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 $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 $g$, $fapprox g $ means $lim_{nto infty}frac{ln f(n)}{ln g(n)}=1$.
Now I don't really see why this holds. It might have to do something with the gambler's ruin problem, but the sets that we want to hit are random here ...
and I don't have more ideas on how to approach this.
Any help would be greatly appreciated!
brownian-motion stopping-times
asked Dec 24 '18 at 15:28
John Doe
163
add a comment |
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=1$ and $B^2(0)=-1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 $g$, $fapprox g $ means $lim_{nto infty}frac{ln f(n)}{ln g(n)}=1$.
Now I don't really see why this holds. It might have to do something with the gambler's ruin problem, but the sets that we want to hit are random here ...
and I don't have more ideas on how to approach this.
Any help would be greatly appreciated!
brownian-motion stopping-times
asked Dec 24 '18 at 15:28
John Doe
163
add a comment |
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=1$ and $B^2(0)=-1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 $g$, $fapprox g $ means $lim_{nto infty}frac{ln f(n)}{ln g(n)}=1$.
Now I don't really see why this holds. It might have to do something with the gambler's ruin problem, but the sets that we want to hit are random here ...
and I don't have more ideas on how to approach this.
Any help would be greatly appreciated!
brownian-motion stopping-times
asked Dec 24 '18 at 15:28
John Doe
163
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=1$ and $B^2(0)=-1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 $g$, $fapprox g $ means $lim_{nto infty}frac{ln f(n)}{ln g(n)}=1$.
Now I don't really see why this holds. It might have to do something with the gambler's ruin problem, but the sets that we want to hit are random here ...
and I don't have more ideas on how to approach this.
Any help would be greatly appreciated!
brownian-motion stopping-times
brownian-motion stopping-times
asked Dec 24 '18 at 15:28
John Doe
163
asked Dec 24 '18 at 15:28
John Doe
163
edited Dec 27 '18 at 13:42
asked Dec 24 '18 at 15:28
John Doe
163
asked Dec 24 '18 at 15:28
John Doe
163
asked Dec 24 '18 at 15:28
John Doe
163
163
add a comment |
add a comment |
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