91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from an exponential distribution for which the value of the parameter β is unknown. We need to calculate the M.L.E. of median

Let m be the median of the distribution then from property of exponential distribution we know that,

\(m = \frac{{\ln 2}}{\beta }\)

The likelihood function is

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

Taking logarithm, differentiating with respect to \(\beta \) and equate it with 0, we get

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

Hence the M.L.E is

\(\begin{array}{c}\hat m = \frac{{\ln 2}}{{\frac{1}{{{{\bar x}_n}}}}}\\ = {{\bar x}_n}\ln 2\end{array}\)