Random walk : probability of return in $\leq N$ steps, equivalent as $N\to\infty$

Question

It is well known that in dimension 1, the probability of at least one return to the origin in $N$ steps or less is $$ q_{N}=1-{1 \over {2^{2N}}}{\binom {2N}{N}}\,.$$ Using Stirling's formula one has the equivalent $$ q_{N}\sim 1 - {1 \over {\sqrt {\pi N}}} $$ which explains that the convergence is slowish, and is consistent with the fact that the expected number of steps (which is actually $\sum 1-q_N$) needed to return to the origin is not finite.

I was wondering if we could obtain a similar equivalent in dimension 2 ? I expect convergence to be much slower, maybe involving $\frac 1{\log(N)}$ or even $\frac 1{\log(\log(N))}$ ?

If one writes $u_n$ for the probability to be at the origin at step $n$ and $f_n$ for the probability of a first return at step $n$, then the sequences $u$ and $f$ are linked through a generating series, as already pointed out in Tom Boardman's answer here Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1

In the document Tom Boardman linked http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter12.pdf, pages 3-8 the $f_n$ are computed in dimension $1$ but not $2$. Knowing the $f_n$, we would just have to compute an equivalent of their sum from $1$ to $N$ as $N\to \infty$ (the series $w$ in the document), but I'm not sure these computations are "easily" doable.

Does anybody know if this has already been done, or if there is an easier way to answer my question in bold ?

Thanks a lot

Answer 1

With notations from the ebook cited in the OP: $$ U^{(2)}(x)\ =\ 1 + \sum_{n=1}^\infty \underbrace{\frac1{4^{2n}}\binom{2n}n^2}_{u_{2n}^{(2)}} x^{2n}, $$ and what you are asking about is $1 - q_N = \sum_{n=N+1}^{+\infty} f_n^{(2)}$, where: $$ F^{(2)}(x)\ =\ \sum_{n=0}^{+\infty} f_n^{(2)} x^n\ =\ 1 - \frac1{U^{(2)}(x)}. $$

By Stirling's formula, we obtain the following approximation as $x \to 1^-$: $$ U^{(2)}(x) \ \sim\ \sum_{n=1}^\infty \frac{x^{2n}}{\pi n} \ =\ -\frac1\pi \ln(1-x^2). $$ Moreover $0 \leq f_n^{(2)} \leq u_n^{(2)} = O(\frac 1n)$. So we can expect that some sort of Tauberian Theorem applies, from which we would get: $$ 1-q_N \sim \frac1{U^{(2)}(e^{-\frac1N})} \sim \frac{\pi}{\ln N} $$

A complete proof can be found in G. Lawler and V. Limic, Random Walk: A Modern Introduction, Proposition 4.2.4 (in a more general situation).