m507b-hw3soln-s08

m507b-hw3soln-s08 - MATH 507b ASSIGNMENT 3 SOLUTIONS SPRING...

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MATH 507b ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (1.2) 1 2 in two steps means 1 3 2, so p 2 (1 , 2) = p (1 , 3) p (3 , 2) = ( . 9)( . 4) = . 36. 2 3 in 3 steps means 2 1 3 3 or 2 2 1 3 or 2 1 1 3, so p 3 (2 , 3) = ( . 7)( . 9)( . 6) + ( . 3)( . 7)( . 9) + ( . 7)( . 1)( . 9) = . 63 . (1.3) We claim that for all μ and all n 0, P μ ( X n = 0) = β α + β + (1 - α - β ) n ± μ (0) - β α + β ² . Equivalently, subtracting from 1, P μ ( X n = 1) = α α + β - (1 - α - β ) n ± μ (0) - β α + β ² . For n = 0 the first equality says P μ ( X n = 0) = μ (0) which is true by definition. Suppose the claim is true for some n . Then P μ ( X n +1 = 0) = P ( X n +1 = 0 | X n = 0) P μ ( X n = 0) + P ( X N =1 = 1 | X n = 1) P μ ( X n = 1) = (1 - α ) ³ β α + β + (1 - α - β ) n ± μ (0) - β α + β ²´ + β ³ α α + β - (1 - α - β ) n ± μ (0) - β α + β ²´ = β α + β + (1 - α - β ) n +1 ± μ (0) - β α + β ² , so the claim is true for n + 1. (1.5) Since just one son and daughter from each generation are mated, and any other sib- lings are irrelevant, we may assume that each generation contains exactly one male and one female. The son effectively chooses a gene at random from each parent, independently. The daughter does the same, independently of the son. If, for example, the parents are Aa,Aa then the son and daughter are (without regard to order): 1
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AA,AA with probability 1 4 · 1 4 = 1 16 , AA,Aa with probability 2 · 1 4 · 1 2 = 1 4 , AA,aa with probability 2 · 1 4 · 1 4 = 1 8 , Aa,Aa with probability 1 2 · 1 2 = 1 4 , Aa,aa with probability 2 · 1 4 · 1 2 = 1 4 , aa,aa
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m507b-hw3soln-s08 - MATH 507b ASSIGNMENT 3 SOLUTIONS SPRING...

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