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Let $B$ and $W$ be independent Brownian motions, let $\tau$ be a stopping time adapted to $\mathcal{F}^{W}$, do we always have $E[\int_{0}^{\tau}B_{s}dW_{s}]=0$?

I know that $\int_{0}^{t}B_{s}dW_{s}$ is a square integrable martingale, and I know that if the integrand were nonrandom, then the answer would be no. But I get stuck in the original question. I tried to use the tower property, but I don't know how to play with this Ito stochastic integral.

Thanks for your help!

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What do you know? What have you tried? –  Sasha Aug 17 '12 at 21:02
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What happened to your accept rate? –  Did Aug 17 '12 at 21:05
    
Nicely done! Now to @Sasha's question... –  Did Aug 18 '12 at 8:16
    
@did Could you please comment on this question? –  user7762 Aug 28 '12 at 0:37
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2 Answers

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By the integration by parts formula $\int B_sdW_s = B_tW_t-\int W_s dB_s$ . Get rid of the first term by agreeing that we'll only look at stopping times where $W_{\tau} = 0$. Conditioning on all of $\mathcal F_W$ the r.h.s is $N(0, \int^{\tau}W^2(s) ds)$. Choose $\tau$ so that unconditioning yields a non $\mathbb L_1$ function. More specifically, let $X = e^{e^{W_1}}$ and choose a stopping time $\tau$ satisfying 3 conditions: $\tau > 1, W_{\tau} = 0, \int^{\tau}W^2(s) ds) \ge X$. The second condition can be satisfied because $W_t = 0$ for some arbitrarily large $t$, and the third can be satisfied because the integral increases to $\infty$. By the first remarks the integral we're looking at can be represented as $\sqrt{\int^{\tau}W^2(s) ds)} Z$ where $Z \sim N(0,1)$ and is independent of $\int^{\tau}W^2(s) ds)$. Since you have the distribution fairly explicitly, you can calculate $\mathbb E(\vert W_{\tau} \vert)$ by first conditioning on $\mathcal F_W$ and find that it is the same as $\sqrt{\frac 2 {\pi}} \mathbb E \sqrt{\int^{\tau}W^2(s) ds)} \ge \sqrt{\frac 2 {\pi}} \mathbb E \vert X \vert = \infty$ so the r.v. is not $\mathbb L_1$.

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Could you explain what do you mean by saying "choose $\tau$ so that unconditioning yields a non L1 function"? –  user7762 Aug 28 '12 at 0:42
    
tried to do that. all r.v.'s of your form are symmetric so they are not going to have non-zero expectation, non $\mathbb L _1$ is the onl way you can fail to have expectation $0$. –  mike Aug 28 '12 at 11:26
    
Could you explain why $\mathbb{E}[\int_{0}^{\tau}W_{s}dB_{s}|\mathcal{F}^{W}]$ has the same distribution as $\sqrt{\int_{0}^{\tau}W^{2}(s)ds}Z$ ? I don't know how to apply independence lemma to an ito integral. Thanks! –  user7762 Aug 28 '12 at 23:09
    
When you condition on all of $\mathcal F_X$ the integral term is constant because it is $\mathcal F_W$ measurable, and by independence $B_t$ is the same brownian motion it was originally. So the condition distribution can be gotten from the distribution of terms like $\int f(t) dB_t$ where $f(t)$ is a deterministic fctn of $t$, and that is gaussian with mean $0$ and variance $\int f^2(t) dt$. –  mike Aug 29 '12 at 11:39
    
I know heuristically this is how the independence lemma works. But I'm having trouble giving a rigorous proof. For example, what is an explicit formula for the r.v. $Z$? –  user7762 Aug 29 '12 at 13:30
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Edit: Oops, I missed the $\mathcal{F}^W$ in the original problem. So this doesn't answer the question, but I'll leave it anyway.

If you allow arbitrary stopping times (or even just those adapted to the filtration $\mathcal{F}^{B,W}$), the answer is no.

Let $X_t = \int_0^t B_s\,dW_s$. We note that the quadratic variation of $\newcommand{\X}{\langle X\rangle}X_t$ is $\X_t = \int_0^t B_s^2\,ds$. In particular, we have $\X_t \uparrow \infty$ almost surely.

For $a<0<b$, let $\tau_{a,b} = \inf\{t : X_t \notin (a,b)\}$. Also let $\tau_a = \inf\{t : X_t = a\}$. Since $X_t^2 - \X_t$ is a martingale, for any $a,b$ we have $$E[\X_{t \wedge \tau_{a,b}}] = E[X_{t \wedge \tau_{a,b}}^2] \le (|a| \vee |b|)^2.$$ But since $\X_t \uparrow \infty$, we must have $\tau_{a,b} < \infty$ almost surely.

Now $X_{t \wedge \tau_{a,b}}$ is a bounded martingale, so by optional stopping we have $$0 = E[X_{\tau_{a,b}}] = a P(\tau_a < \tau_b) + b (1-P(\tau_a < \tau_b)).$$ Thus $P(\tau_a < \tau_b) = b/(b-a)$. Letting $b \to \infty$, we have $P(\tau_a < \infty) = 1$. But then $E[X_{\tau_a}] = a \ne 0$.

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