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Unformatted text preview: MATH 56A: FALL 2006 HOMEWORK AND ANSWERS 2. Math 56a: Homework 2 p. 35 #1.1, 1.2, 1.3 1.1. Every morning a newspaper is added to a pile. With probability 1/3 the pile is emptied emptied. But if the pile has 5 newspapers, it is always emptied. Make this into a Markov chain. This is a Markov process with states 0, 1, 2, 3, 4 representing the number of newspapers in the pile in the evening. The transition matrix is 1/3 2/3 0 0 0 1/3 0 2/3 0 0 1/3 0 0 2/3 0 P = 1/3 0 0 0 2/3 1 0 0 0 0 1.2. Given that P = 1/3 2/3 3/4 1/4 What is the probability that X3 = 1 given that X0 = 0? The answer is the (0, 1) entry of P 3 . Instead of doing this in the straightforward boring method I will use right eigenvectors: P vi = i vi . The eigenvalues of P are 1, 5/12 with right eigenvectors v0 = (1, 1) and v1 = (8, 9): e0 = 9 1 v0 + v1 17 17 5 12 3 1 v1 17 9 P e0 = v0 + 17
3 whose 0th coordinate is 9 p3 (0, 1) = + 17 1 5 12 3 8 109 = 17 216 2 HW AND ANSWERS 2006 1.3. Given that .4 .2 .4 P = .6 0 .4 .2 .5 .3 what is the long term probability of being in state 1? The answer is 25/66 = 0.378787879 1) By directly raising this matrix to a high power. By squaring the matrix 4 times you get 0.378787879 0.257575758 0.363636364 P 16 = 0.378787879 0.257575758 0.363636364 0.378787879 0.257575758 0.363636364 So, the invariant probability distribution is (0.378787879, 0.257575758, 0.363636364) The first coordinate is the long term probability of being in state 1. 2) By directly computing the invariant distribution as a left eigenvector. The invariant probability distribution is given by the equation: (x, y, z)P = (x, y, z) with solution 25 17 4 = (x, y, z) = , , 66 66 11 = (0.378787879, 0.257575758, 0.363636364) ...
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 Spring '11
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