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For a Brownian motion $B_t$ there are many results about its $t \rightarrow + \infty$ or $t \rightarrow 0+$ behavior, such as $$\limsup_{t\rightarrow \infty} \frac{B_t}{(2t\log\log t)^{1/2}} = 1 \quad a.s.\\ \liminf_{t\rightarrow \infty} \frac{B_t}{(2t\log\log t)^{1/2}} = -1 \quad a.s.$$

I interpret the above result as follows: as $t \rightarrow +\infty$ the paths of the process $B_t$ will a.s. grow by the rate $(2t\log\log t)^{1/2}$. Another way I think of this is if you scale $B_t$ by $(2t\log\log t)^{-1/2}$ then the resulting process will a.s. be within $[-1, 1]$ in the limit $t \rightarrow +\infty$.

Assuming the above interpretations are correct, I am having a hard time coming up with a similar interpretation to the $t \rightarrow 0+$ limit, for example: $$\liminf_{t\rightarrow 0+} \frac{B_t}{(2t\log\log \frac1t)^{1/2}} = -1 \quad a.s.$$ A Brownian motion by definition starts at $0$ a.s., and if we assume a continuous version of the Brownian motion I would expect that the $t\rightarrow 0+$ limit is always zero and not -1. So what is the above saying exactly?

CBBAM
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We can use time-inversion Show that $X(t)=t W(1/t)$ is a Brownian motion if $W(t)$ is a Brownian motion. i.e. $X_{t}:=tB_{1/t}$ is a BM and so

$$\liminf_{t\rightarrow 0+} \frac{B_t}{(2t\log\log \frac1t)^{1/2}} \stackrel{d}{=}\liminf_{t\rightarrow 0+} \frac{tB_{1/t}}{(2t\log\log \frac1t)^{1/2}}$$

$$=\liminf_{t\rightarrow +\infty} \frac{\frac{1}{t}B_{t}}{(2\frac{1}{t}\log\log t)^{1/2}}$$

$$=\liminf_{t\rightarrow +\infty} \frac{B_{t}}{(2t\log\log t)^{1/2}}=-1.$$

Thomas Kojar
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  • Thank you for your answer, I understand the proof but I'm having some trouble understanding what's really going on visually/intuitively. In the $t \rightarrow 0+$ limit does this result say that the Brownian motion never goes below $-1$ a.s.? If so, doesn't this already follow from the definition of Brownian motion (i.e. $B_0 = 0$ a.s.)? – CBBAM Mar 03 '24 at 01:53
  • @CBBAM this gives you the particular envelope of approaching zero/infinity for all $0\leq t\leq \epsilon$ or for all $t\geq \frac{1}{\epsilon} $ respectively. – Thomas Kojar Mar 03 '24 at 02:01
  • I see, so the bound is more about what happens in the time interval $(0, \epsilon)$? as opposed to at $t = 0$? – CBBAM Mar 03 '24 at 03:25