Question

6. Suppose that the proportion 0 of defective items in large shipment is unknown and that the prior distribution of 0 is the beta distribution with parameters 1 and 10. Assume in a random sample of 20 items that 1 item is found to be defective. (a) What is the expected value and variance of the prior distribution? (b) What is the posterior distribution? (c) What is the Bayes estimator for 0 if one uses the quadratic loss function? (d) Find the M.L.E. for 0. Is it the same as the Bayes estimator? (e) Suppose that you change the sampling plan and will keep on sampling until you find 3 defective items. Let X be the number of non-defective items until this happens. Derive the M.L.E. and Bayes estimator again.

Answer 1

Let X denote the random variable representing the number of defective items in the random sample of 20 items.

Now, we are given that the prior distribution of $\theta$ is the beta distribution with parameters 1 and 10

=> $\theta$ ~ Beta(α=1, β=10)

Moreover, since $\theta$ is the proportion of defective items in the random sample of 20 items, we can conclude that:

X|$\theta$ ~ Binomial(n=20, $\theta$ )

(a)

Now, the expected value and variance of prior distribution is given by:

E(0) = +8 Var(@) = 1 + 10 [ANSWER] a 8 (a + b)2(a + 8 + 1) 1 * 10 (1 + 10)2 *(1 + 10 + 1) 10 112 * 12 5 726 ANSWER

(b)

To find the posterior distribution, we use the following relation:

Posterior $\propto$ Prior*Likelihood ................(1)

Now, we are given the following details about the prior:

$\theta$ ~ Beta(α=1, β=10)

Prior = fe(0) * 99-1(1-0)8-1 Bla,B)* B(1.10) *" 10) *01-(1 - 0) 10-1 B(1.10) *(1-0)9

Also, we are given that X|$\theta$ ~ Binomial(n=20, $\theta$ ) and that we observed X = 1 (since in a random sample of 20 items we found 1 defective item). Thus:

Likelihood = Pxle(X = 1) = (1°)041 – 0320-1 [Using the formula for PMF of a Binomial Distribution = 20*4* (1 - 0) 19

Now, substituting the value of Prior and Likelihood in equation (1), we get:

Posterior, fox(@) & ( 20 10 *(1 – 0)*) * (20 * 8 * (1 – 0)19) 20 B(1,10) * *** (1 - 0)28 *(1 - 0)28 [Ignoring the constant terms] Now, we recognize the above expression as the kernel of a Beta distribution with parameters 2 and 29 # Posterior, fex (0) = * 02-1(1 - 0)29-1 ;0<o<1 (ANSWER 1

0X Beta(a=2,3 = 29) ANSWER

(c)

The Bayes estimator for $\theta$ under the quadratic loss function is just the mean of the Posterior disribution, thus we get:

Bayes estimator for under quadratic loss = E[0|X] = E[Betala = 2,8 = 29)] at B 2+29 = 31 ANSWER = 0.064516 ANSWER

(d)

The likelihood function is given by:

Likelihood, L = Pxel X = 1) = (20)*(1 – 6)20-1 [Using the formula for PMF of a Binomial Distribution] = 20**(1 - 0) 19 Taking logs both sides, we get: In(L) = ln(20) +In() + 19 * ln(1 - 0) Differentiating both sides w.r.t. 0, we get: d (In(L)) 1 19 der ā -1- ............ (2) Setting the above expression equal to zero, we get:

一日 | 19 1 -8 -1-8 199 200

To check that the last expression for $\theta$ is indeed a point of maxima for the likelihood, we again differentiate expression (2) w.r.t. $\theta$ :

of age) – $ 65 - ".) 1 19 = 82 (1 - 0)2 < 0 Maxima

Thus, the MLE of $\theta$ is:

Ô= = 0.05 [ANSWER]

From part (c) and the last expression we see that the Bayes estimator under quadratic loss and the M.L.E. for $\theta$ are not the same. [ANSWER]