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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 (1u 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 (1u ) x1 udu = 1 / [ x ( x + 1)]...
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This note was uploaded on 10/24/2010 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.
 Spring '08
 Chisholm

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