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Suppose I define $X$ to be normally distributed with $\mu = 0, \sigma^2 = 1$, so that $X$ has the pdf

$f_{X}(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}, \quad -\infty < x <\infty.$

Let discrete variable $Y$ be defined by the following: $P(Y = -\sqrt{3})= P(Y = \sqrt{3}) = \frac{1}{6}$ and $P(Y=0) = \frac{2}{3}$.

How can I show that $\mathrm{E}[X^d] = \mathrm{E}[Y^d]$ for $d = \{1, 2, 3, 4, 5\}?$

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  • $\begingroup$ Is this a homework question? If so, please add the self-study tag, read it’s wiki, and say what progress you’ve made. $\endgroup$ – Dave Oct 4 at 15:44
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    $\begingroup$ The odd moments are easy and the even moments are not much harder (especially if you know the excess kurtosis of a standard normal is $0$) $\endgroup$ – Henry Oct 4 at 15:45
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  1. Begin with the following fact:

$E[X^m] = \begin{cases} 0, \quad \quad \quad \quad m\ \mathrm{is\ odd} \\ 2^{-m/2}\frac{m!}{(m/2)!}, \quad m\ \mathrm{is\ even} \end{cases}$

So, $E[X] = E[X^3] = E[X^5] = 0, E[X^2] = 1,$ and $E[X^4] = 3$.

  1. For $Y$, use the formula $E[Y] = \sum y P(Y)$
  • Ex: $E[Y] = \sum y P(Y) = (-\sqrt{3})(1/6) + (\sqrt{3})(1/6) + (0)(⅔) = 0$
  • Continue these steps for each of the other moments.
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