6
$\begingroup$

I have two possibly related questions. Let $\tau:=\min\{t\geq0:B_t=1\}$, where $B_t$ is a standard Brownian motion.

  1. I am supposed to derive the fact that $\mathbf{E}\tau=\infty$ by applying some properties of stochastic integrals to $\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dB_t$. I have tried this: $\mathbf{E}\tau=\mathbf{E}\int^{\tau}_0dt=\mathbf{E}\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dt=\mathbf{E}\left(\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dB_t\right)^2$ And I'm stuck (not even sure if going the right way).

  2. Is Ito's lemma false with stopping times as upper limits of the integrals? For example, is $\int^{\tau}_0dB_t=B_{\tau}$ false? How can I see that?

EDIT: A discussion on possible proofs of $\mathbf{E}\tau=\infty$ is presented in the answers here. I am just curious about the particular suggestion.

And the stochastic integral with a hitting time as an upper limit is defined as follows $\int^{\tau}_0dB_t:=\int^\infty_0\mathbf{1}_{[0,\tau]}(t)dB_t:=L^2-\lim_{N\to\infty}\int^N_0\mathbf{1}_{[0,\tau]}(t)dB_t$ Then I guess the question 2. would be equivalent to asking if Ito's lemma is true for integrals with infinite upper limit.

  • 0
    *Curious about*... what, exactly?2012-03-25

3 Answers 3

4

Here is a solution to (1). Not sure if it strictly falls under the requirements of the problem, but it doesn't use anything too fancy.

Let $\tau_r$ be the first time Brownian motion hits either $1$ or $-r$ and $f(x)=x^2$. Since the infinitesimal generator of Brownian motion (restricted to smooth functions) is $1/2 \Delta$, we have by Dynkin's formula that

$\mathbf{E}[f(B_{\tau_r})]=\mathbf{E}[B_{\tau_r}^2]=f(0)+\mathbf{E}\left[ \int_0^{\tau_r} 1 ds \right]=\mathbf{E}[\tau_r] .$

Using the Gambler's ruin estimate, we have $\mathbf{E}[B_{\tau_r}^2]=\frac{r^2}{r+1}+\frac{r}{r+1}.$ Letting $r \rightarrow \infty$, we see that $\mathbf{E}[\tau_r] \rightarrow \infty$ as $r \rightarrow \infty$. It follows that $\mathbf{E}[\tau]=\infty$.

  • 0
    @DidierPiau: I would like to accept your answer. A few words on the second question would also be appreciated.2012-03-26
3

To answer your second question first, you should be careful to determine exactly how you define a stochastic integral with a stopping time in the limit. But any reasonable definition should result in $\int_0^\tau dB_t = B_\tau$.

Edit: I take it back. Using your definition, $\int_0^\tau dB_t$ does not exist when $\tau = \inf\{t \ge 0 : B_t = 1\}$. For we have $\int_0^N 1_{[0, \tau]} dB_t = B_{\tau \wedge N}$, and this does not converge in $L^2$ as $N \to \infty$ (if it did, it would also converge in $L^1$, but we have $E[B_{\tau \wedge N}]=0$ while $E[B_\tau] = 1$). Of course, it does converge almost surely.

For your main question, here's an argument possibly in the spirit of your question, though it seems too complicated still.

Let $\tau_k = \inf\{t \ge 0 : B_t = k\}$ be the hitting time of $k$. Note that, $0 = \tau_0 \le \tau_1 \le \tau_2 \le \dots$ by continuity of Brownian motion, and by recurrence $\tau_k < \infty$. Let $\sigma_k = \tau_k - \tau_{k-1}$ be the time needed to get from $k-1$ to $k$. By the strong Markov property and translation invariance, the $\sigma_k$ are iid. Thus we have $E[\tau_k] = E[\sigma_1 + \dots + \sigma_k] = k E[\sigma_1].\quad(1)$

Now for any $k,n$, we have $E \int_0^\infty |1_{[0, \tau_k \wedge n]}(t)|^2\,dt = E[\tau_k \wedge n] \le n < \infty$, so the Itô isometry gives $E[\tau_k \wedge n] = E\left[\left(\int_0^\infty 1_{[0, \tau_k \wedge n]}(t)\,dB_t\right)^2\right] = E[B_{\tau_k \wedge n}^2]\quad(2)$ as in your argument. (Another way to get this identity is to note that $X_t = B_t^2 - t$ is a martingale, hence so is $X_{\tau_k \wedge t}$.) By monotone convergence, $E[\tau_k \wedge n] \uparrow E[\tau_k]$ as $n \to \infty$. And by Fatou's lemma, $\liminf_{n \to \infty} E[B_{\tau_k \wedge n}^2] \ge E[B_{\tau_k}^2] = k^2$. So passing to the limit, we have $E[\tau_k] \ge k^2.\quad(3)$

Combining (1) and (3), we see $E[\sigma_1] \ge k$. Since $k$ was arbitrary, we must have $E[\sigma_1] = \infty$, which completes the proof.

Note the essential observation here is that the strong Markov property suggests that $E[\tau_k]$ scales linearly with $k$, as in (1), while (2) suggests quadratic scaling. The only way out of this paradox is to have $E[\tau_k] = \infty$.

  • 0
    @nokiddn: And the integrals converge to the right thing as well.2012-03-26
2

Hi a direct and analytic proof consists in using the density of $\tau$ and to calclulate $E[\tau]$ directly. $\tau$'s density is given by :

$f(t)_\tau=\frac{1}{\sqrt{2\pi.t^3}}e^{-\frac{1}{2.t}}$ (many books give this e.g. Karlin Taylor, "A First course in Stochastic Processes")

From this you can show your assertion.

Anyway, I think that the solution given by ShawnD is more in the spirit of your problem, and essentially is an application of Doob's Optimal Sampling Theorem for martingales.

Best regards