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$ 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.

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    $\begingroup$ For $C^2$ everything is nice: en.wikipedia.org/wiki/Dynkin%27s_formula $\endgroup$
    – Steve Huntsman
    Commented Jul 31, 2010 at 16:02
  • $\begingroup$ If $f$ is bounded, can't you use the dominated convergence theorem to get continuity all the time? $\endgroup$
    – weakstar
    Commented Jul 31, 2010 at 18:19
  • $\begingroup$ 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 $\tau$. $\endgroup$
    – kenneth
    Commented Aug 1, 2010 at 3:18

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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.

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  • $\begingroup$ It is quite tricky one. It helps my understanding. $\endgroup$
    – kenneth
    Commented Aug 2, 2010 at 5:32
  • $\begingroup$ Hi, there is a little typo it's $f(B_0)$ not $f(B_1)$ I think Regards $\endgroup$
    – The Bridge
    Commented Aug 2, 2010 at 7:16
  • $\begingroup$ 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 $x$. $\endgroup$
    – user6096
    Commented Aug 2, 2010 at 14:29
  • $\begingroup$ That's right I misread your defintion of $\tau$ $\endgroup$
    – The Bridge
    Commented Aug 2, 2010 at 15:26
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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.

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  • $\begingroup$ 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 $\partial D$. $\endgroup$
    – kenneth
    Commented Aug 1, 2010 at 2:38
  • $\begingroup$ (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 $\endgroup$
    – kenneth
    Commented Aug 1, 2010 at 2:39
  • $\begingroup$ 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. $\endgroup$
    – user6096
    Commented Aug 1, 2010 at 20:20
  • $\begingroup$ Byron, Actually, the stopping time in your example above is deterministic (measurable by $\mathcal{F}_0$), with discontinuity w.r.t. topology of $C[0,\infty)$. $\endgroup$
    – kenneth
    Commented Aug 2, 2010 at 11:44
  • $\begingroup$ 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. $\endgroup$
    – user6096
    Commented Aug 2, 2010 at 14:32
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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

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  • $\begingroup$ 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 $P^x$. $\endgroup$
    – kenneth
    Commented Aug 1, 2010 at 2:36
  • $\begingroup$ Hi, Bridge, $B_\tau = B(\tau, \omega)$, and I guess your expression is true only if $\tau$ is not correlated to $\omega$, i.e. $P^{x}(d \tau \otimes d \omega) = P^{x}(d \tau) P^{x}(d \omega)$. $\endgroup$
    – kenneth
    Commented Aug 1, 2010 at 5:22
  • $\begingroup$ That's right, sorry for that. $\endgroup$
    – The Bridge
    Commented Aug 1, 2010 at 11:26

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