fix: Amend the orthogonal projection scheme

.gitattributes

@@ -1 +1,7 @@
 main.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/regression_plan_3D.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/vector_orthogonality.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/base_plan.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/coordinates_systems.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/ordinary_least_squares.pdf filter=lfs diff=lfs merge=lfs -text
+figures/schemes/orthogonal_projection.pdf filter=lfs diff=lfs merge=lfs -text

content/chapters/part1/1.tex

@@ -129,6 +129,13 @@ $\X \hat{\beta}$ 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}[$R^2$]
     \[
@@ -139,7 +146,7 @@ If $H$ is the projection matrix of the subspace generated by $\X$, $X\Y$ is the
 \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 $\norm{\Y - \hat{\beta}\X}^2$}
+    \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}
 
@@ -149,3 +156,165 @@ If $H$ is the projection matrix of the subspace generated by $\X$, $X\Y$ is the
     \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 

definitions.tex

@@ -0,0 +1,6 @@
+\DeclareMathOperator{\VVar}{\mathbb{V}} % variance
+\DeclareMathOperator{\One}{\mathbf{1}}
+\DeclareMathOperator{\Cor}{\mathrm{Cor}}
+\newcommand{\M}[1][]{\ensuremath{\ifstrempty{#1}{\mathcal{M}}{\mathbb{M}_{#1}}}}
+\newcommand{\X}{\ensuremath{\mathbf{X}}}
+\newcommand{\Y}{\ensuremath{\mathbf{Y}}}

figures/schemes/orthogonal_projection.tex

@@ -12,21 +12,31 @@
     \tkzDefParallelogram(A,B,C)
     \tkzGetPoint{D}
     \tkzDrawPolygon[fill=gray!25!white](A,B,C,D)
-    \draw[decorate,decoration={brace,
+
-    amplitude=8pt},xshift=0pt,very thin,gray] (2,0) -- ++(-1,-0.5) node [black,midway,xshift=0.5em,yshift=-1em] {\color{blue}$a$};
   \end{scope}
-  \begin{scope}[canvas is xz plane at y=0]
+  % Draw the rectangle triangle scheme
+  \begin{scope}[canvas is xz plane at y=1]
     \draw[thick,fill=white,fill opacity=0.7,nodes={opacity=1}]
-    (2,0) node[bullet,label=below right:{$\mathbf{X}$}] {}
+    (2,0) node[bullet,label=right:{$\bar{\mathbf{Y}}$}] (Y_bar) {}
-    -- (0,0) node[bullet] {}
+    -- (0,-0.5) node (B) {}
-    -- (0,3) node[bullet,label=above:{$\mathbf{Y}$}] {} -- cycle;
+    -- (0,3) node[label=above:{$\mathbf{Y}$}] (Y) {} -- cycle;
-    \draw (0.25,0) -- (0.25,0.25) -- (0,0.25);
+    % Right angle annotation
+    \tkzPicRightAngle[draw,
+      angle eccentricity=.5,angle radius=2mm](Y,B,Y_bar)
+    % epsilon: Y - X \hat{\beta} curly brackets annotations
     \draw[decorate,decoration={brace,
-    amplitude=8pt},xshift=0pt,very thin,gray] (0,0) -- (0,3) node [black,midway,xshift=-1.25em,yshift=0em] {\color{blue}$b$};
+      amplitude=8pt},xshift=0pt,very thin,gray] (B) -- (Y) node [black,midway,xshift=-1.25em,yshift=0em] {\color{blue}$b$};
+    % X\hat{\beta} - \hat{Y}
+    \draw[decorate,decoration={brace,
+      amplitude=8pt},xshift=0pt,very thin,gray] (Y_bar) -- (B) node [black,midway,xshift=0.5em,yshift=-1em] {\color{blue}$a$};
+    % 
+    \draw[decorate,decoration={brace,
+      amplitude=8pt},xshift=0pt,very thin,gray] (Y) -- (Y_bar) node [black,midway,xshift=1em,yshift=1em] {\color{blue}$c$};
   \end{scope}
+  % Coordinate system
   \begin{scope}[canvas is xy plane at z=0]
-    \draw[->] (2,0) -- ++(-0.75,0.75) node [left] {$\mathbf{1}$};
+    \draw[->] (2,1) -- node [above] {$\mathbf{1}$} ++(-1,0) ;
-    \draw[->] (2,0) -- ++(-1,-0.5);
+    \draw[->] (2,1) -- ++(-0.45,-1) node [right] {$X_1$};
   \end{scope}
 \end{tikzpicture}
 \end{document}

main.tex

@@ -28,9 +28,6 @@
 \definecolor{myblue}{HTML}{5654fa}
 \colorlet{primary}{myblue}
 
-\input{definitions}
-\input{preamble}
-
 \hypersetup{
   pdftitle={Course - Multivariate Statistics},
   pdfauthor={Samuel Ortion},
@@ -51,6 +48,7 @@
 \input{glossary}
 \input{definitions}
 
+
 \makeindex%
 \makeglossary%
 \begin{document}