Probability that a Brownian Motion in R2 hits a line through the origin.

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Let $B_t=(B_t^1,B_t^2)$ be a Brownian motion in $mathbbR^2$. I wish to show that $P[B_t^2=0, tin(0,epsilon)]=1$ for all $epsilon$; that is, a Brownian motion in $mathbbR^2$ will always hit a line through the origin in some short time. For context, this is part of an attempted solution for Exercise 9.7c in Oksendal's SDE book, 6th edition.



My reason for believing this is that I know that one-dimensional Brownian motion equals zero infinitely many times in a short time interval, but I'm not sure how to adapt this.



Any help would be appreciated. Thanks!




probability stochastic-processes stochastic-calculus brownian-motion




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  • 3




    If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
    – Henry
    Aug 1 at 23:37






  • 1




    Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
    – E-A
    Aug 1 at 23:41










  • Duh! Little embarrassed that I didn't realize this myself. Thanks!
    – liplo
    Aug 1 at 23:44














up vote
1
down vote

favorite












Let $B_t=(B_t^1,B_t^2)$ be a Brownian motion in $mathbbR^2$. I wish to show that $P[B_t^2=0, tin(0,epsilon)]=1$ for all $epsilon$; that is, a Brownian motion in $mathbbR^2$ will always hit a line through the origin in some short time. For context, this is part of an attempted solution for Exercise 9.7c in Oksendal's SDE book, 6th edition.



My reason for believing this is that I know that one-dimensional Brownian motion equals zero infinitely many times in a short time interval, but I'm not sure how to adapt this.



Any help would be appreciated. Thanks!




probability stochastic-processes stochastic-calculus brownian-motion




share|cite|improve this question















  • 3




    If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
    – Henry
    Aug 1 at 23:37






  • 1




    Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
    – E-A
    Aug 1 at 23:41










  • Duh! Little embarrassed that I didn't realize this myself. Thanks!
    – liplo
    Aug 1 at 23:44












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $B_t=(B_t^1,B_t^2)$ be a Brownian motion in $mathbbR^2$. I wish to show that $P[B_t^2=0, tin(0,epsilon)]=1$ for all $epsilon$; that is, a Brownian motion in $mathbbR^2$ will always hit a line through the origin in some short time. For context, this is part of an attempted solution for Exercise 9.7c in Oksendal's SDE book, 6th edition.



My reason for believing this is that I know that one-dimensional Brownian motion equals zero infinitely many times in a short time interval, but I'm not sure how to adapt this.



Any help would be appreciated. Thanks!




probability stochastic-processes stochastic-calculus brownian-motion




share|cite|improve this question











Let $B_t=(B_t^1,B_t^2)$ be a Brownian motion in $mathbbR^2$. I wish to show that $P[B_t^2=0, tin(0,epsilon)]=1$ for all $epsilon$; that is, a Brownian motion in $mathbbR^2$ will always hit a line through the origin in some short time. For context, this is part of an attempted solution for Exercise 9.7c in Oksendal's SDE book, 6th edition.



My reason for believing this is that I know that one-dimensional Brownian motion equals zero infinitely many times in a short time interval, but I'm not sure how to adapt this.



Any help would be appreciated. Thanks!





probability stochastic-processes stochastic-calculus brownian-motion





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 23:34









liplo

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  • 3




    If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
    – Henry
    Aug 1 at 23:37






  • 1




    Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
    – E-A
    Aug 1 at 23:41










  • Duh! Little embarrassed that I didn't realize this myself. Thanks!
    – liplo
    Aug 1 at 23:44












  • 3




    If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
    – Henry
    Aug 1 at 23:37






  • 1




    Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
    – E-A
    Aug 1 at 23:41










  • Duh! Little embarrassed that I didn't realize this myself. Thanks!
    – liplo
    Aug 1 at 23:44







3




3




If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
– Henry
Aug 1 at 23:37




If $B_t^1$ and $B_t^2$ are independent, can you just ignore $B_t^1$ and reduce this to the one-dimensional case?
– Henry
Aug 1 at 23:37




1




1




Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
– E-A
Aug 1 at 23:41




Henry's comment + the fact that the whole thing is rotationally invariant will give you the answer. (i.e. you can construct another BM from these that has the same distributions but one direction is perpendicular to your line (and the other parallel))
– E-A
Aug 1 at 23:41












Duh! Little embarrassed that I didn't realize this myself. Thanks!
– liplo
Aug 1 at 23:44




Duh! Little embarrassed that I didn't realize this myself. Thanks!
– liplo
Aug 1 at 23:44















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