91Ó°ÊÓ

An unknown parameter\(\theta \) needs to be estimated using a random sample \({X_1}...{X_n}\) of size n from the distribution for which \(\theta \) is the parameter.

Let \(\xi \left( \theta \right)\) be the prior distribution \(\theta \) and \(f\left( {x|\theta } \right)\) be the joint pdf of \({X_1}...{X_n}\).

The conditional pdf of \(\theta \) given \({X_1} = {x_1},...{X_n} = {x_n}\) is called the posterior distribution of \(\theta \) and is denoted by \(\xi \left( {\theta |{X_1}...{X_n}} \right)\). Then \(\xi \left( {\theta |{X_1}...{X_n}} \right) \propto {f_n}\left( {X|\theta } \right)\xi \left( \theta \right)\)

Let\(\theta \) be the proportion of defective items in a large manufactured lot.

Let prior distribution of\(\theta \) being uniform on the interval\(\left( {0,1} \right)\). A random sample of n = 8

times yielded exactly 3 defectives.

Let \({X_i} = 0\) if the inspected item is defective and \({X_i} = 1\) is the item is not defective.

Comparing the situation with Bernoulli distribution with parameter\(\theta \). Further, as each of the inspected items is independent of the others.

Hence joint pdf of \({X_1}...{X_n}\) is given by:

\(\begin{aligned}{f_n}\left( {X|\theta } \right) = \prod\limits_{i = 1}^8 {{\theta ^{\left( {{x_i}} \right)}}{{\left( {1 - \theta } \right)}^{\left( {1 - {x_i}} \right)}}} \\ = {\theta ^3}{\left( {1 - \theta } \right)^5}\end{aligned}\)

The prior distribution of\(\theta \) is the uniform distribution on the interval \(\xi \left( \theta \right) = 1\) for \(0 < \theta < 1\)

We know that:

\(\begin{aligned}\xi \left( {\theta |{X_1}...{X_n}} \right) \propto {f_n}\left( {X|\theta } \right)\xi \left( \theta \right)\\ = {\theta ^3}{\left( {1 - \theta } \right)^5} \times 1\\ = {\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{6 - 1}}\end{aligned}\)

Comparing the above expression with beta pdf with parameters \(\alpha = 4\) \(\beta = 6\), it s clear that the conditional distribution of\(\theta \) is proportional to the beta distribution. That is \(\frac{{\left| \!{\overline {\, {\left( {\alpha + \beta } \right)} \,}} \right. }}{{\left| \!{\overline {\, \alpha \,}} \right. \left| \!{\overline {\, \beta \,}} \right. }}\)

If multiplied with the constant term, the posterior pdf \(\theta \) will be beta with \(\alpha = 4\) and\(\beta = 6\). Thus, the posterior pdf of\(\theta \) is given by

\(\xi \left( {\theta |{X_1}...{X_n}} \right) = \frac{{\left| \!{\overline {\, {\left( {4 + 6} \right)} \,}} \right. }}{{\left| \!{\overline {\, 4 \,}} \right. \left| \!{\overline {\, 6 \,}} \right. }}{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{6 - 1}}\)