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I have been trying to prove the following conjecture for a while, but so far to no avail. Would be very grateful for some tips!

(I edited the entire thing to make it clearer)

The conjecture is the following;

Let $\bar{x}\in\mathbb{R}^n$ and $\bar{t}\in\mathbb{R}$. Let B be an $n$-dimensional Brownian motion with $B_{t_0}=x_0$, i.e. the motion starts at time $t_0$ at location $x_0$. Let $\eta>0$, and suppose that $(x_0,t_0)$ is in the interior of the $\eta$-ball with center $(\bar{x},\bar{t})$. Define the following stopping time.

$$\tau(x_0,t_0):=\inf \lbrace t : \sqrt{|B_t-\bar{x}|^2+|t-\bar{t}|^2 }\geq \eta \rbrace$$

Let a continuous function $f$ be defined on the boundary of the $\eta$-ball with center $(\bar{x},\bar{t})$. Let $U$ be the interior of the $\eta$-ball with center $(\bar{x},\bar{t})$. Then define the function $F:U\rightarrow\mathbb{R}$ by

$$F(x_0,t_0):=\int_{\omega} f(B_{\tau(x_0,t_0)}(\omega))dW(\omega)$$

The conjecture is that $F$ is differentiable at the point $(\bar{x},\bar{t})$.

I know from the literature on harmonic functions that the conjecture is true if the stopping time is instead given by $\tau(x_0,t_0):=\inf[t : \sqrt{|B_t-\bar{x}|^2}\geq \eta ]$ and $f$ does not depend on $t$, because in this case $F$ is a harmonic function. I have not found how to use the techniques from that literature in the present case, however.

It would be awesome if someone could give me a hint as to how to go about this!

Or, if someone knows of another stochastic process that has this property that would also be great!

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  • $\begingroup$ Can you write more clearly? To start with, I do not see how the set $A$ you introduced changes with time. It looks like it is the outside of a fixed ball (the fact that that ball is in $\mathbb R^2$ though the BM is in $\mathbb R^{n+1}$ only adds to the general confusion). I'm pretty sure you meant something else but figuring out what exactly that was is beyond my abilities. $\endgroup$
    – fedja
    Commented Nov 1, 2012 at 3:02
  • $\begingroup$ I've entirely rewritten the question, it should be much clearer now. $\endgroup$
    – Sandro
    Commented Nov 1, 2012 at 5:15
  • $\begingroup$ To add curly brackets, use \lbrace and \rbrace $\endgroup$ Commented Nov 1, 2012 at 11:09
  • $\begingroup$ I have found a way to prove continuity of $F$, but I'm still struggling with differentiability. I'd still be extremely glad about a useful hint! $\endgroup$
    – Sandro
    Commented Nov 27, 2012 at 1:48

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