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Next: Expectation and Variance Up: Notes about the Chi-Squared Previous: Moment Generating Function

Percentage Points

The number \( \mathrm{\chi}_{P\%}^2 ( \nu ) \) is defined as the number such that if \( X \ensuremath{\sim \mathrm{\chi^2}(\nu)}\), then

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

and is referred to as the $P\%$ point of a \( \mathrm{\chi}^2 (\nu)
\) 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,

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

So that

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

Therefore we need to solve

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

for $x$ in order to find the $P\%$ point of the distribution. For example, for \( \nu = 2 \),

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

and

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

Hence,

\begin{displaymath}
\ensuremath{\mathrm{e}^{-\frac{x}{2}}}= \frac{P}{100}
\end{displaymath}

so that

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

The procedure is very different for other values of $\nu$. Consider the case \( \nu = 4 \). Here

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

and

\begin{displaymath}
\mathrm{\Gamma}\bigl(\frac{\nu}{2}\bigr) = 1! = 1
\end{displaymath}

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 \( y \ensuremath{\mathrm{e}^{y}}= x \) then \( y = \mathrm{W}(x) \), where \(
\mathrm{W}(x) \) is Lambert's W function,

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

and so

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

\fbox{\parbox[t]{9cm}{\textbf{A note to those people reading this}, who are bett...
... know where I screwed up, \textit{please} tell me - I'm very
curios! Thanks.}}

next up previous
Next: Expectation and Variance Up: Notes about the Chi-Squared Previous: Moment Generating Function
Alexander Frolkin 2001-02-01