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to take the ball down the court and then they are on o↵ense.
To ﬁnd the stationary distribution we
0
32
B0
5
⇡0 ⇡1 ⇡2 ⇡3 B
@1
0
6
0 want to solve
1
1
1
5
1C
C= 0
2.5 1A
0
1 0 0 1 The answer can be found by reading the fourth row of the inverse of the matrix:
10
29 4
29 12
29 3
29 Thus in the long run the chain spends a fraction 4/29 in state 1, and 6/29 in
state 3. To translate this into a more useful statistic, we note that the average
time in state 1 is 1/5 and the average time in state 3 is 1/6, so the number of
baskets per minute for the two teams are
4/29
20
=
= 0.6896
1/5
29 3/29
18
=
= 0.6206
1/6
29 Multiplying by 2 points per basket and 40 minutes per game yields 55.17 and
49.65 points per game respectively.
Detailed balance condition. Generalizing from discrete time we can
formulate this condition as:
⇡ (k )q (k, j ) = ⇡ (j )q (j, k ) for all j 6= k (4.16) The reason for interest in this concept is
Theorem 4.5. If (4.16) holds, then ⇡ is a stationary distribution.
Why is this tru...
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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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