symmetric r.v. raised to an odd power

Question

My prof claims that raising a symmetric r.v., like N(0,1), to an odd power gives a distribution with expectation 0. What's the best way to see this?

Answer 1

The assertion that a symmetric random variable raised to an odd power has mean 0 is not true in general.

(However, it is true that a standard normal raised to an odd power has mean 0.)

To make a general statement, first we'll need to restrict the original variable to have mean zero. That is, our statement would start "a symmetric random variable with mean 0, raised to an odd power..."

Then, we also need the expectation of that odd power to be finite; this is not the case in general. Consider for example the 5th power of a t-distributed random variable (with $\nu=3$, say$).

[Edit: I've only just noticed that whuber offers the same distribution as a counter example (but as a class of distributions, not just a single combination of power and df).]

The integral doesn't converge to a finite number. Since we have symmetry, if $\int_{0}^\infty x^p f(x) dx$ is finite (equal to $a$, say), then the mean of the odd power will be 0 by symmetry:

$E(X^p)=\int_{-\infty}^\infty x^p f(x)\, dx\:$ (law of the unconscious statistician)

$\qquad\quad=\int_{-\infty}^0 x^p f(x)\, dx +\int_{0}^\infty x^p f(x)\, dx$

$\qquad\quad=\int_{-\infty}^0 x^p f(-x)\, dx +\int_{0}^\infty x^p f(x)\, dx\:$ (by symmetry of $f$)

Then let $y=-x$, $dy=-dx$ (watch the negatives carefully, you get three come into it)

$\qquad\quad=-\int_{0}^{\infty} x^p f(x) dx +\int_{0}^\infty x^p f(x)\, dx$

$\qquad\quad=-a+a=0$