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The simple random walk is one starting at $0$ with steps of $-1$ and $1$ with equal probability. Is there a proof not involving (too much) probability - preferably number-theoretic - of why this walk must diverge (i.e. can get arbitrarily large)? Additionally, is there a non-probabilistic proof that the "speed" of this random walk is on the order of $\sqrt{n}$? From the wikipedia page:

$\text{In fact}^{\text{[citation needed]}}:$ $$\lim_{n\rightarrow\infty}E(|S_n|)/\sqrt{n}=\sqrt{2/\pi}$$

Can someone fill in that [citation needed]?

ant11
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2 Answers2

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You can get the asymptotic speed of the random walk by combining the exact result $$\mathbb{E}\left|{S_{2n}\over 2n}\right|={2n\choose n}{1\over 4^n},$$ with these estimates.

Community
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After $n$ steps, the probability that $S_n$ is at a given site is at most $2^{-n}\max\limits_k{n\choose k}\sim c/\sqrt{n}$ hence $P(S_n\in B)\to0$ when $n\to\infty$, for every finite $B$. Thus $(S_n)$ is almost surely unbounded.

Regarding the "order" of $S_n$, recall that $U_n=S_n/\sqrt{n}$ converges in distribution to a standard normal random variable $Z$. Additionally, $E(U_n^2)=1$ for every $n$, thus $(U_n)$ is bounded in $L^2$, in particular $E(|U_n|)\to E(|Z|)=\sqrt{2/\pi}$ by uniform integrability.

Are these non-probabilistic? Dunno...

Did
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