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
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.
For every set of values \( y_i,x_i \) we have thus the corresponding set of equations $$ \begin{align*} y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ \dots & \dots \\ y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\ \end{align*} $$
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 $$ \hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. $$
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 $$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$
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?
We have defined the matrix \( \hat{X} \) $$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$
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 $$ Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). $$
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 $$ \frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). $$
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 $$ \hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. $$
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.
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.
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 \( b_i = y_i/\sigma_i \).
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 $$ \hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. $$
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 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 $$ \frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. $$
For a linear fit we don't need to invert a matrix!! Defining $$ \gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, $$ $$ \gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, $$ $$ \gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), $$ $$ \gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, $$ $$ \gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, $$ we obtain $$ \beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, $$ $$ \beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. $$
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.
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]])])
# 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.
We can repeat the above algorithm using scikit-learn as follows
# 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()
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.
# 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()
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.
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.
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 $$ \chi^2 = \frac{1}{n} \sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, $$
where \( \sigma_i^2 \) is the variance (to be defined later) of the entry \( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves however the aim of scaling the equations and make the cost function dimensionless.
Minimizing the cost function is a central aspect of our discussions to come. Finding its minima as function of the model parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring theme in these series of lectures. Essentially all machine learning algorithms we will discuss center around the minimization of the chosen cost function. This depends in turn on our specific model for describing the data, a typical situation in supervised learning. Automatizing the search for the minima of the cost function is a central ingredient in all algorithms. Typical methods which are employed are various variants of gradient methods. These will be discussed in more detail later. Again, you'll be surprised to hear that many practitioners minimize the above function ''by the eye', popularly dubbed as 'chi by the eye'. That is, change a parameter and see (visually and numerically) that the \( \chi^2 \) function becomes smaller.
There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define the relative error as $$ \epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. $$ We can modify easily the above Python code and plot the relative error instead
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.random.rand(100,1)
y = 5*x+0.01*np.random.randn(100,1)
linreg = LinearRegression()
linreg.fit(x,y)
ypredict = linreg.predict(x)
plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
plt.axis([0,1.0,0.0, 0.5])
plt.xlabel(r'$x$')
plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
plt.title(r'Relative error')
plt.show()
Depending on the parameter in front of the normal distribution, we may have a small or larger relative error. Try to play around with different training data sets and study (graphically) the value of the relative error.
As mentioned above, scikit-learn has an impressive functionality. We can for example extract the values of \( \alpha \) and \( \beta \) and their error estimates, or the variance and standard deviation and many other properties from the statistical data analysis.
Here we show an example of the functionality of scikit-learn.
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
x = np.random.rand(100,1)
y = 2.0+ 5*x+0.5*np.random.randn(100,1)
linreg = LinearRegression()
linreg.fit(x,y)
ypredict = linreg.predict(x)
print('The intercept alpha: \n', linreg.intercept_)
print('Coefficient beta : \n', linreg.coef_)
# The mean squared error
print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
# Explained variance score: 1 is perfect prediction
print('Variance score: %.2f' % r2_score(y, ypredict))
# Mean squared log error
print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
# Mean absolute error
print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
plt.plot(x, ypredict, "r-")
plt.plot(x, y ,'ro')
plt.axis([0.0,1.0,1.5, 7.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Linear Regression fit ')
plt.show()
The function coef gives us the parameter \( \beta \) of our fit while intercept yields \( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as $$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, $$
The smaller the value, the better the fit. Ideally we would like to have an MSE equal zero. The attentive reader has probably recognized this function as being similar to the \( \chi^2 \) function defined above.
The r2score function computes \( R^2 \), the coefficient of determination. It provides a measure of how well future samples are likely to be predicted by the model. Best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of \( \hat{y} \), disregarding the input features, would get a \( R^2 \) score of \( 0.0 \).
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 $$ \bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. $$
Another quantity will meet again in our discussions of regression analysis is mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error. The MAE is defined as follows $$ \text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. $$ Finally we present the squared logarithmic (quadratic) error $$ \text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, $$
where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error estimate is best to use when targets having exponential growth, such as population counts, average sales of a commodity over a span of years etc.
We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn.
import matplotlib.pyplot as plt
import numpy as np
import random
from sklearn.linear_model import Ridge
from sklearn.preprocessing import PolynomialFeatures
from sklearn.pipeline import make_pipeline
from sklearn.linear_model import LinearRegression
x=np.linspace(0.02,0.98,200)
noise = np.asarray(random.sample((range(200)),200))
y=x**3*noise
yn=x**3*100
poly3 = PolynomialFeatures(degree=3)
X = poly3.fit_transform(x[:,np.newaxis])
clf3 = LinearRegression()
clf3.fit(X,y)
Xplot=poly3.fit_transform(x[:,np.newaxis])
poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
plt.plot(x,yn, color='red', label="True Cubic")
plt.scatter(x, y, label='Data', color='orange', s=15)
plt.legend()
plt.show()
def error(a):
for i in y:
err=(y-yn)/yn
return abs(np.sum(err))/len(err)
print (error(y))
Similarly, using R, we can perform similar studies. (more details on R will be inserted later).
# Importing various packages
from math import exp, sqrt
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]
theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(theta_linreg)
theta = np.random.randn(2,1)
eta = 0.1
Niterations = 1000
m = 100
for iter in range(Niterations):
gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
theta -= eta*gradients
print(theta)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(theta)
ypredict2 = xbnew.dot(theta_linreg)
plt.plot(xnew, ypredict, "r-")
plt.plot(xnew, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Random numbers ')
plt.show()
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDRegressor
x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)
xb = np.c_[np.ones((100,1)), x]
theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(theta_linreg)
sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
sgdreg.fit(x,y.ravel())
print(sgdreg.intercept_, sgdreg.coef_)
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
m = 100
x = 2*np.random.rand(m,1)+4.
y = 4+3*x*x+ +x-np.random.randn(m,1)
xb = np.c_[np.ones((m,1)), x]
theta = 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(theta)
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()
import matplotlib.pyplot as plt
import numpy as np
from sklearn import linear_model
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error, r2_score
#creating data with random noise
x=np.arange(50)
delta=np.random.uniform(-2.5,2.5, size=(50))
np.random.shuffle(delta)
y =0.5*x+5+delta
#arranging data into 2x50 matrix
a=np.array(x) #inputs
b=np.array(y) #outputs
#Split into training and test
X_train=a[:37, np.newaxis]
X_test=a[37:, np.newaxis]
y_train=b[:37]
y_test=b[37:]
print ("X_train: ", X_train.shape)
print ("y_train: ", y_train.shape)
print ("X_test: ", X_test.shape)
print ("y_test: ", y_test.shape)
print ("------------------------------------")
print ("Ordinary Least Squares")
#Add Ordinary Least Squares fit
reg=LinearRegression()
reg.fit(X_train, y_train)
pred=reg.predict(X_test)
print ("Prediction Shape: ", pred.shape)
print('Coefficients: \n', reg.coef_)
# The mean squared error
print("Mean squared error: %.2f"
% mean_squared_error(y_test, pred))
# Explained variance score: 1 is perfect prediction
print('Variance score: %.2f' % r2_score(y_test, pred))
#plot
plt.scatter(X_test,y_test,color='green', label="Training Data")
plt.plot(X_test, pred, color='black', label="Fit Line")
plt.legend()
plt.show()
print ("------------------------------------")
print ("Ridge Regression")
ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
ridge.fit(X_train,y_train)
print ("Ridge Coefficient: ",ridge.coef_)
print ("Ridge Intercept: ", ridge.intercept_)
#Look into graphing with Ridge fit
print ("------------------------------------")
print ("Lasso")
lasso=linear_model.Lasso(alpha=0.1)
lasso.fit(X_train,y_train)
predl=lasso.predict(X_test)
print("Lasso Coefficient: ", lasso.coef_)
print("Lasso Intercept: ", lasso.intercept_)
plt.scatter(X_test,y_test,color='green', label="Training Data")
plt.plot(X_test, predl, color='blue', label="Lasso")
plt.legend()
plt.show()
How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then $$ \hat{A} = \hat{U}\hat{\Sigma}\hat{V} $$
Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue
Discuss the mathematics here
Add examples on classification problems