Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.