91Ó°ÊÓ

A random sample \({X_1},{X_2},...,{X_n}\) of size n is taken from exponential distribution with unknown parameter\(\theta \).

Let prior distribution of \(\theta \)be gamma distribution with parameters \(\alpha > 0\) and\(\beta > 0\).

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

So,

\(E\left[ {L\left( {\theta ,\delta *\left( x \right)} \right)|x} \right] = \mathop {\min }\limits_a E\left[ {L\left( {\theta ,a} \right)|x} \right]\)

Consider a random sample of size n taken from a exponential distribution with unknown parameter\(\theta \).

Consider that prior distribution of\(\theta \)is gamma with parameters\(\alpha \)and\(\beta \).

Thus, the 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\(\alpha + n\)and\(\beta + \sum\limits_{i = 1}^n {{x_i}} \).

So, mean of posterior distribution is:

\({\mu _1} = \frac{{\alpha + n}}{{\beta + \sum\limits_{i = 1}^n {{x_i}} }}\)

\(\begin{array}{l} = \frac{{{\raise0.7ex\hbox{${\left( {\alpha + n} \right)}$} \!\mathord{\left/

{\vphantom {{\left( {\alpha + n} \right)} n}}\right.\kern-\nulldelimiterspace}

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

{\vphantom {{\left( {\beta + \sum\limits_{i = 1}^n {{x_i}} } \right)} n}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$n$}}}}\\ = \frac{{1 + {\raise0.7ex\hbox{$\alpha $} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$n$}}}}{{{{\overline x }_n} + {\raise0.7ex\hbox{$\beta $} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$n$}}}}\end{array}\) ..……………………………………. (2)

Let’s apply limit\(n \to \infty \)to Bayes estimator of\(\theta \)computed in equation (2).

We understand that as\(n \to \infty \),\({\raise0.7ex\hbox{$\alpha $} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$n$}} \to 0\)and\({\raise0.7ex\hbox{$\beta $} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$n$}} \to 0\)

So,

\(\mathop {\lim }\limits_{n \to \infty } {\mu _1} = \frac{1}{{{{\overline x }_n}}}\) …………………………………….. (3)

From law of large number we know that\(\frac{1}{{{{\overline x }_n}}}\)converges in probability to\(\theta \)as mean of exponential distribution is\(\frac{1}{\theta }\).

So, from equation (3) it follows those Bayes estimators for\(n = 1,2,...\)form a consistent sequence of estimators of\(\theta \)when the loss function is squared error function.