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Given $Z$ is a standard normal random variable with mean 0 and variance 1 ($Z \sim N(0,1)$), could anyone provide an explanation for why $E\left[ Z^4 \right] = 3$?

I know that:

$$ \begin{aligned} E \left[ Z \right] &= \mu \\ &= 0 \end{aligned} $$

and

$$ \begin{aligned} 1 &= Var(Z) \\ &= E[Z^2] - E[Z]^2 \\ &= E[Z^2] - 0^2 \\ &= E[Z^2] \end{aligned} $$

But I'm not sure the rationale behind $$ E[Z^4] = 3 $$

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    $\begingroup$ What is wrong with using the normal MGF? $\endgroup$ – Sycorax Apr 13 '16 at 19:52
  • $\begingroup$ @C11H17N2O2SNa Thanks! I'll work through it and make sure that I get the correct answer from the MGF as well. I remembered the professor giving a very simplistic explanation in class and not pulling out the MGF, so I was wondering if another intuitive explanation existed. $\endgroup$ – Chris C Apr 13 '16 at 19:59
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    $\begingroup$ This is answered at stats.stackexchange.com/a/176814/919 using the moment generating function. The elementary way substitutes $y=-z^2/2$ in the Normal integral to obtain $$\mathbb{E}(Z^{2k})=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}z^{2k} \exp(-z^2/2)dz=\frac{2^k}{\sqrt{\pi}}\int_0^\infty y^{k-1/2}\exp(-y)dy = \frac{2^k}{\sqrt{\pi}}\Gamma\left(k+\frac{1}{2}\right),$$ which is an equivalent way to express the same results. $\endgroup$ – whuber Apr 13 '16 at 20:41
  • $\begingroup$ Thank you @whuber, that's a very comprehensive answer and I learned a lot by reading it. $\endgroup$ – Chris C Apr 13 '16 at 20:52
  • $\begingroup$ Possible duplicate of How to calculate the expected value of a standard normal distribution? $\endgroup$ – kjetil b halvorsen Apr 8 '19 at 6:40
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You can show the result using moment generating functions or my direct integration (which maybe more difficult). I find the following to be satisfying.

Let $X = Z^2$. Then $X \sim \chi^2_1$. Expected value of a $\chi^2_1$ is 1 and the variance is 2.

Thus we can find the second moment of $X$. \begin{align*} E[X^2]& = Var(X) + E[X]^2\\ & = 2 + 1 = 3 \end{align*}

But $E[Z^4] = E[X^2] = 3$.

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  • $\begingroup$ Thank you! I'll try to work through the method with the moment generating function myself as well just to make sure that I get the same answer, but this method makes a lot of sense to me. $\endgroup$ – Chris C Apr 13 '16 at 19:57
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As a note, the moments of a standard normal random variable have an elegant recursive relation: $$E[Z^{n + 1}] = n E[Z^{n - 1}], n = 1, 2, 3, \ldots. \tag{1}$$

From $(1)$ and $E[Z] = 0, E[Z^2] = 1$, it is easy to see that \begin{align} & E[Z^{2n}] = (2n - 1)!! \\ & E[Z^{2n - 1}] = 0. \end{align}

To show $(1)$, consider integration by parts.

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