The third equation is redundant since if we add up

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Unformatted text preview: p = ⇡ says ⇡1 ⇡2 ⇡3 0 .7 @.3 .2 which translates into three equations .2 .5 .4 1 .1 .2A = ⇡1 .4 .7⇡1 + .3⇡2 + .2⇡3 .2⇡1 + .5⇡2 + .4⇡3 .1⇡1 + .2⇡2 + .4⇡3 = ⇡1 = ⇡2 = ⇡3 ⇡2 ⇡3 20 CHAPTER 1. MARKOV CHAINS Note that the columns of the matrix give the numbers in the rows of the equations. The third equation is redundant since if we add up the three equations we get ⇡1 + ⇡2 + ⇡3 = ⇡1 + ⇡2 + ⇡3 If we replace the third equation by ⇡1 + ⇡2 + ⇡3 = 1 and subtract ⇡1 from each side of the first equation and ⇡2 from each side of the second equation we get .3⇡1 + .3⇡2 + .2⇡3 = 0 .2⇡1 .5⇡2 + .4⇡3 = 0 ⇡1 + ⇡2 + ⇡3 = 1 (1.10) At this point we can solve the equations by hand or using a calculator. By hand. We note that the third equation implies ⇡3 = 1 substituting this in the first two gives .2 = .5⇡1 ⇡1 ⇡2 and .1⇡2 .4 = .2⇡1 + .9⇡2 Multiplying the first equation by .9 and adding .1 times the second gives 2.2 = (0.45 + 0.02)⇡1 or ⇡1 = 22/47 Multiplying the first equation by .2 and adding 0.16...
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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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