Intersection-exponent for one-dimensional Brownian motion 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! 
 A: For the lower bound, the following pretty naive estimate works:
$$\mathbb{P}(B^{1}[0,T_{n}^{1}]\cap{B^{2}[0,T_{n}^{2}]}=\emptyset)\geq{\mathbb{P}(\{B^{1} \hspace{3pt}\text{hits $-n$ before hitting $0$}\}\cap\{B^{2} \hspace{3pt}\text{hits $n$ before hitting $0$}\})}= \\ =\mathbb{P}(B^{1} \hspace{3pt}\text{hits $-n$ before hitting $0$})\cdot\mathbb{P}(B^{2} \hspace{3pt}\text{hits $n$ before hitting $0$}\})=\frac{1}{n^{2}}$$
Also, $\frac{1}{n^{2}}$ is larger than your upper bound of $\frac{1}{(n+1)^{2}}$ which suggests that you probably made a calculation error somewhere. For what it's worth, you can get an upper bound by the same reasoning as the one I used for the lower bound above:
$$\mathbb{P}(B^{1}[0,T_{n}^{1}]\cap{B^{2}[0,T_{n}^{2}]}=\emptyset)\leq{\mathbb{P}(\{B^{1} \hspace{3pt}\text{hits $-n$ before hitting $1$}\}\cap\{B^{2} \hspace{3pt}\text{hits $n$ before hitting $-1$}\})}= \\ =\mathbb{P}(B^{1} \hspace{3pt}\text{hits $-n$ before hitting $1$})\cdot\mathbb{P}(B^{2} \hspace{3pt}\text{hits $n$ before hitting $-1$}\})=\frac{4}{(n+1)^{2}}$$
