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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
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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
    
Oh, I got it! we write this as $$e^{\frac{1}{2}t^2}= \sum_{k=0}^{\infty}\frac{t^{2k}}{k!2^k}.$$ Then we derivate $(2k)$ times and the only non-zero term (since $t=0) will be $\frac{(2k)!}{k!2^k}$!! thanks for the help! :* –  Marie. P. May 6 '12 at 16:15
    
I strongly recommend you write up your solution neatly, post it as an answer to your own question (not just allowed but encouraged on math.SE). You can even accept your own answer as the best answer. –  Dilip Sarwate May 6 '12 at 21:41
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3 Answers 3

up vote 3 down vote accepted

So, here is the solution:

We know the MGF, $$M_X(t)=E[e^{tX}],$$ and if we find the derivative w.r.t. $t$ of both sides at $t=0$, we get that $$M^{(k)}_X(0)=E[X^k \cdot e^0].$$ So we try to investigate the function $e^{\frac{t^2}{2}}$ (and its derivatives) at $t=0$, and we use the fact that $$e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}.$$ Hence $$e^{\frac{t^2}{2}}=\sum_{j=0}^{\infty}\frac{ \left( \frac{t^2}{2} \right)^j}{j!}=\sum_{j=0}^{\infty} \frac{t^{2j}}{j!\cdot 2^j}.$$ Now we find the $(2k)^{th}$ derivative, and evaluate at $t=0$. But since $t=0$, we have that all summands that contain factor $t$ are zero, which leaves only one: The coefficient of $t^0$ in the $(2k)^{th}$ derivative is what we want. But this is equal to $(2k)!$ times the coefficient of $t^{2k}$ in the above sum, hence $\frac{(2k)!}{k!\cdot 2^k},$ as desired.

All the best! marie :)

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Use gamma function to evaluate the integral

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$$ \begin{align} \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty x^{2k}\,e^{-x^2/2}\,\mathrm{d}x &=\frac2{\sqrt{2\pi}}\int_0^\infty x^{2k-1}\,e^{-x^2/2}\,\mathrm{d}x^2/2\\ &=\frac2{\sqrt{2\pi}}\int_0^\infty (2x)^{k-1/2}\,e^{-x}\,\mathrm{d}x\\ &=\frac{2^{k+1/2}}{\sqrt{2\pi}}\Gamma(k+1/2)\\ &=\frac{(2k)!}{2^kk!} \end{align} $$

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