# another version of doob inequality

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!

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Thans for the help of Pascal Maillard. I find the following proof:

Let $T$ the hitting time of $2\lambda$, then

$$E[M_t1_{\{M_t\geq\lambda\}}]\geq E[M_t1_{\{M_t\geq\lambda, T\leq t\}}]=E[M_t1_{\{T\leq t\}}]-E[M_t1_{\{M_t\geq\lambda, T\leq t\}}]$$

For the first term by Doob inequality we have

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

For the second term we have

$$E[M_t1_{\{M_t\geq\lambda, T\leq t\}}]\leq E[\lambda 1_{\{M_t\geq\lambda, T\leq t\}}]\leq \lambda P(S_t>2\lambda)$$

which gives

$$E[M_t1_{\{M_t\geq\lambda\}}]\geq E[M_t1_{\{T\leq t\}}]-E[M_t1_{\{M_t\geq\lambda, T\leq t\}}]\geq 2\lambda P(S_t>2\lambda)-\lambda P(S_t>2\lambda)$$

That is just to be proved!

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