545454 0272727 0181818 converting the answer to

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Unformatted text preview: 6⇡2 + .3⇡3 ⇡1 + ⇡2 + ⇡3 = ⇡1 = ⇡2 =1 Subtracting ⇡1 from both sides of the first equation and ⇡2 from both sides of the second, this translates into ⇡ A = (0, 0, 1) with 0 1 .2 .1 1 .4 1A A = @ .2 .3 .3 1 Note that here and in the previous example the first two columns of A consist of the first two columns of the transition probability with 1 subtracted from the diagonal entries, and the final column is all 1’s. Computing the inverse and reading the last row gives (0.545454, 0.272727, 0.181818) Converting the answer to fractions using the first entry in the MATH menu gives (6/11, 3/11, 2/11) To check this we note that 6/11 = ✓ 0 .8 3/11 2/11 @.2 .3 4.8 + .6 + .6 11 1 .1 .2A .4 .1 .6 .3 .6 + 1.8 + .6 11 .6 + .6 + .8 11 ◆ Example 1.20. Basketball (continuation of 1.10). To find the stationary matrix in this case we can follow the same procedure. A consists of the first three columns of the transition matrix with 1 subtracted from the diagonal, and a final column of all 1’s. 1/4 0 2/3 0 1 /4 0 1...
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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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