I quote Jacod-Protter (2004):
Doob's First Martingale Inequality Let $M=\left(M_n\right)_{n\ge0}$ be a martingale or a positive submartingale. Set $M^*_n=\sup_{j\le n}|M_j|$. Then
$$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|\right\}}{\alpha}\tag{1}$$
Does $(1)$ imply that for all $p\ge1$:
$$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|^{\color{red}{p}}\right\}}{\alpha^{\color{red}{p}}}\tag{2}$$?
If so, why? How could one show that for all $p\ge1$: $$(1)\implies (2)\hspace{1cm}\text{ for all }p\ge1$$?
Best Answer
The inequality (2) you state $$\mathbb{P}\left(M_n^*\ge \alpha^p\right)\le\frac{\mathbb{E}\left\{|M_n|^{\color{red}{p}}\right\}}{\alpha^{\color{red}{p}}}\tag{2}$$ is wrong for $\alpha\in (0,1)$ and $p>1$: Consider $M_n=\alpha^p$ for all $n$. Then the LHS of (2) is 1 and the RHS is $\alpha^p<1$. The correct form of (2) is $$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|^{\color{red}{p}}\right\}}{\alpha^{\color{red}{p}}}\tag{3}$$ and this follows from (1) since for $p>1$ the sequence $M_n^p$ is a submartingale by convexity of $x \to x^p$ and Jensen's inequality. Applying (1) to this sequence yields (3), since $$\mathbb{P}\left(M_n^*\ge \alpha\right) =\mathbb{P}\left((M_n^p)^*\ge \alpha^p\right) . $$