So I have a Markov chain on the nonnegative integers such that, starting from $x$, the chain goes to $x+1$ with probability $p$, $0<p<1$, and goes to state $0$ with probability $1-p$.

I'm trying to show that this chain is recurrent namely, $\sum\limits_{n=1}^\infty P_x(T_x = n) =1$ for all $x$ in the state space. After some workings, I've worked out $P_x(T_x = n)$ to be $\sum\limits_{k=1}^{n-x} {n-x-1 \choose k-1} \cdot (1-p)^k \cdot p^{n-k}$.

Now, this is the part where I am stuck, I need to show that $\sum\limits_{n=1}^\infty (\sum\limits_{k=1}^{n-x} {n-x-1 \choose k-1} \cdot (1-p)^k \cdot p^{n-k}) =1$

It'd be great if anyone could lead me from here or even better, give a more intuitive approach.

  • $\begingroup$ What is $P_x$ and $T_x$? I guess $P_x(X_n \in A) = \mathbb{P}(X_n \in A | X_0 = x)$ and then $T_x$ is the first return time to state $x$, given that we start in state $x$? Clearly, this chain is stable since we have catastrophes where the process returns to state $0$, which can occur from every state in the state space. $\endgroup$ – Ritz Sep 18 '14 at 12:52

I don't know how much you know about Markov Chains but you can simplify your problem by using the concept of irreductibility. We can notice that $$\forall (i,j) \in \mathbb{N}^2, \exists n\geq 0, \quad P(X_n=j|X_0=i)>0 $$ (i.e. we can go from any $i$ to any $j$). The chain is therefore irreductible.

A property of irreductible chains (or subchains) is that every state withing share the same properties of reccurence or transience. So now you only need to pick a single state and show it is either recurrent or transient. We'll pick state $0$.

We can see that $$ P_0(T_0 = n) = p^{n-1} (1-p)$$ because to get to $0$ starting from $0$ in exactly $n$ steps means going $i \rightarrow (i+1)$ up to state $n-1$ then going to $0$.

Therefore $T_0 \sim \mathcal{G}(1-p)$ a geometric law of parametre $(1-p)$ so $$E[T_0] = \frac{1}{1-p}<\infty$$

So state $0$ is positive recurrent, therefore it is recurrent (+irrectibility of the chain) means that $(X_n)$ is recurrent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.