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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