5102-Solution-Set-02

5102-Solution-Set-02 - SOLUTIONS TO PROBLEM SET 2 STAT...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS TO PROBLEM SET 2 STAT 5102-004 (1) Suppose we are interested in the proportion of defective items in a large population. Consider taking a sample from that population. If the population is large compared to any sample we might see, we can safely think of the observations as independent Bernoulli trials with common probability that a defective item is drawn. Problem 6.2.6 describes sampling a fixed number n of items from the population and counting the number of defectives. This sampling is described by a binomial distribution. The distribution of the number of defective items Y has density f ( y | ) = n y y (1- ) n- y (0 , 1) n N y { ,...,n } . Now, assume that we have a uniform prior for the proportion of defectives. The posterior density is proportional to the likelihood times the prior density. In this case, lik( | y ) y (1- ) n- y and ( ) 1, so that the posterior density is proportional to ( y +1)- 1 (1- ) ( n- y +1)- 1 (0 , 1) . We recognize this as the kernel of a beta density with = y + 1 and = n- y + 1. Since y = 3 of the n = 8 observations are defectives, the posterior for is Beta(4 , 6). Problem 6.2.9 describes sampling from the population until a fixed number r of defective items are observed. This sampling is described by a negative binomial distribution. The distribution of the number of non-defective items X has density f ( x | ) = x + r- 1 x r (1- ) x (0...
View Full Document

Page1 / 3

5102-Solution-Set-02 - SOLUTIONS TO PROBLEM SET 2 STAT...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online