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

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