1
$\begingroup$

We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=-1$ and $B^2(0)=1$ and $T_n^i=\inf\{t\geq0:|B^i(t)|=n\}$.

In Gregory Lawler's: Hausdorff Dimension of Cut-Points for Brownian Motion, it is claimed, that

$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\} \approx n^{-2}$

where for functions $f$ and $g$, $f\approx g $ means $\lim_{n\to \infty}\frac{\ln f(n)}{\ln g(n)}=1$.

Now I think, that in this case even

$\lim_{n\to\infty}\frac{\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}}{n^{-2}}=1\quad$ without the logarithms holds true.

Note that

$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}=\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}$,

and we get an upper bound by the application of the Optional Stopping Theorem:

$\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}\leq \textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq 1; \min\limits_{t\in[0,T_n^2]}{B^2(t)}\leq -1\}=(\frac{1}{n+1})^2$.

However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=\min\limits_{0\leq s\leq t}{B(s)}$, but I was not successful so far.

Any help would be greatly appreciated!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.