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+%% -*- mode: latex; mode: reftex; mode: flyspell; coding: utf-8; tex-command: "pdflatex.sh" -*-
+
+%% Any copyright is dedicated to the Public Domain.
+%% https://creativecommons.org/publicdomain/zero/1.0/
+%% Written by Francois Fleuret <francois@fleuret.org>
+
+\documentclass[11pt,a4paper,oneside]{article}
+\usepackage[paperheight=15cm,paperwidth=8cm,top=2mm,bottom=15mm,right=2mm,left=2mm]{geometry}
+%\usepackage[a4paper,top=2.5cm,bottom=2cm,left=2.5cm,right=2.5cm]{geometry}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath,amssymb,dsfont}
+\usepackage[pdftex]{graphicx}
+\usepackage[colorlinks=true,linkcolor=blue,urlcolor=blue,citecolor=blue]{hyperref}
+\usepackage{tikz}
+\usetikzlibrary{arrows,arrows.meta,calc}
+\usetikzlibrary{patterns,backgrounds}
+\usetikzlibrary{positioning,fit}
+\usetikzlibrary{shapes.geometric,shapes.multipart}
+\usetikzlibrary{patterns.meta,decorations.pathreplacing,calligraphy}
+\usetikzlibrary{tikzmark}
+\usetikzlibrary{decorations.pathmorphing}
+\usepackage[round]{natbib}
+\usepackage[osf]{libertine}
+\usepackage{microtype}
+
+\usepackage{mleftright}
+
+\newcommand{\setmuskip}[2]{#1=#2\relax}
+\setmuskip{\thinmuskip}{1.5mu} % by default it is equal to 3 mu
+\setmuskip{\medmuskip}{2mu} % by default it is equal to 4 mu
+\setmuskip{\thickmuskip}{3.5mu} % by default it is equal to 5 mu
+
+\setlength{\parindent}{0cm}
+\setlength{\parskip}{1ex}
+%\renewcommand{\baselinestretch}{1.3}
+%\setlength{\tabcolsep}{0pt}
+%\renewcommand{\arraystretch}{1.0}
+
+\def\argmax{\operatornamewithlimits{argmax}}
+\def\argmin{\operatornamewithlimits{argmin}}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\def\given{\,\middle\vert\,}
+\def\proba{\operatorname{P}}
+\newcommand{\seq}{{S}}
+\newcommand{\expect}{\mathds{E}}
+\newcommand{\variance}{\mathds{V}}
+\newcommand{\empexpect}{\hat{\mathds{E}}}
+\newcommand{\mutinf}{\mathds{I}}
+\newcommand{\empmutinf}{\hat{\mathds{I}}}
+\newcommand{\entropy}{\mathds{H}}
+\newcommand{\empentropy}{\hat{\mathds{H}}}
+\newcommand{\ganG}{\mathbf{G}}
+\newcommand{\ganD}{\mathbf{D}}
+\newcommand{\ganF}{\mathbf{F}}
+
+\newcommand{\dkl}{\mathds{D}_{\mathsf{KL}}}
+\newcommand{\djs}{\mathds{D}_{\mathsf{JS}}}
+
+\newcommand*{\vertbar}{\rule[-1ex]{0.5pt}{2.5ex}}
+\newcommand*{\horzbar}{\rule[.5ex]{2.5ex}{0.5pt}}
+
+\def\positionalencoding{\operatorname{pos-enc}}
+\def\concat{\operatorname{concat}}
+\def\crossentropy{\LL_{\operatorname{ce}}}
+
+\begin{document}
+
+\vspace*{0ex}
+
+\begin{center}
+{\Large On Random Variables}
+
+Fran\c cois Fleuret
+
+\today
+
+\vspace*{1ex}
+
+\end{center}
+
+\underline{Random variables} are central to any model of a random
+process, but their mathematical definition is unclear to most. This is
+an attempt at giving an intuitive understanding of their definition
+and utility.
+
+\section{Modeling randomness}
+
+To formalize something ``random'', the natural strategy is to define a
+distribution, that is, in the finite case, a list of values /
+probabilities. For instance, the head / tail result of a coin flipping
+would be
+%
+\[
+\{(H, 0.5), (T, 0.5)\}.
+\]
+
+This is perfectly fine, until you have several such objects. To model
+two coins $A$ and $B$, it seems intuitively okay: they have nothing to
+do with each other, they are ``independent'', so defining how they
+behave individually is sufficient.
+
+\section{Non-independent variables}
+
+The process to generate two random values can be such that they are
+related. Consider for instance that $A$ is the result of flipping a
+coin, and $B$ as *the inverse value of $A$*.
+
+Both $A$ and $B$ are legitimate RVs, a both have the same distribution
+(H, 0.5) (T, 0.5). So where is the information that they have a
+relation?
+
+With models of the respective distributions of $A$ and $B$, this is
+nowhere. This can be fixed in some way by specifying the distribution
+of the pair $(A, B)$. That would be here
+%
+\[
+\{(H/H, 0.0), (H/T, 0.5), (T/H, 0.5), (T/T, 0.0)\}.
+\]
+
+The distribution of $A$ and $B$ individually are called the
+\underline{marginal} distributions, and this is the \underline{joint}
+distribution.
+
+Note that the joint is a far richer object than the two marginals, and
+in general many different joints are consistent with given marginals.
+Here for instance, the marginals are the same as if $A$ and $B$ where
+two independent coins, even though they are not.
+
+Even though this could somehow work, the notion of a RV here is very
+unclear: it is not simply a distribution, and every time a new one is
+defined, it require the specification of the joint with all the
+variables already defined.
+
+\section{Random Variables}
+
+The actual definition of a RV is a bit technical. Intuitively, in some
+way, it consists of defining first ``the source of all randomness'',
+and then every RV is a deterministic function of it.
+
+Formally, it relies first on the definition of a set $\Omega$ such
+that its subsets can be measured, with all the desirable properties,
+such as $\mu(\Omega)=1, \mu(\emptyset)=0$ and $A \cap B = \emptyset
+\Rightarrow \mu(A \cup B) = \mu(A) + \mu(B)$.
+
+There is a technical point: for some $\Omega$ it may be impossible to
+define such a measure on all its subsets due to tricky
+infinity-related pathologies. So the set $\Sigma$ of
+\underline{measurable} subsets is explicitly specified and called a
+$\sigma$-algebra. In any practical situation this technicality does
+not matter, since $\Sigma$ contains anything needed.
+
+The triplet $(\Omega, \Sigma, \mu)$ is a \underline{measured set}.
+
+Given such a measured set, an \underline{random variable} $X$ is a
+mapping from $\Omega$ into another set, and the
+\underline{probability} that $X$ takes the value $x$ is the measure of
+the subset of $\Omega$ where $X$ takes the value $x$:
+%
+\[
+P(X=x) = \mu(X^{-1}(x))
+\]
+
+You can imagine $\Omega$ as the square $[0,1]^2$ in $\mathbb{R}^2$
+with the usual geometrical area for $\mu$.
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+For instance if the two coins $A$ and $B$ are flipped independently, we
+could picture possible random variables with the proper distribution
+as follows:
+
+\nopagebreak
+
+\begin{tikzpicture}[scale=0.8]
+\draw[pattern=north east lines] (0,0) rectangle ++(0.5,0.5);
+\draw (0,0) rectangle ++(1,0.5);
+\node at (2.5,0.2) {$A=\text{head}/\text{tail}$};
+
+\draw[fill=red!50] (4.5, 0) rectangle ++(0.5,0.5);
+\draw (4.5,0) rectangle ++(1,0.5);
+\node at (7.0,0.2) {$B=\text{head}/\text{tail}$};
+\end{tikzpicture}
+%
+
+\nopagebreak
+
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50,draw=none] (0, 0) rectangle (2, 4);
+\draw[draw=none,pattern=north east lines] (0, 0) rectangle (4,2);
+\draw (0,0) rectangle (4,4);
+
+%% \draw[draw=green,thick] (0,0) rectangle ++(2,2);
+%% \draw[draw=green,thick] (0.1,2.1) rectangle ++(1.8257,1.8257);
+%% \draw[draw=green,thick] (2.1,0.1) rectangle ++(0.8165,0.8165);
+
+\end{tikzpicture}
+%
+\hspace*{\stretch{1}}
+%
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50,draw=none] (0, 0) rectangle ++(1, 4);
+\draw[fill=red!50,draw=none] (1.5, 0) rectangle ++(1, 4);
+\draw[draw=none,pattern=north east lines] (0, 0.25) rectangle ++(4,0.5);
+\draw[draw=none,pattern=north east lines] (0, 1.25) rectangle ++(4,0.5);
+\draw[draw=none,pattern=north east lines] (0, 2.) rectangle ++(4,0.5);
+\draw[draw=none,pattern=north east lines] (0, 2.5) rectangle ++(4,0.5);
+\draw (0,0) rectangle (4,4);
+\end{tikzpicture}
+%
+\hspace*{\stretch{1}}
+%
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50,draw=none] (0, 0) rectangle (2, 2);
+\draw[fill=red!50,draw=none] (0, 4)--(2,4)--(4,2)--(2,2)--cycle;
+\draw[draw=none,pattern=north east lines] (0.5, 4)--(1.5,4)--(3.5,2)--(2.5,2)--cycle;
+\draw[draw=none,pattern=north east lines] (3, 3) rectangle (4,4);
+\draw[draw=none,pattern=north east lines] (0,4)--(1,3)--(0,2)--cycle;
+\draw[draw=none,pattern=north east lines] (2.25,0) rectangle (3.25,2);
+\draw[draw=none,pattern=north east lines] (0, 0) rectangle (2,1);
+\draw (0,0) rectangle (4,4);
+\end{tikzpicture}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+And if $A$ is flipped and $B$ is the inverse of $A$, possible RV would
+be
+
+\nopagebreak
+
+\begin{tikzpicture}[scale=0.8]
+%% \node at (3.2, 1) {Flip A and B = inverse(A)};
+
+\draw[pattern=north east lines] (0,0) rectangle ++(0.5,0.5);
+\draw (0,0) rectangle ++(1,0.5);
+\node at (2.5,0.2) {$A=\text{head}/\text{tail}$};
+
+\draw[fill=red!50] (4.5, 0) rectangle ++(0.5,0.5);
+\draw (4.5,0) rectangle ++(1,0.5);
+\node at (7.0,0.2) {$B=\text{head}/\text{tail}$};
+\end{tikzpicture}
+
+\nopagebreak
+
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50] (0,0) rectangle (4,4);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 0) rectangle (2,4);
+\draw (0,0) rectangle (4,4);
+\end{tikzpicture}
+%
+\hspace*{\stretch{1}}
+%
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50] (0,0) rectangle (4,4);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 0) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (1, 0) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (3, 0) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 1) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (2, 1) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 2) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (1, 3) rectangle ++(1,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (2, 3) rectangle ++(1,1);
+\draw (0,0) rectangle (4,4);
+\end{tikzpicture}
+%
+\hspace*{\stretch{1}}
+%
+\begin{tikzpicture}[scale=0.600]
+\draw[fill=red!50] (0,0) rectangle (4,4);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 0)--(1,1)--(3,1)--(3,4)--(0,1)--cycle;
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (0, 3) rectangle ++(2,1);
+\draw[preaction={fill=white},draw=none,pattern=north east lines] (3,0) rectangle ++(1,1);
+%% \draw (0,0) grid (4,4);
+\draw (0,0) rectangle (4,4);
+\end{tikzpicture}
+
+%% Thanks to this definition, additional random variables can be defined
+%% with dependency structures. For instance, if $A$ and $B$ are two
+%% separate coin flipping, and then a third variable $C$ is defined by
+%% rolling a dice and taking the value of $A$ if it gives $1$ and the
+%% value of $B$ otherwise.
+
+\end{document}