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Homesymmetric random variable

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# symmetric random variable

Let $(\Omega,\mathcal{F},P)$ be a probability space and $X$ a real random variable defined on $\Omega$. $X$ is said to be *symmetric* if $-X$ has the same distribution function as $X$. A distribution function $F:\mathbb{R}\to[0,1]$ is said to be *symmetric* if it is the distribution function of a symmetric random variable.

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

$F(-x)=P(X\leq-x)=P(-X\leq-x)=P(X\geq x)=1-P(X\leq x)=1-F(x),$ |

so that $F(x)+F(-x)=1$. This also shows that if $X$ has a density function $f(x)$, then $f(x)=f(-x)$.

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.

## Mathematics Subject Classification

60E99*no label found*60A99

*no label found*

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## Comments

## expectation of symmetric variables

The remark according to which if a random variable $X$ is symmetric

^{}, then $E[X]$ exists, does not seem to be correct. For example, if a random variable has a Cauchy–Lorentz distribution^{}, i.e., its probability density function iswith location parameter $x_{0}=0$ and scale parameter $\gamma>0$, then it is symmetric, but its mean is undefined.

I think that the correct statement is that if $X$ is symmetric *and* $E[X]$ exists, then $E[X]=0$.