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.



Wiener measure of hitting sets A,B but not C (or easier hitting A but not C)
This is a variation on [Brownian motion and hitting a Quadrilateral](https://mathoverflow.net/questions/180360) 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.




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*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.](http://www.rle.mit.edu/cehv/documents/16-AmericanJournalofPhysics441132-11341976.pdf) (The result we need from that paper is the total induced charge on each plane.) In this way one obtains 
$$P\_{A|C}=\frac{c}{a+c}.$$


