91Ó°ÊÓ

Arandom sample of size n is taken from a Poisson distribution for which the value of the mean θ is unknown and the prior distribution of θ is a gamma distribution for which the mean is\({\mu _0}\)

Consider that prior distribution of\(\theta \)is gamma with parameters\(\alpha \)and \(\beta \) . Thus, mean of the prior distribution is:

\({\mu _0} = \frac{\alpha }{\beta }\)……………………………………………… (1)

Due to conjugate pairs of posterior and prior distributions, the posterior distribution of\(\theta \),\(\xi \left( {\theta |{x_1},{x_2},...,{x_n}} \right)\), is also gamma with parameters\(\alpha + \sum\limits_{i = 1}^n {{x_i}} \)and\(\beta + n\).

So, mean of posterior distribution is:

\(\begin{array}{c}{\mu _1} = \frac{{\alpha + \sum\limits_{i = 1}^n {{x_i}} }}{{\beta + n}}\\ = \frac{\alpha }{{\beta + n}} + \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\beta + n}}\\ = \frac{{{\raise0.7ex\hbox{${\beta \alpha }$} \!\mathord{\left/

{\vphantom {{\beta \alpha } \beta }}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$\beta $}}}}{{\beta + n}} + \frac{{{\raise0.7ex\hbox{${n\sum\limits_{i = 1}^n {{x_i}} }$} \!\mathord{\left/

{\vphantom {{n\sum\limits_{i = 1}^n {{x_i}} } n}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$n$}}}}{{\beta + n}}\end{array}\)

\( = \left( {\frac{\beta }{{\beta + n}}} \right)\frac{\alpha }{\beta } + \left( {\frac{n}{{\beta + n}}} \right){\overline X _n}\)

Substitute from equation (1)

\({\mu _1} = \left( {1 - \frac{n}{{\beta + n}}} \right){\mu _0} + \left( {\frac{n}{{\beta + n}}} \right){\overline x _n}\)

Thus, it is proved that the mean of posterior distribution of \(\theta \) is a weighted average of the \({\gamma _n}{\overline X _n} + \left( {1 - {\gamma _n}} \right){\mu _0}\)where\({\gamma _n} = \frac{n}{{\beta + n}}\)

Apply limit\(n \to \infty \)for\({\gamma _n}\),

\(\begin{array}{c}\mathop {\lim }\limits_{n \to \infty } = \mathop {\lim }\limits_{n \to \infty } \frac{\beta }{{\beta + n}}\\ = 1\end{array}\)

Thus, it is proved that as \(n \to \infty \), \({\gamma _n} \to 1\)