Continuity in intial state of Brownian Motion - MathOverflow

Continuity in intial state of Brownian Motion

2010-07-31T15:33:48Z

$ B = (B_t, \mathcal{F}_t; t\ge 0 ) $ is a 1-d Brownian family on a measurable space $(\Omega, \mathcal{F})$ with a family of probability measures ${\mathbb{P}^x}$, i.e. $\mathbb{P}^x(B_0 = x) = 1$, and $B$ is 1-d BM starting from $x$ under $\mathbb{P}^x$.

Let $\tau$ be a given stopping time w.r.t. underlying filtration, $f$ be a given continuous bounded real function. Consider $V(x) = \mathbb{E}^x [f(B_\tau)]$, where $\mathbb{E}^x$ is the expectation under $\mathbb{P}^x$.

[Question] Is $V(\cdot)$ continuous for any given stopping time $\tau<\infty$? If not, is there any counter example? Or does continuity hold with further conditions?

If $\tau$ is deterministic, then $V$ has no doubt to be continuous. I am not sure, even if the problem is well formulated with the extension to stopping time $\tau$. Thanks for any of your comments.

Answer by Byron Schmuland for Continuity in intial state of Brownian Motion

2010-07-31T22:55:10Z

Edit: I just noticed that the OP asked about a 1-d Brownian motion. The constriction below only works in three or more dimensions. Back to the drawing board....

Your function $V$ is not necessarily continuous. Its continuity properties depend not only on the function $f$, but also the nature of the random time $\tau$.

A classic counterexample is found by letting $\tau$ to be the hitting time of the complement of a bounded, open region $D$ with an irregular point (as defined in Newtonian potential theory). For instance, you could choose a region with a "Lebesgue spine".

http://en.wikipedia.org/wiki/Lebesgue_spine

Then $E_x[\tau]<\infty$ for all $x$, and for any continuous $f$ your function $V(x):=E_x[f(B_\tau)]$ is the Perron-Wiener-Brelot solution to the Dirichlet problem with data $(D,f)$. That is, $V$ is harmonic on $D$ and $\lim_{D\ni x\to z}V(x)=f(z)$ at all regular points $z\in \partial D$.

However, if the point $z$ is irregular, then choosing $f$ with $f(z)=1$ and $f(y)<1$ otherwise, we have $\liminf_{D\ni x\to z}V(x)$<1. On the other hand, $V(y)=f(y)$ for all $y\not\in \bar D$ so approaching the tip of the spine from outside of $\bar D$, the function $V$ has limit 1.

Thus, $V$ fails to be continuous at $z$. Note that $f$ can be as smooth as you like.

Intuitively, the reason why $V$ is discontinuous is that the spine is so sharp that Brownian motion fails to see it, even as the starting point approaches the tip of the spine from within $D$.

One nice treatment of these questions of probabilistic potential theory is found in Kai Lai Chung's "Lectures from Markov Processes to Brownian Motion". The lim inf result above is Theorem 3 (p.164) in section 4.4 of this book.

Answer by The Bridge for Continuity in intial state of Brownian Motion

2010-08-01T00:35:20Z

Well I am not sure about it, but it could worth a try to start with this to show continuity:

$E^x[f(B_\tau)]=E^0[\int_0^{+\infty}f(x+W_t)dP^{\tau}(t)]=\int_{\mathbb{R}}\int_0^{+\infty}C(t)f(x+y)e^{-\frac{y^2}{2.t}}dP^{\tau}(t)dy$

(where $W_t$ is a BM starting from $0$, $C(t)$ is a normalising constant, and $P^{\tau}(t)$ is the cdf of $\tau$)

Your move now Kenneth

Regards

Answer by Byron Schmuland for Continuity in intial state of Brownian Motion

2010-08-01T15:16:33Z

Here is a simpler example that I hope convinces you that $V$ need not be continuous, even in the one dimensional case.

Take one dimensional Brownian motion $(B_t)$ and define the stopping time $\tau(\omega)=1_{(B_0(\omega)<0)}$. Then, for any bounded measurable $f$, we have $$V(x)=E_x[f(B_1)]1_{(-\infty,0)}(x)+f(x) 1_{[0,\infty)}(x).$$

The function $V$ can be made discontinuous at zero by choosing $f$ to have a strict maximum at $x=0$, since then $E_x[f(B_1)] < f(0)$.

Comment: You really cannot expect the function $V$ to be continuous in general. The values of a typical stopping time $\tau$ are intimately tied up with the sample paths of the Brownian motion; in your words $\tau$ is "strongly correlated'' with $\omega$. It's in the definition of stopping time.

The only stopping times that are independent of the Brownian motion are the deterministic ones.