feat: Add some basic (but probably uncomplete) schemes

content/chapters/part1/1.tex

@@ -13,7 +13,7 @@ with $g$ being
 
 \subsection{Penalized Regression}
 
-When the number of variables is large, e.g, when the number of explicative variable is above the number of observations, if $p >> n$ ($p$: the number of explicative variable, $n$ is the number of observations), we cannot estimate the parameters.
+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.
@@ -21,8 +21,7 @@ Lasso regression, Elastic Net, etc.
 \subsection{Simple Linear Model}
 
 \begin{align*}
-    \Y &= \X & \beta & + & \varepsilon.\\
+    \Y &= \X \beta + \varepsilon \\
-    n \times 1 & n \times 2 & 2 \times 1 & + & n \times 1 \\
     \begin{pmatrix}
         Y_1 \\
         Y_2 \\
@@ -35,11 +34,11 @@ Lasso regression, Elastic Net, etc.
         \vdots & \vdots \\
         1 & X_n
     \end{pmatrix}
-    & \begin{pmatrix}
+    \begin{pmatrix}
         \beta_0 \\
         \beta_1
     \end{pmatrix}
-    & + & 
+    + 
     \begin{pmatrix}
         \varepsilon_1 \\
         \varepsilon_2 \\
@@ -61,8 +60,35 @@ Lasso regression, Elastic Net, etc.
     \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}
@@ -87,26 +113,26 @@ We want to minimize the distance between $\X\beta$ and $\Y$:
     \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{red}(\X^T \X)^{-1}} \X^\T \Y = {\color{red}(\X^T \X)^{-1}} (\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 subspace define by the explanatory variables $\X$
-This formula comes from the orthogonal projection of $\Y$ on the subspace define by the explicative 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{Coefficient of Determination: \texorpdfstring{$R^2$}{R\textsuperscript{2}}}
-\section{Coefficient of Determination: $R^2$}
 \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 $\Y$ explicated by the model.
+    \] 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 $\norm{\Y - \hat{\beta}\X}^2$}
+    \label{fig:scheme-orthogonal-projection}
+\end{figure}

content/chapters/part1/2.tex

@@ -15,8 +15,12 @@ Let $u = \begin{pmatrix}
         v_n
     \end{pmatrix}$
 
+\begin{definition}[Scalar Product (Dot Product)]
     \begin{align*}
-    \langle u, v\rangle & = \left(u_1, \ldots, u_v\right) \begin{pmatrix}
+        \scalar{u, v} & = \begin{pmatrix}
+            u_1, \ldots, u_v
+        \end{pmatrix}
+        \begin{pmatrix}
             v_1    \\
             \vdots \\
             v_n
@@ -24,6 +28,15 @@ Let $u = \begin{pmatrix}
     & = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n
     \end{align*}
 
+    We may use $\scalar{u, v}$ or $u \cdot v$ notations.
+\end{definition}
+\paragraph{Dot product properties}
+\begin{itemize}
+    \item $\scalar{u, v} = \scalar{v, u}$
+    \item $\scalar{(u+v), w} = \scalar{u, w} + \scalar{v, w}$
+    \item $\scalar{u, v}$
+    \item $\scalar{\vec{u}, \vec{v}} = \norm{\vec{u}} \times \norm{\vec{v}} \times \cos(\widehat{\vec{u}, \vec{v}})$
+\end{itemize}
 
 \begin{definition}[Norm]
     Length of the vector.
@@ -41,9 +54,7 @@ Let $u = \begin{pmatrix}
 \end{definition}
 
 \begin{definition}[Orthogonality]
-    \[
+
-        u \perp v \Leftrightarrow \scalar{u, v} = 0
-    \]
 \end{definition}
 
 \begin{remark}
@@ -55,13 +66,12 @@ Let $u = \begin{pmatrix}
     \]
 \end{remark}
 
-Scalar product properties:
+\begin{figure}
-\begin{itemize}
+    \centering
-    \item $\scalar{u, v} = \scalar{v, u}$
+    \includestandalone{figures/schemes/vector_orthogonality}
-    \item $\scalar{(u+v), w} = \scalar{u, w} + \scalar{v, w}$
+    \caption{Illustration for the scalar product of two orthogonal vectors.}
-    \item $\scalar{u, v}$
+    \label{fig:scheme-orthogonal-scalar-product}
-    \item $\scalar{\vec{u}, \vec{v}} = \norm{\vec{u}} \times \norm{\vec{v}} \times \cos(\widehat{\vec{u}, \vec{v}})$
+\end{figure}
-\end{itemize}
 
 \begin{align*}
     \scalar{v-u, v-u} & = \scalar{v, v} + \scalar{u, u} - 2 \scalar{u, v} \\
@@ -74,10 +84,14 @@ Scalar product properties:
     \norm{u + v}^2 & = \norm{u}^2 + \norm{v}^2 + 2 \scalar{u,v}
 \end{align*}
 
+\begin{proposition}[Scalar product of orthogonal vectors]
+\[
+    u \perp v \Leftrightarrow \scalar{u, v} = 0
+\]
+\end{proposition}
 
-If $u \perp v$, then $\scalar{u, v} = 0$
 \begin{proof}[Indeed]
-    $\norm{u-v}^2 = \norm{u+v}^2$,
+    $\norm{u-v}^2 = \norm{u+v}^2$, as illustrated in \autoref{fig:scheme-orthogonal-scalar-product}.
     \begin{align*}
         \Leftrightarrow & -2 \scalar{u, v} = 2 \scalar{u, v} \\
         \Leftrightarrow & 4 \scalar{u, v} = 0                \\
@@ -146,7 +160,7 @@ The scalar product between $z$ and (?) is zero.
             x_2
         \end{pmatrix}        \\
         & = \begin{pmatrix}
-                   a x_1 + b_x2 \\
+                a x_1 + b x_2 \\
                 c x_1 + d x_2
             \end{pmatrix}
     \end{align*}
@@ -182,31 +196,8 @@ The number of columns has to be the same as the dimension of the vector to which
     \end{pmatrix}$
 \end{definition}
 
-\begin{example}
+\begin{figure}
-    \begin{align*}
+    \centering
-        Y & = X \beta + \varepsilon \\
+    \includestandalone{figures/schemes/coordinates_systems}
-        \begin{pmatrix}
+    \caption{Coordinate systems}
-            y_1 \\
+\end{figure}
-            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}
-    \end{align*}
-\end{example}

figures/schemes/base_plan.tex

@@ -0,0 +1,16 @@
+\documentclass[margin=0.5cm]{standalone}
+\usepackage{tikz}
+\usepackage{tkz-euclide}
+
+\begin{document}
+\usetikzlibrary{3d}
+\begin{tikzpicture}
+  \tkzDefPoint(-2,-2){A}
+  \tkzDefPoint(10:3){B}
+  \tkzDefShiftPointCoord[B](1:5){C}
+  \tkzDefShiftPointCoord[A](1:5){D}
+  \tkzDrawPolygon(A,...,D)
+  \tkzDrawPoints(A,...,D)
+  \node at (A) {A};
+\end{tikzpicture}
+\end{document}

figures/schemes/coordinates_systems.tex

@@ -1,12 +1,23 @@
 \documentclass[tikz]{standalone}
 \usepackage{tikz}
-\usepackage{tkz-euclide}
 
 \begin{document}
-\begin{tikzpicture}
+\usetikzlibrary{3d}
-    \tkzInit[xmax=5,ymax=5,xmin=-5,ymin=-5]
+% 1D axis
-    \tkzGrid
+\begin{tikzpicture}[->]
-    \tkzAxeXY
+    \begin{scope}[xshift=0]
-    \draw[thick, latex-latex] (-1,4) -- (4,-6) node[anchor=south west] {$a$};
+    \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$}; 
+    \end{scope}
+% 2D coordinate system
+    \begin{scope}[xshift=50]
+    \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$};
+    \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1,angle=90) node[above] {$y$};
+    \end{scope}
+% 3D coordinate systems
+\begin{scope}[xshift=100]
+    \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$};
+    \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1,angle=90) node[above] {$y$}; 
+    \draw (0, 0, 0) -- (xyz cylindrical cs:z=1) node[below left] {$z$};
+\end{scope}
 \end{tikzpicture}
 \end{document}

figures/schemes/orthogonal_projection.tex

@@ -0,0 +1,32 @@
+% ref. https://tex.stackexchange.com/a/523362/235607
+\documentclass[tikz]{standalone}
+\usepackage{tikz-3dplot}
+\usepackage{tkz-euclide}
+\usepackage{mathtools}
+\begin{document}
+\tdplotsetmaincoords{50}{0}
+\begin{tikzpicture}[tdplot_main_coords,bullet/.style={circle,inner
+        sep=1pt,fill=black,fill opacity=1}]
+  \begin{scope}[canvas is xy plane at z=0]
+    \tkzDefPoints{-2/-1/A,3/-1/B,4/2/C}
+    \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[thick,fill=white,fill opacity=0.7,nodes={opacity=1}]
+    (2,0) node[bullet,label=below right:{$\mathbf{X}$}] {}
+    -- (0,0) node[bullet] {}
+    -- (0,3) node[bullet,label=above:{$\mathbf{Y}$}] {} -- cycle;
+    \draw (0.25,0) -- (0.25,0.25) -- (0,0.25);
+    \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}$b$};
+  \end{scope}
+  \begin{scope}[canvas is xy plane at z=0]
+    \draw[->] (2,0) -- ++(-0.75,0.75) node [left] {$\mathbf{1}$};
+    \draw[->] (2,0) -- ++(-1,-0.5);
+  \end{scope}
+\end{tikzpicture}
+\end{document}

figures/schemes/vector_orthogonality.tex

@@ -0,0 +1,27 @@
+\documentclass[margin=0.5cm]{standalone}
+\usepackage{tikz}
+\usepackage{tkz-euclide}
+\usepackage{mathtools}
+
+\begin{document}
+\begin{tikzpicture}
+    \coordinate (A) at (0.5, 1) {};
+    \coordinate (B) at (-0.5, -1) {};
+    \coordinate (C) at (1.25, -0.70) {};
+    \coordinate (0) at (0, 0) {};
+
+    % left angle
+    \tkzMarkRightAngle[draw=black,size=0.1](A,0,C);
+    \draw[lightgray,very thin] (A) -- (C);
+    % Curly brace annotation for ||u-v||
+    \draw[decorate,decoration={brace,
+    amplitude=10pt},xshift=0pt,yshift=4pt,very thin] (A) -- (C) node [black,midway,xshift=27pt,yshift=0.5em] {$\lVert u-v \rVert$};
+    \draw[lightgray,very thin] (B) -- (C);
+
+    % axis lines
+    \draw[->] (0) -- (A) node[above] {$u$};
+    \draw[->] (0) -- (B) node[below] {$-u$};
+    \draw[->] (0) -- (C) node[right] {$v$};
+
+\end{tikzpicture}
+\end{document}

preamble.tex

@@ -1,3 +1,6 @@
 \usepackage{pgffor}
 \usetikzlibrary{math}
 \usepackage{standalone}
+\usepackage{tikz-3dplot}
+\usepackage{tkz-euclide}
+\usepackage{mathtools}