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SolutionsIII

# SolutionsIII - ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT III...

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ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT III Spring 2007 1. Markov chains are not only extraordinarily useful but also fun to play with. (a) The transition matrix P is defined by [ P ] ij = Prob { i j in one step } , so P = 2 4 1 / 2 1 / 6 1 / 3 1 / 3 1 / 2 1 / 6 2 / 3 1 / 6 1 / 6 3 5 . The stationary distribution π * is the unique row vector [ π * i π * 2 π * 3 ] with nonnega- tive entries summing to 1 that satisfies π * P = π * . Simple algebra (four equations in three unknowns, etc.) yields π * = ˆ 1 / 2 1 / 4 1 / 4 ˜ . (b) The indicated Markov chain, which is a “random walk”-type chain, will have transition matrix Q = 2 4 q 11 q 12 0 q 21 q 22 q 23 0 q 32 q 33 3 5 . To guarantee that this chain has stationary distribution π * from (a), it suffices to make sure that the q ’s satisfy the detailed-balance conditions π * i q ij = π * j q ji for all i and j . There are many ways to make this happen. The conditions hold if q 12 /q 21 = 1 / 2 and q 23 = q 32 . So let’s find α and β so that Q = 2 4 1 - α α 0 2 α 1 - 2 α - β β 0 β 1 - β 3 5 ; we need to make sure all the entries in Q are nonnegative as well. One choice is α = 1 / 4 and β = 1 / 2, which means that Q = 2 4 3 / 4 1 / 4 0 1 / 2 1 / 4 1 / 4 0 1 / 4 3 / 4 3 5 . You can check that π * Q = π * for our given π * . 2. It’s easy to do a complete analysis for a trivial example such as this one, but I dare you to try an example with longer strings and bigger populations. (a) The possible populations are { 0 , 0 } , { 0 , 1 } , and { 1 , 1 } . Let’s be boring and call them 1, 2, and 3, respectively. (b) See the accompanying diagram. (c) For Population 1, the parents for the next generation are always 0 and 0, so mutation is the only thing that can change Population 1 into another. 1 changes to 3 if and only if both parents mutate, which occurs with probability . 0001. 1 changes to 2 if and only if exactly one of the parents mutates, which occurs with probability 2( . 99)( . 01) = . 0198. 1 goes to 1 if and only if neither parent mutates, which happens with probability . 99 2 = . 9801. The situation with Population 3 is similar. The parents are always 1 and 1, mutation is the only change agent, 1

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and 3 goes to 1 with probability . 0001, to 2 with probability . 0198, and to 3 with probability . 9801. As for Population 2, any pair of parents for the next generation is possible. Let’s call A the set of parents 0 and 0; B the set of parents 0 and 1; and C the set of parents 1 and 1. Because of fitness-proportional selection, A occurs with probability 1 / 36, B with probability 10 / 36, and C with probability 25 / 36. Using the standard notation, denote by p 21 the probability that Population 1 arises as the next generation. Then p 21 = (1 / 36)Prob(1 | A ) + (10 / 36)Prob(1 | B ) + (25 / 36)Prob(1 | C ) , where, e.g., the term Prob(1 | A ) means the probability that Population 1 arises when the parents are A . We’ve computed Prob(1 | A ) = . 9801 and Prob(1 | C ) = . 0001 already. And Prob(1 | B ) = . 0099, since 1 arises from B if and only if the 1 mutates and the 0 does not. Accordingly, p 21 = . 0301 approximately. Similarly, p 22 = (1 / 36)Prob(2 | A ) + (10 / 36)Prob(2 | B ) + (25 / 36)Prob(2 | C ) , We’ve computed Prob(2 | A ) = Prob(2 | C ) = . 0198 already. And Prob(2 | B ) = . 9801, since 3 (like 1) arises from B with probability . 0099, and 2 arises from B if and only if 1 doesn’t and 3 doesn’t. Accordingly, p
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