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Unformatted text preview: ECE 496 MIDTERM SOLUTIONS Spring 2007 1. The answer is (b). In fact, the matrix in (b) is the only one that could be the onestep transitionprobability matrix for a Markov chain, since it’s the only one whose rows each add up to 1. 2. The answer is (c). You can test each candidate answer by checking whether π * P = π * , where P is your answer to the previous problem. Answer (c) is the only one that works with any of the matrices, by the way. 3. The answer is (a). In fact, even if you got Problem 1 wrong, you would have to pick (a) here. P (1 , 4) = 0 for all four choices of matrix in Problem 1 (and in the picture there’s no arrow from 1 to 4), so you can’t have light 1 flash and then light 4 flash on the next step. 4. The answer is (c). To determine the probability that light 2 flashes at t = 3, all you need to know is what light flashed at t = 2. The entire past history is irrelevant. Since light 3 flashes at t = 2, the probability that light 2 will flash at t = 3 is P (3 , 2) = 1 / 6. (You can read this off the picture if you’re unsure about your answer to Problem 1.) 5. The answer is (a). The key idea is that the limiting value of the indicated sum is merely the average of the “payoff function” with respect to the stationary distribution π * . (Note: the information that light 2 flashes at t = 0 is irrelevant.) You found π * in Problem 2, so the limit is equal to 3 × 1 9 + 5 × 2 9 + 7 × 1 3 + 11 × 1 3 + 20 × 0 + 37 × 0 = 67 9 ....
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 Spring '07
 DELCHAMPS
 Algorithms, Probability, Quantification, Universal quantification, Existential quantification, speciﬁc copy

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