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Stat333_Assign1_SOLNS.09W

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

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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 unaffected). Formally 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 fixed 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 fill in positions 2 and 3 (order within does not matter) is ( 2 2 ) ( m +2 2 ) = 2 ( m + 1)( m + 2) 5.(b) Here, since the renewal property does not hold and there are not too many doors, it is simplest
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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)]...
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