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I'm reading about the Law of Iterated Logarithm for standard Brownian motion.

As a Corollary of this we have the next:

Suppose that $\{B(t)\}_{t\geq 0}$ is a standard Brownian Motion. Then almost surely $$\displaystyle\limsup_{t\rightarrow 0}\frac{|B(t)|}{\sqrt{2t\log\log\frac{1}{t}}}=1.$$

The proof is as follows:

The process $\{X(t)\}_{t\geq 0}$ defined by $X(t)=tB(1/t)$ is a Brownian Motion for $t>0.$ Then, for LIL we have $$\displaystyle\limsup_{t\rightarrow 0}\frac{|B(t)|}{\sqrt{2t\log\log\frac{1}{t}}}=\displaystyle\limsup_{h\rightarrow\infty}\frac{|X(h)|}{\sqrt{2h\log\log h}}=1.$$

I don't understand it. To apply L.I.L. we need that $|X(h)|$ be Brownian Motion. So I suppose that $|X(h)|$ it is,but I can't see it.

Any kind of help is thanked in advanced.

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  • $\begingroup$ $|X(h)|$ is certainly not a Brownian motion and neither is $t B(t)$. You have to consider the process $X_t := t B(\color{red}{1/t})$. $\endgroup$ – saz May 21 '18 at 7:10
  • $\begingroup$ Thanks @saz. I wrote wrong the process $X(t).$ I edit it immediately. So, I know that $X(t)=tB(1/t)$ is Brownian motion. Then why $|X(t)|$ also is B.M? $\endgroup$ – Squird37 May 21 '18 at 7:14
  • $\begingroup$ Note that a non-negative process cannot be a Brownian motion...in particular, neither $|B|$ nor $|X|$ are Brownian motions . You have to apply the law of iterated logarithm directly to the process $X$ (and not to its absolute value). $\endgroup$ – saz May 21 '18 at 7:26
  • $\begingroup$ I have your point Thanks a lot. $\endgroup$ – Squird37 May 21 '18 at 7:40

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