# The speed of a.s. convergence in SLLN

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 $$\epsilon$$)?

• 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? Nov 21 '19 at 3:56
• General, but I'm most interested in this one. Nov 21 '19 at 3:57
• I do not know how to answer myself, but it is an interesting question. Nov 21 '19 at 4:03

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 $$\mathbb{P}\left(\tau > t\right)=\sum_{\substack{n > 0,\\ n{\rm\ odd}}}\frac{4}{n\pi}(-1)^{(n-1)/2}\exp\left(-\frac18n^2\pi^2t\right)$$