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Let $B$ and $W$ be two independent Brownian motions on $(\mathcal{F}_{t})$. Is $\tau$ defined as: $$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}$$ almost surely finite ? Bounded ?

I would say that it can not be finite due to Brownian motion and therefore can't be bounded. But I don't know how to prove it. What do you think guys ?

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The proof in the other answer shows that $\tau$ is not bounded, but we actually do have that $\tau < \infty$ almost surely. To show that, first note that by Levy's characterization of Brownian motion, $\hat B_t := \frac{1}{\sqrt 2} (B_t - W_t)$ is a Brownian motion. Then $\tau = \inf\{ t : B_t \ge W_t + e^{-t} \} = \inf\{ t : \hat B_t \ge \frac{1}{\sqrt 2} e^{-t}\}$, so we have

\begin{align*} P(\tau = \infty) &= P(\hat B_t < \frac{1}{\sqrt{2}} e^{-t} \text{ for all }t \ge 0) \\ &\le P(\hat B_t < \frac{1}{\sqrt{2}} \text{ for all }t \ge 0). \end{align*}

By the recurrence property of Brownian motion, this final probability is $0$.

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  • $\begingroup$ Nice answer. In fact, it has prompted me to ask another question here $\endgroup$ – Jan Stuller Nov 30 '20 at 16:45
  • $\begingroup$ Wow, I doubt there is some brownian property we should use but I I never would thought this is Levy's characterization. $\endgroup$ – Magneto49_3 Nov 30 '20 at 20:05
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Edit 02/12/2020: upon reflection, I believe my original proof was incorrect, albeit it led to the correct result. I present an amended proof below, and leave the original proof in the appendix for reference.

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Supposing that Doob's Optional Stopping Theorem holds, we should have that:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[B_{0}]$$

Now, because of continuity of Brownian motion, and because $e^{0}=1$, we must have that:

$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}=\inf \{ t \ge 0: B_{t} = W_{t} + e^{-t} \}$$

Therefore:

$$B_{\tau}=W_{\tau}+e^{-\tau}$$

So:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[W_{\tau}+e^{-\tau}]=\mathbb{E}[W_\tau] + e^{-\tau} $$

But:

$$\mathbb{E}[B_{0}]=0$$

So for Doob's theorem to hold, we must have:

$$0=\mathbb{E}[W_{\tau}]+e^{-\tau}$$

Or:

$$\mathbb{E}[W_{\tau}]=-e^{-\tau}$$

Given that $W_0=0$, we have $\mathbb{E}[W_t|W_0]=0$ $\forall t \in \mathbb{R}$, so that $\mathbb{E}[W_{\tau}|W_0]_{=0}=-e^{-\tau}$ is not true for any $\tau\in \mathbb{R}$. Therefore we have a contradiction and Doob's optional stopping theorem doesn't hold, meaning that $\mathbb{E}[\tau]$ is not finite.

For completeness: Doob's optional stopping theorem states three conditions, any of which is sufficient for some stochastic process $X_{\tau}$ to be well defined and for the following to be true: $\mathbb{E}[X_{\tau}]=\mathbb{E}[X_0]$.

One of these three conditions is that $\mathbb{E}[\tau]<\infty$.

Conversely, if one can show that $\mathbb{E}[X_{\tau}]\neq\mathbb{E}[X_0]$, then none of the three conditions of the stopping theorem are true, including the one that stipulates $\mathbb{E}[\tau]<\infty$.

PS: most problems do not require proving the stopping theorem from scratch. In most problems, one just needs to apply the theorem (so the trick is to memorize the conditions of the theorem and the statement $\mathbb{E}[X_{\tau}]=\mathbb{E}[X_0]$, and then just using the statement to prove the conditions or using the conditions to prove the statement).

If you also need to prove the Stopping theorem, it's a bit more difficult, but the proof can be found here, for example.

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Appendix with original proof: Supposing that Doob's Optional Stopping Theorem holds, we should have that:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[B_{0}]$$

Now, because of continuity of Brownian motion, and because $e^{0}=1$, we must have that:

$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}=\inf \{ t \ge 0: B_{t} = W_{t} + e^{-t} \}$$

Therefore:

$$B_{\tau}=W_{\tau}+e^{-\tau}$$

So:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[W_{\tau}+e^{-\tau}]=\mathbb{E}[W_\tau] + e^{-\tau} $$

But:

$$\mathbb{E}[B_{0}]=0\neq \mathbb{E}[W_0]+e^{0}=1$$

(the line above is wrong, because the statement doesn't need to be true for Doob's theorem to hold: counterexample would be: $\tau:=inf \left\{t>0: B_t = 1-t\right\}$. With $B_t$ being a standard Brownian motion, clearly $\mathbb{E}[\tau]$ is finite (and in fact equals to 1), but we don't have $\mathbb{E}[B_0]=1$)

So that:

$$\mathbb{E}[B_{\tau}]\neq \mathbb{E}[B_0]$$

And we have a contradiction (again, this conclusion is wrong, because it cannot be reached just based on the statements above)

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  • $\begingroup$ Thank you Jan, I'll take your advices into consideration ! $\endgroup$ – Magneto49_3 Nov 30 '20 at 20:02
  • $\begingroup$ @Magneto49_3: I think I made a logical mistake in my original proof, I amended my answer with a corrected version. $\endgroup$ – Jan Stuller Dec 2 '20 at 8:57

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