Consider two independent copies of IID random walk on ${\bf Z}$ starting from $0$, and let $N_1(t)$ (resp. $N_2(t)$) denote the number of times, up to time $t$, that the first (resp. second) walker has returned to $0$. We know that $N_1(t),N_2(t) \rightarrow \infty$ almost surely and that $N_1(t)/t,N_2(t)/t \rightarrow 0$ almost surely. Is it the case that $N_1(t)/N_2(t) \rightarrow 1$ almost surely? What if we replace random excursions from $0$ in ${\bf Z}$ by random excursions from $(0,0)$ in ${\bf Z}^2$?
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asked Jan 11, 2023 at 23:54
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1$\begingroup$ I think the answer should be yes by the Hopf ergodic theorem. Unfortunately the theorem doesn't apply absolutely verbatim because the function appearing in the denominator needs to be strictly positive. I suspect this can be fixed by taking a very fast decaying but strictly positive function. I will try to write down full details. $\endgroup$– Anthony QuasCommented Jan 12, 2023 at 2:20
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5$\begingroup$ I think one can show that as $\lambda \to \infty$, the processes $\lambda^{-1/2} N_j(\lambda t)$ converge in law (at least in the sense of finitely dimensional distributions) to the inverse $1/2$-stable subordinators $Z_j(t)$. And this implies that $N_1(t)/N_2(t)$ converges, in the sense of distributions, to a non-trivial random variable $Z_1(1)/Z_2(1)$. $\endgroup$– Mateusz KwaśnickiCommented Jan 12, 2023 at 8:16
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$\begingroup$ @MateuszKwaśnicki Do you still believe this, or has Anthony's solution convinced you otherwise? $\endgroup$– James ProppCommented Jan 12, 2023 at 23:21
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$\begingroup$ I take it back. My functions are not in $L^1$. :( $\endgroup$– Anthony QuasCommented Jan 13, 2023 at 0:43
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2$\begingroup$ @JamesPropp: Unfortunately I had no time to carefully read Anthony's answer, but I am pretty much convinced about the convergence. By Donsker's theorem for stable processes, the inverse functions of $\lambda^{-1} N_j(\lambda t)$ converge as $\lambda \to \infty$ to $1/2$-stable subordinators. Convergence of original functions should now follow by some straightforward calculation, but I was too busy recently to try and work out the details. $\endgroup$– Mateusz KwaśnickiCommented Jan 13, 2023 at 15:55
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1 Answer
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You can easily derive everything from the first principles by the following back of envelope computation:
If we have a random walk starting anywhere, then after $t$ steps the expected number $EN(t)$ of returns is at most $\sum_{m=1}^t \frac 1{\sqrt m}\approx \sqrt t$ and $E[N(t)^2]\le\sum_{1\le m< M\le t} \frac 1{\sqrt m}\frac 1{\sqrt{M-m}}+\sum_{1\le m\le t}\frac 1{\sqrt m}\approx t$ with approximate equality attained if we start at $0$. Note that it implies that $P(N(t)>c\sqrt t)>c$ for some fixed $t>0$ if we start at $0$. On the other hand, if we start above $C\sqrt t$ with large $C>0$, then the probability to ever reach $0$ is as small as we want, so the trivial Cauchy-Schwarz yields $EN(t)\le \sqrt{P(\exists \tau X(\tau)=0)E[N(t)^2]}<\frac{c^2}4\sqrt t$ if $C$ is large enough so $P(N(t)>\frac c2\sqrt t)\le c/2$.
Now just take $T>0$ and let $t$ be the first time after $T$ such that $X_1(t)=0$. Then the probability that $|X_2(t)|>C\sqrt t$ is some constant $p_0$, say. From there and until the time $2t$ with probability $\ge c-\frac c2=\frac c2$, $X_1$ acquires $>c\sqrt t$ returns but $X_2$ less than $\frac c2\sqrt t$ returns. So, with fixed probability ($p_0c/2$) they have noticeably different increments on the scale $\sqrt t$. Also, the probability that $N_2(t)$ is much larger than $\sqrt t$ is small (we condition only on $X_1$, so $X_2$ still has the standard distribution). Thus, for every $M$, with constant probability, there exists $t>M$ such that $|N_1(t)/N_2(t)-1|+|N_1(2t)/N_2(2t)-1|>\delta>0$ for some fixed absolute $\delta$. That definitely rules out convergence to $1$ even in distribution, forget about a.s.
I tried to avoid any high-tech here and just answered the question as posed. I leave more advanced considerations to someone else.
In $\mathbb Z^2$, the situation is the same only with $\log t$ instead of the $\sqrt t$ everywhere.
I hope I haven't written any nonsense (lately I'm more prone to it than I should be).
