Stochastic

28 consider a markov chain with nite state space s

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: g (1) = 1 + 0.6(1.25) 1.75 = = 2.3333 0.75 0.75 Example 1.46. Tennis. In Example 1.40 we formulated the last portion of the game as a Markov chain in which the state is the di↵erence of the scores. The state space was S = {2, 1, 0, 1, 2} with 2 (win for server) and 2 (win for opponent). The transition probability was 2 1 0 -1 -2 1 0 0 .6 0 0 2 1 .6 0 0 0 0 0 .4 0 .6 0 -1 0 0 .4 0 0 -2 0 0 0 .4 1 Let g (x) be the expected time to complete the game when the current state is x. By considering what happens on one step X g (x) = 1 + p(x, y )g (y ) y Since g (2) = g ( 2) = 0, if we let r(x, y ) be the restriction of the transition probability to 1, 0, 1 we have X g (x) r(x, y )g (y ) = 1 y Writing 1 for a 3 ⇥ 1 matrix (i.e., column vector) with all 1’s we can write this as (I r)g = 1 50 CHAPTER 1. MARKOV CHAINS so g = (I r) 1 1. There is another way to see this. If N (y ) is the number of visits to y at times n 0, then from (1.12) Ex N (y ) = 1 X rn (x, y ) n=0 To see that this is (I r) 1 (x, y )...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

Ask a homework question - tutors are online