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
On
$$
\begin{align}
\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty x^{2k}\,e^{-x^2/2}\,\mathrm{d}x
&=\frac2{\sqrt{2\pi}}\int_0^\infty x^{2k-1}\,e^{-x^2/2}\,\mathrm{d}x^2/2\tag{1}\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty (2x)^{k-1/2}\,e^{-x}\,\mathrm{d}x\tag{2}\\
&=\frac{2^{k+1/2}}{\sqrt{2\pi}}\Gamma(k+1/2)\tag{3}\\
&=\frac{(2k)!}{2^kk!}\tag{4}
\end{align}
$$
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 $x\mapsto\sqrt{2x}$
$(3)$: $\Gamma(\alpha)=\int_0^\infty x^{\alpha-1}e^{-x}\,\mathrm{d}x$
$(4)$: $\begin{align}\frac{\Gamma(k+1/2)}{\sqrt\pi} =\frac{\Gamma(k+1/2)}{\Gamma(1/2)} =(k-1/2)(k-3/2)\cdots1/2 =\frac1{2^k}(2k-1)!! =\frac{(2k)!}{4^kk!}\end{align}$
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 :)