# $X$ standard normal distribution, $E[X^k]=?$

I'm stuck with a homework problem where we are supposed to prove that the expected value $E[X^k]$, if $X$ has standard normal distribution, is equal to: $$E[X^{2k}]=\frac{(2k)!}{k!\cdot2^k}.$$ But I cannot think of the correct approach. Can anyone help me?

all the best :)

Marie

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Hint: compute the moment generating function for the normal distribution. All odd moments are zero, so you should probably have $X^{2k}$ there instead of $X^k$. – Chris Janjigian May 6 '12 at 12:19
Hint: If you don't want to compute the moment-generating function, write $$E[X^{2k}] = \int_{-\infty}^{\infty}x^{2k}\phi(x)\,\mathrm dx = 2\int_0^{\infty}x^{2k}\phi(x)\,\mathrm dx$$ where $\phi(x)$ is the standard normal density, and make a change of variables $y = x^2$ to convert the integral into one that is the definition of the Gamma function and use $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ to evaluate it. You will need to know the value of $\Gamma(\frac{1}{2})$. – Dilip Sarwate May 6 '12 at 13:35
hmm, okay, now I have computed the MGF, $M_X(t)=E[e^{tX}]$, and if we find the derivative of both sides at $t=0$, we get that $$M_X^{(k)}(0)=E[X^k\cdot e^{0}].$$ So I try to investigate $[e^{\frac{1}{2}t^2}]'$ at $0$, which is something like $[e^{\frac{1}{2}t^2}]^{(k)}=\sum_{i=0}^{k} a_i^k\cdot t^i\cdot e^{0.5t^2}$, but I cannot find a connection to $\frac{(2k)!}{k!2^k}$, i.e. cannot find a closed-form expression for $a_0^{2k}$. How do I get a formula for the first coeff.? I noticed that the derivate has also only even or odd powers of $t$, but I cannot see the picture yet. :) anyone help? – Marie. P. May 6 '12 at 16:05