91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from a Poisson distribution for which the mean is unknown. We need to calculate the M.L.E. of the standard deviation of the distribution

\(f\left( {\lambda |x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\)

The likelihood function is

\(\begin{array}{c}L\left( {\lambda |x} \right) = \prod\limits_{i = 1}^n {f\left( {\lambda |x} \right)} \\ = \frac{{{e^{ - n\lambda }}{\lambda ^{\sum\limits_{i = 1}^n {{x_i}} }}}}{{\prod\limits_{i = 1}^n {{x_i}!} }}\end{array}\)

Taking logarithm and equate the equation with 0, we get

\(\begin{array}{c}\frac{{\partial \log L\left( {\lambda |x} \right)}}{{\partial \lambda }} = 0\\ - n\lambda + \sum\limits_{i = 1}^n {{x_i}\ln \lambda } - \ln \left( {\prod\limits_{i = 1}^n {{x_i}} } \right) = 0\\ - n + \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{\lambda } = 0\\\lambda = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\\lambda = {{\bar x}_n}\end{array}\)

The M.L.E of \(\lambda \) is \({\bar x_n}\) .The standard deviation of the distribution is \(\sqrt \lambda \) .So, the required M.L.E. is \(\sqrt {{{\bar x}_n}} \)