I have a homework problem in which I'm not certain where to start:
Let $X$ be a random variable with $N (0, 1)$ distribution. Show that
$E(X^n) =\left\lbrace{\begin{array}{cc} 0 & \text{if $n$ is odd} \\ (n-1)(n-3)....3.1 & \text{if $n$ is even} \end{array}}\right. $
Any pointers much appreciated.
The pdf of $X \sim N(0, 1)$ is $f(x) = \dfrac{1}{\sqrt {2\pi}} e^{-x^2/2}$. Therefore,
$\text{E}(X^n) = \dfrac{1}{\sqrt{2\pi}}\displaystyle\int_{-\infty}^{\infty}x^n e^{-x^2/2} \, dx$.
Now, for odd $n$, the integrand is an odd function, therefore the value of the integral is $0$ (as the integration is over a symmetric interval). For even $n$, the integrand is even, so we can write the integral as
$\text{E}(X^n) = \dfrac{1}{\sqrt{2\pi}} 2 \times \displaystyle\int_0^{\infty}x^n e^{-x^2/2} \, dx = \sqrt{\dfrac{2}{\pi}} \displaystyle\int_0^{\infty}x^n e^{-x^2/2} \, dx$
Let $t = x^2/2$. Then $dt = x\,dx \Rightarrow dx = dt/x = dt/\sqrt t$, and $0 < t < \infty$.
$\begin{align} \text{E}(X^n) & = \sqrt{\dfrac{2}{\pi}} \displaystyle\int_0^{\infty}(2t)^{n/2} e^{-t} \dfrac{1}{\sqrt t} \, dt\\ & = 2^{n/2} \sqrt{\dfrac{2}{\pi}} \displaystyle\int_0^{\infty}t^{(n - 1)/2} e^{-t} \, dt\\ & = 2^{n/2} \sqrt{\dfrac{2}{\pi}} \Gamma\left(\dfrac{n - 1}{2} + 1\right)\\ & = 2^{n/2} \sqrt{\dfrac{2}{\pi}} \left(\dfrac{n - 1}{2}\right) \Gamma\left(\dfrac{n - 1}{2}\right)\\ & = 2^{n/2} \sqrt{\dfrac{2}{\pi}} \left(\dfrac{n - 1}{2}\right) \left(\dfrac{n - 3}{2}\right) \cdots \left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{1}{2}\right)\\ & = 2^{n/2} \sqrt{\dfrac{2}{\pi}} \dfrac{(n - 1)(n - 3) \cdots 1}{2^{n/2}}\sqrt{\dfrac{\pi}{2}}\\ & = \boxed{(n - 1)(n - 3) \cdots 1} \end{align}$