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 {\bf Question}

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

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

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

\medskip

 {\bf Answer}

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