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QUESTION
 Find the mgf of the square of a standard normal variable. Hence
 find the mgf of a $\chi^2$-distribution with $\nu$ degrees of
 freedom. Hence show that the $\chi^2$-distribution is a
 particular form of a gamma distribution and state the parameters
 of gamma. What form does the $\chi^2$-distribution take if
 $\nu$=2.

ANSWER
\begin{eqnarray*}
  M_{x^2}(t)&=&E(e^{X^2t})\\
  &=&\int_{-\infty}^\infty
  \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}e^{x^2t}\,dx\\
  &=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
  e^{-\frac{1}{2}x^2(1-2t)}\,dx\\
  &=&\frac{1}{(1-2t)^{\frac{1}{2}}}
  \end{eqnarray*}
  (from $N(\mu,\sigma^2)$ we know that
  $\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}\,dx=1$.
  let $\mu=0,\ \sigma^2=\frac{1}{1-2t}$).\\
  $M_\nu^2=X_1^2+X_2^2+\ldots +X_\nu^2$ where each $X_i$ has an
  independent N(0,1) distribution hence
  $M_{X^2_\nu}(t)=[M_{X^2}(t)]^\nu=\frac{1}{(1-2t)^\frac{\nu}{2}}$\\
  For$\varrho (\lambda)\
  M(t)=\frac{\lambda}{\lambda-t}=\frac{1}{1-\frac{t}{\lambda}}$\\
  For Gamma m, $\lambda\ \
  M(t)=(\frac{1}{1-\frac{t}{\lambda}})^2$(this follows for integer
  m and in fact works for a general m)\\
  Hence by uniqueness $X_\nu^2$ is Gamma $m=\nu\ \
  \lambda=\frac{1}{2}$. If $\nu=2\ \ M(t)=\frac{1}{1-2t}$ i.e.
  $\varrho(\frac{1}{2})$

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