prove that $E(X^{n}) = 0 $

Question

Prove that for any odd $n \in \mathbb{N}$ and $X \sim \mathcal{N}(0,\sigma^{2})$, $E (X^n) = 0$.

I started with the fact that: $$E(X^{n}) = {{1}\over{\sqrt{2\pi} \sigma}} {\int_{-\infty}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx = {{2}\over{\sqrt{2\pi} \sigma}} {\int_{0}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx.$$ Then I used the substitution $$t={{x}\over{\sigma}} \to {{2\sigma^n}\over{\sqrt{2\pi}}} {\int_{0}^{\infty} t^{n} e^{{-t^2}/{2}}} dt.$$ But I have a problem how to solve it. Does anyone know how this equality can be proved? The task tip says to use the Gamma function in the calculation, but I don't see it..

Answer 1

You can use the moment generating function, express it in tayor series and take the derivative.

$$M_X(t)=e^{\frac{\sigma^2t^2}{2}}$$

and

$$\mathbb{E}[X^n]=\frac{d^n}{dt^n}e^{\frac{\sigma^2t^2}{2}}\Bigg|_{t=0}$$

Expanding MGF in taylor series you get

$$e^{\frac{\sigma^2t^2}{2}}=\sum_{n=0}^{\infty}\frac{\Big(\frac{\sigma^2t^2}{2}\Big)^n}{n!}=\sum_{n=0}^{\infty}\frac{\sigma^{2n}\cdot t^{2n}}{2^n\cdot n!}$$

Thus it is evident that, for every integer $n\geq 0$

$$\mathbb{E}[X^{2n}]=\frac{\sigma^{2n}(2n)!}{2^n\cdot n!}$$

and

$$\mathbb{E}[X^{2n+1}]=0$$

Answer 2

In fact, for any symmetric distribution $X$, $E[X^{2n+1}] = 0$ for all $n \in \mathbb{N}$. Simply split into positive and negative parts and observe that the probabilistic density function $f_X$ is symmetric on $\mathbb{R}$. $$\begin{aligned} E\left[X^{2n+1}\right] &= \int_{\mathbb{R}} x^{2n+1}f_X(x) \,\mathrm{d}x \\ &= \int_{-\infty}^0 x^{2n+1} f_X(x) \,\mathrm{d}x + \int_0^{+\infty} x^{2n+1} f_X(x) \,\mathrm{d}x \\ &= 0 \end{aligned}$$

In the first integral, make the substitution $y = -x$, and you'll see that it cancels the second integral.

$$\int_{-\infty}^0 x^{2n+1} f_X(x) \,\mathrm{d}x = \int_{0}^{+\infty} -y^{2n+1} f_X(y) \,\mathrm{d}y$$