Calculate $E(X^{2n})$ where $X$ is normal (0,1)

Question

I need help proving the following:

Let $X$ be normally distributed with parameters $\sigma=0$ and $\mu=1$. Let $n$ be a positive integer. Show that:

$$E(X^{2n})=\frac{(2n)!}{2^nn!}=:(2n-1)!!$$

I've tried the change of variables $Y=X^2$ and then calculating the integral, but got nowhere. Any hints?

Thanks in advance!

does it help to use the definition of $\Gamma$ function? :-) or you need a full solution from scratch on your own using simple calculus rules? — Math-fun, May 23 '16 at 15:21
This is related to the higher-order moments of the normal distribution. — hardmath, May 23 '16 at 15:28
Related: http://math.stackexchange.com/questions/1669735/expectation-of-a-standard-normal-random-variable/1669837#1669837 — heropup, May 23 '16 at 15:35

score 2 · Accepted Answer · answered May 23 '16 at 15:33

2

$$\int_{-\infty}^{\infty}x^{2n}e^{-\frac{1}{2}x^{2}}dx=2\int_{0}^{\infty}x^{2n}e^{-\frac{1}{2}x^{2}}dx=-2\int_{0}^{\infty} x^{2n-1}de^{-\frac{1}{2}x^{2}}=$$$$2\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}dx^{2n-1}-2\left[x^{2n-1}e^{-\frac{1}{2}x^{2}}\right]_{0}^{\infty}=2\left(2n-1\right)\int_{0}^{\infty}x^{2n-2}e^{-\frac{1}{2}x^{2}}dx$$

answered May 23 '16 at 15:33

drhab

153,781

You are welcome. – drhab May 23 '16 at 17:04

score 1 · Answer 2 · answered May 23 '16 at 15:24

1

Hint :

Do a proof by induction using integration by parts by writing:

$x^{2n+2} \exp(-\frac{x^2}{2}) = (-x^{2n+1})(-x\exp(-\frac{x^2}{2}))$

answered May 23 '16 at 15:24

Dark

1,610

score 0 · Answer 3 · answered May 23 '16 at 15:36

You can prove this by induction on $n$, starting at $n=0$. Hint: If $N\ge0$ then integration by parts shows that $$\int_{-\infty}^\infty t^Ne^{-t^2/2}\,dt=\frac 1{N+1}\int_{-\infty}^\infty t^{N+2}e^{-t^2/2}\,dt.$$

Now that that's been done for you, you should derive a similar formula for $\Bbb E[|X|^{2n+1}]$.

Calculate $E(X^{2n})$ where $X$ is normal (0,1)

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