then define the correlation matrix for the two vectors \( \boldsymbol{x} \) and \( \boldsymbol{y} \) as $$ \boldsymbol{K}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{corr}[\boldsymbol{y},\boldsymbol{x}] & 1 \\ \end{bmatrix}, $$
In the above example this is the function we constructed using pandas.
In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression we defined the design/feature matrix \( \boldsymbol{X} \) as $$ \boldsymbol{X}=\begin{bmatrix} x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ \dots & \dots & \dots & \dots \dots & \dots \\ x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ \end{bmatrix}, $$ with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors/features \( p \) refering to the column numbers and the entries \( n \) being the row elements. We can rewrite the design/feature matrix in terms of its column vectors as $$ \boldsymbol{X}=\begin{bmatrix} \boldsymbol{x}_0 & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \dots & \dots & \boldsymbol{x}_{p-1}\end{bmatrix}, $$ with a given vector $$ \boldsymbol{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. $$
With these definitions, we can now rewrite our \( 2\times 2 \) correaltion/covariance matrix in terms of a moe general design/feature matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \) covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \) $$ \boldsymbol{C}[\boldsymbol{x}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_1] & \mathrm{var}[\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ \dots & \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & \mathrm{var}[\boldsymbol{x}_{p-1}]\\ \end{bmatrix}, $$ and the correlation matrix $$ \boldsymbol{K}[\boldsymbol{x}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_0] & 1 & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ \dots & \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & 1\\ \end{bmatrix}, $$
The Numpy function np.cov calculates the covariance elements using the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have the exact mean values. The following simple function uses the np.vstack function which takes each vector of dimension \( 1\times n \) and produces a \( 2\times n \) matrix \( \boldsymbol{W} \) $$ \boldsymbol{W} = \begin{bmatrix} x_0 & y_0 \\ x_1 & y_1 \\ x_2 & y_2\\ \dots & \dots \\ x_{n-2} & y_{n-2}\\ x_{n-1} & y_{n-1} & \end{bmatrix}, $$
which in turn is converted into into the \( 2\times 2 \) covariance matrix \( \boldsymbol{C} \) via the Numpy function np.cov(). We note that we can also calculate the mean value of each set of samples \( \boldsymbol{x} \) etc using the Numpy function np.mean(x). We can also extract the eigenvalues of the covariance matrix through the np.linalg.eig() function.
# Importing various packages
import numpy as np
n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
W = np.vstack((x, y))
C = np.cov(W)
print(C)
The previous example can be converted into the correlation matrix by simply scaling the matrix elements with the variances. We should also subtract the mean values for each column. This leads to the following code which sets up the correlations matrix for the previous example in a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the \( 2\times 2 \) correlation matrix (since we have only two vectors).
import numpy as np
n = 100
# define two vectors
x = np.random.random(size=n)
y = 4+3*x+np.random.normal(size=n)
#scaling the x and y vectors
x = x - np.mean(x)
y = y - np.mean(y)
variance_x = np.sum(x@x)/n
variance_y = np.sum(y@y)/n
print(variance_x)
print(variance_y)
cov_xy = np.sum(x@y)/n
cov_xx = np.sum(x@x)/n
cov_yy = np.sum(y@y)/n
C = np.zeros((2,2))
C[0,0]= cov_xx/variance_x
C[1,1]= cov_yy/variance_y
C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
C[1,0]= C[0,1]
print(C)
We see that the matrix elements along the diagonal are one as they should be and that the matrix is symmetric. Furthermore, diagonalizing this matrix we easily see that it is a positive definite matrix.
The above procedure with numpy can be made more compact if we use pandas.
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
import numpy as np
import pandas as pd
n = 10
x = np.random.normal(size=n)
x = x - np.mean(x)
y = 4+3*x+np.random.normal(size=n)
y = y - np.mean(y)
X = (np.vstack((x, y))).T
print(X)
Xpd = pd.DataFrame(X)
print(Xpd)
correlation_matrix = Xpd.corr()
print(correlation_matrix)
We expand this model to the Franke function discussed above.
# Common imports
import numpy as np
import pandas as pd
def FrankeFunction(x,y):
term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
return term1 + term2 + term3 + term4
def create_X(x, y, n ):
if len(x.shape) > 1:
x = np.ravel(x)
y = np.ravel(y)
N = len(x)
l = int((n+1)*(n+2)/2) # Number of elements in beta
X = np.ones((N,l))
for i in range(1,n+1):
q = int((i)*(i+1)/2)
for k in range(i+1):
X[:,q+k] = (x**(i-k))*(y**k)
return X
# Making meshgrid of datapoints and compute Franke's function
n = 4
N = 100
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)
Xpd = pd.DataFrame(X)
# subtract the mean values and set up the covariance matrix
Xpd = Xpd - Xpd.mean()
covariance_matrix = Xpd.cov()
print(covariance_matrix)
We note here that the covariance is zero for the first rows and columns since all matrix elements in the design matrix were set to one (we are fitting the function in terms of a polynomial of degree \( n \)).
This means that the variance for these elements will be zero and will cause problems when we set up the correlation matrix. We can simply drop these elements and construct a correlation matrix without these elements.
We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix \( \boldsymbol{X} \) as $$ \boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}]. $$
To see this let us simply look at a design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{2\times 2} \) $$ \boldsymbol{X}=\begin{bmatrix} x_{00} & x_{01}\\ x_{10} & x_{11}\\ \end{bmatrix}=\begin{bmatrix} \boldsymbol{x}_{0} & \boldsymbol{x}_{1}\\ \end{bmatrix}. $$
If we then compute the expectation value $$ \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}=\begin{bmatrix} x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ \end{bmatrix}, $$ which is just $$ \boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \\ \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \\ \end{bmatrix}, $$ where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
More material will be added here.
The exercises here are meant to prepare you for work with project 1. The first exercise is a follow-up of exercise 2 from week 36 August 30-September 3).
This exercise is a continuation of exercise 2 from exercise set 1 (week 36, August 30-September 3)). We will use the same function to generate our data set, still staying with a simple function \( y(x) \) which we want to fit using linear regression, but now extending the analysis to include the Ridge and the Lasso regression methods.
We will thus again generate our own dataset for a function \( y(x) \) where \( x \in [0,1] \) and defined by random numbers computed with the uniform distribution. The function \( y \) is a quadratic polynomial in \( x \) with added stochastic noise according to the normal distribution \( \cal{N}(0,1) \).
The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points).
x = np.random.rand(100)
y = 2.0+5*x*x+0.1*np.random.randn(100)
Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 3. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
The code here allows you to perform your own Ridge calculation and perform calculations for various values of the regularization parameter \( \lambda \). This program can easily be extended upon.
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.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(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)
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# matrix inversion to find beta
OLSbeta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(OLSbeta)
# and then make the prediction
ytildeOLS = X_train @ OLSbeta
print("Training R2 for OLS")
print(R2(y_train,ytildeOLS))
print("Training MSE for OLS")
print(MSE(y_train,ytildeOLS))
ypredictOLS = X_test @ OLSbeta
print("Test R2 for OLS")
print(R2(y_test,ypredictOLS))
print("Test MSE OLS")
print(MSE(y_test,ypredictOLS))
# Repeat now for Ridge regression and various values of the regularization parameter
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 20
MSEPredict = np.zeros(nlambdas)
MSETrain = np.zeros(nlambdas)
lambdas = np.logspace(-4, 1, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
Ridgebeta = np.linalg.inv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
# and then make the prediction
ytildeRidge = X_train @ Ridgebeta
ypredictRidge = X_test @ Ridgebeta
MSEPredict[i] = MSE(y_test,ypredictRidge)
MSETrain[i] = MSE(y_train,ytildeRidge)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSETrain, label = 'MSE Ridge train')
plt.plot(np.log10(lambdas), MSEPredict, 'r--', label = 'MSE Ridge Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
Repeat the above but using the functionality of Scikit-Learn. Compare your code with the results from Scikit-Learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
Finally, using Scikit-Learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as $$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, $$ and the \( R^2 \) score function. If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as $$ R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, $$ where we have defined the mean value of \( \hat{y} \) as $$ \bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. $$ Discuss these quantities as functions of the variable \( \lambda \) in the Ridge and Lasso regression methods.
This is a longer exercise and the aim is to study in more detail various regression methods, including the Ordinary Least Squares (OLS) method, Ridge regression and finally Lasso regression. This exercise forms a part of project 1.
We will study how to fit polynomials to a specific two-dimensional function called Franke's function. This is a function which has been widely used when testing various interpolation and fitting algorithms.
The Franke function, which is a weighted sum of four exponentials reads as follows $$ \begin{align*} f(x,y) &= \frac{3}{4}\exp{\left(-\frac{(9x-2)^2}{4} - \frac{(9y-2)^2}{4}\right)}+\frac{3}{4}\exp{\left(-\frac{(9x+1)^2}{49}- \frac{(9y+1)}{10}\right)} \\ &+\frac{1}{2}\exp{\left(-\frac{(9x-7)^2}{4} - \frac{(9y-3)^2}{4}\right)} -\frac{1}{5}\exp{\left(-(9x-4)^2 - (9y-7)^2\right) }. \end{align*} $$
The function will be defined for \( x,y\in [0,1] \). Our first step will be to perform an OLS regression analysis of this function, trying out a polynomial fit with an \( x \) and \( y \) dependence of the form \( [x, y, x^2, y^2, xy, \dots] \). We will fit a function (for example a polynomial) of \( x \) and \( y \). Thereafter we will repeat much of the same procedure using the Ridge and Lasso regression methods, introducing thus a dependence on the bias (penalty) \( \lambda \).
The Python fucntion for the Franke function is included here (it performs also a three-dimensional plot of it)
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import numpy as np
from random import random, seed
fig = plt.figure()
ax = fig.gca(projection='3d')
# Make data.
x = np.arange(0, 1, 0.05)
y = np.arange(0, 1, 0.05)
x, y = np.meshgrid(x,y)
def FrankeFunction(x,y):
term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
return term1 + term2 + term3 + term4
z = FrankeFunction(x, y)
# Plot the surface.
surf = ax.plot_surface(x, y, z, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
# Customize the z axis.
ax.set_zlim(-0.10, 1.40)
ax.zaxis.set_major_locator(LinearLocator(10))
ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
# Add a color bar which maps values to colors.
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()
We will generate our own dataset for a function \( \mathrm{FrankeFunction}(x,y) \) with \( x,y \in [0,1] \). The function \( f(x,y) \) is the Franke function. You should explore also the addition an added stochastic noise to this function using the normal distribution \( \cal{N}(0,1) \).
Write your own code (using either a matrix inversion or a singular value decomposition from e.g., numpy ) or use your code and perform a standard least square regression analysis using polynomials in \( x \) and \( y \) up to fifth order. You can use scikit-learn as well.
Evaluate the Mean Squared error (MSE) $$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, $$
and the \( R^2 \) score function. If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as $$ R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, $$
where we have defined the mean value of \( \hat{y} \) as $$ \bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. $$
You should split your data in train and test and also consider scaling the data.
To set up the design matrix, the following code can be used
def FrankeFunction(x,y):
term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
return term1 + term2 + term3 + term4
def create_X(x, y, n ):
if len(x.shape) > 1:
x = np.ravel(x)
y = np.ravel(y)
N = len(x)
l = int((n+1)*(n+2)/2) # Number of elements in beta
X = np.ones((N,l))
for i in range(1,n+1):
q = int((i)*(i+1)/2)
for k in range(i+1):
X[:,q+k] = (x**(i-k))*(y**k)
return X
# Making meshgrid of datapoints and compute Franke's function
n = 5
N = 1000
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)
Write then your own code for the Ridge method or use Scikit-Learn. Perform the same analysis as you did for ordinary Least Squares (for the same polynomials) but now for different values of \( \lambda \). Compare and analyze your results with those obtained with ordinary Least Squares. Study the dependence on \( \lambda \).
This part is essentially a repeat of the previous ones, but now with Lasso regression. Write either your own code or use the functionalities of Scikit-Learn (recommended). Give a critical discussion of the three methods and a judgement of which model fits the data best.