Editing Law of the iterated logarithm (section)

==Discussion==
The law of iterated logarithms operates "in between" the [[law of large numbers]] and the [[central limit theorem]]. There are two versions of the law of large numbers — [[weak law of large numbers|the weak]] and [[strong law of large numbers|the strong]] — and they both state that the sums ''S''<sub>''n''</sub>, scaled by ''n''<sup>−1</sup>, converge to zero, respectively [[convergence of random variables#Convergence in probability|in probability]] and [[convergence of random variables#Almost sure convergence|almost surely]]:
: <math>
    \frac{S_n}{n} \ \xrightarrow{p}\ 0, \qquad
    \frac{S_n}{n} \ \xrightarrow{a.s.} 0, \qquad \text{as}\ \ n\to\infty.
  </math>

On the other hand, the central limit theorem states that the sums ''S''<sub>''n''</sub> scaled by the factor ''n''<sup>−1/2</sup> converge in distribution to a standard normal distribution. By [[Kolmogorov's zero–one law]], for any fixed ''M'', the probability that the event
<math> \limsup_n \frac{S_n}{\sqrt{n}} \geq M </math>
occurs is 0 or 1.
Then

: <math> \Pr\left( \limsup_n \frac{S_n}{\sqrt{n}} \geq M \right) \geqslant \limsup_n \Pr\left( \frac{S_n}{\sqrt{n}} \geq M \right) = \Pr\left( \mathcal{N}(0, 1) \geq M \right) > 0</math>

so

:<math>\limsup_n \frac{S_n}{\sqrt{n}}=\infty \qquad \text{with probability 1.}</math>

An identical argument shows that

:<math> \liminf_n \frac{S_n}{\sqrt{n}}=-\infty \qquad \text{with probability 1.}</math>

This implies that these quantities cannot converge almost surely. In fact, they cannot even converge in probability, which follows from the equality

:<math>\frac{S_{2n}}{\sqrt{2n}}-\frac{S_n}{\sqrt{n}} = \frac1{\sqrt2}\frac{S_{2n}-S_n}{\sqrt{n}} - \left (1-\frac1\sqrt2 \right )\frac{S_n}{\sqrt{n}}</math>

and the fact that the random variables

:<math>\frac{S_n}{\sqrt{n}}\quad \text{and} \quad \frac{S_{2n}-S_n}{\sqrt{n}}</math>

are independent and both converge in distribution to <math>\mathcal{N}(0, 1).</math>

The ''law of the iterated logarithm'' provides the scaling factor where the two limits become different:
: <math>
    \frac{S_n}{\sqrt{2n\log\log n}} \ \xrightarrow{p}\ 0, \qquad
    \frac{S_n}{\sqrt{2n\log\log n}} \ \stackrel{a.s.}{\nrightarrow}\ 0, \qquad \text{as}\ \ n\to\infty.
  </math>

Thus, although the absolute value of the quantity <math>S_n/\sqrt{2n\log\log n}</math> is less than any predefined ''ε''&nbsp;> 0 with probability approaching one, it will nevertheless almost surely be greater than ''ε'' infinitely often; in fact, the quantity will be visiting the neighborhoods of any point in the interval (-1,1) almost surely.
[[File:LimitTheoremsExhibition.png|thumb|Exhibition of Limit Theorems and their interrelationship]]