Rogers and Williams I.15 shows Kolmogorov's test. Paraphrasing, for a Brownian motion and some $h:\mathbb{R}^+\to\mathbb{R}$ with $t^{-1/2}h(t)\uparrow$ as $t\downarrow 0$, then if

$$\int_0^\epsilon t^{-3/2}h(t)e^{-\frac{h(t)^2}{2t}}dt$$

diverges, then $\mathbb{P}(B_t\leq h(t) \text{ near 0})= \mathbb{P}(\cup_{\delta > 0}\{B_t\leq h\text{ along }[0,\delta]\})=0$, whereas if it converges then that probability is 1.

They claim that the law of iterated logs follows, that is:

$$\mathbb{P}\left[\limsup_{t\downarrow 0}\frac{B_t}{\sqrt{2t\log\log 1/t}}=1\right]=1$$ and they ask the reader to demonstrate this.

It is easy to check that taking $h(t)=\sqrt{2t\log\log 1/t}$ then the integral in Kolmogorov's test diverges. All that is left to do is to demonstrate that the limsup is bounded above by 1 as $t\to 0$. Here is an attempt:

If we take $x=1/t$ then:

\begin{align} \frac{B_t}{\sqrt{2t\log\log 1/t}} =& \frac{B_{1/x}}{\sqrt{2\frac{1}{x}\log\log x}} \\ =& \frac{xB_{1/x}}{\sqrt{2x\log\log x}} \\ \overset{\text{in d}}{=}&\frac{B_x}{\sqrt{2x\log\log x}} \end{align} which is almost surely $\leq 1$ eventually since for large $x$, $P(B_x\geq \sqrt{2x\log\log x})$ is small and because Brownian motion is strongly Markov.

I am uncomfortable with the last step. Can someone make it more precise?



If we define

$$h_k(t) := \sqrt{1+ \frac{1}{k}} h(t) = \sqrt{\left( 1+ \frac{1}{k} \right) 2t \log \log \frac{1}{t}} $$

for fixed $k \in \mathbb{N}$, then

$$\begin{align*} \int_0^{\epsilon} t^{-3/2} h_k(t) e^{-h_k(t)^2/2t} \, dt &= \sqrt{2+\frac{2}{k}} \int_0^{\epsilon} \frac{1}{t} \frac{1}{(\log 1/t)^{1+1/k}} \log \log \frac{1}{t} \, dt<\infty. \end{align*}$$

(The finiteness of the integral on the right-hand side follows e.g. from Cauchy's condensation test and the integral test for convergence.) Applying Kolmogorov's test, we find that

$$\mathbb{P} \left( B_t \leq \sqrt{1+\frac{1}{k}} h(t) \, \, \text{for small $t$} \right) = 1$$

for all $k \in \mathbb{N}$ and therefore

$$\mathbb{P} \left( B_t \leq h(t) \, \, \text{for small $t$}\right) = 1,$$ i.e.

$$\mathbb{P} \left( \limsup_{t \to 0} \frac{B_t}{h(t)} \leq 1 \right)=1.$$


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