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# HW 4 - Homework Set#4 Solutions IE 336 Spring 2011 1(a Let...

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Homework Set #4 Solutions IE 336 Spring 2011 1. (a) Let { 1, 2, 3 } represent the states of the Markov chain. Figure 1: Transition Diagram for Problem 1a (b) Determine p , q , and r . 2 r + 3 p + q 3 = 1 q 8 + r + 3 q 8 + 2 r = 1 r + 6 p + q 6 = 1 q 2 = 6 r Solving the equations simultaneously yields p = 1 / 9, q = 1, and r = 1 / 6. Thus: P = 1 2 3 1 1 / 3 1 / 3 1 / 3 2 1 / 8 13 / 24 2 / 6 3 1 / 6 2 / 3 1 / 6 (c) Determine the walk probability. p 11331121232312323 = p 11 p 13 p 33 p 31 p 11 p 12 p 21 p 12 p 23 p 31 p 12 p 23 p 32 p 23 = p 2 11 · p 3 12 · p 13 · p 21 · p 3 23 · p 3 31 · p 32 · p 33 = 1 3 2 · 1 3 3 · 1 3 · 1 8 · 2 6 3 · 1 6 3 · 2 3 · 1 6 = 3 . 2668 × 10 - 9 1

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2. (a) See Example 2.14 (Section 2.12.11) in the text. Let S = { aa, ab, bb } . To find the second row of the transition matrix, use the law of total probability, conditioning on the gene pair of the father. Also, assume that the father’s pair of chromosomes is binomial, where p is the probability that a randomly chosen eye-color gene is a . P ( X 1 = aa | X 0 = ab ) = X k S P ( X 1 = aa | X 0 = ab, father = k ) · P (father = k ) = P ( X 1 = aa | X 0 = ab, father = aa ) · P (father = aa ) + P ( X 1 = aa | X 0 = ab, father = ab ) · P (father = ab ) + P ( X 1 = aa | X 0 = ab, father = bb ) · P (father = bb ) = 1 2 · p 2 + 1 4 · 2 p (1 - p ) + 0 · (1 - p ) 2 = p 2 2 + p 2 - p 2 2 = p 2 P ( X 1 = ab | X 0 = ab ) = X k S P ( X 1 = ab | X 0 = ab, father = k ) · P (father = k ) = P ( X 1 = ab | X 0 = ab, father = aa ) · P (father = aa ) + P ( X 1 = ab | X 0 = ab, father = ab ) · P (father = ab ) + P ( X 1 = ab | X 0 = ab, father = bb ) · P (father = bb ) = 1 2 · p 2 + 1 2 · 2 p (1 - p ) + 1 2 · (1 - p ) 2 = p 2 2 + p - p 2 + 1 2 - p + p 2 2 = 1 2 P ( X 1 = bb | X 0 = ab ) = X k S P ( X 1 = bb | X 0 = ab, father = k ) ·
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HW 4 - Homework Set#4 Solutions IE 336 Spring 2011 1(a Let...

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