Percentage Points

The number is defined as the number such that if , then

$$\ensuremath{\mathrm{Pr}(X > \mathrm{\chi}_{P\%}^2 ( \nu ))}= \frac{P}{100}$$

and is referred to as the $P\%$ point of a distribution. In other words, the $P\%$ point of a distribution is the point such that $P\%$ of the distribution lies to the right of it.

From the above definition,

$$1 - \mathrm{F}(\mathrm{\chi}_{P\%}^2 ( \nu )) = \frac{P}{100}$$

So that

$$\frac{\mathrm{\Gamma}(\frac{1}{2}\nu,\frac{1}{2} x)}{\mathrm{\Gamma}(\frac{1}{2}\nu)} = \frac{P}{100}$$

Therefore we need to solve

$$\mathrm{\Gamma}\bigl(\frac{\nu}{2}, \frac{x}{2}\bigr) = \frac{P \mathrm{\Gamma}(\frac{1}{2}\nu)}{100}$$

for $x$ in order to find the $P\%$ point of the distribution. For example, for ,

$$\mathrm{\Gamma}\bigl(\frac{\nu}{2}, \frac{x}{2}\bigr) = \int... ...nsuremath{\mathrm{e}^{-t}}\d{t}= \ensuremath{\mathrm{e}^{-x/2}}$$

and

$$\mathrm{\Gamma}\bigl(\frac{\nu}{2}\bigr) = \int_0^{+\infty} \ensuremath{\mathrm{e}^{-t}}\d{t}= 1$$

Hence,

$$\ensuremath{\mathrm{e}^{-\frac{x}{2}}}= \frac{P}{100}$$

so that

$$\mathrm{\chi}_{P\%}^2 ( 2 ) = 2\ln{\frac{100}{P}}$$

The procedure is very different for other values of $\nu$. Consider the case . Here

$$\mathrm{\Gamma}\bigl(\frac{\nu}{2}, \frac{x}{2}\bigr) = \int... ...suremath{\mathrm{e}^{-\frac{x}{2}}}\bigl(\frac{x}{2} + 1\bigr)$$

and

$$\mathrm{\Gamma}\bigl(\frac{\nu}{2}\bigr) = 1! = 1$$

Hence,

$$\begin{eqnarray*} \ensuremath{\mathrm{e}^{-\frac{x}{2}}}\bigl(\frac{x}{2} + 1\bi... ...}{2} + 1\bigr) & = & \frac{-P \ensuremath{\mathrm{e}^{-1}}}{100} \end{eqnarray*}$$

Since if then , where is Lambert's W function,

$$-\bigl(\frac{x}{2} + 1\bigr) = \mathrm{W}\bigl(\frac{-P \ensuremath{\mathrm{e}^{-1}}}{100}\bigr)$$

and so

$$\mathrm{\chi}_{P\%}^2 ( 4 ) = -2\bigl(\mathrm{W}\bigl(\frac{-P \ensuremath{\mathrm{e}^{-1}}}{100}\bigr) + 1\bigr)$$