next up previous
Next: Percentage Points Up: Distribution Functions Previous: Cumulative Distribution Function

Moment Generating Function

The m.g.f. is defined as

\begin{displaymath}
\mathrm{M}(t) := \mathrm{E}(\ensuremath{\mathrm{e}^{tX}}) = ...
...}\Bigr)^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-x/2}}\d{x}\end{displaymath}

To evaluate \( I := \int_{0}^{+\infty}
(\frac{x}{2})^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-\frac{x}{2} (1 - 2t)}}\d{x}\), we substitute \( u = \frac{x}{2} (1 - 2t) \) so that \( \d{u}=
\frac{1}{2} (1 - 2t) \d{x}\), \( \frac{x}{2} = \frac{u}{1 - 2t} \) and \( \d{x}= \frac{2\d{u}}{1 - 2t} \). Hence,

\begin{eqnarray*}
I & = & \frac{2}{1 - 2t} \int_{0}^{+\infty} \Bigl(\frac{u}{1 -...
...rac{2 \mathrm{\Gamma}(\frac{1}{2}\nu)}{(1 -
2t)^{\frac{\nu}{2}}}
\end{eqnarray*}



We can now deduce the m.g.f. as follows.

\begin{eqnarray*}
\mathrm{M}(t) & = & \frac{1}{2 \mathrm{\Gamma}(\frac{1}{2} \nu...
...)}{(1 - 2t)^{\frac{\nu}{2}}} \\
& = & (1 - 2t)^{-\frac{\nu}{2}}
\end{eqnarray*}



Since we have made the substitution \( u = \frac{x}{2} (1 - 2t) \), $u$ must be positive to avoid the limits of integration being changed. Hence, the m.g.f. is only valid for \( 1 - 2t > 0 \) i.e. \( t <
\frac{1}{2} \). We now have

\begin{eqnarray*}
\mathrm{M}'(t) & = & (-\frac{\nu}{2})(-2)(1 - 2t)^{-\frac{\nu}{2} - 1}
\\
& = & \frac{\nu}{(1 - 2t)^{\frac{\nu}{2} + 1}}
\end{eqnarray*}



and

\begin{eqnarray*}
\mathrm{M}''(t) & = & \frac{-2 \nu (-\frac{\nu}{2} - 1)}{(1 -
...
... 2}} \\
& = & \frac{\nu^2 + 2\nu}{(1 - 2t)^{\frac{\nu}{2} + 2}}
\end{eqnarray*}



Since \( \mathrm{M}'(0) = \mathrm{E}(X) \) and \( \mathrm{M}''(0) =
\mathrm{E}(X^2) \), we have

\begin{displaymath}
\mathrm{E}(X) = \mathrm{M}'(0) = \nu
\end{displaymath}

and

\begin{displaymath}
\mathrm{E}(X^2) = \mathrm{M}''(0) = \nu^2 + 2\nu
\end{displaymath}

and so

\begin{displaymath}
\mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2 = \nu^2 + 2\nu -
\nu^2 = 2\nu
\end{displaymath}



Alexander Frolkin 2001-02-01