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I have $X\sim N(0, \sigma^2)$ where $\sigma^2$ is unknown. The HW question asks me to find the expected values of $X^2$ and $X^4$.

Attempt:

My guess is to use MGFs. I know that the MGF of $X$ is $M_X(t)=e^{\frac{1}{2}\sigma^2t^2}$, but don't exactly know how to use this properly. I thought $t$ corresponded to the $t^{th}$ moment but then $M_X(2)=e^{2\sigma^2}$. This is the second moment that I thought was supposed to equal the variance, but clearly it doesn't.

If anyone has any pointers or know of any good resources to point me in the direction of, I would very much appreciate it.

StubbornAtom
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Wivaviw
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  • In this answer at https://math.stackexchange.com/questions/1669735/expectation-of-a-standard-normal-random-variable/1669837#1669837 the moments of a standard normal random variable are derived. Yours would simply consider the variable $X = \sigma Z$, hence $\operatorname{E}[X^{2k}] = \sigma^{2k} \operatorname{E}[Z^{2k}].$ – heropup Nov 04 '20 at 03:24

1 Answers1

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To find the $n^{\rm th}$ moments, take the $n^{\rm th}$ differential of the moment generating function, $\mathsf M_{\small X}(t)$, with respect to $t$ and then evaluate this at $t=0$.

The second moment is thus the second differential evaluated at $t=0$. $$\begin{align}\left.\dfrac{\mathrm d^2~\mathsf M_{\small X}(t)}{\mathrm d t~^2}\right\rvert_{t=0} &= \left.\dfrac{\mathrm d^2~\mathrm e^{\sigma^2 t^2/2}}{\mathrm d t~^2}\right\rvert_{t=0}\\[1ex]&=\left.\dfrac{\mathrm d~\sigma^2t\mathrm e^{\sigma^2t^2/2}}{\mathrm d t}\right\rvert_{t=0}\\[1ex]&=\left[\sigma^2(\sigma^2t^2+1)\mathrm e^{\sigma^2t^2/2}\right]_{t=0}\\&=\sigma^2\end{align}$$

Which tells you that $\mathsf E(X^2)=\sigma^2$

Now do the fourth moment.

Graham Kemp
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