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Probability Density Function

If \( X \ensuremath{\sim \mathrm{\chi^2}(\nu)}\), then

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

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

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

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

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

Hence,

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

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

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

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

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

with \( \alpha = \frac{\nu}{2} \). Hence,

\begin{displaymath}
2 \mathrm{\Gamma}\Bigl(\frac{\nu}{2}\Bigr) = \frac{1}{A_\nu}
\end{displaymath}

Therefore,

\begin{displaymath}
A_\nu = \frac{1}{2\mathrm{\Gamma}(\frac{1}{2} \nu)}
\end{displaymath}

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*}





Alexander Frolkin 2001-02-01