Markov Chain with $n+1$ states

Question

Let a Markov Chain with $|I|=n+1$ states $I=\{0,1,...,n\}$.

For state $i\neq 0$ the probability to move in the next step to $0$ is $p_i\gt 0$, and with probability $1-p_i$ to stay in $i$.

When we are in state $0$ the probability to move in the next step to $i\neq 0$ is $a_i$ such that:

$$\sum_{i\in I\backslash \{0\}} a_i=1$$

Suppose the experiment starts in state $0$, what is the expected value of the number of moves until the next visit to zero?

For convenience let's call this random varible $S$.

What is the stationary distribution?

My approach:

For the expected value of $S$: $$\mathbb{E}[S]=1+\sum_{i\in I\backslash \{0\}} \frac{a_i}{p_i}$$ For the stationary distribution:

I have noticed that if there is some state $i\neq 0$ such that $p_i\lt1$ than the chain is ergodic and otherwise it has a period $2$. But I don't know how to put it all together.

Answer 1

S = no of visits before returning to state 0 Range (S) ={1,2,...}=N

Now,to find E[S]

Apply the lemma E[s] = $\sum_{k=1}^{\inf} P[S\geq k]$

Now, P[$S\geq k$] = p[$\cup_i X_n= i \forall 1 \leq n \leq k-1]$

=$\sum_{i=1}^{n+1} a_i(1-p_i)^{k-1}$(Since,disjoint union)

So, E[S] = $\sum_{k=1}^{\inf}[\sum_{i=1}^{n+1} a_i(1-p_i)^{k-1}]$

=$\sum_{i=1}^{n+1}a_i[\sum_{k=1}^{\inf}(1-p_i)^{k-1}]$(By fubinni 's exchange)

= $\sum_{i=1}^{n+1}[a_i/(p_i)]$