91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from a distribution with the p.d.f. given in Exercise 10 of Sec. 7.5. We need to calculate the M.L.E. of \({e^{ - \frac{1}{\theta }}}\)

\(\begin{array}{c}f\left( {x|\theta } \right) = \theta {x^{\theta - 1}}\,\,\,\,\,\,0 < x < 1\\ = 0\,\,\,\,\,{\rm{otherwise}}\end{array}\)

Now, \(L\left( \theta \right) = \log f\left( {x|\theta } \right)\) If we differentiate the function and equate it with 0 we get\(\begin{array}{l}\frac{{\partial L\left( \theta \right)}}{{\partial \theta }} = 0\\ \Rightarrow \frac{\partial }{{\partial \theta }}\left[ {n\log \theta + \left( {\theta - 1} \right)\sum\limits_{i = 1}^n {\log {x_i}} } \right] = 0\\ \Rightarrow \frac{n}{\theta } + \sum\limits_{i = 1}^n {\log {x_i}} = 0\\ \Rightarrow \hat \theta = - \frac{n}{{\sum\limits_{i = 1}^n {\log {x_i}} }}\end{array}\)

Hence the M.L.E. of \({e^{ - \frac{1}{\theta }}}\)is as follows:

\(\begin{array}{c}{e^{ - \frac{1}{\theta }}} = \exp \left[ { - \frac{1}{{ - \frac{n}{{\sum\limits_{i = 1}^n {\log {x_i}} }}}}} \right]\\ = \exp \left[ {\log {{\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)}^{\frac{1}{n}}}} \right]\\ = {\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{\frac{1}{n}}}\end{array}\)

Hence the M.L.E. is \({\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{\frac{1}{n}}}\)