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Is it true that if the probability density function of a continuous random variable is an even function, then the continuous random variable is symmetric?

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  • $\begingroup$ You get that this is just saying an even function is symmetric around zero? The only connection to probability is the definition of a continuous random variable being symmetric (which is that its pdf is symmetric at some point - which in this case happens to be zero). $\endgroup$ Feb 12 at 13:28

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Yes, this is true. Consider a probability density function $p(x)$ that is even:

By definition, the probability density function is symmetric if there exists $x_0$ s.t $ p(x_0+δ)=p(x_0-δ)$ for all real numbers δ. But take $x_0=0$. Then it follows immediately that if p is even then $p(δ)=p(-δ)$ for all real $δ$. So clearly the probability density function is symmetric.

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  • $\begingroup$ Strictly speaking you showed that for $x_0\not=0$ the PDF is not even. $\endgroup$
    – Kurt G.
    Feb 12 at 10:52
  • $\begingroup$ @KurtG. The statement is still true right? $\endgroup$
    – adisnjo
    Feb 12 at 10:59
  • $\begingroup$ Hmmm, how so ? If I assume that p(x) is even (I have added that for clarity), does that not show the symmetry property or is this answer missing something? $\endgroup$
    – J.Dmaths
    Feb 12 at 10:59
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    $\begingroup$ Sorry I got it wrong. All correct: if $p$ is even then RV is symmetric. I had the converse implication in mind. $\endgroup$
    – Kurt G.
    Feb 12 at 11:01

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