# Asymptotic behavior for the return to zero of a simple random walk

I got stuck today trying to understand an argument of the Frank den Hollander Book's. The problem is described below.

Let $S_n=\sum_{i=1}^n X_i$ be the simple random walk in $\mathbb{Z}^d$, that is $$\mathbb{P}(X_i=x)= \left\{ \begin{array}{ll} \frac{1}{2d}&\text{if}\ \|x\|=1;\\ &&\\ 0&\text{otherwise.} \end{array}\right.$$ I would like to know how to prove that $$\mathbb{P}(S_{2n}=0)\sim 2\left(\frac{d}{4\pi n}\right)^{\frac{d}{2}}, \qquad n\to\infty.$$

I learn from the Gregory Lawler book's that this is a consequence of the Local Central Limit Theorem. But I would like to know if one can prove this fact without use this result. I tried to Taylor Expand $$\hat{p}(k)=\frac{1}{d}\sum_{j=1}^d \cos k_j$$ $k=(k_1,\dots,k_d)\in [-\pi,\pi)^d$ and use that $$\mathbb{P}(S_{2n}=0)=\left(\frac{1}{2\pi}\right)^d\int_{[-\pi,\pi)^d} [\hat{p}(k)]^{2n} dk.$$ But It is not working. Any help or reference is welcome. Thanks.

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## 1 Answer

The approach is taken on pages 78 and 79 of Principles of Random Walk (2nd edition) by Frank Spitzer. I was able to see these pages using Google Books.

Spitzer first translates $[-\pi,\pi)^d$ by the vector $(\pi/2,\pi/2,\dots,\pi/2)$ which doesn't change the value of the integral. Then he argues that the bulk of the integral is concentrated at two points, the origin and $(\pi,\pi,\dots,\pi)$ both contributing the same value asymptotically.

The Taylor's series expansion ${1\over d}\sum_{j=1}^d \cos k_j\approx \exp(-|k|^2/2d)$ near the origin finishes the result.

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Yes, I think that's probably the most direct method. Generally for integrating some function $f$ raised to a large power $m$, you can identify the maximum points of $\vert f\vert$ and expand to second order around these points, giving a Gaussian integral. – George Lowther Sep 28 '11 at 23:51
Hi Byron, thank you very much for the reference. – Simone Sep 29 '11 at 1:19