Expectation

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Expectation

The expectation for this distribution is given by

$\begin{displaymath} \mathrm{E}(X) = \frac{1}{2 \mathrm{\Gamma}(\frac{1}{2} \nu)}... ...{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-\frac{x}{2}}}\d{x}\end{displaymath}$

Upon substitution, yet again, $t = \frac{x}{2}$ , we obtain

$\begin{displaymath} \mathrm{E}(X) = \frac{4}{2 \mathrm{\Gamma}(\frac{1}{2} \nu)}... ...mma}(\frac{\nu}{2} + 1)}{\mathrm{\Gamma}(\frac{\nu}{2})} = \nu \end{displaymath}$

since $\mathrm{\Gamma}(\alpha + 1) = \alpha \mathrm{\Gamma}(\alpha)$ and so $\mathrm{\Gamma}(\frac{\nu}{2} + 1) = \frac{\nu}{2} \mathrm{\Gamma}(\frac{\nu}{2})$ .

Alexander Frolkin 2001-02-01