Counting returns in null-recurrent random walk 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 ${\bf Z}^2$?
 A: 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).
