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Solutions_practice_theoretical_part

Solutions_practice_theoretical_part - IEOR E4701 Practice...

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Unformatted text preview: IEOR E4701 Practice Midterm - Theoretical Solutions Summer 2011 Question 1 - gambler's ruin Let us recall the gambler's ruin problem. We have two players, A and B. A coin will be tossed independently at each time. The coin comes up heads with probability p ∈ (0 , 1) and tails with probability q = 1- p . If the coin comes up heads at time n ≥ 1 , player A receives one dollar from B. Otherwise, if the tails come up, player B receives 1 dollar from player A. The initial wealth of A is i and the total wealth of the players equals N > i . We denote by X n the wealth of player A right after the n-th toss has performed and the money has been transferred. As we indicated earlier, X = i . We de ne T k = min { n ≥ 0 : X n = k } for any k = 0 , 1 ,...,N . a) Let M be the maximum wealth that player A ever reaches during the duration of the game and de ne g ( i ) = E i ( M · I ( T < T N )) . Provide a recursion that g ( i ) satis es, together with corresponding boundary conditions. Solution: g ( i ) = iP i ( T < T i +1 ) + g ( i + 1) P i ( T i +1 < T ) subject to g (0) = 0 = g ( N ) . Recall that P i ( T N < T ) = 1- ( q/p ) i 1- ( q/p ) N . b) Provide an explicit expression for the quantity P i ( M > k,T < T N ) for each value of k = 0 , 1 ,...,N . Solution: First notice that, for i = 0 , N , P i ( M > k, T < T N ) = 0 , regardless of the value of k . It is also the case that the probability above is zero for k = N, N- 1 regardless of the value of i . We now need to divide to cases with respect to k < N- 1 for i 6 = 0 , N : 1 IEOR E4701 Practice Midterm - Theoretical Solutions Summer 2011 • ≤ k < i : Since X = i ≤ M , we have P i ( M > k, T < T N ) = P...
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