hw7sols - ISyE 3232 H. Ayhan Stochastic Manufacturing and...

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

View Full Document Right Arrow Icon
Solutions to Homework 7 October 24, 2011 1. According to the fact that the row summation of transition matrix P equals one, we can fill the missing values in P: 1 5 4 5 0 2 5 1 2 1 10 0 1 10 9 10 . And α is the initial distribution, so the summation of its entries equals one too. So we have α = ( 1 5 , 2 5 , 2 5 ) . a) Pr( x 1 = 0 | x 0 = 1 ) = P 1 , 0 = 2 5 . b) We let α i denote the distribution of X i (so α 0 = α ). Then we have α i = α 0 P i . So X 1 ’s distribution α 1 = αP = ( 1 5 , 2 5 , 2 5 ) 1 5 4 5 0 2 5 1 2 1 10 0 1 10 9 10 = ( 1 5 , 2 5 , 2 5 ) . c) α 15 = αP 15 . We find from b) α 1 = αP = ( 1 5 , 2 5 , 2 5 ) = α . So αP i +1 = αP i = ... = α . Let i = 14 ; we have α 15 = αP 15 = α . d) It’s irreducible. Every state can communicate with each other. e) State 0 can reach itself in one step. So the Markov Chain is aperiodic. f) Yes, it’s postively recurrent. Because it’s a irreducible and finite Markov Chain. 1 ISyE 3232 H. Ayhan Stochastic Manufacturing and Service Systems Fall 2011
Background image of page 1

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

View Full DocumentRight Arrow Icon
2. (a) In order to show that { X n ,n
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

hw7sols - ISyE 3232 H. Ayhan Stochastic Manufacturing and...

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