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On a book I'm reading (Stochastic processes by Bass. R.F.) after he proves the law of iterated algorithm for a brownian motion $W$, namely that

$$\limsup_{t\rightarrow \infty} \frac{|W_t|}{\sqrt{2t\log{\log{t}}}}=1\text{ a.s. and } \limsup_{t\rightarrow 0} \frac{|W_t|}{\sqrt{2t\log{\log{\frac{1}{t}}}}}=1\text{ a.s.}$$

he continues by claiming that

the law of iterated logarithm show that the paths of $W_t$ are not differentiable at time $0$, a.s.

and I'm not sure I understand correctly why this is so. From the second $\limsup$, can we say that $$\limsup_{t\rightarrow 0} \frac{|W_t|}{t}=\limsup_{t\rightarrow 0}\sqrt{\frac{2\log{\log{\frac{1}{t}}}}{t}}=+\infty$$ and therefore the derivative at $0$ does not exist since $\limsup_{t\rightarrow 0}|\frac{W_t-W_0}{t-0}|=\infty$? I mean, since the $\limsup$ of the ratio is finite, and the denominator goes to infinity, it would appear logical to me that the numerator goes to infinity as well, but I'm not sure it is that easy or that I have formulated it well.

John Doe
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1 Answers1

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Indeed, if some (deterministic) functions $x$ and $y$ and some (finite or infinite) number $t_0$ are such that $\limsup\limits_{t\to t_0}x(t)=1$ and $y(t)$ converges to some (finite or infinite) limit when $t\to t_0$, then $$\limsup\limits_{t\to t_0}x(t)y(t)=\lim\limits_{t\to t_0}y(t).$$ Apply this to $t_0=0$ and $$ x(t)=\frac{|W_t(\omega)|}{\sqrt{2t\log\log(1/t)}},\qquad y(t)=\frac{\sqrt{2t\log\log(1/t)}}t. $$ Two limsups would not be enough to conclude but one limsup and one limit are.

Did
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  • Thanks a lot for telling me about this property. Just two small questions: a) In the proposed $x$ and $y$ shouldn't it be $\log\log\frac{1}{t}$? I mean that's when we get the $\limsup$ for $t\rightarrow 0$ in the law of iterated logarithm, and we check differentiability at $0$. b) I suppose this holds true for $t\rightarrow a$ in general, not for $0$ only, right? Infinity as well? – John Doe Feb 16 '14 at 16:16
  • Yes, twice. $ $ – Did Feb 16 '14 at 16:17