# Intuition of law of iterated logarithm?

Let $X_i$ be iid random variables with $EX_i = 0$ and $Var X_i=1$ and $S_n=X_1+\cdots+X_n$. Then the law of the iterated logarithm says almost everywhere we have

$$\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} = \sqrt{2}$$

On the other hand the central limit theorem says

$$\frac{S_n}{\sqrt{n}} \to N(0,1)$$

Can anyone explain why dividing by an extra $\sqrt{\log{\log{n}}}$ should go from giving $N(0,1)$ to something bounded by the constant $\sqrt{2}$?

To try to understand I considered the simple case when each $X_n$ is $N(0,1)$ so that $S_n/\sqrt{n}$ is also normally distributed as $N(0,1)$. Then $S_n/\sqrt{n\log{\log{n}}}$ is distributed as $N(0,1/\log{\log{n}})$. Then it would seem to me that to even have just $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ requires either

$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) < \infty$$

or if

$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) = \infty$$

then to achieve $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ the sets {$\omega : \frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}$} cannot for example cover the probability space over and over infinitely forever. I don't know the value of $\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right)$ but since it is the sum of the probability of the tail ends of a bunch of normal distributions you would expect there to be no closed form even for partial sums.

In the other direction for $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}}$ to not have a value lower than $\sqrt{2}$ isn't it necessary that something like the following holds

$$\sum_{n=3}^\infty P\left(\sqrt{2}-\epsilon < \frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}\right) = \infty$$

Can anyone explain why this number $\sqrt{2}$ should pop up?

I already asked the above on math.stackexchange (link) but apparently moving it here was impossible hence the duplicate post.

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The proof goes via lots of Borel-Cantelli. Heuristically if you believe the central limit theorem, $S_n$ should be normal with mean 0 and variance $n$, so that $S_n/\sqrt n$ is approximately $N(0,1)$. The appearance of the $\sqrt{2\log\log n}$ is roughly because $\mathbb P(N>\sqrt{2\log\log n})$ is on the cusp of summability. So that $\mathbb P(N>\sqrt{2.000001\log\log n})$ is summable, so happens finitely many times (this is the easier part), whereas $\mathbb P(N>\sqrt{1.999999\log\log n})$ is not summable and so [quite a lot of annoying technical details skipped] happens infinitely often. –  Anthony Quas May 31 '13 at 6:09
To supplement Anthony's comment slightly: recall that the normal distribution in the central limit theorem (with variance $1$) is $\frac{1}{ \sqrt{2\pi}} e^{\frac{x^2}{2 }}$. Roughly speaking, it's the 2 in the denominator of the exponent that ultimately gives rise to the $\sqrt{2}$ in the law of the iterated logarithm. –  Mark Lewko May 31 '13 at 6:40
@Anthony: Are you saying that with $S_n$ having mean $0$ and variance $n$ that it is in fact true that $$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n}} > \sqrt{(2+\epsilon)\log{\log{n}}}\right) < \infty$$ and $$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n}} > \sqrt{(2-\epsilon)\log{\log{n}}}\right) = \infty$$ –  user16557 May 31 '13 at 7:38
@unknown: that's clearly false, but I think it's true if the summand is divided by n. –  George Lowther May 31 '13 at 13:30
Actually, for Borel-Cantelli to work in this situation you need to sample time points along a geometric progression, not linearly spaced. i.e., for any $q > 1$, $$\sum_{n\in\lbrace q^r\colon r\in\mathbb{N}\rbrace}\mathbb{P}\left(\frac {S_n}{\sqrt{n}} > \sqrt{A\log\log n}\right)$$ is finite for $A > 2$ and infinite for $A < 2$. This follows from $\mathbb{P}(S_n/\sqrt{n} > K)\sim (2\pi)^{-1/2}K^{-1}\exp(-K^2/2)$. –  George Lowther Jun 1 '13 at 0:52
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