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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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    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.2012-05-06

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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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$ \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\tag{1}\\ &=\frac2{\sqrt{2\pi}}\int_0^\infty (2x)^{k-1/2}\,e^{-x}\,\mathrm{d}x\tag{2}\\ &=\frac{2^{k+1/2}}{\sqrt{2\pi}}\Gamma(k+1/2)\tag{3}\\ &=\frac{(2k)!}{2^kk!}\tag{4} \end{align} $ Explanation:
$(1)$: the integrand is even, so we can double the integral over $(0,\infty)$.
$\phantom{(1)\text{:}}$ Furthermore, $x^{2k-1}\,\color{#C00000}{\mathrm{dx^2/2}} =x^{2k-1}\,\color{#C00000}{x\,\mathrm{d}x}=x^{2n}\,\mathrm{d}x$
$(2)$: Substitute $x\mapsto\sqrt{2x}$
$(3)$: $\Gamma(\alpha)=\int_0^\infty x^{\alpha-1}e^{-x}\,\mathrm{d}x$
$(4)$: $\begin{align}\frac{\Gamma(k+1/2)}{\sqrt\pi} =\frac{\Gamma(k+1/2)}{\Gamma(1/2)} =(k-1/2)(k-3/2)\cdots1/2 =\frac1{2^k}(2k-1)!! =\frac{(2k)!}{4^kk!}\end{align}$

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    After a fair amount of Googling, I have read that for a general normal distribution, the formula is $\frac{(2k)!\sigma^k}{2^k k!}$. Is that consistent with your understanding?2016-01-17
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Use gamma function to evaluate the integral