Probability Density Function

If , then

$$\ensuremath{\mathrm{Pr}(X = x)}= \mathrm{f}(x) = A_\nu \Bigl... ...igr)^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-\frac{x}{2}}}$$

for $x > 0$. Here, $\nu$ is an integral valued parameter known as the number of degrees of freedom. $A_\nu$ is a constant depending on $\nu$. For $\mathrm{f}(x)$ to be a p.d.f., is must satisfy

$$\int_{-\infty}^{+\infty} \mathrm{f}(x) \d{x}= 1$$

Since $\mathrm{f}(x)$ is only defined for $x > 0$, it is sufficient for it to satisfy

$$\int_{0}^{+\infty} \mathrm{f}(x) \d{x}= 1$$

Hence,

$$\int_{0}^{+\infty} \Bigl(\frac{x}{2}\Bigr)^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-\frac{x}{2}}}\d{x}= \frac{1}{A_\nu}$$

We substitute $t = \frac{x}{2}$, so that $\d{x}= 2 \d{t}$, and

$$2 \int_{0}^{+\infty} t^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-t}}\d{t}= \frac{1}{A_\nu}$$

We notice that this is the Euler gamma function, defined by

$$\mathrm{\Gamma}(\alpha) := \int_{0}^{+\infty} \ensuremath{\mathrm{e}^{-x}}x^{\alpha - 1} \d{x}$$

with . Hence,

$$2 \mathrm{\Gamma}\Bigl(\frac{\nu}{2}\Bigr) = \frac{1}{A_\nu}$$

Therefore,

$$A_\nu = \frac{1}{2\mathrm{\Gamma}(\frac{1}{2} \nu)}$$

We can now make the definition

$$\begin{eqnarray*} \mathrm{f}(x) & = & \frac{1}{2\mathrm{\Gamma}(\frac{1}{2} \nu)... ...nu)} x^{\frac{\nu}{2} - 1} \ensuremath{\mathrm{e}^{-\frac{x}{2}}}\end{eqnarray*}$$