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 {\bf Question}

 Suppose that a r.v. $X$ has the mgf

$$m(t)=e^{t^2+3t}\ \ {\rm for}\ \ -\infty<t<\infty.$$

 Find $E\{[x-E(X)]^r\}$, the $r$th central moment of $X$,
for $r=1,2, \cdots$.

Does $X$ have a normal distribution?  Give your reasoning.


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 {\bf Answer}

It is known that $X \sim N(\mu,\sigma^2)$ if and only if $M_X(t) =
e^{\mu t+\frac{\sigma^2t^2}{2}}$

Therefore $M_X(t) = e^{3t+t^2}$ is the mgf of
$N(\mu=3,\sigma^2=2)$

Let $Y=X-\mu$

Therefore $E\left\{[X-E(X)]^r\right\}=E(Y^r)$

But \begin{eqnarray*} M_Y(t)& = & e^{-t\mu}e^{\mu t+t^2}\\ & = &
e^{t^2}\\ & = & \sum_{k=0}^{\infty} \frac{t^{2k}}{k!}\\ & = &
\sum_{k=0}^{\infty} \frac{(2k)!}{k!} \cdot \frac{t^{2k}}{(2k)!}
\end{eqnarray*}

Therefore $E(Y^r) = \left\{ \begin{array}{cl} 0 & {\rm if}\ r\
{\rm is\ odd}\\ \frac{(2k)!}{k!} & {\rm if}\ r=2k
\end{array} \right.$

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