1
$\begingroup$

I am doing research on the hitting probability of various sets (eg. 3D convex) and specifically how changes in perimeter/surface area change the hitting probability.

By hitting probability I mean $P(B[0,t]\cap A\neq \varnothing$ for some t).

Question 1) (Answered below) Say we have two bounded planes in $\mathbb{R}^{3}$, one with greater surface area than the other and equidistant from the origin. Then given a brownian motion starting from the origin, I want to get the hitting probability of each plane. enter image description here

I suppose the larger surface area plane will have a great hitting probability. But what would be a rigorous way of proving that?

Thanks that is answered below.

Question 2) Also, given the exact coordinates of one of the bounded planes A above, can we compute $P(T_{A}<\infty)$? How can I go about it?

$\{T_{A} <\infty\}=\{B_{1}(t)=a, |B_{2}(t)|\geq b,|B_{3}(t)|<c$ for some $t>0\}$. Here by a,b,c I mean the distance from origin, length and width of the square $A$. So I have to compute: $P_{0}\{(B_{1}(t)=a)\cap (|B_{2}(t)|\geq b)\cap (|B_{3}(t)|<c)$ for some $t>0\}$=

Also, can you provide some books/papers that expose Brownian motion and surface area for more general sets?

Thnx

$\endgroup$
2
  • $\begingroup$ A plane with a greater area than another plane? Brrr... $\endgroup$
    – Did
    Sep 6, 2014 at 8:01
  • $\begingroup$ @Did OP did specify bounded planes. $\endgroup$
    – Todd Trimble
    Sep 6, 2014 at 15:57

1 Answer 1

1
$\begingroup$

I suppose the larger surface area plane will have a greater hitting probability. But what would be a rigorous way of proving that?

Depends on whether for each plane $A$, the center of $A$ is the nearest point of $A$ to the origin.

If yes, then you could, by symmetry, assume the smaller plane is a subset of the larger plane, to get your result.

If no, then the result could fail.

$\endgroup$
5
  • $\begingroup$ Thanks. Also, given the exact coordinates of bounded plane A in 3d, can we compute $P(T_{A}<\infty)$? $\endgroup$
    – TKM
    Sep 7, 2014 at 21:28
  • $\begingroup$ @TKM I guess that's another question $\endgroup$ Sep 7, 2014 at 22:12
  • $\begingroup$ We usually do 1 question per question $\endgroup$ Sep 7, 2014 at 22:13
  • $\begingroup$ okay, I will make a new one. $\endgroup$
    – TKM
    Sep 8, 2014 at 15:46
  • $\begingroup$ mathoverflow.net/questions/180360 $\endgroup$ Sep 9, 2014 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.