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 (p01−p00000⋯p101−p1000⋯p2001−p200⋯⋮⋮⋮⋮⋮⋮⋮pr0⋯⋯⋯1−pr0⋯⋮⋮⋮⋮⋮⋮⋮) This transition matrix consiste of a column vector (p0,p1,⋯,pr,⋯) and a diagonal matrix diag(1−p0,1−p1,⋯,1−pr,⋯).
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−(1−p0)f00(2)=(1−p0)p1=(1−p0)[1−(1−p1)]=(1−p0)−(1−p0)(1−p1)f00(3)=(1−p0)(1−p1)p2=(1−p0)(1−p1)[1−(1−p2)]=(1−p0)(1−p1)−(1−p0)(1−p1)(1−p2)⋯⋯⋯⋯f00(n)=n−2∏i=0(1−pi)−n−1∏i=0(1−pi) Then we have m+1∑n=1f00(n)=m+1∑n=1[n−2∏i=0(1−pi)−n−1∏i=0(1−pi)]=1−m∏i=0(1−pi) 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.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
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.
Theorem 6.2 If i\leftrightarrow j, then
- i and j have the same period
- i is transient if and only if j is transient. (Equivalently, i is persistent if and only if j is persistent.)
- 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.