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See here for a PDF version of this vignette.
This document assumes basic familiarity with Markov chains and linear algebra.
In this note, we illustrate one way of analytically obtaining the stationary distribution for a finite discrete Markov chain.
Assume our probability transition matrix is: \[P = \begin{bmatrix} 0.7 & 0.2 & 0.1 \\ 0.4 & 0.6 & 0 \\ 0 & 1 & 0 \end{bmatrix}\]
Since every state is accessible from every other state, this Markov chain is irreducible. Every irreducible finite state space Markov chain has a unique stationary distribution. Recall that the stationary distribution \(\pi\) is the row vector such that \[\pi = \pi P\].
Therefore, we can find our stationary distribution by solving the following linear system: \[ \begin{aligned} 0.7\pi_1 + 0.4\pi_2 &= \pi_1 \\ 0.2\pi_1 + 0.6\pi_2 + \pi_3 &= \pi_2 \\ 0.1\pi_1 &= \pi_3 \end{aligned} \] subject to \(\pi_1 + \pi_2 + \pi_3 = 1\). Putting these four equations together and moving all of the variables to the left hand side, we get the following linear system: \[ \begin{aligned} -0.3\pi_1 + 0.4\pi_2 &= 0 \\ 0.2\pi_1 + -0.4\pi_2 + \pi_3 &= 0 \\ 0.1\pi_1 - \pi_3 &= 0 \\ \pi_1 + \pi_2 + \pi_3 &= 1 \end{aligned} \]
We will define the linear system in matrix notation: \[\underbrace{\begin{bmatrix} -0.3 & 0.4 & 0 \\ 0.2 & -0.4 & 1 \\ 0.1 & 0 & -1 \\ 1 & 1 & 1 \end{bmatrix}}_A \begin{bmatrix} \pi_1 \\ \pi_2 \\ \pi_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}}_b \\ A\pi^T = b\]
The stationary distribution, which is usually represented by a row vector, is transposed with \(\pi^T\).
Since this linear system has more equations than unknowns, it is an
overdetermined system. Overdetermined systems can be solved using a QR
decomposition, so we use that here. (In brief, qr.solve
works by finding the QR decomposition of \(A\), \(A=QR\) with \(Q'Q=I\) and \(R\) an upper triangular matrix. Then if
\(A\pi^T = b\) it must be the case that
\(QR\pi^T=b\) which implies \(R\pi^T = Q'b\), and this can be solved
easily because \(R\) is
triangular.)
A <- matrix(c(-0.3, 0.2, 0.1, 1, 0.4, -0.4, 0, 1, 0, 1, -1, 1 ), ncol=3,nrow=4)
b <- c(0,0,0, 1)
pi <- qr.solve(A,b)
names(pi) <- c('state.1', 'state.2', 'state.3')
pi
# state.1 state.2 state.3
# 0.54054054 0.40540541 0.05405405We find that: \[ \pi_1 \approx 0.54, \pi_2 \approx 0.41, \pi_3 \approx 0.05 \]
Therefore, under proper conditions, we expect the Markov chain to spend more time in states 1 and 2 as the chain evolves.
Recall that we are attempting to find a solution to \[\pi = \pi P\] such that \(\sum_i \pi_i =1\). First we rearrange the expression above to get: \[ \begin{aligned} \pi - \pi P &= 0 \\ \pi (I - P) &= 0 \\ (I - P)^T\pi^T &= 0 \end{aligned} \]
One challenge though is that we need the constrained solution which
respects that \(\pi\) describes a
probability distribution (i.e. \(\sum \pi_i =
1\)). Luckily this is a linear constraint that is easily
represented by adding another equation to the system. So as a small
trick, we need to add a row of all 1’s to our \((I-P)^T\) (call this new matrix \(A\)) and a 1 to the last element of the
zero vector on the right hand side (call this new vector \(b\)). Now we want to solve \(A\pi = b\) which is over-determined so we
solve it as above using qr.solve.
The function stationary below implements the general
approach, and we test it with the worked example above.
stationary <- function(transition) {
stopifnot(is.matrix(transition) &&
nrow(transition)==ncol(transition) &&
all(transition>=0 & transition<=1))
p <- diag(nrow(transition)) - transition
A <- rbind(t(p),
rep(1, ncol(transition)))
b <- c(rep(0, nrow(transition)),
1)
res <- qr.solve(A, b)
names(res) <- paste0("state.", 1:nrow(transition))
return(res)
}
stationary(matrix(c(0.7, 0.2, 0.1, 0.4, 0.6, 0, 0, 1, 0),
nrow=3, byrow=TRUE))
# state.1 state.2 state.3
# 0.54054054 0.40540541 0.05405405
sessionInfo()
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