From 2ea3ddb978f549400ea7f74127db3760f7f15d05 Mon Sep 17 00:00:00 2001 From: Eduardo Horta Date: Fri, 23 Jul 2021 14:16:32 -0300 Subject: [PATCH] Update 60E99-SymmetricRandomVariable.tex The previous version of the text claimed that symmetric random variables are always integrable. This assertion is false, take for example a standard Cauchy random variable. --- 60E99-SymmetricRandomVariable.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/60E99-SymmetricRandomVariable.tex b/60E99-SymmetricRandomVariable.tex index 30dd414..ef14b20 100644 --- a/60E99-SymmetricRandomVariable.tex +++ b/60E99-SymmetricRandomVariable.tex @@ -39,7 +39,7 @@ Let $(\Omega,\mathcal{F},P)$ be a probability space and $X$ a real random variable defined on $\Omega$. $X$ is said to be \emph{symmetric} if $-X$ has the same distribution function as $X$. A distribution function $F:\mathbb{R}\to [0,1]$ is said to be \emph{symmetric} if it is the distribution function of a symmetric random variable. -\textbf{Remark}. By definition, if a random variable $X$ is symmetric, then $E[X]$ exists ($<\infty$). Furthermore, $E[X]=E[-X]=-E[X]$, so that $E[X]=0$. Furthermore, let $F$ be the distribution function of $X$. If $F$ is continuous at $x\in\mathbb{R}$, then $$F(-x)=P(X\le -x)=P(-X\le -x)=P(X\ge x)=1-P(X\le x)=1-F(x),$$ so that $F(x)+F(-x)=1$. This also shows that if $X$ has a density function $f(x)$, then $f(x)=f(-x)$. +\textbf{Remark}. If a random variable $X$ is symmetric and integrable (that is, $E|X|$ is finite), then $E[X]=E[-X]=-E[X]$, so that $E[X]=0$. Furthermore, let $F$ be the distribution function of $X$. If $F$ is continuous at $x\in\mathbb{R}$, then $$F(-x)=P(X\le -x)=P(-X\le -x)=P(X\ge x)=1-P(X\le x)=1-F(x),$$ so that $F(x)+F(-x)=1$. This also shows that if $X$ has a density function $f(x)$, then $f(x)=f(-x)$. Notice that symmetry does not imply integrability, e.g. a standard Cauchy random variable is symmetric but not integrable. There are many examples of symmetric random variables, and the most common one being the normal random variables centered at $0$. For any random variable $X$, then the difference $\Delta X$ of two independent random variables, identically distributed as $X$ is symmetric. %%%%%