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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

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

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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.

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

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