So I know that a connected Markov chain has a stationary distribution $\pi$ that satisfies

$$\lim_{t \rightarrow \infty} a_t = \pi,$$

where $a_t$ is the average probability distribution at time $t$. So this average converges to our stationary distribution. However, does that mean that $\lim_{t \rightarrow \infty} p_t = \pi$ (where $p_t$ is the probability vector of where the Markov chain is after $t$ steps)?

Just going off of the average definition, couldn't it be that the Markov chain bounces around two or more different probability distributions, so that the average is still $\pi$, but $\pi$ itself is never a probability distribution for the states of the Markov chain?

For example, say we have 3 states and $\pi = (0.2, 0.3, 0.5)$, then $p_{(1)} = (0.1, 0.2, 0.7)$ and $p_{(2)} = (0.3, 0.4, 0.3)$. Say for even $i$, $p_i = p_{(1)}$, and for uneven $i$, $p_i = p_{(1)}$, which means that the distribution of the process oscillates between $p_{(1)}$ and $p_{(2)}$. So the average $\lim_{t \rightarrow \infty} a_t$ is indeed $\pi$, but not $\lim_{t \rightarrow \infty} p_t = \pi$, since $p_t$ always switches between $p_{(1)}$ and $p_{(2)}$.

This is obviously a simple example, but I hope you understand my general question.

  • 1
    $\begingroup$ I think you need to add aperiodic to guarantee stationarity. Now if there exists a stationarity distribution, it is necessarily unique. $\endgroup$
    – Xi'an
    Jul 31, 2023 at 16:05

1 Answer 1


Here is a fairly general statement from Meyn & Tweedie (1994) that addresses the question: convergence to the invariant probability measure holds for all Harris recurrent aperiodic chains (without aperiodicity it fails!)

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with an additional result about the unicity of the stationary measure provided by recurrence only

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and another one about the convergence of the averages

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where$$S_n(f)=\sum_{i=1}^n f(\it\Phi_i)$$


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