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Hi everyone, please consider the following problem:

Let $(M_t)_{t\geq 0}$ be a continuous and positive submartingale and $S_t=\sup_{0\leq s\leq t}M_s$. Please prove that for any $\lambda>0$ we have

$$\lambda P(S_t>2\lambda)\leq E[M_t1_{{M_t>\lambda}}]$$

This inequality makes me remember the Doob inequality

$$\lambda P(S_t>\lambda)\leq E[M_t1_{{S_t>\lambda}}]$$

So it is enough to show that

$$E[M_t1_{{S_t>\lambda}}]\leq 2E[M_t1_{{M_t>\lambda}}]$$

But I have no idea to deal about the term $1_{{M_t>\lambda}}$, even by introducing a stopping time $T_{\lambda}$. Could someone help me prove this inequality or give some idea? Many thanks!

edited Oct 24 at 21:22

Will Jagy
12.5k●1●26●45

asked Oct 24 at 20:36

Higgs88
43●4

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There's a typo in your title: Dood -> Doob. – HW Oct 24 at 20:38

How do you know that this inequality holds? – Alekk Oct 24 at 21:59

Pascal Maillard · Accepted Answer · 2012-10-24 22:52:06Z UTC

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Nice little exercise, it should go to math.stackexchange.com, though. I'll give you a hint: Let $T$ be the hitting time of $2\lambda$. Then, $E[M_t 1_{M_t\ge \lambda}] \ge E[M_t 1_{M_t\ge \lambda,\ T\le t}] \ge E[M_t1_{T\le t}]- E[M_t1_{M_t\le \lambda}1_{T\le t}].$

answered Oct 24 at 22:52

Pascal Maillard
331●1●7

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Thanks so much for the hint of Pascal Maillard. Here is a proof: By Doob inequality: $$E[M_t1_{\{T\leq t\}}]\geq 2\lambda P(S_t>2\lambda)$$ and $$E[M_t1_{\{M_t\leq\lambda, T\leq t\}}]\leq E[\lambda 1_{\{M_t\leq\lambda, T\leq t\}}]\leq\lambda P(S_t>\lambda)$$ which gives $$E[M_t1_{\{T\leq t\}}]-E[M_t1_{\{M_t\leq\lambda, T\leq t\}}]\geq 2\lambda P(S_t>\lambda)-\lambda P(S_t>\lambda)=\lambda P(S_t>\lambda)$$ That is what we need! – Higgs88 Oct 25 at 13:32

modification of Doob inequality

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