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Let $G_1, \dots, G_n$ be iid random variables, such that $G_1 \sim \mathcal N(0,1)$

Let $$S_n = \sum_{i=1}^n G_i\quad \text{and} \quad\tilde{S}_n = \frac{1}{\sqrt{2n\log\log n}}S_n$$

It is easy to see that $\tilde{S}_n \sim N(0,\sigma_n)$, where $\sigma_n = \frac{1}{\sqrt{2\log\log n}}$.

By the iterated logarithm law, $$\limsup_{n \rightarrow \infty} \ \tilde{S}_n = 1 \quad \text{a.s} $$

So what is troubling for me, is that I cannot prove by direct means that if $(Y_n)_{n \in \mathbb N}$ are indepedent random variables such that $Y_n \sim \mathcal N(0,\sigma_n)$ then it holds that $$\limsup_{n \rightarrow \infty} \ Y_n = 1 \quad \text{a.s} $$

For instance, only trying to prove that : a.s, $Y_n > 1+\epsilon$ occurs only a finite number of times :

Using Borell-Cantelli we want to prove that $$\sum_{n} \mathbb P(Y_n>1+\epsilon) <\infty $$ However, Gaussian concentration only gives : $$\mathbb P(Y_n>x) = P(\sigma_nG>x) \leq e^{-\frac{x^2}{2\sigma_n^2}} $$

One can check it is far from summable. So where do we loose information ? Is the inequality too loose ?

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  • $\begingroup$ The $\tilde{S}_n$ are not independent. So why are your $Y_n$ independent? $\endgroup$ – Tobias Fritz Jun 28 '18 at 16:21
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In fact, $\limsup_{n \to \infty} Y_n = \infty\ne1$ a.s. More specifically, \begin{equation} \limsup_{n \to \infty} \frac{Y_n}{\sqrt{\ln n/\ln\ln n}} = 1 \end{equation} a.s. or, equivalently, \begin{equation} \limsup_{n \to \infty} \frac{Z_n}{\sqrt{2\ln n}} = 1 \tag{*} \end{equation} a.s., where $Z_n:=Y_n/\sigma_n\sim N(0,1)$. Indeed, $\P(Z_1>z)=\exp\{-z^2/(2+o(1))\}$ as $z\to\infty$. So, for any $\ep\in(0,1)$, \begin{equation} \sum_n\P(Z_n>\sqrt{(2+\ep)\ln n})= \sum_n\exp\{-(2+\ep)\ln n/(2+o(1))\}<\infty \end{equation} and \begin{equation} \sum_n\P(Z_n>\sqrt{(2-\ep)\ln n})= \sum_n\exp\{-(2-\ep)\ln n/(2+o(1))\}=\infty. \end{equation} So, (*) follows by the Borel--Cantelli lemma.

(There is no paradox here, since your $Y_n$'s are independent, and the $\tilde{S}_n$'s are rather strongly dependent.)

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