91Ó°ÊÓ

Given 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 Poisson, \({X_i} \sim P\left( \lambda \right)\).

The pmf is:

\(f\left( x \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}},x = 0,1, \ldots \)

In the method of moments method,

\(\begin{array}{l}{\mu _j}\left( \theta \right) = E\left( {X_i^j} \right) \ldots \left( 1 \right)\\{m_j} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^j} \ldots \left( 2 \right)\\Equating\,\,\left( 1 \right)\,\,and\,\,\left( 2 \right)\\{\mu _j} = {m_j}\end{array}\)

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

\(\begin{array}{l}{\mu _1}\left( \theta \right) = E\left( {X_i^1} \right) \ldots \left( 1 \right)\\{m_1} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^1} \ldots \left( 2 \right)\end{array}\)

Equating (1) and (2)

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

Let’s calculate MLE to estimate the \(\lambda \) parameter of an exponential distribution.

\(L\left( \lambda \right) = \prod\limits_{i = 1}^n {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \)

Applying log likelihood,

\(\begin{array}{c}\ln L\left( \lambda \right) = \ln \prod\limits_{i = 1}^n {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{{x_i}!}}} \\ = \ln \left( {{e^{ - n\lambda }}} \right) + \ln {\lambda ^{\sum\limits_{i = 1}^n {{x_i}} }} - \ln \prod\limits_{i = 1}^n {{x_i}} \\ = - n\lambda + \sum\limits_{i = 1}^n {{x_i}} \ln \lambda - \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} \end{array}\)

Differentiating it to find the value of \(\lambda \)that maximises the log likelihood function.

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

Equating to zero,

\(\begin{array}{c} - n + \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\hat \lambda }} = 0\\n = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\hat \lambda }}\\\hat \lambda = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\end{array}\)

Finally, the M.L.E. is

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

Hence, the method of moments estimator for the parameter of a Poisson distribution is the M.L.E.