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I understand that

  • if all states of an irreducible Markov chain are non-null recurrent, then the MC has a unique stationary distribution

  • if an irreducible Markov chain is finite, then all of its states are non-null recurrent.

Therefore, an irreducible finite Markov chain with transition probabilities $P$ has a unique stationary distribution, let's call it $\pi$.

What are the conditions under which $\pi$ is also a limiting distribution?

Is it true that if there exists a distribution $x$ such that the sequence of distributions $x, xP, xP^2, xP^3, \ldots$ does not converge to $\pi$, then the chain must be periodic?

Another way of asking this is: If $\pi$ is a stationary distribution but not a limiting distribution, we know that some initial distributions will converge to $\pi$, but what will the others do? Do they necessarily converge to a "periodic stationary distribution", can they just bounce around crazily,...?

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Your statements are slightly confusing, since a transient chain could be considered "non null recurrent" (after all, it is not null recurrent). So you should replace your statements with "positive recurrent":

  • if all states of an irreducible Markov chain are positive recurrent, then the MC has a unique stationary distribution

  • if an irreducible Markov chain is finite, then all of its states are positive recurrent.

You can also say that if a Markov chain is irreducible and positive recurrent, then the (unique) stationary distribution has strictly positive components.


Every finite state irreducible Markov chain $\{M(t)\}_{t=0}^{\infty}$ has a unique stationary distribution $\pi =(\pi_i)_{i \in S}$ (where $S$ denotes the finite state space). When you simulate, with probability 1, the sample path fractions of time converge to this distribution, so that: $$ \lim_{T\rightarrow\infty} \frac{1}{T}\sum_{t=0}^{T-1} 1\{M(t)=i\} = \pi_i \quad, \forall i \in S \quad \mbox{(with prob 1)}$$ regardless of the initial state $M(0)$. Taking expectations of both sides and using the bounded convergence theorem together with $E[1\{M(t)=i\}]=P[M(t)=i]$ we also get: $$ \lim_{T\rightarrow\infty} \frac{1}{T}\sum_{t=0}^{T-1} P[M(t)=i] = \pi_i \quad, \forall i \in S$$


If the chain is finite state, irreducible, and also aperiodic, you can further say $$ \lim_{n\rightarrow\infty} P[M(t)=i|M(0)=j] = \pi_i \quad, \forall i \in S$$ regardless of the initial state $j \in S$. So if the chain is finite state, irreducible, but limiting probabilities do not converge, then the chain cannot be aperiodic.

If $M(t)$ is finite state, irreducible, and periodic with period $d>1$, then limiting probabilities cannot converge (assuming we start in a particular state with probability 1). This is because: \begin{align*} \lim_{k\rightarrow\infty} P[M(kd)=i|M(0)=i] > 0 \\ \lim_{k\rightarrow\infty} P[M(kd+1)=i|M(0)=i] = 0 \end{align*} This is because the Markov chain $\{Z(k)\}_{k=0}^{\infty}$ defined by $Z(k)=M(kd)$ is irreducible and aperiodic (over an appropriately reduced state space) and so all states it can reach have positive steady state values.

With this reasoning, it can be shown that $P[M(t)=i|M(0)=j]$ converges (as $t\rightarrow\infty$) to a periodic function with period $d$. The particular $d$-periodic function it converges to depends on the intial state.

This should be pretty clear if you do a few matlab examples: Plot $\vec{\pi}(t) = \vec{\pi}(0)P^t$ versus $t \in \{0, 1, 2, ...\}$ for some examples with $\vec{\pi}(0)=[1 , 0, 0, ...]$ or $\vec{\pi}(0)=[0, 1, 0, 0, ...]$.

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  • $\begingroup$ Thanks, this clears things up. Is your objection to my use of "non-null recurrent" the fact that it may be confused with "non-(null recurrent)"? The former meaning recurrent, but non-null... I thought this was standard usage. Besides this point, the main idea I wanted to gain an intuition for is the fact hat under the stated conditions, the Markov chain always heads towards some sort of regularity (even if that regularity is periodic). It's sort of surprising to me that the probability vector never behaves "chaotically", for lack of a better term. Right? $\endgroup$ – theQman Nov 29 '17 at 15:59
  • $\begingroup$ Yes, at first I wrote your statements were "slightly incorrect" since I interpreted as non-(null recurrent). I then realized it could be interpreted as (non-null) recurrent, so I changed to "slightly confusing." I personally prefer "positive recurrent." There is no chaos, the probabilities converge to a $d$-periodic function. There is only one probability vector that satisfies $\pi = \pi P$ and so that acts like an "attractor" for the time averages (which must satisfy this equation since the number of times we arrive to state $i$ is always within 1 of the number of times we depart). $\endgroup$ – Michael Nov 29 '17 at 17:00

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