A Markov chain $X_n$ $(n\ge1)$ with state space S=0,1,2,3,... has the following transition probabilities:
$$p_{i,i+1}=\frac{(i+1)^2}{2i^2+2i+1},\ p_{i,i-1}=\frac{i^2}{2i^2+2i+1}, \ p_{0,1}=1, \ and \ i\ge 1$$
I am trying to find if it is recurrent or transient. What I have is the following:
Obviously, $p_{i,i+1}$ and $p_{i,i-1}$ depends on $i$, it means that each state will have a different probability. For example, when $i=1$, then $p_{1,2}=4/5$ and $p_{1,0}=1/5$. So, it seems that $p_{i,i+1}$ is decreasing as $p_{i,i-1}$ increases on each step.
$$T=\pmatrix{0&1&0&0&...&0&0&0\\1/5&0&4/5&0&...&0&0&0\\0&4/13&0&9/13&...&0&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\0&0&0&0&...&1/2&0&1/2}_{nxn}$$
According to wiki, a state i is recurrent if and only if the expected number of visits to i is infinite:
$$ \sum_{n=0}^{\infty}p_{ii}^{(n)}=\infty\\ p_{ii}=Pr(X_n=i|X_0=0)\ for\ n\ge 1$$
for a recurrent state i, the mean hitting time is defined as:
$$M_i=E\left[ T_i \right]=\sum_{n=1}^{\infty}nf_{ii}^{(n)}$$
State i is positive recurrent (or non-null persistent) if Mi is finite; otherwise, state i is null recurrent (or null persistent).
If this Markov chain had a finite state space:
I think that I should be positive recurrent because it is possible to get to any state from any state.
How does the infinite state space change this fact? I haven't found good information about it.
What about this?
If we start from zero, it seems that the MC may return to zero almost surely with probability 1.
Thank you!
Adapting my answer from here, which asks basically the same question.
Let $h_n=\mathbb P_n(\text{hit }0)$, so $h_0=1$. Then $$h_n=p_{n,n-1}h_{n-1}+p_{n,n+1}h_{n+1}\implies \frac{h_{n+1}-h_n}{h_n-h_{n-1}}=\frac{n^2}{(n+1)^2}.$$ Telescoping, we get $h_{n+1}-h_n=\frac{1}{(n+1)^2}(h_1-h_0)$. So summing over $0\leq n\leq m-1$, we obtain $$h_m=1+(h_1-1)\sum_{k=1}^m\frac{1}{k^2}.$$ As $\sum_{k=1}^m\frac{1}{k^2}\to\frac{\pi^2}{6}$, the fact that $(h_m)_{m=0}^\infty$ is the smallest non-negative solution to this equation implies that $h_1=1-\frac{6}{\pi^2}<1$. This is enough to show that the chain is transient.