91Ó°ÊÓ

has a chi-square distribution with m degree of freedom\({\chi ^2}\)

\(\)\(f\left( x \right) = {2^{ - m/2}}\Gamma \left( {m/2} \right){x^{m/2 - 1}}{e^{ - x/2}}\,;\,x > 0\)

To find the mode of the distribution we need to differentiate log f(x) with respect to x, and need to equate it with 0.

\(\)\(\begin{align}f\left( x \right) &= {2^{ - m/2}}\Gamma \left( {m/2} \right){x^{m/2 - 1}}{e^{ - x/2}}\,\\\log \,f\left( x \right) &= - \frac{m}{2}\log 2 - \log \Gamma \left( {\frac{m}{2}} \right) + \left( {\frac{m}{2} - 1} \right)\log x - \frac{x}{2}\\\frac{{d\log f\left( x \right)}}{{dx}} &= \left( {\frac{m}{2} - 1} \right)\frac{1}{x} - \frac{1}{2}\end{align}\)

\(\begin{align}Now,\,\,\frac{{d\log f\left( x \right)}}{{dx}} &= 0\\\left( {\frac{m}{2} - 1} \right)\frac{1}{x} - \frac{1}{2} &= 0\\\left( {\frac{m}{2} - 1} \right)\frac{1}{x} &= \frac{1}{2}\\\left( {\frac{{m - 2}}{2}} \right)\frac{1}{x} &= \frac{1}{2}\\x &= m - 2\end{align}\)

So the mode of the chi-square distribution is m-2 for \(m > 2\)