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hw4 - solutions

hw4 - solutions - IE 336 Operations Research Stochastic...

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IE 336: Operations Research - Stochastic Models Homework Set #4 Solutions Spring 2008 1. Let P= p q 0 3 q 0 r + q 0 0 1 be the transition matrix of a Markov chain (a) Determine transition diagram: 1 2 3 1 3q p q r+q (b) If q > p 2 , is it possible that p = 2 3 ? Explain It is impossible that p = 2 3 . Let assume that p = 2 3 , then q > ( 2 3 ) 2 = 4 9 . p + q > 2 3 + 4 9 = 10 9 (*) But P is a transition matrix p + q = 1 (**) Then (*) and (**) p must be different from 2 3 (c) If r = 1 5 , determine the values of p and q P is a transition matrix braceleftbigg p + q = 1 4 q + r = 1 Then q = 1 5 and p = 4 5 2. Consider the ”eye color” example from class (section 2.21 in the book). Show that transition matrix P = aa ab bb aa p 1 - p 0 ab p 2 1 2 1 - p 2 bb 0 p 1 - p For the first row of the transition matrix P { X 1 = aa | X 0 = aa } = P { X 1 = aa | X 0 = aa,father = aa } P { father = aa } + P { X 1 = aa | X 0 = aa,father = ab } P { father = ab } + P { X 1 = aa | X 0 = aa,father = bb } P { father = bb } 1
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Since P { X 1 = aa | X 0 = aa,father = aa } = 1, P { X 1 = aa | X 0 = aa,father = ab } = 1 2 , P { X 1 = aa | X 0 = aa,father = bb } = 0, P { father = aa } = p 2 , P { father = ab } = 2 p (1 - p ) P { X 1 = aa | X 0 = aa } = p Clearly, P { X 1 = bb | X 0 = aa } = 0 P { X 1 = aa | X 0 = aa } + P { X 1 = ab | X 0 = aa } = 1 P { X 1 = ab | X 0 = aa } = 1 - p For the second row of the transition matrix P { X 1 = aa | X 0 = ab } = 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 } Since P { X 1 = aa | X 0 = ab,father = aa } = 1 2 , P { X 1 = aa | X 0 = ab,father = ab } = 1 4 , P { X 1 = aa | X 0 = ab,father = bb } = 0, P { father = aa } = p 2 , P { father = ab } = 2 p (1 - p ) P { X 1 = aa | X 0 = ab } = 1 2 p 2 + 1 4 (2 p (1 - p )) = p 2 P { X 1 = ab | X 0 = ab } = 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 } Since P { X 1 = ab | X 0 = ab,father = aa } = 1 2 , P { X 1 = ab | X 0 = ab,father = ab } = 1 2 , P { X 1 =
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