91Ó°ÊÓ

Let the random variables be i.i.d and defined as\({X_1} \ldots {X_n}\).Every one of these random variables is assumed to be a sample from the same exponential, \({X_i} \sim \exp \left( \beta \right)\)

The pdf is:

\(f\left( x \right) = \beta {e^{ - \beta x}},x > 0\)

Let us calculate MLE to estimate the\(\beta \)parameter of an exponential distribution.

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

Applying log likelihood and differentiating it to find the value of\(\beta \)that maximises the log likelihood function, we equate it with 0.

\(\begin{array}{c}\frac{\partial }{{\partial \beta }}{\rm{log}}L\left( \beta \right) = \frac{{\partial \ln \left( {{\beta ^n}{e^{ - \beta \sum\limits_{i = 1}^n x }}} \right)}}{{\partial \beta }}\\ = \frac{n}{\beta } - \sum\limits_{i = 1}^n {{x_i}} \end{array}\)

Finally, the MLE is

\(\beta = \frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }}\)

Since the mean of the exponential distribution is \(\frac{1}{\beta }\), MLE estimate becomes \(\frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\).

Since,

\(\begin{array}{l}{X_i} \sim Exp\left( \beta \right)\\ \Rightarrow \sum\limits_{i = 1}^n {{X_i}} \sim Gamma\left( {n,\beta } \right)\end{array}\)

Let,

\(Z = \sum\limits_{i = 1}^n {{X_i}} \)

\(\begin{array}{c}E\left[ {\frac{n}{{\sum\limits_{i = 1}^n {{X_i}} }}} \right] = nE\left[ {\frac{1}{Z}} \right]\\ = n\int\limits_0^\infty {\frac{1}{z}} \times \frac{1}{{\Gamma n}}{\beta ^n}{z^{n - 1}}{e^{ - \beta z}}dz\\ = \frac{{n\beta \Gamma \left( {n - 1} \right)}}{{\Gamma \left( n \right)}}\int\limits_0^\infty {\frac{1}{{\Gamma \left( {n - 1} \right)}}{\beta ^{n - 1}}{z^{n - 2}}{e^{ - \beta z}}} dz\\ = \frac{{n\beta \left( {n - 2} \right)!}}{{\left( {n - 1} \right)!}}\\ = \frac{{n\beta }}{{n - 1}}\end{array}\)

As\(n \to \infty \), this converges to\(\beta \).

Hence, it is a consistent estimator.