TITLE: Odd order moments of a symmetrical distribution
QUESTION [17 upvotes]: Is it true that for every symmetrical distribution all odd-order moments are equal to zero?
If yes, how would I be able to prove such a thing?

REPLY [15 votes]: I think it's easier than arguments about integrals and sums.  Saying a random variable $X$ has a symmetric distribution means that $X$ and $-X$ have the same distribution.  This means that for any function $f$, if $E[f(X)]$ exists then so does $E[f(-X)]$ and they are equal.  So take $f(x) = x^n$ with $n$ odd.  If $X$ has an $n$th moment, then $E[X^n]$ exists and so $$E[X^n] = E[(-X)^n] = E[(-1)^n X^n] = E[-(X^n)] = -E[X^n]$$ which means $E[X^n] = 0$.