Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$ 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.

share|improve this question
1  
For $C^2$ everything is nice: en.wikipedia.org/wiki/Dynkin%27s_formula –  Steve Huntsman Jul 31 '10 at 16:02
    
If $f$ is bounded, can't you use the dominated convergence theorem to get continuity all the time? –  weakstar Jul 31 '10 at 18:19
    
Hi, Weakstar, By DCT, we have to deal with $\lim_{x\to 0} \tau^x \to \tau^0$, and it is not possible without further info on $\tau$. –  kenneth Aug 1 '10 at 3:18
add comment

3 Answers

up vote 5 down vote accepted

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.

share|improve this answer
    
It is quite tricky one. It helps my understanding. –  kenneth Aug 2 '10 at 5:32
    
Hi, there is a little typo it's $f(B_0)$ not $f(B_1)$ I think Regards –  The Bridge Aug 2 '10 at 7:16
    
I think it's OK. The way I've defined $\tau$, when the initial state $B_0(\omega)$ is negative, the value of $\tau(\omega)$ is 1, so $B_\tau=B_1$. Otherwise $B_\tau=B_0$, which is $x$. –  Byron Schmuland Aug 2 '10 at 14:29
    
That's right I misread your defintion of $\tau$ –  The Bridge Aug 2 '10 at 15:26
add comment

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.

share|improve this answer
    
Dear Byron, Thank you very much for your clear explanation. I am still thinking of in which condition continuity holds. You mentioned: BM fails to see it. Do you mean $\tau$ is not predictable? More precisely, if $\Omega = C[0,\infty)$ is canonical continuous process space, then $\tau:\Omega \to [0,\infty)$ is not continuous with sharp $\partial D$. –  kenneth Aug 1 '10 at 2:38
    
(Cont. from above) Here is my digestion of the above. Think of $\mathbb{P}^x \Rightarrow \mathbb{P}^0$ (weak convergence) as $x\to 0$, this implies $\mathbb{P}^x F \to \mathbb{P}^0 F$ as $x\to 0$ for all continuous bounded $F:\Omega\to \mathbb{R}$. So, continuity of $V$ holds, if $F= f(B_\tau): \Omega \to \mathbb{R}$ is continuous, and it holds if $\tau$ is continuous. For instance, if $\partial D$ is smooth in some sense, such that $\tau$ is continuous, then continuity of $V$ holds. Thank you –  kenneth Aug 1 '10 at 2:39
    
kenneth, You are quite right that if $\tau$ is continuous, then so is $V$. But most of the stopping times, for example hitting times, used in stochastic analysis are not continuous. Except the deterministic times, I mean. –  Byron Schmuland Aug 1 '10 at 20:20
    
Byron, Actually, the stopping time in your example above is deterministic (measurable by $\mathcal{F}_0$), with discontinuity w.r.t. topology of $C[0,\infty)$. –  kenneth Aug 2 '10 at 11:44
    
A random variable is usually called deterministic when it is equal to a constant a.s. So <em> deterministic </em> and <em> measurable with respect to ${\cal F}_0$ </em> are very different concepts. –  Byron Schmuland Aug 2 '10 at 14:32
show 2 more comments

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

share|improve this answer
    
Hi, Bridge, Thank you very much for your suggestions. One confuse thing is that, the cdf of $\tau$ depends on starting point $x$, i.e. $\tau$ has different distribution under $P^x$. –  kenneth Aug 1 '10 at 2:36
    
Hi, Bridge, $B_\tau = B(\tau, \omega)$, and I guess your expression is true only if $\tau$ is not correlated to $\omega$, i.e. $P^{x}(d \tau \otimes d \omega) = P^{x}(d \tau) P^{x}(d \omega)$. –  kenneth Aug 1 '10 at 5:22
    
That's right, sorry for that. –  The Bridge Aug 1 '10 at 11:26
    
Thank you for your suggestions. That's helpful. –  kenneth Aug 1 '10 at 13:02
    
Thank you for your suggestions. That's helpful. –  kenneth Aug 1 '10 at 13:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.