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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?

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  • $\begingroup$ What does U.I. mean? Uniformly integrable? $\endgroup$ – Math1000 Dec 22 '14 at 2:04
  • $\begingroup$ That is what I think, Math1000. $\endgroup$ – ncmathsadist Dec 22 '14 at 2:22
  • $\begingroup$ 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? $\endgroup$ – Did Dec 22 '14 at 6:10
  • $\begingroup$ yeah, I was just thinking about proving uniform integrability by some other methods. I overlooked the possibility of proving boundedness. $\endgroup$ – Richard Dec 23 '14 at 0:04
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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.

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