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Consider a Markov chain $\{X_t\}$ on a finite state $\mathcal{S} = \{1,\dots, S\}$ space whose transition matrix $P$ is populated by elements of the form $$ p_{ij} = P(X_{t+1} = j | X_t = i)$$ and we know that the $S$ vector $p_i = (p_{i,1},\dots, p_{i,S})$ follows a Dirichlet distribution with concentration parameter $(\alpha_{i1},\dots,\alpha_{iS})$, and this is true for $i \in \mathcal{S}$. Does the notion of an associated stationary distribution make sense in this context? And if so, can it be described in terms of a Dirichlet distribution?

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  • $\begingroup$ @Xi'an No, but you only know the distribution of the elements, not the realizations $\endgroup$
    – Dan.phi
    May 6, 2021 at 14:14

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The transition matrix $P$ is almost surely strongly irreducible, hence produces a Markov chain with a stationary distribution $\pi$ associated with the eigenvalue $\lambda_1=1$ of $P$, i.e., $\pi P=\pi$. According to this paper, when $S$ grows to infinity, $\pi$ converges to the uniform distribution in total variation.

When considering the case $S=2$, the stationary distribution associated with$$P=\left(\begin{matrix} 1-a &a\\ b &1-b\end{matrix} \right)$$ is$$\pi=(b\ \ a)\big/(a+b)$$where $a\sim\mathcal B(\alpha_1,\alpha_2)$ and $b\sim\mathcal B(\alpha_2,\alpha_1)$, which shows $\pi$ is not a Dirichlet variate.

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  • $\begingroup$ What about with fixed $S$? $\endgroup$
    – Dan.phi
    May 6, 2021 at 16:26
  • $\begingroup$ Right, thanks, I accepted the answer. May I ask about the case before where a new realization was drawn at each $t$? $\endgroup$
    – Dan.phi
    May 6, 2021 at 16:30
  • $\begingroup$ This is a very interesting case, but while there is a stabilisation due to the finite state space, I thought the chain cannot be stationary in the Markov sense, since the transition varies at each time. See however this answer on math overflow that shows the opposite, by integrating out the Markov matrix. $\endgroup$
    – Xi'an
    May 6, 2021 at 16:54

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