\chapter{Linear Model}

\section{Simple Linear Regression}

\[
    Y_i = \beta_0 + \beta_1 X_i  + \varepsilon_i
\]
\[
    \Y = \X \beta + \varepsilon.
\]
\[
    \begin{pmatrix}
        Y_1 \\
        Y_2 \\
        \vdots \\
        Y_n
    \end{pmatrix}
    = 
    \begin{pmatrix}
        1 & X_1 \\
        1 & X_2 \\
        \vdots & \vdots \\
        1 & X_n 
    \end{pmatrix}
    \begin{pmatrix}
        \beta_0 \\
        \beta_1
    \end{pmatrix}
    +
    \begin{pmatrix}
        \varepsilon_1 \\
        \varepsilon_2 \\
        \vdots
        \varepsilon_n
    \end{pmatrix}
\]

\paragraph*{Assumptions}
\begin{enumerate}[label={\color{primary}{($A_\arabic*$)}}]
    \item $\varepsilon_i$ are independent;
    \item $\varepsilon_i$ are identically distributed;
    \item $\varepsilon_i$ are i.i.d $\sim \Norm(0, \sigma^2)$ (homoscedasticity).
\end{enumerate}

\section{Generalized Linear Model}

\[
    g(\EE(Y)) = X \beta
\]
with $g$ being
\begin{itemize}
    \item Logistic regression: $g(v) = \log \left(\frac{v}{1-v}\right)$, for instance for boolean values,
    \item Poisson regression: $g(v) = \log(v)$, for instance for discrete variables. 
\end{itemize}

\subsection{Penalized Regression}

When the number of variables is large, e.g, when the number of explanatory variable is above the number of observations, if $p >> n$ ($p$: the number of explanatory variable, $n$ is the number of observations), we cannot estimate the parameters.
In order to estimate the parameters, we can use penalties (additional terms).

Lasso regression, Elastic Net, etc.

\subsection{Statistical Analysis Workflow}

\begin{enumerate}[label={\bfseries\color{primary}Step \arabic*.}]
    \item Graphical representation;
    \item ...
\end{enumerate}
\[
    Y = X \beta + \varepsilon,
\]
is noted equivalently as
\[
        \begin{pmatrix}
            y_1 \\
            y_2 \\
            y_3 \\
            y_4
        \end{pmatrix}
        = \begin{pmatrix}
                1 & x_{11} & x_{12} \\
                1 & x_{21} & x_{22} \\
                1 & x_{31} & x_{32} \\
                1 & x_{41} & x_{42}
            \end{pmatrix}
        \begin{pmatrix}
            \beta_0 \\
            \beta_1 \\
            \beta_2
        \end{pmatrix} +
        \begin{pmatrix}
            \varepsilon_1 \\
            \varepsilon_2 \\
            \varepsilon_3 \\
            \varepsilon_4
        \end{pmatrix}.
\]
\section{Parameter Estimation}

\subsection{Simple Linear Regression}

\subsection{General Case}

If $\X^T\X$ is invertible, the OLS estimator is:
\begin{equation}
\hat{\beta} = (\X^T\X)^{-1} \X^T \Y
\end{equation}

\subsection{Ordinary Least Square Algorithm}

We want to minimize the distance between $\X\beta$ and $\Y$:
\[
    \min \norm{\Y - \X\beta}^2
\]
(See \autoref{ch:elements-of-linear-algebra}).
\begin{align*}
    \Rightarrow& \X \beta = proj^{(1, \X)} \Y\\
    \Rightarrow& \forall v \in w,\, vy = v proj^w(y)\\
    \Rightarrow& \forall i: \\
    & \X_i \Y = \X_i X\hat{\beta} \qquad \text{where $\hat{\beta}$ is the estimator of $\beta$} \\
    \Rightarrow& \X^T \Y = \X^T \X \hat{\beta} \\
    \Rightarrow& {\color{gray}(\X^T \X)^{-1}} \X^T \Y = {\color{gray}(\X^T \X)^{-1}} (\X^T\X) \hat{\beta} \\
    \Rightarrow& \hat{\beta} = (\X^T\X)^{-1} \X^T \Y
\end{align*}

This formula comes from the orthogonal projection of $\Y$ on the vector subspace defined by the explanatory variables $\X$

$\X \hat{\beta}$ is the closest point to $\Y$ in the subspace generated by $\X$.

If $H$ is the projection matrix of the subspace generated by $\X$, $X\Y$ is the projection on $\Y$ on this subspace, that corresponds to $\X\hat{\beta}$.

\section{Sum of squares}

$\Y - \X \hat{\beta} \perp \X \hat{\beta} - \Y \One$ if $\One \in V$, so
\[
    \underbrace{\norm{\Y - \bar{\Y}\One}}_{\text{Total SS}} = \underbrace{\norm{\Y - \X \hat{\beta}}^2}_{\text{Residual SS}} + \underbrace{\norm{\X \hat{\beta} - \bar{\Y} \One}^2}_{\text{Explicated SS}}
\]

\section{Coefficient of Determination: \texorpdfstring{$R^2$}{R\textsuperscript{2}}}
\begin{definition}[$R^2$]
    \[
        0 \leq R^2 = \frac{\norm{\X\hat{\beta} - \bar{\Y}\One}^2}{\norm{\Y - \bar{\Y}\One}^2} = 1 - \frac{\norm{\Y - \X\hat{\beta}}^2}{\norm{\Y - \bar{\Y}\One}^2} \leq 1
    \] proportion of variation of $\Y$ explained by the model.
\end{definition}

\begin{figure}
    \centering
    \includestandalone{figures/schemes/orthogonal_projection}
    \caption{Orthogonal projection of $\Y$ on plan generated by the base described by $\X$. $\color{blue}a$ corresponds to $\norm{\X\hat{\beta} - \bar{\Y}}^2$ and $\color{blue}b$ corresponds to $\hat{\varepsilon} = \norm{\Y - \hat{\beta}\X}^2$} and $\color{blue}c$ corresponds to $\norm{Y - \bar{Y}}^2$.
    \label{fig:scheme-orthogonal-projection}
\end{figure}

\begin{figure}
    \centering
    \includestandalone{figures/schemes/ordinary_least_squares}
    \caption{Ordinary least squares and regression line with simulated data.}
    \label{fig:ordinary-least-squares}
\end{figure}




\begin{definition}[Model dimension]
    Let $\M$ be a model.
    The dimension of $\M$ is the dimension of the subspace generated by $\X$, that is the number of parameters in the $\beta$ vector.

    \textit{Nb.} The dimension of the model is not the number of parameter, as $\sigma^2$ is one of the model parameters.
\end{definition}

\section{Gaussian vectors}


\begin{definition}[Normal distribution]

\end{definition}


\begin{definition}[Gaussian vector]
    A random vector $\Y \in \RR[n]$ is a gaussian vector if every linear combination of its component is ...
\end{definition}

\begin{property}
    $m = \EE(Y) = (m_1, \ldots, m_n)^T$, where $m_i = \EE(Y_i)$


    ...

    \[
        \Y \sim \Norm_n(m, \Sigma)
    \]
    where $\Sigma$ is the variance-covariance matrix!
    \[
        \Sigma = \E\left[(\Y -m)(\Y - m)^T\right].
    \]



\end{property}

\begin{remark}
    \[
        \Cov(Y_i, Y_i) = \Var(Y_i)
    \]
\end{remark}

\begin{definition}[Covariance]
    \[
        \Cov(Y_i, Y_j) = \EE\left((Y_i-\EE(Y_j))(Y_j-\EE(Y_j))\right)
    \]
\end{definition}


When two variable are linked, the covariance is large.

If two variables $X, Y$ are independent, $\Cov(X, Y) = 0$.

\begin{definition}[Correlation coefficient]
    \[
        \Cor(Y_i, Y_j) = \frac{\EE\left((Y_i-\EE(Y_j))(Y_j-\EE(Y_j))\right)}{\sqrt{\EE(Y_i - \EE(Y_i)) \cdot \EE(Y_j - \EE(Y_j))}}
    \]
\end{definition}

Covariance is really sensitive to scale of variables. For instance, if we measure distance in millimeters, the covariance would be larger than in the case of a measure expressed in metters. Thus the correlation coefficient, which is a sort of normalized covariance is useful, to be able to compare the values.

\begin{remark}
    \begin{align*}
        \Cov(Y_i, Y_i) &= \EE((Y_i - \EE(Y_i)) (Y_i - \EE(Y_i))) \\
        &= \EE((Y_i - \EE(Y_i))^2) \\
        &= \Var(Y_i)
    \end{align*}
\end{remark}

\begin{equation}
    \Sigma = \begin{pNiceMatrix}
        \VVar(Y_1) & & & &\\
        & \Ddots & & & \\
        &  \Cov(Y_i, Y_j) & \VVar(Y_i) & & \\
        & & & \Ddots & \\
        & & & & \VVar(Y_n) 
    \end{pNiceMatrix}
\end{equation}

\begin{definition}[Identity matrix]
    \[
        \mathcal{I}_n = \begin{pNiceMatrix}
            1 & 0 & 0  \\
            0 & \Ddots & 0\\
            0 & 0 & 1
        \end{pNiceMatrix}
    \]
    
\end{definition}


\begin{theorem}[Cochran Theorem (Consequence)]
    Let $\mathbf{Z}$ be a gaussian vector: $\mathbf{Z} \sim \Norm_n(0_n, I_n)$.

    \begin{itemize}
        \item If $V_1, V_n$ are orthogonal subspaces of $\RR[n]$ with dimensions $n_1, n_2$ such that
        \[
            \RR[n] = V_1 \overset{\perp}{\oplus} V_2.
        \]
        \item If $Z_1, Z_2$ are orthogonal of $\mathbf{Z}$ on $V_1$ and $V_2$ i.e. $Z_1 = \Pi_{V_1}(\mathbf{Z}) = \Pi_1 \Y$ and $Z_2 = \Pi_{V_2} (\mathbf{Z}) = \Pi_2 \Y$... 
        (\textcolor{red}{look to the slides})
    \end{itemize}
\end{theorem}

\begin{definition}[Chi 2 distribution]
    If $X_1, \ldots, X_n$ i.i.d. $\sim \Norm(0, 1)$, then;,
    \[
        X_1^2 + \ldots X_n^2 \sim \chi_n^2
    \]
\end{definition}

\subsection{Estimator's properties}


\[
    \Pi_V = \X(\X^T\X)^{-1} \X^T
\]

\begin{align*}
    \hat{m} &= \X \hat{\beta} = \X(\X^T\X)^{-1} \X^T \Y \\
    \text{so} \\
    &= \Pi_V \Y 
\end{align*}

According to Cochran theorem, we can deduce that the estimator of the predicted value $\hat{m}$ is independent $\hat{\sigma}^2$

All the sum of squares follows a $\chi^2$ distribution:
\[
   ...
\]

\begin{property}

\end{property}

\subsection{Estimators consistency}

If $q < n$,
\begin{itemize}
    \item $\hat{\sigma}^2 \overunderset{\PP}{n\to\infty} \sigma^{*2}$.
    \item If $(\X^T\X)^{-1}$...
    \item ...
\end{itemize}

We can derive statistical test from these properties.


\section{Statistical tests}

\subsection{Student $t$-test}


\[
    \frac{\hat{\theta}-\theta}{\sqrt{\frac{\widehat{\VVar}(\hat{\theta})}{n}}} \underset{H_0}{\sim} t
\]

where 
