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# Markov Chain Analysis and Stationary Distribution

This example shows how to derive the symbolic stationary distribution of a trivial Markov chain by computing its eigen decomposition.

The stationary distribution represents the limiting, time-independent, distribution of the states for a Markov process as the number of steps or transitions increase.

Define (positive) transition probabilities between states A through F as shown in the above image.

syms a b c d e f cCA cCB positive;

Add further assumptions bounding the transition probabilities. This will be helpful in selecting desirable stationary distributions later.

assumeAlso([a, b, c, e, f, cCA, cCB] < 1 & d == 1);

Define the transition matrix. States A through F are mapped to the columns and rows 1 through 6. Note the values in each row sum up to one.

P = sym(zeros(6,6));
P(1,1:2) = [a 1-a];
P(2,1:2) = [1-b b];
P(3,1:4) = [cCA cCB c (1-cCA-cCB-c)];
P(4,4) = d;
P(5,5:6) = [e 1-e];
P(6,5:6) = [1-f f];
P
P =

$\left(\begin{array}{cccccc}a& 1-a& 0& 0& 0& 0\\ 1-b& b& 0& 0& 0& 0\\ \mathrm{cCA}& \mathrm{cCB}& c& 1-\mathrm{cCA}-\mathrm{cCB}-c& 0& 0\\ 0& 0& 0& d& 0& 0\\ 0& 0& 0& 0& e& 1-e\\ 0& 0& 0& 0& 1-f& f\end{array}\right)$

Compute all possible analytical stationary distributions of the states of the Markov chain. This is the problem of extracting eigenvectors with corresponding eigenvalues that can be equal to 1 for some value of the transition probabilities.

[V,D] = eig(P');

Analytical eigenvectors

V
V =

Analytical eigenvalues

diag(D)
ans =

$\left(\begin{array}{c}1\\ 1\\ c\\ d\\ a+b-1\\ e+f-1\end{array}\right)$

Find eigenvalues that are exactly equal to 1. If there is any ambiguity in determining this condition for any eigenvalue, stop with an error - this way we are sure that below list of indices is reliable when this step is successful.

ix = find(isAlways(diag(D) == 1,'Unknown','error'));
diag(D(ix,ix))
ans =

$\left(\begin{array}{c}1\\ 1\\ d\end{array}\right)$

Extract the analytical stationary distributions. The eigenvectors are normalized with the 1-norm or sum(abs(X)) prior to display.

for k = ix'
V(:,k) = simplify(V(:,k)/norm(V(:,k)),1);
end
Probability = V(:,ix)
Probability =

The probability of the steady state being A or B in the first eigenvector case is a function of the transition probabilities a and b. Visualize this dependency.

fsurf(Probability(1), [0 1 0 1]);
xlabel a
ylabel b
title('Probability of A');

figure(2);
fsurf(Probability(2), [0 1 0 1]);
xlabel a
ylabel b
title('Probability of B');

The stationary distributions confirm the following (Recall states A through F correspond to row indices 1 through 6 ):

• State C is never reached and is therefore transient i.e. the third row is entirely zero.

• The rest of the states form three groups, { A , B }, { D } and { E , F } that do not communicate with each other and are recurrent.

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