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Hmwk7solution

# Hmwk7solution - ISyE 3232 Stochastic Manufacturing and...

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ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 J. Dai and H. Ayhan Solutions to Homework 7 1. Since α is a distribution, the sum of entries should be 1. α = (1 / 5 , 2 / 5 , 2 / 5) . The transition matrix P has the property that elements in each row of P sum up to one. Therefore we have P = 1 / 5 4 / 5 0 2 / 5 1 / 2 1 / 10 0 1 / 10 9 / 10 (a) Pr( X 1 = 0 | X 0 = 1) = P 1 , 0 = 2 / 5 . (b) The row vector αP = (0 . 2 , 0 . 4 , 0 . 4) describes distribution of X 1 . For i = 0 , 1 , 2 Pr( X 1 = i ) = Pr( X 1 = i | X 0 = 0) Pr( X 0 = 0) + Pr( X 1 = i | X 0 = 1) Pr( X 0 = 1) + Pr( X 1 = i | X 0 = 2) Pr( X 0 = 2) = α 0 P 0 ,i + α 1 P 1 ,i + α 2 P 2 ,i = i th element of row vector αP. Note that αP = α, so α is actually a stationary distribution. (c) The row vector αP 15 = (0 . 2 , 0 . 4 , 0 . 4) describes distribution of X 15 . For i = 0 , 1 , 2 Pr( X 15 = i ) = Pr( X 15 = i | X 0 = 0) Pr( X 0 = 0) + Pr( X 15 = i | X 0 = 1) Pr( X 0 = 1) + Pr( X 15 = i | X 0 = 2) Pr( X 0 = 2) = α 0 P 15 0 ,i + α 1 P 15 1 ,i + α 2 P 15 2 ,i = i th element of row vector αP 15 . 2. In the solutions below we use the following formulas: for two events A and B P ( A | B ) = P ( A T B ) P ( B ) and for two random variables Y and Z and a constant k E [ Y | Z = k ] = X n =0 n P { Y = n | Z

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Hmwk7solution - ISyE 3232 Stochastic Manufacturing and...

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