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Let's assume X is a random variable with an even pdf.

To show that X and -X are identically distributed, we need to prove that $F_X(x)=F_-X(x)$. We also know that X having an even pdf means $f_X(x)=-f_X(x)$. What I did here is that I took the integral of $f_X(x)$ and $-f_X(x)$ both from negative infinity to x and set them equal to each other. When we integrate this we should get $F_X(x)=F_-X(x)$, right?

I'm not so sure how to show that $M_x(t)$, or the moment is even. Do we just do $M_x(t)=-M_x(-t)$ and infer our results from there?

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    $\begingroup$ you mean $f_X(x) = f_X(-x)$ $\endgroup$
    – Ant
    Sep 27, 2015 at 17:16

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It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:

\begin{eqnarray*} M_X(-t) &=& \; E(e^{-tx}) \\ &=& \; E(e^{t(-x)}) \\ &=& \; M_{-X}(t) \\ &=& \; M_X(t)\qquad\qquad\text{since $X$ and $-X$ are identically distributed.} \end{eqnarray*}

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