Recall the Gambler’s ruin problem from the last session. Suppose someone plays the following gambling game. They bet \(1\) pound on a fair flip of a coin and win \(1\) pound if it comes up heads, and lose if come up tails. They start with \(1\) pound, and stop playing once they have reached \(5\) pounds or have no money left. Express this problem up as a Markov chain. In particular:
Consider the transition matrix \(P\) on the states \(\{1,2,3,4\}\) given by:
P <- matrix(c(0, 1/2, 1/2, 0, 0, 1/2, 0,0,0,1/2, 1/2,0,0, 1/2, 0, 0,0,0,1/2, 1/2, 1/3, 1/3, 0, 0, 1/3), nrow =5)
P <-t(P)
P
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0000000 0.5000000 0.5 0.0 0.0000000
## [2,] 0.5000000 0.0000000 0.0 0.0 0.5000000
## [3,] 0.5000000 0.0000000 0.0 0.5 0.0000000
## [4,] 0.0000000 0.0000000 0.0 0.5 0.5000000
## [5,] 0.3333333 0.3333333 0.0 0.0 0.3333333
Compute powers of \(P\) until you see that all the entries are positive.
What does it mean if all the entries are positive?
Recall that \(\lambda\) is a probability measure if
Show that \[\rho = (1-t)\lambda + t \mu\] is a probability measure on \(\Omega\) for all \(t \in [0,1]\).
Let \(P\) be a transition matrix on a state space \(S\). Show that if \(\lambda\) and \(\mu\) are stationary measures of \(P\), then any convex combination of \(\lambda\) and \(\mu\) is a stationary measure for \(P\).
Consider the transition matrix on four states given by \[ \left( \begin{array}{cccc} 1/2 & 1/2 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1/3 & 2/3 \\ 0 & 0 & 1/2 & 1/2 \\ \end{array} \right) \]
Find three stationary measures for \(P\).
Does \(P\) have finitely many stationary measures?
Consider the transition matrix \(P\) on the states \(\{1,2,3\}\) given by:
P <- matrix( c( 0,0,1,0, 0,0,0,1, 1/2,1/2,0,0, 1,0,0,0), nrow =4)
P <- t(P)
P
## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 1 0
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 1.0 0.0 0 0