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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 = 


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

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

Analytical eigenvectors

V = 

(b-1a-100-c-dcCB-bcCA-bcCB+ccCAσ1b-1a-d00-10100-c-dcCA-acCA-acCB+ccCBσ110010000-c-dc+cCA+cCB-10000001010100000f-1e-1000-f-1d-e0-101-d1000101)where  σ1=c+cCA+cCB-1a+b-ac-bc+c2-1

Analytical eigenvalues

ans = 


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'));
ans = 


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);
Probability = V(:,ix)
Probability = 

(b-1a-1σ10σ5σ21σ10σ6σ200-c-1σ3c+cCA+cCB-1011σ3000000)where  σ1=b-12a-12+1  σ2=σ3c+cCA+cCB-1a+b-c-1  σ3=c-12c+cCA+cCB-12+σ62σ4+σ52σ4+1  σ4=c+cCA+cCB-12a+b-c-12  σ5=cCB-bcCA-bcCB+ccCA  σ6=cCA-acCA-acCB+ccCB

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');

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.