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Next: Moment Generating Function Up: Distribution Functions Previous: Probability Density Function

Cumulative Distribution Function

Having defined the p.d.f. of the distribution, we can now deduce the c.d.f. -- the function $\mathrm{F}(x)$, such that

\begin{displaymath}
\ensuremath{\mathrm{Pr}(X \leqslant x)}= \mathrm{F}(x) = \int_{-\infty}^{x}
\mathrm{f}(t) \d{t}\end{displaymath}

Since the p.d.f. is defined only for $x > 0$,

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

Again, we substitute \( u = \frac{t}{2} \) to obtain

\begin{displaymath}
\mathrm{F}(x) = \frac{1}{\mathrm{\Gamma}(\frac{1}{2}\nu)} \int_0^{x/2}
u^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-u}}\d{u}\end{displaymath}

since \( u = 0 \) when \( t = 0 \) and \( u = \frac{x}{2} \) when \( t
= x \). Further, we notice that

\begin{displaymath}
\int_0^{x/2} u^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-...
...\infty}
u^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-u}}\d{u}\end{displaymath}

We recognise that the second integral on the R.H.S. is the incomplete gamma function,

\begin{displaymath}
\mathrm{\Gamma}(\alpha, a) := \int_a^{+\infty} \ensuremath{\mathrm{e}^{-t}}t^{\alpha - 1}
\d{t}\end{displaymath}

Hence,

\begin{displaymath}
\mathrm{F}(x) = \frac{1}{\mathrm{\Gamma}(\frac{1}{2} \nu)} \...
...c{1}{2}
\nu, \frac{1}{2} x)}{\mathrm{\Gamma}(\frac{1}{2} \nu)}
\end{displaymath}



Alexander Frolkin 2001-02-01