Can someone explain why odd moments vanish for centered, symmetric-about-0 random variables?

Question

(as in the title). I'm particularly interested in whether there's an intuitive way to understand this. I've done a handful of calculations with moments, but the concept is still a bit new to me.

Answer 1

If $-X$ has the same distribution as $X$, then $-X^n = (-X)^n$ has the same distribution as $X^n$. Then $E[X^n] = E[-X^n] = -E[X^n]$ so...

Answer 2

My intuition about this is in two parts. First, because of the symmetry about $0$, the average of your random variable (the first moment) is $0$. If it were positive, then by symmetry it should be negative, and vice versa.

Second, what about higher odd moments, $E(X^k)$? Well, $X^k$ has the same symmetry property as $X$. Reversing the sign of $X$ also reverses the sign of $X^k$. So the previous paragraph applies to the random variable $X^k$ and shows that its expectation is also $0$.

(Note: All this is under the assumption that the moments exist.)