Let $X_1, X_2, \ldots\sim \mathcal{N}(0, 1)$ be i.i.d, and $S_n = X_1 + X_2 + \cdots + X_n$. The strong law of large numbers states that $\frac{S_n}{n} \to 0$ almost surely.

Fix $\epsilon > 0$. Define an integer valued random variable

$$N = \sup\{n: |\frac{S_n}{n}| > \epsilon\}$$

i.e the "last time the deviation is large". By SLLN, $N$ is finite almost surely. What is its distribution (it will depend on $\epsilon$)?

  • $\begingroup$ Just out of curiosity, are you only interested in the case where the random variables are normally distributed, or are you interested in the general case? $\endgroup$
    – Math1000
    Nov 21, 2019 at 3:56
  • $\begingroup$ General, but I'm most interested in this one. $\endgroup$ Nov 21, 2019 at 3:57
  • $\begingroup$ I do not know how to answer myself, but it is an interesting question. $\endgroup$
    – Math1000
    Nov 21, 2019 at 4:03

1 Answer 1


It can be shown that $\epsilon^2N$ approaches a limiting distribution as $\epsilon\to0$.

Letting $W$ be a Weiner process, the process $W_n$ over positive integer $n$ has the same distribution as $S_n$. Hence, $N$ can be expressed as the maximum $n$ satisfying $\lvert W_n\rvert/n > \epsilon$. However, $\epsilon tW_{\epsilon^{-2}t^{-1}}$ is also a Wiener process (it is Gaussian, and has the same covariances as $W_t$). So $N$ has the same distribution as $\tilde N_\epsilon$, which I am using to denote the maximum $n$ such that $$ \lvert\epsilon n W_{\epsilon^{-2}n^{-1}}\rvert/n > \epsilon $$ or, equivalently, $\lvert W_{\epsilon^{-2}n^{-1}}\rvert > 1$. Let $\tau$ be the first time that $\lvert W\rvert$ hits $1$. $$ \tau=\inf\left\{t\in\mathbb R^+\colon\lvert W_t\rvert\ge1\right\}. $$ We clearly have $\tilde N_\epsilon < \epsilon^{-2}\tau^{-1}$. Also, for any $\delta > 0$, the process $W_t$ will exceed $1$ with probability 1 in the interval $(\tau,\tau+\delta)$. Hence, as the sequence $\epsilon^{-2}n^{-1}$ becomes dense in the limit $\epsilon\to0$, we will have $\epsilon^{-2}\tilde N_\epsilon^{-1} < \tau+\delta$ for sufficiently small $\epsilon$. This shows that $$ \epsilon^2N\stackrel{d}=\epsilon^2\tilde N_\epsilon\to\tau^{-1}. $$ The distribution of $\tau$ can be computed as an infinite sum (in various ways). See my answer to a previous question on the probability that Brownian motion remains within a bound which, with a little rearranging, gives $$ \mathbb{P}\left(\tau > t\right)=\sum_{\substack{n > 0,\\ n{\rm\ odd}}}\frac{4}{n\pi}(-1)^{(n-1)/2}\exp\left(-\frac18n^2\pi^2t\right) $$


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