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 {\bf Question}

 Suppose that $X$ has the standard normal distribution,
i.e.

$$f(x)=\frac{1}{\sqrt2\pi}e^{-\frac{1}{2}x^2},\ \
-\infty<x<\infty.$$

 Derive the moment generating function of $Y=X^2$. What
distribution does $Y$ follow?

\medskip

 {\bf Answer}

\begin{eqnarray*} M_X(t) & = & E(e^{tY})\\ & = & E\{e^{tX^2}\}\\ &
= & \int_{-\infty}^{\infty} e^{tx^2}
\frac{e^{-\frac{1}{2}x^2}}{\sqrt{2\pi}} \,dx\\ & = &
\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{2}
x^2(1-2t)}}{\sqrt{2\pi}} \,dx\\ & = & \int_{-\infty}^{\infty}
\frac{e^{-\frac{1}{2} \frac{x^2}{b^2}}}{\sqrt{2\pi}} \,dx = b\\ &
& {\rm where}\ b^2=\frac{1}{1-2t}\ \ ({\rm if}\ 1-2t>0 \Rightarrow
t<\frac{1}{2})\\ & = & \left(\frac{1}{1-2t}\right)^{\frac{1}{2}}\
{\rm if}\ t<\frac{1}{2}.
\end{eqnarray*}

Since the above is the mgf of the $\chi^2$ distribution with 1
degree of freedom we can conclude that $Y=X^2$ follows the
$\chi^2$ distribution with 1 degree of freedom.

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