** Chapter 11 Question 3, Page 414 **
Example 11.5:
Example 11.5 Each time a certain horse runs in a three-horse race, he has proba- bility 1/2 of winning, 1/4 of coming in second, and 1/4 of coming in third, indepen- dent of the outcome of any previous race. We have an independent trials process, but it can also be considered from the point of view of Markov chain theory. The transition matrix is
P = matrix(c(.5, .25, .25,
.5, .25, .25,
.5, .25, .25), nrow = 3, ncol = 3, byrow=T)
answer
P2 = P %*% P
print(P2)
## [,1] [,2] [,3]
## [1,] 0.5 0.25 0.25
## [2,] 0.5 0.25 0.25
## [3,] 0.5 0.25 0.25
P3 = P2 %*% P
print(P3)
## [,1] [,2] [,3]
## [1,] 0.5 0.25 0.25
## [2,] 0.5 0.25 0.25
## [3,] 0.5 0.25 0.25
# To visualize PN for very large N:
P100 = P
for (i in 1:99)
P100 = P100 %*% P
print(P100)
## [,1] [,2] [,3]
## [1,] 0.5 0.25 0.25
## [2,] 0.5 0.25 0.25
## [3,] 0.5 0.25 0.25
From the above we can see that \(P^n\) is the same as P.