The above symmetric property isn't referring to the double stochastic property

\begin{bmatrix} 0.2&0.8&0&0&0\\ 0.2&0.2&0.6&0&0\\ 0&0.4&0.2&0.4&0\\ 0&0&0.6&0.2&0.2\\ 0&0&0&0.8&0.2 \end{bmatrix}

The stationary distribution is given as (1/16,1/4,3/8,1/4,1/16).

Is it true that for an irreducible and aperiodic markov chain, if we can label the first row to last row (state 1 to 5) and also alternatively (state 5 to 1) and we get the same transition matrix, we will have a symmetric stationary distribution i.e. (a,b,c,b,a)?


1 Answer 1


More generally let $\pi$ be a permutation matrix. The act of permuting rows an columns is equivalent to $\pi P \pi$. Let $\mu$ be the stationary distribution of $P$, and suppose $\pi P\pi=P$. Define $\mu'=\pi^T\mu$. Then:

$$\mu^T=\mu^T P = \mu^T \pi P\pi=\mu'^TP\pi$$

Multiply both sides on the right by $\pi^{-1}=\pi^T$ ($\pi$ is orthonormal afterall). This gives:


Since the Markov chain $P$ is assumed to be irreducible and aperiodic, it has a unique stationary distribution, which allows us to conclude $\mu'=\mu$. Thus if $P$ is left invariant under permutations of its rows and columns by $\pi$, this implies $\mu=\pi\mu$, i.e. $\mu$ is invariant under $\pi$.

If you look at the individual coefficients, you get $\mu_i=\mu_{\pi^{-1}(i)}$.

  • $\begingroup$ did that mean my inference that we will have a "symmetric" stationary distribution is true for all Markov chain P when it is assumed to be irreducible and aperiodic $\endgroup$
    – Joseph
    Nov 1, 2017 at 8:01
  • $\begingroup$ @Joseph: Right, I added a line to clarify that. In your case you have the reverse permutation, which implies $\mu_i=\mu_{n-i}$, since $\pi^{-1}(i)=n-i$. $\endgroup$
    – Alex R.
    Nov 1, 2017 at 17:26

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