# Showing that a discrete random variable has the same moments as a Normal Distribution

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 $$d = \{1, 2, 3, 4, 5\}?$$

• Is this a homework question? If so, please add the self-study tag, read it’s wiki, and say what progress you’ve made. – Dave Oct 4 at 15:44
• 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 $0$) – Henry Oct 4 at 15:45

$$E[X^m] = \begin{cases} 0, \quad \quad \quad \quad m\ \mathrm{is\ odd} \\ 2^{-m/2}\frac{m!}{(m/2)!}, \quad m\ \mathrm{is\ even} \end{cases}$$
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: $$E[Y] = \sum y P(Y) = (-\sqrt{3})(1/6) + (\sqrt{3})(1/6) + (0)(⅔) = 0$$