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?
Best Answer
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.