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Pre-requisites

This document assumes basic familiarity with Markov chains and linear algebra.

Overview

In this note, we illustrate one way of analytically obtaining the stationary distribution for a finite discrete Markov chain.

3x3 example

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 vector such that \[\pi = \pi P\].

Therefore, we can find our stationary distribution by solving the following linear system: \[\begin{align*} 0.7\pi_1 + 0.4\pi_2 &= \pi_1 \\ 0.2\pi_1 + 0.6\pi_2 + \pi_3 &= \pi_2 \\ 0.1\pi_1 &= \pi_3 \end{align*}\] 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: \[\begin{align*} -0.3\pi_1 + 0.4\pi_2 &= 0 \\ 0.2\pi_1 + -0.4\pi_2 + \pi_3 &= 0 \\ 0.1\pi_1 - \pi_3 &= 0 \\ \pi_1 + \pi_2 + \pi_3 &= 1 \end{align*}\]

We will define the linear system in matrix notation: \[\underbrace{\begin{bmatrix} -0.3 & 0.4 & 0 \\ 0.2 & -0.4 & 1 \\ 0.1 & 0 & -1 \\ 1 & 1 & 1 \end{bmatrix}}_A \begin{bmatrix} \pi_1 \\ \pi_2 \\ \pi_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}}_b \\ A\pi = b\]

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 \(A\pi = b\) it must be the case that \(QR\pi=b\) which implies \(R\pi = Q'b\), and this can be solved easily because \(R\) is triangular.)

library(Matrix)
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 0.05405405 

We find that: \[\begin{align*} \pi_1 \approx 0.54, \pi_2 \approx 0.41, \pi_3 \approx 0.05 \end{align*}\]

Therefore, under proper conditions, we expect the Markov chain to spend more time in states 1 and 2 as the chain evolves.

The General Approach

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: \[\begin{align} \pi - \pi P &= 0 \\ (I - P)\pi &= 0 \end{align}\]

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 \((I-P)\) (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 overdetermine so we solve it as above using qr.solve.


sessionInfo()
R version 4.0.5 (2021-03-31)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS Big Sur 10.16

Matrix products: default
BLAS:   /Library/Frameworks/R.framework/Versions/4.0/Resources/lib/libRblas.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.0/Resources/lib/libRlapack.dylib

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] Matrix_1.3-2

loaded via a namespace (and not attached):
 [1] Rcpp_1.0.6        whisker_0.4       knitr_1.32        magrittr_2.0.1   
 [5] workflowr_1.6.2   lattice_0.20-41   R6_2.5.0          rlang_0.4.10     
 [9] fansi_0.4.2       stringr_1.4.0     tools_4.0.5       grid_4.0.5       
[13] xfun_0.22         utf8_1.2.1        git2r_0.28.0      htmltools_0.5.1.1
[17] ellipsis_0.3.2    rprojroot_2.0.2   yaml_2.2.1        digest_0.6.27    
[21] tibble_3.1.1      lifecycle_1.0.0   crayon_1.4.1      later_1.2.0      
[25] vctrs_0.3.8       fs_1.5.0          promises_1.2.0.1  glue_1.4.2       
[29] evaluate_0.14     rmarkdown_2.7     stringi_1.5.3     compiler_4.0.5   
[33] pillar_1.6.0      httpuv_1.5.5      pkgconfig_2.0.3  

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