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feat: Add some basic (but probably uncomplete) schemes

This commit is contained in:
Samuel Ortion 2023-09-22 23:48:46 +02:00
parent a92a13d354
commit ae51323390
12 changed files with 191 additions and 85 deletions
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62
content/chapters/part1/1.tex
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@ -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.\\
n \times 1 & n \times 2 & 2 \times 1 & + & n \times 1 \\
\Y &= \X \beta + \varepsilon \\
\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 explicative variables $\X$
This formula comes from the orthogonal projection of $\Y$ on the subspace define 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{Coefficient of Determination: $R^2$}
\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 $\Y$ 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}

111
content/chapters/part1/2.tex
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@ -15,15 +15,28 @@ Let $u = \begin{pmatrix}
v_n
\end{pmatrix}$
\begin{align*}
\langle u, v\rangle & = \left(u_1, \ldots, u_v\right) \begin{pmatrix}
v_1 \\
\vdots \\
v_n
\end{pmatrix} \\
& = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n
\end{align*}
\begin{definition}[Scalar Product (Dot Product)]
\begin{align*}
\scalar{u, v} & = \begin{pmatrix}
u_1, \ldots, u_v
\end{pmatrix}
\begin{pmatrix}
v_1 \\
\vdots \\
v_n
\end{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}
@ -55,18 +66,17 @@ Let $u = \begin{pmatrix}
\]
\end{remark}
Scalar 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{figure}
\centering
\includestandalone{figures/schemes/vector_orthogonality}
\caption{Illustration for the scalar product of two orthogonal vectors.}
\label{fig:scheme-orthogonal-scalar-product}
\end{figure}
\begin{align*}
\scalar{v-u, v-u} & = \scalar{v, v} + \scalar{u, u} - 2 \scalar{u, v} \\
& = \norm{v}^2 + \norm{u}^2 \\
& = -2 \scalar{u, v}
& = \norm{v}^2 + \norm{u}^2 \\
& = -2 \scalar{u, v}
\end{align*}
\begin{align*}
@ -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 \\
@ -138,17 +152,17 @@ The scalar product between $z$ and (?) is zero.
Then,
\begin{align*}
Ax & = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} \\
& = \begin{pmatrix}
a x_1 + b_x2 \\
c x_1 + d x_2
\end{pmatrix}
& = \begin{pmatrix}
a x_1 + b x_2 \\
c x_1 + d x_2
\end{pmatrix}
\end{align*}
Similarly,
@ -164,9 +178,9 @@ The scalar product between $z$ and (?) is zero.
x_3 \\
x_4
\end{pmatrix}
& = \begin{pmatrix}
a x_1 + b x_2 + c x_3 \ldots
\end{pmatrix}
& = \begin{pmatrix}
a x_1 + b x_2 + c x_3 \ldots
\end{pmatrix}
\end{align*}
\end{example}
@ -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{align*}
Y & = X \beta + \varepsilon \\
\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}
\end{align*}
\end{example}
\begin{figure}
\centering
\includestandalone{figures/schemes/coordinates_systems}
\caption{Coordinate systems}
\end{figure}

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figures/schemes/base_plan.pdf Normal file
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figures/schemes/base_plan.tex Normal file
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@ -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}

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figures/schemes/coordinates_systems.pdf
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figures/schemes/coordinates_systems.tex
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@ -1,12 +1,23 @@
\documentclass[tikz]{standalone}
\usepackage{tikz}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}
\tkzInit[xmax=5,ymax=5,xmin=-5,ymin=-5]
\tkzGrid
\tkzAxeXY
\draw[thick, latex-latex] (-1,4) -- (4,-6) node[anchor=south west] {$a$};
\usetikzlibrary{3d}
% 1D axis
\begin{tikzpicture}[->]
\begin{scope}[xshift=0]
\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}

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figures/schemes/orthogonal_projection.pdf Normal file
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figures/schemes/orthogonal_projection.tex Normal file
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@ -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}

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figures/schemes/vector_orthogonality.pdf Normal file
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figures/schemes/vector_orthogonality.tex Normal file
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@ -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}

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main.pdf
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5
preamble.tex
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@ -1,3 +1,6 @@
\usepackage{pgffor}
\usetikzlibrary{math}
\usepackage{standalone}
\usepackage{standalone}
\usepackage{tikz-3dplot}
\usepackage{tkz-euclide}
\usepackage{mathtools}