Lecture 14. The law of iterated logarithm

Posted on August 23, 2012 by Fabrice Baudoin

We already observed that as a consequence of Kolmogorov’s continuity theorem, the Brownian paths are \gamma-Hölder continuous for every \gamma \in \left(0,\frac{1}{2}\right). The next proposition, which is known as the law of iterated logarithm shows in particular that Brownian paths are not \frac{1}{2}-Hölder continuous.

Theorem. Let (B_t)_{t\ge 0} be a Brownian motion. For s \ge 0,

\mathbb{P}\left( \lim \inf_{t \rightarrow 0} \frac{B_{t+s}-B_s}{\sqrt{2t \ln \ln \frac{1}{t}}} =-1 , \lim \sup_{t \rightarrow 0} \frac{B_{t+s}-B_s}{\sqrt{2t\ln \ln \frac{1}{t}}} =1 \right)=1.

Proof

Thanks to the symmetry and invariance by translation of the Brownian motion, it suffices to show that:

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t\ln \ln \frac{1}{t}}} =1 \right)=1.

Let us first prove that

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t \ln \ln \frac{1}{t}}} \le 1 \right)=1.

Let us denote

h(t)=\sqrt{2t \ln \ln \frac{1}{t}}.

Let \alpha, \beta >0, from Doob’s maximal inequality applied to the martingale \left(e^{\alpha B_t-\frac{\alpha^2}{2}t} \right)_{t \ge 0}, we have for t \ge 0:

\mathbb{P} \left( \sup_{0 \le s \le t} \left( B_s-\frac{\alpha}{2}s\right)>\beta \right)=\mathbb{P} \left( \sup_{0 \le s \le t} e^{\alpha B_s -\frac{\alpha^2}{2}s} >e^{\alpha \beta}\right) \le e^{-\alpha \beta}.

Let now \theta , \delta \in (0,1). Using the previous inequality for every n \in \mathbb{N} with

t=\theta^n, \alpha=\frac{(1+\delta)h(\theta^n)}{\theta^n}, \beta=\frac{1}{2} h (\theta^n),

yields when n \rightarrow +\infty,

\mathbb{P} \left( \sup_{0 \le s \le \theta^n} \left( B_s -\frac{(1+\delta)h(\theta^n)}{2\theta^n}s\right)>\frac{1}{2} h (\theta^n) \right)=O\left( \frac{1}{n^{1+\delta}} \right).

Therefore from Borel-Cantelli lemma, for almost every \omega \in \Omega, we may find N(\omega)\in \mathbb{N} such that for n \ge N(\omega),

\sup_{0 \le s \le \theta^n} \left( B_s(\omega)-\frac{(1+\delta)h(\theta^n)}{2\theta^n} s\right) \le \frac{1}{2}h (\theta^n).

But,

\sup_{0 \le s \le \theta^n} \left( B_s(\omega) -\frac{(1+\delta)h(\theta^n)}{2\theta^n} s\right) \le \frac{1}{2} h (\theta^n)

implies that for \theta^{n+1} \le t\le \theta^n,

B_t (\omega) \le \sup_{0 \le s \le \theta^n} B_s(\omega) \le \frac{1}{2} (2+\delta)h (\theta^n) \le \frac{(2+\delta)h(t)}{2\sqrt{\theta}}.

We conclude:

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t\ln \ln \frac{1}{t}}} \le \frac{2+\delta}{2\sqrt{\theta}}\right)=1.

Letting now \theta \rightarrow 1 and \delta \rightarrow 0 yields

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t \ln \ln \frac{1}{t}}} \le 1\right)=1.

Let us now prove that

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t\ln \ln \frac{1}{t}}} \ge 1 \right)=1.

Let \theta \in (0,1). For n \in \mathbb{N}, we denote

A_n=\left\{ \omega, B_{\theta^n}(\omega)-B_{\theta^{n+1}}(\omega) \ge (1-\sqrt{\theta})h(\theta^n)\right\}.

Let us prove that
\sum \mathbb{P} (A_n)=+\infty.

The basic inequality
\int_{a}^{+\infty} e^{-\frac{u^2}{2}}du \ge \frac{a}{1+a^2}e^{-\frac{a^2}{2}},

implies

\mathbb{P} (A_n)=\frac{1}{\sqrt{2\pi}}\int_{a_n}^{+\infty}e^{-\frac{u^2}{2}}du \ge \frac{a_n}{1+a_n^2}e^{-\frac{a_n^2}{2}},

with

a_n=\frac{(1-\sqrt{\theta})h(\theta^n)}{\theta^{n/2} \sqrt{1-\theta}}.

When n \to +\infty,

\frac{a_n}{1+a_n^2} e^{-\frac{a_n^2}{2}}=O\left(\frac{1}{n^{\frac{1+\theta-2\sqrt{\theta}}{1-\theta}}}\right),

therefore,

\sum \mathbb{P} (A_n)=+\infty.

As a consequence of the independence of the Brownian increments and of Borel-Cantelli lemma, the event

B_{\theta^n}-B_{\theta^{n+1}} \ge (1-\sqrt{\theta})h(\theta^n)

will occur almost surely for infinitely many n‘s. But, thanks to the first part of the proof, for almost every \omega, we may find N(\omega) such that for n \ge N(\omega),

B_{\theta^{n+1}}>-2h(\theta^{n+1})\ge-2\sqrt{\theta} h(\theta^n).

Thus, almost surely, the event

B_{\theta^n} >h(\theta^n)(1-3\sqrt{\theta})

will occur for infinitely many n‘s. This implies

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t \ln \ln \frac{1}{t}}} \ge 1-3\sqrt{\theta} \right)=1.

We finally get

\mathbb{P}\left( \lim \sup_{t \rightarrow 0} \frac{B_{t}}{\sqrt{2t \ln \ln \frac{1}{t}}} \ge 1 \right)=1.

by letting \theta \to 0 \square

As a straightforward consequence, we may observe that the time inversion invariance property of Brownian motion implies:

Corollary. Let (B_t)_{t\ge 0} be a standard Brownian motion.

\mathbb{P}\left( \lim \inf_{t \rightarrow +\infty} \frac{B_{t}}{\sqrt{2t \ln \ln t}} =-1,\lim \sup_{t \rightarrow +\infty} \frac{B_{t}}{\sqrt{2t \ln \ln t}} =1\right)=1.

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