# Transition probability convergence for Harris Chains

I'm studying theorem 6.8.8 of Durrett - convergence of transition probabilities for Harris chains and I have a (I think) pretty hard question which would help me more than words can describe if one of you nice people could help me solve.

It is the part that $W_m=S_m - T_m$ is a random walk (a sum of iid random variables) and has mean zero and furthermore that it means that $M=\inf\{n\geq 1 | S_m = T_m\}< \infty$ a.s. which I can't understand.

I can only show that $s_m=S_m-S_{m-1}$ and $t_m=T_m-T_{m-1}$ are iid for $m\geq2$. And I know the Chung-Fuch theorem saying that if a Random Walk is one-dimensional and $S_n / n \to 0$ in probability then it is recurrent, true here by Kolmogorovs law of large numbers. This gives the result if $P(W_m=0)>0$, but that I can't realize why should be the case.

Edit3:

I've done some further realizations myself. So the result is true if for $R_m:=S_m-T_m-(S_1-T_1)$ has $R_n=S_1-T_1$ for some $n$. $R_n$ will be a recurrent random walk from the above and I found this theorem P1 saying that for all $x\in \mathbb{R}$ there exists a $N=N(x)$ such that the probability of transitioning from 0 to x in more than n steps is positive. Since recurrent means 0 i.o. we can use Theorem 6.3.3 of Durrett (page 284, 296 in scribd see below) to find that it actually hits any $x\in \mathbb{R}$ with probability 1 and therefore also $S_1-T_1$ from which the result follows. This requires though that I can show $R_m$ is strongly aperiodic, but I guess that follows from the aperiodicity of the original chain.

What do you guys think - does it stick?

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Are $x$ and $\alpha$ meant to be the same? Otherwise the mean wouldn't be $0$, since $x$ could transition to $\alpha$ with probability $1$. (Actually, upon rereading the question and seeing that you got the result for $m\ge2$, I'm thinking that this might be the problem, that you're taking $\alpha$ and $x$ to be different. In that case the statement doesn't hold.) –  joriki Oct 25 '12 at 14:22
Thanks very much for your reply! I figured out how to simply add the text, would you mind affirming that the result simply isn't true as it is stated? –  Henrik Oct 25 '12 at 15:42
Here's the entire book, with hyperlinked theorems :-) –  joriki Oct 25 '12 at 15:55

Apologies for my incorrect first answer. This seems to be just a question of careful indexing. Indeed the times $S_0$ and $T_0$ need not have the same mean, but that just means that the random walks start at different points. The steps $s_m$ and $t_m$ that you defined are actually i.i.d. for $m\ge1$, not just $m\ge2$, since once $\alpha$ has been reached at $S_0$ and $T_0$, respectively, there's no memory of how it was reached and the remaining return times are identically distributed.
In any case, I don't think the proof relies on these details, since all it wants to conclude from this is that $\inf\{m\ge1\colon S_m=T_m\}\lt\infty$, and that would hold even if any finite number of initial steps had different distributions or means.
P.S.: Reading through your question again, I'm wondering whether part of the problem is that you write "mean zero" where the text has "mean $0$ steps" – it's only the steps that have mean zero, and this is also only what's needed for the conclusion drawn.
$\alpha$ definitely plays an important role. But I think only it is for $x\in A$ that it is $\varepsilon$, or am I misunderstanding you? –  Henrik Oct 25 '12 at 17:21
Thank you, i think i'm starting to get there :) What is the difference between mean zero and mean zero steps? And how do we get the conclusion that $\inf \{ m \geq 1 : S_m = T_m \}< \infty$ ? –  Henrik Oct 25 '12 at 18:34
@Henrik: I would take "mean zero" to mean that $S_m-T_m$ has mean zero, which apparently it doesn't, whereas "mean zero steps" just means that the increments $(S_m-T_m)-(S_{m-1}-T_{m-1})$ have mean zero. So $S_m$ and $T_m$ may start out with different means, but that difference stays constant. We can conclude that $\inf\{m\geq1:S_m=T_m\}\lt\infty$ a.s. because a one-dimensional random walk with mean zero steps almost surely hits every point; hence $S_m-T_m$ hits zero almost surely (independent of whether it started with mean zero or not). –  joriki Oct 25 '12 at 18:41