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I am trying to formulate the measure of event

$E=\{B[0\infty)\cap A,B \neq \varnothing$ and $B[0\infty)\cap C= \varnothing\}$,

where $B[0\infty)$ is a Brownian path and $A,B,C$ are pairwise disjoint compact non-empty sets.

Q1: I am looking for $\mu_{W}(E)$. Any solutions?

One guess is:

$\mu_{W}(E)=\int_{t_{1}}^{\infty}\int_{0}^{\infty} \int_{B} \int_{A} \int_{\mathbb{R}^{d}/(A\cup B)}p(x_{1},x_{2},t_{1})p(x_{2},x_{3},t_{2}) dx_{1}dx_{2}dx_{3}dt_{1}dt_{2}+ $

$+\int_{t_{1}}^{\infty}\int_{0}^{\infty} \int_{A} \int_{B} \int_{\mathbb{R}^{d}/(A\cup B)}p(x_{1},x_{2},t_{1})p(x_{2},x_{3},t_{2}) dx_{1}dx_{2}dx_{3}dt_{1}dt_{2}-$

$-\int_{0}^{\infty}\int_{C} \int_{\mathbb{R}^{d}/C} p(x_{1},x_{2},t_{1}) dx_{1}dx_{2}dt_{1}$

The first term is hitting A and then B, the second is the converse and the last term is hitting C.

The inner integral is the startpoint of the Brownian motion. For the first two terms ,the outer integral has $t_{1}$ as the startpoint, to denote the transition from A to B.

Q2:Expressing hitting set A but not C, in terms of their individual hitting probabilities $P(H(A))$ and $P(H(C))$.

I guess $P_{\mathbb{R}^{d}/\bar{C}}(H(A))-P_{\partial A}(H(C))$.

thanks Ilya. Still I would appreciate if someone can give a precise answer.

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    $\begingroup$ I think if you express each of the integrals as a probability of a certain event, it would be easier for you to realize whether they correspond to what you want or not. $\endgroup$
    – SBF
    Jul 17, 2014 at 19:30
  • $\begingroup$ that's what I was doing. I updated it. $\endgroup$
    – TKM
    Jul 17, 2014 at 19:37
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    $\begingroup$ I'm pretty sure there is a couple of neat PDEs that the solution satisfies, though likely there is also a direct approach - not that I know of unfortunately. Let's take a look at the integrals. If $p(x,y,t)$ is a density of lending at $y$ from $x$ in time $t$, then the last integral is not even a probability. $$ \int_C p(x,y,t)\mathrm dy = P(x,C,t) $$ that is a probability of landing in a set $C$ from $x$ in time $t$. Your latter integral is integrating this probability over $x\in \Bbb R$ and $t\in \Bbb R_+$ - that's not a probability of hitting $C$. $\endgroup$
    – SBF
    Jul 17, 2014 at 19:50
  • $\begingroup$ i want x and t to vary. I am not just looking at paths starting from fixed point x and hitting C at a fixed time t, which is what you wrote. But I will review it anyhow. $\endgroup$
    – TKM
    Jul 17, 2014 at 19:55
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    $\begingroup$ My point was: if you take a look at the latter integral, you can't say it's less than $1$, so it's not a probability of anything - in particular not of hitting $C$. On a separate note, if you know what's the probability of hitting $B$ without touching $C$, compute this probability for initial conditions on the boundary of $A$. $\endgroup$
    – SBF
    Jul 17, 2014 at 20:25

1 Answer 1

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This is a variation on Brownian motion and hitting a Quadrilateral and can be reduced in a similar way to the solution of a problem in electrostatics. Let me assume that the Brownian motion starts at the origin and ask for the probability $P_{A|C}$ that the particle eventually hits the boundary $A$ without first hitting the boundary $C$.

In the equivalent electrostatic problem the boundaries $A$ and $C$ are grounded (potential $\Phi=0$) and the probability $P_{A|C}$ is obtained by integrating (minus) the electric field $\nabla\Phi$ over the boundary of $A$:

$$\nabla^2\Phi(\vec{r})=-\delta(\vec{r}),\;\;\Phi(\vec{r})=0\;\;\text{for}\;\;\vec{r}\in A,C$$

$$P_{A|C}=\int_A \frac{\partial\Phi}{\partial\vec{r}}\cdot\hat{n}\;dS,$$

with $\hat{n}$ a unit vector normal to $A$ and pointing outward.


Example: $A$ and $C$ are two infinite parallel planes, $A$ is at $z=-a$ and $C$ is at $z=c$, with $a+c=d$ the separation of the planes. The particle starts at $z=0$. The potential in this case can be obtained by the method of image charges, as an infinite alternating series. Summing this series to obtain $P_{A|C}$ is a bit tricky, a reliable way to do this is described by Marcus Zahn in Point charge between two parallel grounded planes. (The result we need from that paper is the total induced charge on each plane.) In this way one obtains $$P_{A|C}=\frac{c}{a+c}.$$

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