18 chapter 1 markov chains example 115 consider the

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Unformatted text preview: r q of initial probabilities. If there are k states, then pn (x, y ) is a k ⇥ k matrix. So to make the matrix multiplication work out right, we should take q as a 1 ⇥ k matrix or a “row vector.” 18 CHAPTER 1. MARKOV CHAINS Example 1.15. Consider the weather chain (Example 1.3) and suppose that the initial distribution is q (1) = 0.3 and q (2) = 0.7. In this case ✓ ◆ .6 .4 .3 .7 = .32 .68 .2 .8 since .3(.6) + .7(.2) = .32 .3(.4) + .7(.8) = .68 Example 1.16. Consider the social mobility chain (Example 1.4) and suppose that the initial distribution: q (1) = .5, q (2) = .2, and q (3) = .3. Multiplying the vector q by the transition probability gives the vector of probabilities at time 1. 0 1 .7 .2 .1 .5 .2 .3 @.3 .5 .2A = .47 .32 .21 .2 .4 .4 To check the arithmetic note that the three entries on the right-hand side are .5(.7) + .2(.3) + .3(.2) = .35 + .06 + .06 = .47 .5(.2) + .2(.5) + .3(.4) = .10 + .10 + .12 = .32 .5(.1) + .2(.2) + .3(.4) = .05 + .04 + .12 = .21 If qp = q then q is called a stationary distribution. If the distribution at time 0 is the same as the distribution at time 1, then by the Markov property...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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