Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
You ask about having a mod move this to MO. At the moment, mods cannot do this (but I anticipate being able to do it soon as some of the technical sides of MO are being moved). If you want to ask this there, you'll have to do it yourself. But if you do, make sure to link back to this question so that there is no duplication of effort. –  mixedmath May 31 '13 at 4:24
    
I might add that MO is strictly for research level questions, however. I'm not sure about the difficulty of this question (not my field) but from Did's comments it sounds like the answer is well-understood. Maybe you should do a little more research yourself before asking there. –  Alexander Gruber May 31 '13 at 4:29
    
@AlexanderGruber Your estimation of the level of the question is accurate. Definitely not MO-stuff. –  Did May 31 '13 at 5:09
add comment

1 Answer

The main difference is that a convergence is almost sure and the other one is in distribution. To grasp how both can occur simultaneously, consider an independent sequence of random variables $(\xi_n)$ such that $P(\xi_n=\sqrt{2\log\log n})=1/n$ and $P(\xi_n=0)=1-1/n$.

Then $\xi_n\to0$ in distribution and, by Borel-Cantelli lemma, $$ \limsup_{n\to\infty}\frac{\xi_n}{\sqrt{\log\log n}}=\sqrt2\quad \text{almost surely}. $$

share|improve this answer
    
But how can this sort of thing happen with normal distributions? The number $\sqrt{2}$ appears as some arbitrary cut off? –  user782220 May 30 '13 at 21:16
    
Are you asking me to reprove the law of the iterated logarithm? –  Did May 30 '13 at 21:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.