Chapter 6 Period, Ergodic, Communicate (Lecture on 01/21/2021)

We looked at random walk in different dimensions and proved that in dimensions 1 and 2, state 0 is recurrent, but in dimensions more than 3, state 0 is transient. We have seen that states can be classified into transient state and recurrent state. For recurrent state it can be further classified into null recurrent state and non-null recurrent state. We saw one example where we found out whether state 0 is null/non-null recurrent for a one-dimensional random walk.

In that kind of problem we cares about whether f00=1 or f00<1. However, usually it is hard to find f00(n) directly, so we tend to pii(n), which is much easier to calculate, then use Corollary 4.1 to get whether f00=1 or f00<1. That is not always the case. In the following example, we will see that calculate f00(n) directly is actually more easier.

Example 6.1 (Success runs of a Binomial trail) The Markov chain is given by an infinite dimensional transition probability matrix (p01p00000p101p1000p2001p200pr01pr0) This transition matrix consiste of a column vector (p0,p1,,pr,) and a diagonal matrix diag(1p0,1p1,,1pr,).

Question: Provide conditions on p0,,pr,, such that state 0 becomes a recurrent state.

Let f00(n) be the probability of first time visiting state 0 at the n-th time given that x0=0. We then have f00(1)=p0=1(1p0)f00(2)=(1p0)p1=(1p0)[1(1p1)]=(1p0)(1p0)(1p1)f00(3)=(1p0)(1p1)p2=(1p0)(1p1)[1(1p2)]=(1p0)(1p1)(1p0)(1p1)(1p2)f00(n)=n2i=0(1pi)n1i=0(1pi) Then we have m+1n=1f00(n)=m+1n=1[n2i=0(1pi)n1i=0(1pi)]=1mi=0(1pi) and we get f00=lim

Thus, when \sum_{i=0}^{\infty}p_i=\infty, it is equivalent to f_{00}=1-\lim_{m\to\infty}\prod_{i=0}^m(1-p_i)=1, which is also equivalent to 0 is recurrent state.
Result: \lim_{m\to\infty}\prod_{i=0}^m(1-p_i)=0\Longleftrightarrow \sum_{i=0}^{\infty}p_i=\infty.
Definition 6.1 (Period of State) The period d(i) of a state i is defined by d(i)=gcd\{n:p_{ii}(n)>0\}, which means the greatest common divisor of the times at which a return from i to i is possible. We call i periodic if d(i)>1 and aperiodic if d(i)=1.
Example 6.2 (Period of one-dimensional random walk) For a one-dimensional random walk, state 0 is periodic with period d(0)=2. This is because p_{00}(2n+1)=0, so the greatest common divisor of the times at which a return from state 0 to 0 is possible is 2. Indeed, every state of a one dimensional random walk is periodic with period 2.
Definition 6.2 (Ergodic) A state is called Ergodic if it is persistent, non-null recurrent and aperiodic.
For our real life data applications, we use models for which the constructed MCMC is always assumed to be ergodic, meaning that every state of the Markov chain we used is ergodic.

We are now going to do classification of the states in Markov chain.

Definition 6.3 (Communicate and Intercommunicate) We say state i communicates with state j, written as i\to j, if the chain ever visit state j with positive probability starting from state i. That is, i\to j if p_{ij}(m)>0 for some m\geq 0.

We say state i and j intercommunicate if i\to j and j\to i. In which case we write i\leftrightarrow j.

Thus, if i\neq j and i\to j then f_{ij}>0.

Definition 6.4 i\leftrightarrow j is a equivalence relationship if it satisfies:

  • Reflexive: i\leftrightarrow i.

  • Transitive: i\leftrightarrow j and j\leftrightarrow k, then i\leftrightarrow k.

  • Symmetric: i\leftrightarrow j\quad \Longrightarrow\quad j\leftrightarrow i
If one can show i\leftrightarrow j is an equivalent relationship, then it will imply that the state space can be partitioned into intercommunicating subsets.
Theorem 6.1 Intercommunication is a equivalence relationship.

Proof. Firstly, i\leftrightarrow i is trival because p_{ii}(0)=1. Then for symmatricity, by definition it is obvious i\leftrightarrow j implies j\leftrightarrow i. We only need to show transitivity. That is, if i\leftrightarrow j and j\leftrightarrow k, we need to show i\leftrightarrow k.

Since i\to j, by definition, there exist m such that p_{ij}(m)>0 and since j\to k, there exist n such that p_{jk}(n)>0. Using Chapman-Kolmogorov equation (Theorem 3.2), we have \begin{equation} p_{ik}(m+n)=\sum_{r=1}^{m+n}p_{ij}(r)p_{jk}(m+n-r)\stackrel{\text{take}\,r=m}{\geq} p_{ij}(m)p_{jk}(n)>0 \tag{6.5} \end{equation}
which implies i\to k.

Similarly, we can show that k\to i. They together tell us that i\leftrightarrow k.

Theorem 6.2 If i\leftrightarrow j, then

    1. i and j have the same period
    1. i is transient if and only if j is transient. (Equivalently, i is persistent if and only if j is persistent.)
    1. i is (non-)null persistent if and only if j is (non-)null persistent.

Proof. For part (a), let d(i)=gcd\{m:p_{ii}(m)>0\}. Similarly, let d(j)=gcd\{m:p_{jj}(m)>0\}. Since i\leftrightarrow j, there exists h and l such that p_{ij}(h)>0 ad p_{ji}(l)>0. By Chapman-kolmogorov theorem, p_{ii}(h+l)\geq p_{ij}(h)p_{ji}(l)>0. Thus, h+l is divisible by d(i).

Now take any m\in\{n:p_{jj}(n)>0\}, we have p_{jj}(m)>0. By Chapman-Kolmogorov, p_{ii}(h+l+m)\geq p_{ij}(h)p_{jj}(m)p_{ji}(l)>0, which implies that h+l+m is divisible by d(i). Therefore, (h+l+m)-(h+l)=m is also divisible by d(i). Since m is any time such that p_{jj}(m)>0, we have showed that for any m such that p_{jj}(m)>0, it is divisible by d(i). Therefore, gcd\{m:p_{jj}(m)>0\}=d(j) is divisible by d(i).

We can similarly show that d(i) is divisible by d(j). Therefore d(i)=d(j) and part (a) is proved.

Then for part (b), since i\leftrightarrow j, there exist h and l such that p_{ij}(h)>0 and p_{ji}(l)>0. Let \alpha=p_{ij}(h)p_{ji}(l)>0. Now, by Chapman-Kolmogorov equation, p_{ii}(h+l+r)\geq p_{ij}(h)p_{ji}(l)p_{jj}(r). Therefore, p_{ii}(h+l+r)\geq \alpha p_{jj}(r) and \sum_{r}p_{ii}(h+l+r)\geq \alpha \sum_{r}p_{jj}(r). Thus, \sum_{r}p_{jj}(r)<\infty if \sum_{r}p_{ii}(h+l+r)<\infty. Now \sum_{r}p_{ii}(h+l+r)<\infty if i is transient. Therefore, we have proved that j is transient if i is transient.

We can reverse the role of i and j to show that i is transient if j is transient, (b) is proved.

If we work with a real life scenario with all intercommunicating states. In order to show that all states are ergodic, we just need to show that one of these states is ergoric.