Let
\(f\left( {x|\theta } \right) = a\left( \theta \right)b\left( x \right)\exp \left( {c\left( \theta \right)d\left( x \right)} \right)\)
Then
\(\begin{align}\lambda \left( {x|\theta } \right) &= \log \left( {a\left( \theta \right)b\left( x \right)\exp \left( {c\left( \theta \right)d\left( x \right)} \right)} \right)\\ &= \log a\left( \theta \right) + \log b\left( x \right) + c\left( \theta \right)d\left( x \right)\end{align}\)
Differentiating\(\lambda \left( {x|\theta } \right)\)with respect to\(\theta \)as
\({\lambda ^{'}}\left( {x|\theta } \right) = \frac{{{a^{'}}\left( \theta \right)}}{{a\left( \theta \right)}} = {c^{'}}\left( \theta \right)d\left( x \right)\)
Therefore
\(\begin{align}{\lambda ^{'}}\left( {x|\theta } \right) &= \sum\limits_{i = 1}^n {{\lambda ^{'}}\left( {x|\theta } \right)} \\ &= n\frac{{{a^{'}}\left( \theta \right)}}{{a\left( \theta \right)}} + {c^{'}}\left( \theta \right)\sum\limits_{i = 1}^n {d\left( {{x_i}} \right)} \end{align}\)
If we choose,
\(u\left( \theta \right) = \frac{1}{{{c^{'}}\left( \theta \right)}}\) and
\(v\left( \theta \right) = \frac{{n{a^{'}}\left( \theta \right)}}{{a\left( \theta \right){c^{'}}\left( \theta \right)}}\)
Then from the equation of cramer rao inequality, that is
\(Var\left( T \right) \ge \frac{{{{\left( {{m^{'}}\left( \theta \right)} \right)}^{2}}}}{{nI\left( \theta \right)}}\) satisfy with
\(T = \sum\limits_{i = 1}^n {d\left( {{X_i}} \right)} \) . therefore the estimator \(T = \sum\limits_{i = 1}^n {d\left( {{X_i}} \right)} \) is an efficient estimator.
