# solhw43825 - SOLUTIONS FOR HOMEWORK 5, STAT 4382 1. Exerc....

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Unformatted text preview: SOLUTIONS FOR HOMEWORK 5, STAT 4382 1. Exerc. 3.1.1. Well, the probability in question is Pr( X = 0 , X 1 = 1 , X 2 = 2) . This joint probability can be written via conditional ones as follows, Pr( X = 0 , X 1 = 1 , X 2 = 2) = Pr( X = 0)Pr( X 1 = 1 | X = 0)Pr( X 2 = 2 | X 1 = 1 , X = 0) . Then, using Markov property of the chain we can simplify the right side, Pr( X = 0 , X 1 = 1 , X 2 = 2) = Pr( X = 0)Pr( X 1 = 1 | X = 0)Pr( X 2 = 2 | X 1 = 1) . (1) Now we can plug in numbers, Pr( X = 0 , X 1 = 1 , X 2 = 2) = ( . 3)( . 2)(0) = 0 . 2. Exerc. 3.1.2. Well, here we have a similar situation. Write Pr( X 2 = 1 , X 3 = 1 | X 1 = 0) = Pr( X 3 = 1 | X 2 = 1 , X 1 = 0)Pr( X 2 = 1 | X 1 = 0) [and continue using Markov property] = Pr( X 3 = 1 | X 2 = 1)Pr( X 2 = 1 | X 1 = 0) . Now plug in the given values from the transition probability matrix, Pr( X 2 = 1 , X 3 = 1 | X 1 = 0) = ( . 6)( . 2) = . 12 . For the second probability in question the formula is the same, Pr( X 1 = 1 , X 2 = 1 | X = 0) = Pr( X 2 = 1 | X 1 = 1)Pr( X 1 = 1 | X = 0) = . 12 . Remark: Note that the probabilities are the same. It should be clear why: there is just a unit shift in index (time) and, because the Markov chain is stationary — we have this equality. 3. Exerc. 3.1.3. Note that the probability in question is not as it is written in the text but Pr( X = 1 , X 1 = 0 , X 2 = 2 | X = 1) because it is given that X = 1. In the strict sense, the problem is not written correctly in the text. But in any case, now we can continue, using the Markov property of the chain, Pr( X = 1 , X 1 = 0 , X 2 = 2 | X = 1) == Pr( X 2 = 1 | X 1 = 1)Pr( X 1 = 1 | X = 0)Pr( X = 0 | X = 0) [and then we plug in numbers and continue]...
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## This note was uploaded on 02/27/2012 for the course ECONOMICS 2301 taught by Professor Ketsler during the Spring '11 term at University of Texas at Dallas, Richardson.

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solhw43825 - SOLUTIONS FOR HOMEWORK 5, STAT 4382 1. Exerc....

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