**Last updated:** 2019-03-31

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**Knit directory:** `fiveMinuteStats/analysis/`

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Rmd | 3bb3b73 | mbonakda | 2016-02-24 | add two mixture model vignettes + merge redundant info in markov chain vignettes |

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Rmd | 0f93e3c | mbonakda | 2016-01-30 | split simulating discrete markov chains into three separate notes |

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 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. Recall from linear algebra that an overdetermined system is consistent (i.e. we can solve for \(\pi\) exactly) when \(b\) is in the column space of \(A\). We can check this numerically by obtaining the rank of \(A\), then obtaining the rank of an augmented matrix with \(b\) appended as a column of \(A\). (Side note: it can be difficult to find the rank of an arbitrary matrix numerically, especially for large matrices. But for our small example, we are safe.)

```
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)
rank.A <- as.numeric(rankMatrix(A))
rank.Ab <- as.numeric(rankMatrix(cbind(A,b)))
print(paste("The rank of A =", rank.A, "and the rank of the augmented matrix =", rank.Ab))
```

`[1] "The rank of A = 3 and the rank of the augmented matrix = 3"`

We see that \(A\) has full column rank, and that the rank is unchanged when we add \(b\) as a column. Therefore, \(b\) is in the column space of \(A\), and this system is consistent. We can find \(\pi\) by solving the normal equations: \[A^TA\pi = A^Tb\]

We use the solve function in R to solve for the stationary distribution \(\pi\):

```
pi <- drop(solve(t(A) %*% A, t(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.

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 overdetermined, so we solve the normal equations \(A^TA\pi = A^Tb\).

`sessionInfo()`

```
R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.1
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/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_1.2-15
loaded via a namespace (and not attached):
[1] workflowr_1.2.0 Rcpp_1.0.0 lattice_0.20-38 digest_0.6.18
[5] rprojroot_1.3-2 grid_3.5.2 backports_1.1.3 git2r_0.24.0
[9] magrittr_1.5 evaluate_0.12 stringi_1.2.4 fs_1.2.6
[13] whisker_0.3-2 rmarkdown_1.11 tools_3.5.2 stringr_1.3.1
[17] glue_1.3.0 xfun_0.4 yaml_2.2.0 compiler_3.5.2
[21] htmltools_0.3.6 knitr_1.21
```

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