FYS-STK4155/doc/pub/Regression/ipynb/Regression.ipynb

Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis

Morten Hjorth-Jensen, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

Date: Oct 11, 2018

Copyright 1999-2018, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

Regression analysis, overarching aims

Regression modeling deals with the description of the sampling distribution of a given random variable y varies as function of another variable or a set of such variables \hat{x} =[x_0, x_1,\dots, x_p]^T. The first variable is called the dependent, the outcome or the response variable while the set of variables \hat{x} is called the independent variable, or the predictor variable or the explanatory variable.

A regression model aims at finding a likelihood function p(y\vert \hat{x}), that is the conditional distribution for y with a given \hat{x}. The estimation of p(y\vert \hat{x}) is made using a data set with

• n cases i = 0, 1, 2, \dots, n-1

• Response (dependent or outcome) variable y_i with i = 0, 1, 2, \dots, n-1

• p Explanatory (independent or predictor) variables \hat{x}_i=[x_{i0}, x_{i1}, \dots, x_{ip}] with i = 0, 1, 2, \dots, n-1

The goal of the regression analysis is to extract/exploit relationship between y_i and \hat{x}_i in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions .

Regression analysis, overarching aims II

Consider an experiment in which p characteristics of n samples are measured. The data from this experiment are denoted \mathbf{X}, with \mathbf{X} as above. The matrix \mathbf{X} is called the design matrix. Additional information of the samples is available in the form of \mathbf{Y} (also as above). The variable \mathbf{Y} is generally referred to as the response variable. The aim of regression analysis is to explain \mathbf{Y} in terms of \mathbf{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 \mathbf{X} and \mathbf{Y}. This assumption gives rise to the linear regression model where $\beta = (\beta_1, \ldots, \beta_p)^{\top}$ is the regression parameter. The parameter \beta_j, j=1, \ldots, p, represents the effect size of covariate j on the response. That is, for each unit change in covariate j (while keeping the other covariates fixed) the observed change in the response is equal to \beta_j.

General linear models

Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \hat{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 \hat{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_i x_i^j+\epsilon_i,

where \epsilon_i is the error in our approximation.

Rewriting the fitting procedure as a linear algebra problem

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_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\
\end{align*}

Rewriting the fitting procedure as a linear algebra problem, follows

Defining the vectors

\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,

and

\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T,

and

\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,

and the matrix

\hat{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

\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}.

Generalizing the fitting procedure as a linear algebra problem

We are obviously not limited to the above polynomial. We could replace the various powers of x with elements of Fourier series, that is, 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_1x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}

Generalizing the fitting procedure as a linear algebra problem

We redefine in turn the matrix \hat{X} as

\hat{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

\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}.

The left-hand side of this equation forms know. Our error vector \hat{\epsilon} and the parameter vector \hat{\beta} are our unknow quantities. How can we obtain the optimal set of \beta_i values?

Optimizing our parameters

We have defined the matrix \hat{X}

\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_1x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}

Optimizing our parameters, more details

We well use this matrix to define the approximation \hat{\tilde{y}} via the unknown quantity \hat{\beta} as

\hat{\tilde{y}}= \hat{X}\hat{\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 parametrized values \tilde{y}_i, namely

Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right),

or using the matrix \hat{X} as

Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right).

Interpretations and optimizing our parameters

The function

Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right),

can be linked to the variance of the quantity y_i if we interpret the latter as the mean value of for example a numerical experiment. When linking 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 Q(\hat{\beta}) by requiring

\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \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 Q(\hat{\beta})}{\partial \beta_j} = -2\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 Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right).

Interpretations and optimizing our parameters

We can rewrite

\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right),

as

\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta},

and if the matrix \hat{X}^T\hat{X} is invertible we have the solution

\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}.

Interpretations and optimizing our parameters

The residuals \hat{\epsilon} are in turn given by

\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta},

and with

\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0,

we have

\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0,

meaning that the solution for \hat{\beta} is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.

The \chi^2 function

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.

Introducing the standard deviation \sigma_i for each measurement y_i, we define now the \chi^2 function as

\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),

where the matrix \hat{\Sigma} is a diagonal matrix with \sigma_i as matrix elements.

The \chi^2 function

In order to find the parameters \beta_i we will then minimize the spread of \chi^2(\hat{\beta}) by requiring

\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,

which results in

\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,

or in a matrix-vector form as

\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).

where we have defined the matrix \hat{A} =\hat{X}/\hat{\Sigma} with matrix elements a_{ij} = x_{ij}/\sigma_i and the vector \hat{b} with elements b_i = y_i/\sigma_i.

The \chi^2 function

We can rewrite

\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),

as

\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},

and if the matrix \hat{A}^T\hat{A} is invertible we have the solution

\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.

The \chi^2 function

If we then introduce the matrix

\hat{H} =  \left(\hat{A}^T\hat{A}\right)^{-1},

we have then the following expression for the parameters \beta_j (the matrix elements of \hat{H} are h_{ij})

\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}

We state without proof the expression for the uncertainty in the parameters \beta_j as (we leave this as an exercise)

\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,

resulting in

\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!

The \chi^2 function

The first step here is to approximate the function y with a first-order polynomial, that is we write

y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.

By computing the derivatives of \chi^2 with respect to \beta_0 and \beta_1 show that these are given by

\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,

and

\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.

The \chi^2 function

For a linear fit we don't need to invert a matrix!!
Defining

\gamma =  \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2},

4 0

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4 1

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4 2

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\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2},

we obtain

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\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.

This approach (different linear and non-linear regression) suffers often from both being underdetermined and overdetermined in the unknown coefficients \beta_i. A better approach is to use the Singular Value Decomposition (SVD) method discussed below. Or using Lasso and Ridge regression. See below.

Simple regression model

We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the $x$-axis. These numbers define then the value of a function y(x)=4+3x+N(0,1). Thereafter we order the x values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function c\_ arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

In [1]:

import numpy as np
print(np.c_[np.array([1,2,3]), np.array([4,5,6])])
print(np.c_[np.array([[1,2,3]]), 0, 0, np.array([[4,5,6]])])

In [2]:

%matplotlib inline

# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt

x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)

xb = np.c_[np.ones((100,1)), x]
beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(beta)

plt.plot(xnew, ypredict, "r-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Linear Regression')
plt.show()

We see that, as expected, a linear fit gives a seemingly (from the graph) good representation of the data.

Simple regression model, now using scikit-learn

We can repeat the above algorithm using scikit-learn as follows

In [3]:

# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression

x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)
linreg = LinearRegression()
linreg.fit(x,y)
xnew = np.array([[0],[2]])
ypredict = linreg.predict(xnew)

plt.plot(xnew, ypredict, "r-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Random numbers ')
plt.show()

Simple linear regression model using scikit-learn

We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. What follows is a simple Python code where we have defined function y in terms of the variable x. Both are defined as vectors of dimension 1\times 100. The entries to the vector \hat{x} are given by random numbers generated with a uniform distribution with entries x_i \in [0,1] (more about probability distribution functions later). These values are then used to define a function y(x) (tabulated again as a vector) with a linear dependence on x plus a random noise added via the normal distribution.

The Numpy functions are imported used the import numpy as np statement and the random number generator for the uniform distribution is called using the function np.random.rand(), where we specificy that we want 100 random variables. Using Numpy we define automatically an array with the specified number of elements, 100 in our case. With the Numpy function randn() we can compute random numbers with the normal distribution (mean value \mu equal to zero and variance \sigma^2 set to one) and produce the values of y assuming a linear dependence as function of x

y = 2x+N(0,1),

where N(0,1) represents random numbers generated by the normal distribution. From scikit-learn we import then the LinearRegression functionality and make a prediction $\tilde{y} = \alpha + \beta x$ using the function fit(x,y). We call the set of data (\hat{x},\hat{y}) for our training data. The Python package scikit-learn has also a functionality which extracts the above fitting parameters \alpha and \beta (see below). Later we will distinguish between training data and test data.

For plotting we use the Python package matplotlib which produces publication quality figures. Feel free to explore the extensive gallery of examples. In this example we plot our original values of x and y as well as the prediction ypredict (\tilde{y}), which attempts at fitting our data with a straight line.

The Python code follows here.

In [4]:

# Importing various packages
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression

x = np.random.rand(100,1)
y = 2*x+np.random.randn(100,1)
linreg = LinearRegression()
linreg.fit(x,y)
xnew = np.array([[0],[1]])
ypredict = linreg.predict(xnew)

plt.plot(xnew, ypredict, "r-")
plt.plot(x, y ,'ro')
plt.axis([0,1.0,0, 5.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Simple Linear Regression')
plt.show()

Simple linear regression model

This example serves several aims. It allows us to demonstrate several aspects of data analysis and later machine learning algorithms. The immediate visualization shows that our linear fit is not impressive. It goes through the data points, but there are many outliers which are not reproduced by our linear regression. We could now play around with this small program and change for example the factor in front of x and the normal distribution. Try to change the function y to

y = 10x+0.01 \times N(0,1),

where x is defined as before.

Less noise

Does the fit look better? Indeed, by reducing the role of the normal distribution we see immediately that our linear prediction seemingly reproduces better the training set. However, this testing 'by the eye' is obviouly not satisfactory in the long run. Here we have only defined the training data and our model, and have not discussed a more rigorous approach to the cost function.

How to study our fits

We need more rigorous criteria in defining whether we have succeeded or not in modeling our training data. You will be surprised to see that many scientists seldomly venture beyond this 'by the eye' approach. A standard approach for the cost function is the so-called \chi^2 function

\chi^2 = \frac{1}{n}
\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2},

Warning:

Output truncated. This notebook contains too many cells to display efficiently.