91Ó°ÊÓ

The number of defects in a 1200-foot roll of magnetic recording tape has a Poisson distribution for which the value of the mean θ is unknown, and the prior distribution of θ is the gamma distribution with parameters 3 and 1.

Consider an experiment with only two outcomes: Success and Failure.

Let x be the number of success we observe out of nconducted trails of the experiment.

Let\(\theta \)be the true proportion of successes.

If it is assigned a uniform prior distribution over the range (0,1) for\(\theta \)and assume that the likelihood of observing x successes in n trials, given the value of\(\theta \), is a binomial distribution, then the posterior distribution for\(\theta \)is beta with parameters\(\alpha = x + 1\)and\(\beta = n - x + 1\). The Bayesian posterior estimate, variance of the estimate of\(\theta \)and posterior distribution of\(\theta \)and posterior distribution of\(\theta \)are given as below:

\(\begin{array}{l}E\left( {\theta /x} \right) = \frac{\alpha }{{\alpha + \beta }}\\V\left( {\theta /x} \right) = \frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}\left( {\alpha + \beta + 1} \right)}}\end{array}\)

Let’s consider that the number of defects in a 1200-foot roll of magnetic recording tape has a poisson distribution for which the value of the mean\(\theta \)is unknown, a prior distribution\(\theta \), is equal to the gamma distribution with parameters\(\alpha = 3\)and\(\beta = 1\).

When five rolls of this tape are selected at random and inspected, the numbers of defects found on the rolls are 2,2,6,0 and 3.

Since,

If\({X_{1,}}..,{X_n}\)form a random sample from the Poisson distribution with parameter\(\theta \)and the the prior distribution of\(\theta \)is gamma distribution with parameters\(\alpha \)and\(\beta \)given that\({X_i} = {x_i}\left( {i = 1,2,...n} \right)\)is the gamma distribution with parameters\(\alpha + \sum\limits_{i = 1}^n {{x_i}} \)and\(\beta + n\).

Therefore, by using the above theorem the posterior distribution for\(\theta \)is gamma distribution with parameters

\(\begin{array}{c}\alpha + \sum\limits_{i = 1}^n {{x_i}} = 3 + 13\\ = 16\end{array}\)

And

\(\begin{array}{c}\beta + n = 1 + 5\\ = 6\end{array}\)

Therefore, the mean of the posterior distribution is computed as follows

\(\begin{array}{c}E\left( {\theta |{X_i} = {x_i}\left( {i = 1,...,n} \right)} \right) = \frac{{16}}{6}\\ = \frac{8}{3}\end{array}\)