Let's assume a Markov chain has 2 recurrent classes and a transient state from which we can go to either of the recurrent classes. If one of those recurrent classes is periodic, would it effect the existence of steady state distribution of the other (aperiodic) class?

Even if both recurrent classes are aperiodic, how can we say they both converge, (but to different values)? If I start from the transient state and make the transition to one of the recurrent classes, I will never go to the other recurrent class, and in the long run, the steady state probabilities will be zero. But I just read that we can evaluate two classes separately and they can both have convergence. It makes sense to me when there is one recurrent class and some transient states, but I don't see how we can have convergence for both recurrent classes when we have two of them.


The recurrent classes do not affect each other. Once we get into one recurrent class, we never leave, and the structure of the other recurrent class is irrelevant.

Just take an example, say with 4 states and transition probability matrix $(P_{ij})$ given by:

$$ (P_{ij}) = \left[ \begin{array}{cccc} 0 & 1/2 & 1/2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right] $$

State 1 is transient. State 2 forms an aperiodic recurrent class: If we get to state 2 then we always stay there. States 3 and 4 form a periodic recurrent class: If we get to state 3, we bounce around (periodically) between 3 and 4.

-Given we start in state 2: We have a steady state distribution of $\pi=[0,1,0,0]$ (the limiting probabilities converge to this, and the time average fractions of time also converge to this with probability 1).

-Given we start in state 3: The limiting probabilities do not converge (they oscillate depending on even or odd slots), but the time averages converge to $p = [0,0,1/2,1/2]$ with probability 1.

-Given we start in state 1: Then the time averages will certainly converge, but they will not converge to a constant vector with probability 1. Rather, they will converge to a random vector. What they converge to will depend on the outcome of the first transition. If $p=[p_1,p_2,p_3,p_4]$ are the time averages, then given we start in state 1 we get: $$ p = \left\{ \begin{array}{ll} [0, 1, 0, 0] &\mbox{ with prob $1/2$} \\ [0, 0, 1/2, 1/2] & \mbox{ with prob $1/2$} \end{array} \right. $$

In general, for a finite state discrete time Markov chain with $K$ recurrent classes, each recurrent class $k \in \{1, \ldots, K\}$ has a unique probability distribution $\pi_k$ that satisfies $\pi_k = \pi_k P$ (where $P$ is the transition probability matrix) and such that $\pi_k$ has support only on the states of recurrence class $k$. If we start at a state in recurrence class $k$, then with probability 1 the time averages converge to $\pi_k$. If we start in a transient state, then the time averages will converge to a random vector $p$. We will eventually end up in one of the recurrent states (being the one we first visit). We can define $\theta_k$ as the probability we first visit recurrence class $k$ (defined for each $k \in \{1, \ldots, K\}$). Then $p$ is a random vector with $p = \pi_k$ with probability $\theta_k$, for $k \in \{1, \ldots, K\}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thank you, I think what I missed was to think about the distinction between limiting probabilities and the time averages. My only issue is the first point (given we start in state 2)Now thinking about it more carefully, I feel like in the case where we have at least one periodic recurrent class in our chain, we can eliminate the convergence of limiting probabilities, am I wrong? @Michael $\endgroup$ – Selori May 3 '15 at 3:54
  • 1
    $\begingroup$ Well in the above example, we do have one periodic recurrent class (states 3 and 4), but limiting probabilities exist given that we start in state 2 (these would be conditional limiting probabilities given we start in state 2). If we start in state 3 or 4, then limiting probabilities do not exist. $\endgroup$ – Michael May 3 '15 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.