Solve the mgf and expected value of a discrete random variable

Question

This problem is from Casella and Berger

A distribution cannot be uniquely determined by a finite collection of moments, as this example from Romano and Siegel (1986) shows. Let X have the normal distribution, that is, X has pdf

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

Define a discrete random variable Y by

$P(Y=\sqrt{3})=P(Y=-\sqrt{3})=\frac{1}{6}$, $P(Y=0)=\frac{2}{3}$.

Show that

$EX^r=EY^r$ for $r=1, 2, 3, 4, 5$

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I have solved that $EX^r$ are 0, 1, 0 , 3, 0 as $r=1, 2, 3, 4, 5$. How to make the mgf of Y and get the expection?

Answer 1

The MGF of $Y$ is by definition

\begin{align} \text{M}_Y(s) &= E[e^{sY}]\\ &= \sum_y e^{sy}p_Y(y) \end{align}

where $p_Y(y)$ is the PMF of $Y$. You can expand that sum as

$$M_Y(s) = \frac{1}{6}e^{s\sqrt3} + \frac{1}{6}e^{-s\sqrt3} + \frac{2}{3}$$

As usual, the $n$-th moment would be given by

$$E[Y^n] = \frac{d^n}{ds^n}M_Y(s)\bigg|_{s=0}$$