hw10soln

# hw10soln - ORIE 3500/5500 Fall'10 HW 10 Solutions...

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ORIE 3500/5500, Fall ’10 HW 10 Solutions Assignment 10 Solutions Problem 1 Let X be the time in jail and Y be the result of the ﬁrst draw. From the information given, P ( Y = 0) = P ( Y = 1) = P ( Y = 3) = 1 / 3 . Now, we have E ( X ) = E ( E ( X | Y )) = E ( X | Y = 0) P ( Y = 0) + E ( X | Y = 1) P ( Y = 1) + E ( X | Y = 3) P ( Y = 3) = 1 3 ( E ( X | Y = 0) + E ( X | Y = 1) + E ( X | Y = 3) ) . But we know that E ( X | Y = 0) = 0 , E ( X | Y = 1) = 1 + E ( X ) , E ( X | Y = 3) = 3 + E ( X ) . The ﬁrst equation is clear: if the prisoner draws the number zero, he will be released immediately. The second equation is true because if the prisoner draws the number 1, he will stay in jail for one year and then will be at the exact same situation as before he made the ﬁrst draw. His expected additional time until freedom will be just E ( X ). The third equation follows from similar reasoning. Hence we obtain

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## This note was uploaded on 10/29/2011 for the course MATH 3310 at Cornell.

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hw10soln - ORIE 3500/5500 Fall'10 HW 10 Solutions...

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