# Stopped process of Brownian motion

I am baffled about the following problem:

Let $(B_t)$ be a standard Brownian motion. Let $$\tau:= \inf\{ t \geq 0 :B_t = x \} \wedge \inf\{ t \geq 0 :B_t = -y \}$$ be a stopping time, where $x,y >0$. I am eager to know why the stopped process $(B_{t \wedge \tau})_{t \geq 0}$ is U.I..

Moreover, we know that $(\tilde{B}_t := B^2_t -t)_{t \geq 0}$ is a martingale. But the book also claims that $(\tilde{B}_{t \wedge \tau \wedge n})_{t \geq 0}$ is U.I., for any $n \in \mathbb{N}$. Why is that?

• What does U.I. mean? Uniformly integrable? – Math1000 Dec 22 '14 at 2:04
• That is what I think, Math1000. – ncmathsadist Dec 22 '14 at 2:22
• Richard: 1. Which parts of the answer were escaping you when you asked this question? 2. Since the answer does not solve the part about $\tilde B$, I guess that you managed to solve it yourself. Is that correct? – Did Dec 22 '14 at 6:10
• yeah, I was just thinking about proving uniform integrability by some other methods. I overlooked the possibility of proving boundedness. – Richard Dec 23 '14 at 0:04

We know that $\mathbb{E}[\tau] < \infty$ and $B_{t \wedge \tau}$ is a martingale. It is also easy to see that $-y \leq B_{t \wedge \tau} \leq x$. Any bounded martingale is uniform integrable. Hence, $B_{t \wedge \tau}$ is UI martingale.