STAT_333_Assignment_3_Solutions

STAT_333_Assignment_3_Solutions - STAT 333 Assignment 3...

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STAT 333 Assignment 3 SOLUTIONS Type I Problems 1. #10, page 265 of the text, 9 th edition. Gary’s mood is given by an MC with S = {C, S, G} and transition matrix = 5 . 0 3 . 0 2 . 0 3 . 0 4 . 0 3 . 0 1 . 0 4 . 0 5 . 0 P . If we want to examine the probability he is not in a glum mood (state G) in *any* of the next three days, we can modify the matrix to make him stay in that state if he ever enters it. (In other words, make Glum an absorbing state.) = 1 0 0 3 . 0 4 . 0 3 . 0 1 . 0 4 . 0 5 . 0 ' P . Then, if we multiply this matrix by itself 3 times, we will have the 3-step probabilities for this modification. = 1 0 0 561 . 0 220 . 0 219 . 0 415 . 0 292 . 0 293 . 0 ) ' ( 3 P . The third element in the first row (P (3) C,G = 0.415) represent the probability that, given he was Cheerful, he moves to the Glum state any time in the next three days. So we want 1 – 0.415 = 0.585. This can also be done by listing all possible paths of Cheerful and So-So, and summing up. 2. #11, page 265 of the text, 9 th edition. Because the process is a Markov Chain, knowing Gary hasn’t been cheerful in a week is irrelevant if we know the state he was in 4 days ago. (We can ignore the history bringing him to that point.) We again modify the matrix so that Cheerful is an absorbing state. = 5 . 0 3 . 0 2 . 0 3 . 0 4 . 0 3 . 0 0 0 1 ' ' P . Taking the 4 th power of the matrix, = 1885 . 0 1593 . 0 6522 . 0 1593 . 0 1354 . 0 7053 . 0 0 0 1 ) ' ' ( 4 P . We want the probability he goes from Glum to Glum in 4 steps (P (4) G,G = 0.1885), but conditional on the fact that he did not every go to the Cheerful state in those 4 steps (1 - P (4) G,C = 1 - 0.6522 = 0.3478). So the probability is 0.1885/0.3478 = 0.542 This can also be done (similarly to #10) by brute force, though there are more cases. 3. #25, page 267 of the text, 9 th edition. There are two ways to approach this problem. The state space is always {0, 1, …, k}. You can define the Markov Chain such that X n = the number of pairs of shoes at a fixed door (say the
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front door) each day, or you can define Xn = the number of pairs of shoes at the door they happen to leave from each day. Both are Markovian and both will get the correct answer. The second case is slightly easier in terms of calculating the probabilities. For 0 < i < k, P i,i = P i, i-1 = P i, k-i = P i, k-i+1 = ¼ The first probability is the situation where the runner returns to the same door she left from one day, and then leaves from the same door the next day. The second is the situation where she comes home to the opposite door she left from, but then leaves from the original door the next day. Other cases are similarly defined and all equally likely.
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This note was uploaded on 01/16/2011 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

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STAT_333_Assignment_3_Solutions - STAT 333 Assignment 3...

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