Stat333_Assign1_SOLNS.09W

Stat333_Assign1_SOLNS.09W - 5.(b) Here, since the renewal...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Stat 333 Winter 2009 Assign #1 Selected Partial Solutions 1. A is right and the guard is wrong. That is, if we let B * = the guard says B is going free, then P ( A executed) = P ( A executed | B * ) = 1 / 3. Intuitively this occurs because the guard is only allowed to choose between B and C . If he says B , then C ’s probability of execution increases from 1/3 to 2/3 ( A is unaFected). ±ormally you can proceed by Bayes rule: P ( A executed | B * ) = P ( A executed B * ) P ( A executed B * ) + P ( A not executed B * ) = (1 / 3)(1 / 2) (1 / 3)(1 / 2) + (1 / 3)(1) = 1 / 3 (1) noting that [ A not executed B * ] = [ C executed]. 2. Here the key to (c) is the identity [ Y > n ] = [ X 1 = max( X 1 , . . ., X n ] so P ( Y > n ) = 1 /n and P ( Y = n ) = P ( Y > n - 1) - P ( Y > n ) = 1 / [ n ( n - 1)]. 3. To obtain the covariance here, you need to consider m black balls plus two ²xed red balls i and j . These m + 2 balls come out in some random ordering; each ordering is equally likely by symmetry. Thus the probability that red balls i and j ²ll in positions 2 and 3 (order within does not matter) is ( 2 2 ) ( m +2 2 ) = 2 ( m + 1)( m + 2)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.(b) Here, since the renewal property does not hold and there are not too many doors, it is simplest to write down all the paths that lead out. (In fact, using this method you can actually determine the probability mass function of X and directly compute E ( X ) and E ( X 2 ).) 8. (a) Since U and V are independent, U uniform over (0, 1), V uniform over (0, 2), we have P ( V > U ) = i 1 P ( V > U | U = u ) f ( u ) du = i 1 P ( V > u | U = u ) du by substitution and f ( u ) = 1 = i 1 P ( V > u ) du since U and V independent = i 1 (1-u 2 ) du = 3 / 4 (2) (b) note P ( V > U | I V = 1) = 1 while P ( V > U | I V = 0) = P ( V > U | < V < 1) = 1 / 2 by symmetry. Therefore P ( V > U ) = (1)(1 / 2) + (1 / 2)(1 / 2) = 3 / 4 9. P ( X = x ) = I 1 P ( X = x | U = u ) f ( u ) du = I 1 (1-u ) x-1 udu = 1 / [ x ( x + 1)]...
View Full Document

This note was uploaded on 10/24/2010 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

Ask a homework question - tutors are online