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For a doubly stochastic markov process defined by a n by n transition matrix, does the stationary distribution go to p = 1/n for each state? If so, why?

Dan Schwartz
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1 Answers1

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It has been answered several times here. Still for completeness I am answering it

  • we will prove that $[\frac{1}{n} ,\frac{1}{n} ,...,\frac{1}{n} ]$ satisfies $\pi P =\pi$
  • consider arbitrary $j \in \{1,2,..n\}$
  • $ \sum_i \pi_i p_{ij} = \sum_i \frac{1}{n} p_{ij} =\frac{1}{n} \sum_i p_{ij} = \frac{1}{n} . 1 = \pi_j$ (because for doubly stochastic matrix $\sum_i p_{ij} = 1$)
  • since this holds for arbitrary j.We are done
  • your finite markov chain can have multiple stationary distributions. one sufficient condition for unique stationary distribution is irreducibility \begin{bmatrix} 0.5 & 0.5 & 0 & 0 \\ 0.5 & 0.5 & 0 & 0 \\ 0& 0 &0.5 &0.5 \\ 0&0 &0.5 &0.5 \end{bmatrix} there are many stationary distributions for this doubly stochastic matrix
manifold
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