91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from the uniform distribution on the interval \(\left[ {{\theta _1},{\theta _2}} \right]\).We need to find out the M.L.E.’s of\({\theta _1}\)and\({\theta _2}\).

The pdf of X is

\(\begin{array}{c}f\left( x \right) = \frac{1}{{{\theta _2} - {\theta _1}}}\,\,,{\theta _1} \le x \le {\theta _2}\\ = 0\,\,\,,{\rm{otherwise}}\end{array}\)

The likelihood function is

\(\begin{array}{c}L\left( x \right) = \prod\limits_{i = 1}^n {f\left( x \right)} \\ = \frac{1}{{{{\left( {{\theta _2} - {\theta _1}} \right)}^n}}}\end{array}\)

From the property we know that,

\({\theta _1} \le \min \left\{ {{x_1},...,{x_n}} \right\} < \max \left\{ {{x_1},...,{x_n}} \right\} \le {\theta _2}\)

The likelihood function is maximized when \(\left( {{\theta _2} - {\theta _1}} \right)\) is minimized . This implies \({\theta _1}\)is the maximum of \({X_1},...,{X_n}\)and \({\theta _2}\) is the minimum of \({X_1},...,{X_n}\)

Hence \({\hat \theta _1} = \min \left\{ {{x_1},...,{x_n}} \right\}\,\,{\rm{and}}\,\,{\hat \theta _2} = \max \left\{ {{x_1},...,{x_n}} \right\}\)