tea.mathoverflow.net - Discussion Feed (Doob's inequality)

George Lowther comments on "Doob's inequality" (20823)

George Lowther — Fri, 14 Dec 2012 14:31:35 -0800

Yes, I agree that this is not really suitable for mathoverflow, and is more likely to get an answer at math.stackexchange. Also, the inequality mentioned is a special case of the following question (and answer) already asked on math.SE http://math.stackexchange.com/q/88371/1321, and the technique is to use time-change.

HJRW comments on "Doob's inequality" (20821)

HJRW — Fri, 14 Dec 2012 13:55:56 -0800

To clarify, I think MemT is asking if this question could be reopened. As s/he has explained that s/he's revising for an exam, I have explained that this is not on-topic for MO and recommended math.stackexchange.com.

MemT comments on "Doob's inequality" (20818)

MemT — Fri, 14 Dec 2012 13:03:31 -0800

Hey. Could anyone tell me if I have a process $I_t=\int_0^t f_tdB_t,$ where $(f_t,t\ge 0)$ is bounded, $|f_t|\leq M$ almost surely for all $t \ge 0$, how can I show that
$$\mathcal{P}\left[\sup_{0\leq t\leq T}|I_t|>\lambda\right]\leq \exp\left(-\frac{\lambda^2}{2M^2T}\right).$$

First I tried by defining $Y_t^{\alpha}=\exp\left(\alpha I_t-\frac{1}{2}\int_{0}^t f^{2}(s)ds\right)$, where $\alpha\in \mathbb{R}$ to get an upper bound. But I need to know how to show that $Y_t^{\alpha}$ is a mgale. Thank