91Ó°ÊÓ

The prior distribution of θ is as given in Exercise 2, and 20 items are selected at random from the shipment.

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

Let x be the number of successes observed out of n conducted experiment trials and \(\theta \)be the true proportion of successes.

Let 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 parameter\(\alpha = x + 1\)and\(\beta = n - x + 1\).

The Bayesian posterior estimate, the variance of the estimate of\(\theta \)and the posterior distribution of\(\theta \)are given as below

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

Considering the proportion\(\theta \)of defective, is items in a large shipment is unknown, the prior distribution of\(\theta \), is equal to the beta distribution with parameters\(\alpha = 5\)and\(\beta = 10\)

Consider that 20 items are selected at random from the shipment and the number of defectives items in the sample as ‘z,’ then the posterior distribution for\(\theta \)is beta with parameters

\(\alpha = 5 + z\)and

\(\begin{array}{c}\beta = n - z + \beta \\ = 30 - z\end{array}\)

Since for Bayes estimates of\(\theta \), the mean squared error of the estimate is equal to the variance of the posterior distribution. Therefore, the variance of the posterior distribution is computed as follows

\(\begin{array}{c}Var\left( {\theta |x} \right) = \frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}\left( {\alpha + \beta + 1} \right)}}\\ = \frac{{\left( {5 + z} \right) \times \left( {30 - z} \right)}}{{{{\left( {\left( {5 + z} \right) + \left( {30 - z} \right)} \right)}^2} \times \left( {\left( {5 + z} \right) + \left( {30 - z} \right) + 1} \right)}}\\ = \frac{{\left( {5 + z} \right) \times \left( {30 - z} \right)}}{{{{\left( {35} \right)}^2} \times \left( {36} \right)}}\end{array}\)

Now, differentiate the variance concerning ‘z’ and equate the differentiation with zero to get the maximum value of ‘z.’ So, differentiation is as follows:

\(\begin{array}{c}\frac{{dVar\left( {\theta |x} \right)}}{{dz}} = \frac{d}{{dz}}\frac{{\left( {5 + z} \right) \times \left( {30 - z} \right)}}{{{{\left( {35} \right)}^2} \times \left( {36} \right)}}\\ = \frac{1}{{{{\left( {35} \right)}^2} \times \left( {36} \right)}} \times \frac{{d\left( {150 + 25z - {z^2}} \right)}}{{dz}}\\ = \frac{{25 - 2z}}{{{{\left( {35} \right)}^2} \times 36}}\end{array}\)

To compute the maximum value of ‘z,’ equates this derivative concerning zero as follows

\(\begin{array}{c}\\\end{array}\)\(\begin{array}{c}\frac{{dVar\left( {\theta |x} \right)}}{{dz}} = 0\\\frac{{25 - 2z}}{{{{\left( {35} \right)}^2} \times \left( {36} \right)}} = 0\\25 - 2z = 0\end{array}\)

\(z = 12.5\)

Therefore, 12.5 defective items maximize the mean squared error of the Bayes estimates.

From the above derivative\(150 + 25z - {z^2}\)is a quadratic term and the coefficient of\({z^2}\).

Therefore, the value of ‘z’ is minimum at z=0