91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from an exponential distribution. Here the parameter β is unknown (β > 0).We need to calculate of the M.L.E.of β.

\({X_1},...,{X_n}\)form an exponential distribution.

Let s be the sum of the observed values. Then the likelihood function is

\(f\left( {x|\beta } \right) = {\beta ^n}\exp \left( { - \beta s} \right)\)

Let \(L\left( \beta \right) = \log f\left( {x|\beta } \right)\)then

\(\frac{{\partial L\left( \beta \right)}}{{\partial \beta }} = \frac{n}{\beta } - s\)To find maximum value of \(L\left( \beta \right) = \log f\left( {x|\beta } \right)\)then we have

\(\frac{{\partial L\left( \beta \right)}}{{\partial \beta }} = 0\).This implies that

\(\begin{array}{c}\frac{n}{\beta } - s = 0\\\frac{n}{\beta } = s\\\beta = \frac{n}{s}\\\beta = \frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }}\\ = \frac{1}{{\bar x}}\end{array}\)

So, the M.L.E. of β is \(\hat \beta = \frac{1}{{\bar x}}\)