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Let $(M_s)$ a martingale s.t. $\mathbb E[M_t^2]<\infty $. Regarding the proof of $$\mathbb E[\sup_{0\leq s\leq t}M_s]\leq 4\mathbb E[M_t^2],$$

the person who answer the question in this post mention that $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y},$$ and I don't understand why this should be true. The best I know is $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|)}{y},$$ which is Doob inequality. But I don't get how we can get to $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y}.$$ Any idea ?

Bruce
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1 Answers1

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I guess that your martingale is at least right continuous, otherwise your statement is not true. Under this hypothesis, let $\tau=\inf\{s>0\mid |M_s|\geq y\}\wedge t$. The process $(|M_t|)_{t\geq 0}$ is a submartingale, i.e. by Optional Stopping theorem $$\mathbb E|M_\tau|\leq \mathbb E|M_t|\tag{1}.$$ Notice that \begin{align*} |M_\tau|&\geq y\\ &=y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s|\geq y\}}+y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| <y\}}\\ &\geq y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s|\geq y\}}+|M_t|\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| \}}, \end{align*}

and thus, $$\mathbb E|M_\tau|\geq y\mathbb P\left\{\sup_{0\leq s\leq t}|M_s|\geq y\right\}+\mathbb E\Big[|M_t|\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| <y\}}\Big]\tag{2}.$$

Combine $(1)$ and $(2)$ gives the wished result.

Surb
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