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Unformatted text preview: STAT455/855 Fall 2003 Applied Stochastic Processes Final Exam, Brief Solutions 1. (15 marks) (a) (7 marks) The distribution of Y is given by P ( Y = y ) = 1 6 5 6 y 2 for y = 2 , 3 ,... The above follows because each of the rolls 2 ,...,y 1 should be different from the previous roll, which occurs with probability (5 / 6) y 2 , and then the y th roll should be the same as the previous roll, which occurs with probability 1 / 6. If we let N denote the net profit for the game, then E [ N ] = kP ( Y k ) + X y = k +1 ( y k ) P ( Y = y ) = k + X y = k +1 yP ( Y = y ) = k + X y = k +1 y 1 6 5 6 y 2 = k + 5 6 k 1 X y =1 ( k + y ) 1 6 5 6 y 1 = k + 5 6 k 1 ( k + 6) . (b) (8 marks) We now compute the expected net profit for the modified game. Let N mod denote this net profit. Then, conditioning on Y we get E [ N mod ] = X y =2 E [ N mod Y = y ] P ( Y = y ) = k X y =2 E [ N mod Y = y ] P ( Y = y ) + X y = k +1 ( y k ) P ( Y = y ) = k X y =2 E [ N mod Y = y ] P ( Y = y ) + E [ N ] + kP ( Y k ) (from part(a)) STAT 455/855  Final Exam Solutions, 2003 Page 2 of 6 Thus, E [ N mod ] < E [ N ] k X y =2 E [ N mod Y = y ] P ( Y = y ) + kP ( Y k ) < . We now consider E [ N mod Y = y ] for y k . Since we lose k dollars in the original game and we pay an additional 1 dollar to play the second game, our expected net profit will be the expected amount we win in the second play of the game less this amount. Letting Z denote the number of rolls it takes to get two consecutive rolls that are the same, we have E [ N mod Y = y ] = ( k + 1) + X z = k +1 zP ( Z = z ) = ( k + 1) + 5 6 k 1 ( k + 6) , where the second equality follows from part(a) since Z and Y have the same distribution. Note that E [ N mod Y = y ] does not depend on y , so we can cancel...
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This note was uploaded on 09/15/2010 for the course STAT 455/855 taught by Professor Glentakahara during the Fall '09 term at Queens University.
 Fall '09
 GLENTAKAHARA

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