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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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]...
View Full Document

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.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online