Intersection-exponent for one-dimensional Brownian motion
$begingroup$
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=-1$ and $B^2(0)=1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 think, that in this case even
$lim_{ntoinfty}frac{textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}}{n^{-2}}=1quad$ without the logarithms holds true.
Note that
$textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}=textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}$,
and we get an upper bound by the application of the Optional Stopping Theorem:
$textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}leq textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq 1; minlimits_{tin[0,T_n^2]}{B^2(t)}leq -1}=(frac{1}{n+1})^2$.
However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=minlimits_{0leq sleq t}{B(s)}$, but I was not successful so far.
Any help would be greatly appreciated!
probability brownian-motion stopping-times
asked Jan 18 at 12:49
John DoeJohn Doe
263
$endgroup$
add a comment |
$begingroup$
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=-1$ and $B^2(0)=1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 think, that in this case even
$lim_{ntoinfty}frac{textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}}{n^{-2}}=1quad$ without the logarithms holds true.
Note that
$textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}=textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}$,
and we get an upper bound by the application of the Optional Stopping Theorem:
$textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}leq textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq 1; minlimits_{tin[0,T_n^2]}{B^2(t)}leq -1}=(frac{1}{n+1})^2$.
However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=minlimits_{0leq sleq t}{B(s)}$, but I was not successful so far.
Any help would be greatly appreciated!
probability brownian-motion stopping-times
asked Jan 18 at 12:49
John DoeJohn Doe
263
$endgroup$
add a comment |
$begingroup$
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=-1$ and $B^2(0)=1$ and $T_n^i=inf{tgeq0:|B^i(t)|=n}$.
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 think, that in this case even
$lim_{ntoinfty}frac{textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}}{n^{-2}}=1quad$ without the logarithms holds true.
Note that
$textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}=textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}$,
and we get an upper bound by the application of the Optional Stopping Theorem:
$textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}leq textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq 1; minlimits_{tin[0,T_n^2]}{B^2(t)}leq -1}=(frac{1}{n+1})^2$.
However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=minlimits_{0leq sleq t}{B(s)}$, but I was not successful so far.
Any help would be greatly appreciated!
probability brownian-motion stopping-times
asked Jan 18 at 12:49
John DoeJohn Doe
263
$endgroup$
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 think, that in this case even
$lim_{ntoinfty}frac{textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}}{n^{-2}}=1quad$ without the logarithms holds true.
Note that
$textbf{P}{B^1[0,T_n^1]cap B^2[0,T_n^2]=emptyset}=textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}$,
and we get an upper bound by the application of the Optional Stopping Theorem:
$textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq minlimits_{tin[0,T_n^2]}{B^2(t)}}leq textbf{P}{maxlimits_{tin[0,T_n^1]}{B^1(t)}leq 1; minlimits_{tin[0,T_n^2]}{B^2(t)}leq -1}=(frac{1}{n+1})^2$.
However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=minlimits_{0leq sleq t}{B(s)}$, but I was not successful so far.
Any help would be greatly appreciated!
probability brownian-motion stopping-times
probability brownian-motion stopping-times
asked Jan 18 at 12:49
John DoeJohn Doe
263
asked Jan 18 at 12:49
John DoeJohn Doe
263
edited Jan 18 at 13:31
John Doe
asked Jan 18 at 12:49
John DoeJohn Doe
263
asked Jan 18 at 12:49
John DoeJohn Doe
263
asked Jan 18 at 12:49
John DoeJohn Doe
263
263
add a comment |
add a comment |
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