# Continuity of hitting distributions

Hi everybody

Let $U$ be the domain (as shown in the picture) and $\bar{U}$ its closure, further more set $\partial_r U$ to be the reflecting boundary and $\partial_a U$ the absorbing one. The process $(X_t)_{t\geq 0}$ is chosen as described above and $\tau=\inf(t\geq 0, X_t \in \partial_a U)$ is the first hitting time of the absorbing boundary.

Thanks in advance - any kind of advice would be great :)

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The way I understand your question as you move along the absorbing boundary from $A$ to $A^c \cap U_{\alpha}$ the function changes (abruptly) from $0$ to 1. –  mike Jun 3 '12 at 20:27
Tnak you mike, you are obviously right - it should be continuous on $\bar{U}\setminus \partial_a U$ - or I should rather consider the hitting Probability of the entire absorbing boundary $\partial_a U$. I think the latter approach makes more sense - I will edit my initial post. –  Boldwing Jun 3 '12 at 20:38

How about transforming the corner conformally into a part of a line? The process would transform into a time-changed Brownian motion, again with normal reflection (since the map is conformal up to the boundary, with the only exception at that corner point, and I guess the process never hits that point). And then use the same argument with convergence of hitting distribution.

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the process itself never hits the corner point - you are correct. One Point sets are polar for the reflected browmian motion of the type described above. Still I could start my process in the corner and here it would be very helpful to know whether continuity holds -actually it is essential Perhaps your approach could work - do you have anny suggestion for an explicit map that might do the trick ? (I must admit I have never worked with conformal maps before - at least not that explicitly) What is more - as far as I know konformal maps preserve angles thus my corner would still be there ? –  Boldwing Jun 3 '12 at 21:11
en.wikipedia.org/wiki/Riemann_mapping_theorem There exists an essentially unique conformal map from your domain into, say, the upper half-plane. It is conformal inside the domain and at the smooth parts of the boundary, and of course it has no reason to be smooth at angles. Consult any course on complex analysis. Perhaps in Markushevich there must be something about regularity at the boundary. –  Alexander Shamov Jun 3 '12 at 22:45
To straighten out a corner at the origin, use $f(z) = z^c$ (you will need a branch cut). –  Douglas Zare Jun 4 '12 at 0:50
Okey $f(z)=z^c$ should do the trick for the multiplication of complex numbers means adding the respective angles etc. Now the only thing I need to do is finding a branch cut that won't hinder - e.g. one which does not go through the domain I am studying. Choosing $(-\infty,0]$ and just move my domain away from the origin by using a linear map. The branch cut would insure that $z^{-c}$ is also analytic etc. By the way do I actually need a branch cut - as far as I know $z^c$ should be analytic on the entire plain for $c \in \mathbb{N}$ ? –  Boldwing Jun 4 '12 at 6:27
If you are smoothing out an angle of $\pi/n$ you can use $z \mapsto x^n$. In your diagram it looked like you had an angle of $3\pi/2$, so $n=2/3$. Anyway, the branch cut isn't a big deal. –  Douglas Zare Jun 4 '12 at 10:48

First of all I would like to thank all of you for your advice - it really helps to know that there is sombebody out there you can turn to (if the people who are actually supposed to be helping and supporting you are both not avaliable and not interested)

The comformal map argument seems to work. Unfortunately I found out that my reasoning pertaining to the case when a point lies on the flat side of the boundary (a surface ball) doesn't seem to apply. Most approaches I was able to finde in the literature do not deal with the case of a reflecting boundary. The best possible thing I was able to find was in Probabilistic Techniques in Analysis (Richard F. Bass) The book offers a chapter on lipshitz domains and on page 195 the following result ist presented - $\omega(x,A)=\int_A M(x,y)\omega(x_0,d\omega)$ where $\omega(x,A)=\mathbb{P}^x(X_{\tau} \in A)$ and $A\subset \partial D$ with $D$ bein a lipschitz domain.

My question thus boils down to the following: How do I show the continuity of $\mathbb{P}^x(X_{\tau} \in\partial_a U)$ for $x \in \partial_r U$ ?

I have been battling this problem for weeks now :(

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It's good practice to keep at problems for weeks or months at a time. However, for the specific case of a reflecting boundary which is a line segment, can't you just consider a Brownian motion on the region plus the mirror image? –  Douglas Zare Jun 6 '12 at 6:09
In terms of practice experience - you are certainly right one learns how to juggle and keep several ideas present at the same time. My current best gues is to use the fact, that the process will be located within the halfcircle (not the boundary) with Probability one after an infinitisimaly small amount of time. I have also though about using the symmetry like you suggested - but I do not know hot to modell it perfectly - the probabilities would't just double - I will write some of my current thoughts in an edit above :) –  Boldwing Jun 6 '12 at 6:48
I think the probabilities would simply double. –  Douglas Zare Jun 6 '12 at 7:01
perhaps you are right and I am overthinking the problem. Do you think it would suffice to say that the probabilities will just double - without providing a rigorous proof ? (I mean it somehow is obvious - no idea why I discarded the thought in the first place) –  Boldwing Jun 6 '12 at 7:46
I think you should write up the argument carefully as an exercise. Relate the Brownian motion on the double with the Brownian motion on the shape with the reflecting boundary. Afterwards, decide whether to include it. I would lean toward including it unless you decide that it isn't important. –  Douglas Zare Jun 6 '12 at 9:39