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Assume $X_i$ has zero mean and unit variance. Define $S_n = \sum_{i=1}^n X_i$.

In the law of large numbers, the quantity $S_n/n$ means sample mean.

In the central limit theorem, the quantity $S_n/\sqrt{n}$ means normalized sample mean to have zero mean and unit variance.

I was wondering what the meaning of the quantity $S_n/\sqrt{n\log\log n}$ in the law of iterated logarithm is?

Thanks!

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What do you mean when you say "$S_n/n$ means sample mean" and "$S_n/\sqrt{n}$ means normalized sample mean..."? –  Stefan Hansen Oct 18 '12 at 17:06
    
They are sample mean and normalized sample mean for the sample $(X_1, \dots, X_n)$. –  Tim Oct 18 '12 at 17:52
    
Does the normalized sample mean have a special meaning? –  Stefan Hansen Oct 18 '12 at 18:22
    
normalized to have mean zero and variance 1. –  Tim Oct 18 '12 at 18:38

1 Answer 1

The mean of a random walk is expected to be 0 and the variance is expected to grow like $\sqrt{n}$. Similar for iid random varialbe with mean 0 and variance 1.

What happens if we just add them? The sum of the random variables should grow like
$$ \sup \left[ Y_1 + \dots + Y_n\right] \approx \sqrt{ 2 n} \log \log n $$ Intuitively $\log n$ is the number of "bits" or "digits" of $n$.

Another issue is the difference between "almost sure" and "in probability" convergence.

  • For almost sure convergence, for almost every sequence of "coin-flips", the sequence $X_1, \dots, X_n \to X$.
  • For convergence in probability, you measure each $X_n$ individually. $\mathbb{P}\big[|X_n - X |< \epsilon\big] \to 0$

According to Wikipedia, $\sup \frac{1}{\sqrt{2n}} \left[ Y_1 + \dots + Y_n\right] \approx \log \log n$ converges in probability but not almost surely... so although random walks thought to grow like $\sqrt{n}$ the "peak value" is growing slightly faste

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Sorry but I understand almost nothing in this answer (and you consistently write $\log\log$ instead of $\sqrt{\log\log}$). –  Did Oct 18 '12 at 20:10

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