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 $$ \begin{align} y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0 \label{_auto1}\\ y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1 \label{_auto2}\\ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2 \label{_auto3}\\ \dots & \dots \label{_auto4}\\ y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}. \label{_auto5}\\ \label{_auto6} \end{align} $$ Defining the vectors $$ \hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$ $$ \hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, $$ $$ \hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, $$ 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 $$ \begin{align} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0 \label{_auto7}\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1 \label{_auto8}\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1 \label{_auto9}\\ \dots & \dots \label{_auto10}\\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1 \label{_auto11}\\ \dots & \dots \label{_auto12}\\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}. \label{_auto13}\\ \label{_auto14} \end{align} $$ We redefine in turn the matrix \( \hat{X} \) as $$ \hat{X}= \begin{bmatrix} x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ x_{20}& x_{21}^1 &x_{22}2& \dots & \dots &x_{2,n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ x_{n-1,00}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ \end{bmatrix} $$ 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?
$$ \begin{align} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0 \label{_auto15}\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1 \label{_auto16}\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1 \label{_auto17}\\ \dots & \dots \label{_auto18}\\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1 \label{_auto19}\\ \dots & \dots \label{_auto20}\\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}. \label{_auto21}\\ \label{_auto22} \end{align} $$