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

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

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

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

Again, we substitute to obtain

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

since when and when . Further, we notice that

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

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

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

Hence,

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