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 $M_n^{\ast }(\omega ):={\max }_{0\le k\le n}{M}_k(\omega )\forall \omega \in \Omega .$ $(M_n^{\ast }{)}_{n\in \mathbb{N}}$ 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. 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 this expression is clearly in ${\mathcal{F}}_n$, it follows that $\tau$ is a stopping time for the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. For $n\in \mathbb{N}$,

\begin{aligned} \alpha \cdot \mathbb{P}(M_n^* \geq \alpha) &= \alpha \cdot \mathbb{P}(\tau \leq n)\ &= \mathbb{E}\bigl(\alpha \cdot \mathbf{1}{\tau \leq n}\bigr)\ &\leq \mathbb{E}\bigl(M\tau \cdot \mathbf{1}{\tau \leq n}\bigr)\ &= \sum{m=0}^n \mathbb{E}\bigl(M_\tau \cdot \mathbf{1}{\tau=m}\bigr)\ &= \sum{m=0}^n \mathbb{E}\bigl(M_m \cdot \mathbf{1}{\tau=m}\bigr)\ &\leq \sum{m=0}^n \mathbb{E}\bigl(M_n \cdot \mathbf{1}{\tau=m}\bigr)\ &= \mathbb{E}\bigl(M_n \cdot \mathbf{1}{\tau\leq n}\bigr)\ &= \mathbb{E}\bigl(M_n \mathbf{1}_{M_n^* \geq \alpha}\bigr)\ &\leq \mathbb{E}(M_n). \end{aligned}$$

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 $\Vert X{\Vert }_p\le \frac{p}{p-1}\Vert Y{\Vert }_p.$

Proof. We will first show with the additional assumption $X\in L^p$. It is a basic property of the $L^p$-norm that it can be expressed as

$${}^{0}9e0be\Vert X{\Vert }_p^p=p{\int }_0^{\infty }{\lambda }^{p-1}\mathbb{P}(X\ge \lambda )d\lambda .$$

On applying $\alpha \cdot \mathbb{P}(X\ge \alpha )\le \mathbb{E}(Y\cdot 1_{X\ge \alpha })$ and Fubini,

$$\begin{array}{rl}\Vert X{\Vert }_p^p& \le p{\int }_0^{\infty }{\lambda }^{p-2}\mathbb{E}(Y\cdot 1_{\{X\ge \lambda \}})d\lambda \\ & =p{\int }_0^{\infty }{\lambda }^{p-2}{\int }_{\{X\ge \lambda \}}Yd\mathbb{P}d\lambda \\ & ={\int }_{\Omega }Y\left({\int }_0^{X}p{\lambda }^{p-2}d\lambda \right)d\mathbb{P}\\ & =\frac{p}{p-1}{\int }_{\Omega }Y{X}^{p-1}d\mathbb{P}.\end{array}$$

On applying Hölder’s inequality with conjugate exponents $p$ and $p^{\prime }=\frac{p}{p-1},$ we obtain

$$\begin{array}{rl}\Vert X{\Vert }_p^p& \le \frac{p}{p-1}{\left(\mathbb{E}(Y^p)\right)}^{1/p}{\left(\mathbb{E}\left(X^{(p-1)p^{\prime }}\right)\right)}^{1/p^{\prime }}\\ & =\frac{p}{p-1}{\left(\mathbb{E}(Y^p)\right)}^{1/p}{\left(\mathbb{E}(X^p)\right)}^{(p-1)/p}.\end{array}$$

This leads to desired inequality. The lemma will now be shown for general non-negative $X$. For $n\in \mathbb{N}$, define $X_n\in L^p$ to be the truncation $X_n(\omega ):=\min \{X(\omega ),n\},\forall \omega \in \Omega .$ It is easy to see that $\alpha \cdot \mathbb{P}(X\ge \alpha )\le \mathbb{E}(Y\cdot 1_{X\ge \alpha })$ must hold for $X_n$ for each $n\in \mathbb{N}$. It then follows from the previous argument that $\Vert X{\Vert }_p\le \frac{p}{p-1}\Vert Y{\Vert }_p.$ must hold for $X_n$ and $Y$. As $X_n$ converges monotonically pointwise to $X$ as $n\to \infty$, it follows from the monotone convergence theorem that

$$\Vert X{\Vert }_p=\underset{n\to \infty }{\lim }\Vert X_n{\Vert }_p\le \frac{p}{p-1}\Vert Y{\Vert }_p.$$

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 $L^2(\Omega ,\Sigma ,\mathbb{P})$.

Proof. Define, for each $n\in \mathbb{N}$, $d_n:=M_{n+1}-M_n.$ For $j>k$,

$$\begin{array}{rl}\mathbb{E}(d_j{d}_k)& =\mathbb{E}(\mathbb{E}(d_j{d}_k\mid {\mathcal{F}}_{k+1}))\\ & =\mathbb{E}(d_k\mathbb{E}(M_{j+1}-M_j\mid {\mathcal{F}}_{k+1}))\\ & =\mathbb{E}(d_k(M_{k+1}-M_{k+1}))\\ & =0.\end{array}$$

Similarly for $j<k$. It must be shown, for almost every $\omega \in \Omega$, that

$$\underset{n\to \infty }{\mathrm{liminf}}{M}_n(\omega )=\underset{n\to \infty }{\mathrm{limsup}}{M}_n(\omega ).$$

To this end, define for each $a,b\in \mathbb{Q}$ with $a<b$ the set

$$A_{a,b}:=\left\{\omega \in \Omega :\underset{n\to \infty }{\mathrm{liminf}}{M}_n(\omega )\le a<b\le \underset{n\to \infty }{\mathrm{limsup}}{M}_n(\omega )\right\}.$$

Evidently, for any $\omega \in \Omega$, ${\mathrm{liminf}}_{n\to \infty }{M}_n(\omega )\ne {\mathrm{limsup}}_{n\to \infty }{M}_n(\omega )$ if and only if there exist $a,b\in \mathbb{Q}$ with $a<b$ for which $\omega \in A_{a,b}$. Therefore, if it can be shown that the set ${\bigcup }_{a,b\in \mathbb{Q}}{A}_{a,b}$ has measure zero, then it follows that $M_n(\omega )$ will have a limit for almost every $\omega \in \Omega$. This will be achieved by demonstrating that each $A_{a,b}$ has measure zero. Let $m\in \mathbb{N}$ and define $\epsilon :=\left(\frac{b-a}{2}\right).$ $A_{a,b}\subset \{{\sup }_{k\ge m}|M_k-M_m{|}^2\ge {ϵ}^2\},$ Then combine Markov inequality and Doob’s $L^2$ maximal inequality,

$$\begin{array}{rl}\mathbb{P}(A_{a,b})& \le \mathbb{P}\left(\underset{k\ge m}{\sup }|M_k-M_m{|}^2\ge {\epsilon }^2\right)\\ & \le \frac{1}{{\epsilon }^2}\mathbb{E}\left(\underset{k\ge m}{\sup }|M_k-M_m{|}^2\right)\\ & =\frac{1}{{\epsilon }^2}\mathbb{E}\left(\underset{j\in \mathbb{N}}{\sup }|M_{j+m}-M_m{|}^2\right)\\ & \le \frac{4}{{\epsilon }^2}\underset{j\in \mathbb{N}}{\sup }\mathbb{E}\left(|M_{j+m}-M_m{|}^2\right),\end{array}$$

where we used the monotone convergence theorem and the fact that $(|M_{j+m}-M_m|{)}_{j\in \mathbb{N}}$ is a submartingale to obtain the last estimate. On applying $\mathbb{E}(d_j{d}_k)=0$ to eliminate the cross terms,

$$\begin{array}{rl}\mathbb{P}(A_{a,b})& \le \frac{4}{{\epsilon }^2}\underset{j\in \mathbb{N}}{\sup }\mathbb{E}\left({\left|\sum _{l=m}^{m+j-1}{d}_l\right|}^2\right)\\ & =\frac{4}{{\epsilon }^2}\underset{j\in \mathbb{N}}{\sup }\sum _{l=m}^{m+j-1}\mathbb{E}\left(|d_l{|}^2\right)\\ & =\frac{4}{{\epsilon }^2}\sum _{l=m}^{\infty }\mathbb{E}\left(|d_l{|}^2\right).\end{array}$$

Now, for each $n\in \mathbb{N}$, it follows from $\mathbb{E}(d_j{d}_k)=0$ that

$${\left(\sum _{l=0}^n\mathbb{E}\left(|d_l{|}^2\right)\right)}^{1/2}=\mathbb{E}{\left({\left|\sum _{l=0}^n{d}_l\right|}^2\right)}^{1/2}\le 2\underset{l\in \mathbb{N}}{\sup }\Vert M_l{\Vert }_2<\infty .$$

This demonstrates that ${\sum }_{l=0}^{\infty }\mathbb{E}\left(|d_l{|}^2\right)<\infty ,$ and therefore

$$\mathbb{P}(A_{a,b})\le \frac{4}{{\epsilon }^2}\sum _{l=m}^{\infty }\mathbb{E}\left(|d_l{|}^2\right)\to 0$$

as $m\to \infty$. That is, $\mathbb{P}(A_{a,b})=0,$ then ${\mathrm{liminf}}_{n\to \infty }{M}_n={\mathrm{limsup}}_{n\to \infty }{M}_n$, with ${\sup }_{n\in \mathbb{N}}\mathbb{E}(M_n^2)<\infty$, there exists some random variable $M_{\infty }:\Omega \to \mathbb{R}$ for which $M_m\to M_{\infty }$ almost surely as $m\to \infty$. Finally, it will be shown that $M_m\to M_{\infty }$ in $L^2$. For $m\in \mathbb{N}$, by the dominated convergence theorem,

$$\begin{array}{rl}\mathbb{E}\left(|M_{\infty }-M_m{|}^2\right)& =\mathbb{E}\left({\left|\underset{n\to \infty }{\lim }{M}_n-M_m\right|}^2\right)\\ & =\mathbb{E}\left({\left|\underset{n\to \infty }{\lim }\sum _{j=m}^n(M_{j+1}-M_j)\right|}^2\right)\\ & =\mathbb{E}\left({\left|\underset{n\to \infty }{\lim }\sum _{l=m}^n{d}_l\right|}^2\right)\\ & \le \sum _{l=m}^{\infty }\mathbb{E}\left(|d_l{|}^2\right),\end{array}$$

which tends to $0$ as $m\to \infty$.