91Ó°ÊÓ

Let the random variables be IID and defined as\({{\rm{X}}_{\rm{1}}}{\rm{ \ldots }}{{\rm{X}}_{\rm{n}}}\).

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:

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

In the method of moments method, the parameter and the sample estimator moments are equated.

Let the parameter moment for the population be defined as follows:

\({\mu _j}\left( \theta \right) = E\left( {X_i^j} \right)\)

Here the j term varies.

Therefore, for j=1,

\(\begin{array}{c}{\mu _1}\left( \theta \right) = E\left( {X_i^1} \right)\\ = E\left( X \right) \ldots \left( 1 \right)\end{array}\)

This denotes that \({\mu _1}\left( \theta \right)\) is equal to the population mean, that is,\(E\left( {{X_1}} \right)\)

Let the sample moment for the population be be defined as follows:

\({m_j} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^j} \)Here the j term varies.

Therefore, for j=1,

\(\begin{array}{c}{m_1} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^1} \\ = \frac{1}{n}\sum\limits_{i = 1}^n X \ldots \left( 2 \right)\end{array}\)

This denotes that \({m_1}\) is equal to the sample mean, that is,\(\frac{1}{n}\sum\limits_{i = 1}^n X \)

Therefore, for equating the first-order moment with mean,

Equating (1) and (2)

\(\begin{array}{c}{\mu _1} = {m_1}\\ = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \end{array}\)

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

\(\begin{array}{c}{\bf{L}}\left( {\bf{\theta }} \right){\bf{ = \beta }}{{\bf{e}}^{{\bf{ - \beta }}{{\bf{x}}_{\bf{1}}}}}{\bf{ \times \beta }}{{\bf{e}}^{{\bf{ - \beta }}{{\bf{x}}_{\bf{2}}}}}{\bf{ \times \ldots \times \beta }}{{\bf{e}}^{{\bf{ - \beta }}{{\bf{x}}_{\bf{n}}}}}\\ = \prod\limits_{i = 1}^n {\beta {e^{ - \beta {x_i}}}} \\ = {\beta ^n}{e^{ - \beta \sum\limits_{i = 1}^n x }}\end{array}\)

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

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

Equating to zero and obtaining the value of the parameter.

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

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

Hence, the method of moments estimator for the parameter of an exponential distribution is the M.L.E.