Moment Generating Function

The m.g.f. is defined as

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

To evaluate , we substitute so that , and . 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 , $u$ must be positive to avoid the limits of integration being changed. Hence, the m.g.f. is only valid for i.e. . 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 and , we have

$$\mathrm{E}(X) = \mathrm{M}'(0) = \nu$$

and

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

and so

$$\mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2 = \nu^2 + 2\nu - \nu^2 = 2\nu$$