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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).
 Spring '10
 DURRETT
 The Land

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