Martingale and Convergence

Maximal Sequence

Let $(M_n{)}_{n\in \mathbb{N}}$ be a non-negative submartingale with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. For each $n\in \mathbb{N}$, define $M_n^{\ast }:\Omega \to \mathbb{R}$ through $M_n^{\ast }(\omega ):={\max }_{0\le k\le n}{M}_k(\omega )\forall \omega \in \Omega .\text{(2.5)}$ $(M_n^{\ast }{)}_{n\in \mathbb{N}}$ is called the maximal sequence associated with $(M_n{)}_{n\in \mathbb{N}}$.

Doob's Maximal Inequality

Let $(M_n{)}_{n\in \mathbb{N}}$ be a non-negative submartingale and $(M_n^{\ast }{)}_{n\in \mathbb{N}}$ its corresponding maximal sequence. Then for each $n\in \mathbb{N}$ and $\alpha >0$, $\alpha \cdot \mathbb{P}(M_n^{\ast }\ge \alpha )\le \mathbb{E}(M_n{1}_{M_n^{\ast }\ge \alpha })\le \mathbb{E}(M_n).$

Proof First note that for any $n,m\in \mathbb{N}$ with $n\ge m$ and $A\in {\mathcal{F}}_m$, $\mathbb{E}(1_A{M}_m)\le \mathbb{E}(1_A{M}_n)$. This follows from $\mathbb{E}(1_A{M}_m)={\int }_A{M}_{m}d\mathbb{P}\le {\int }_A\mathbb{E}(M_n\mid {\mathcal{F}}_m)d\mathbb{P}=\mathbb{E}(\mathbb{E}(1_A{M}_n\mid {\mathcal{F}}_m))=\mathbb{E}(1_A{M}_n),$ where we applied the submartingale property to obtain the estimate in the second line. \noindent Fix $\alpha >0$ and define $\tau :\Omega \to \mathbb{N}\cup \{\infty \}$ through $\tau (\omega ):=\min \{m\in \mathbb{N}:M_m(\omega )\ge \alpha \}\forall \omega \in \Omega ,$ where we set $\tau (\omega )=\infty$ if the above set is empty. For any $n\in \mathbb{N}$, the set $\{\tau \le n\}$ can be expressed as $\{\tau \le n\}={\bigcup }_{j=0}^n\{M_j\ge \alpha \}.$ As the right-hand side of the above expression is clearly in ${\mathcal{F}}_n$, it follows that $\tau$ is a stopping time for the filtration

Corollary

Let $(M_n{)}_{n\in \mathbb{N}}$ be a non-negative submartingale. Then $\alpha \cdot \mathbb{P}\left({\sup }_{k\in \mathbb{N}}{M}_k\ge \alpha \right)\le {\sup }_{k\in \mathbb{N}}\mathbb{E}(M_k).$

Lemma

Let $X,Y$ be two non-negative random variables with $Y\in L^p(\Omega ,\Sigma ,\mathbb{P})$ for some $p\in (1,\infty )$. Suppose that the estimate $\alpha \cdot \mathbb{P}(X\ge \alpha )\le \mathbb{E}(Y\cdot 1_{X\ge \alpha })$ holds for any $\alpha >0$. Then it must be true that $X\in L^p$ and $\Vert X{\Vert }_p\le \frac{p}{p-1}\Vert Y{\Vert }_p.$

Doob's $L^p$ Inequalities

For $p\in (1,\infty )$ and $n\in \mathbb{N}$, $\Vert M_n^{\ast }{\Vert }_p\le \frac{p}{p-1}\cdot \Vert M_n{\Vert }_p.$

Proof Combining above lemma and Doob’s Maximal Inequality, we’ll get this $L^p$ inequality.

Doob's Martingale Convergence Theorem.

Let $(M_n{)}_{n\in \mathbb{N}}$ be a martingale with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. Suppose that ${\sup }_{n\in \mathbb{N}}\mathbb{E}(M_n^2)<\infty$. Then there must exist some $M_{\infty }\in L^2(\Omega ,\Sigma ,\mathbb{P})$ such that $M_n$ converges to $M_{\infty }$ as $n\to \infty$ both almost surely and in $L^2(\Omega ,\Sigma ,\mathbb{P})$.