Markov chain and martingale: Taking one step to the right or going back to zero

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Consider the Markov chain on nonnegative integers that does the following: from any site $x \geq 0$ it jumps to $x + 1$ with probability $p$ and to 0 with probability $1 − p$. Is this chain transient? Null recurrent? Positive recurrent? You need to prove whatever you claim.

Let $S_0=x$ and $S_n=x+X_1+X_2+\dots +X_n$. $S_n$ is recurrent if $$P(\exists n\geq 1~|~ S_n=0)=1.$$ Furthermore, if the first return time is finite then we have positive recurrence and if it is infinite then we have null recurrence.

I believe that this depends on $p$. I am interested in $$P(\text{return to x} ~|~ \text{started at x})$$ I am having a hard time determining what I should do next and I was hoping for a bit of help.

I want to try and use a martingale. I am wondering if I need to know the mean and the variance of the Markov chain? I am not really sure how to find either of these.

Thank you for any help you can give to assist me.

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I'm not sure how to use martingales in this instance but here is one way:

For any $i,j \in \mathbb{Z}_{\geq 0},$ we have $P^{(j+1)}_{i,j} \geq P_{i,0} \cdot P_{0,1} \cdot P_{1,2} \cdots P_{j-1,j} = (1-p)p^j > 0$. Similarly, $P^{(i+1)}_{j,i}>0$. Then all states $i,j$ communicate and so the MC is irreducible.

Next, note that $P^{(n)}_{0,0} = (1-p)\sum^{n-1}_{k=0} {n-1 \choose k}p^{n-1-k}(1-p)^k = (1-p)(1)^{n-1}=(1-p).$ Therefore,$\ \sum^{\infty}_{n=1}P^{(n)}_{0,0} = \sum^{\infty}_{n=1} (1-p) = \infty$ and so the MC is recurrent.

Finally, let $T_i$ be the waiting time for state $i$ and note that $P(T_0=k\ | \ X_0=0)=p^{k-1}(1-p).$ Then it follows that $E[T_0 \ | \ X=0] = \sum^{\infty}_{k=0}kp^{k-1}(1-p)=\frac{1-p}{p} \sum^{\infty}_{k=0}kp^k= \frac{1}{1-p}< \infty.$ So the MC is positive recurrent.