We know that a simple random walk (walk along one of the $2d$ lattices' directions with the same probability $\frac{1}{2d}$) on $\mathbb{Z}^{d}$ is transient when $d\geq 3$. And the probability $P_{0}(S_{2n}=0)$ can be computed: $P_{0}(S_{2n}=0)=\sum_{l_{1}+\dots+l_{d}=n}\frac{(2n)!}{(l_{1}!\cdots l_{d}!)^{2}}\frac{1}{(2d)^{2n}}$. Where $S_{2n}$ means that the random walk returns the origin after $2n$ times move.

It's easy to imagine that this probability with $n$ fixed is decreasing as $d$ is increased since if $d$ is larger, the random walk need return the origin at all $d$ directions, then the constraint on the random walk is more strict. But how can I prove the result rigorously?

  • $\begingroup$ What is $S_{2n}$ ? Have you try Stirling approximation to show your claim ? $\endgroup$ – Loïc Teyssier Sep 27 '17 at 12:05
  • $\begingroup$ @sbbb88522: would you please explain and (at least grammatically...) clarify the "since the stability is weaken" part of your question? $\endgroup$ – Peter Heinig Sep 27 '17 at 12:07
  • $\begingroup$ Sorry for my vague expression before. I think this problem may seem to be an elementary one but I have tried some analysis and probabilistic methods and have made no progress. I think Stirling approximation is necessary but I can;t figure out how to apply it. Thank you for your comments. $\endgroup$ – sbbb885522 Sep 27 '17 at 14:20