Martingale Inequalities and Convergence — Examples

In the natural filtration, $S_{k-1}$ is known and ${\xi }_k$ is independent of the past with mean zero. Thus F1 gives $\mathbb{E}[S_k\mid {\mathcal{F}}_{k-1}]=S_{k-1}+\mathbb{E}{\xi }_k=S_{k-1},$ so $S$ is a square-integrable martingale.

Expanding the square gives $\mathbb{E}{S}_n^2={\sum }_{k=1}^n\mathbb{E}{\xi }_k^2+2{\sum }_{j<k}\mathbb{E}({\xi }_j{\xi }_k)\mathrm{.}$ For $j<k$, independence and centering give $\mathbb{E}({\xi }_j{\xi }_k)=\mathbb{E}{\xi }_j\mathbb{E}{\xi }_k=0$; hence $\mathbb{E}{S}_n^2={\sum }_k{\sigma }_k^2$.

Apply F3's $L^1$ inequality to the nonnegative submartingale $S_k^2$ at level ${\lambda }^2$. The event is exactly $\{{\max }_{k\le n}\mathrm{\mid }{S}_k\mathrm{\mid }\ge \lambda \}$, so the first bound follows from step 2.1. Apply the $p=2$ inequality to $\mathrm{\mid }{S}_k\mathrm{\mid }$ to get $\mathrm{\parallel }{\max }_{k\le n}\mathrm{\mid }{S}_k\mathrm{\mid }{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{S}_n{\mathrm{\parallel }}_2;$ squaring and using step 2.1 gives the second. AC is used exactly through F4.

Since $M_n\ge 0$, $M_n^{+}=M_n$, and the martingale identity gives $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ for every $n$. Thus F1 applies and gives $M_n\to M_{\mathrm{\infty }}$ almost surely with $M_{\mathrm{\infty }}$ finite and integrable.

Fatou gives $\mathbb{E}{M}_{\mathrm{\infty }}\le {\mathrm{lim\hspace{0.17em}inf}}_n\mathbb{E}{M}_n=\mathbb{E}{M}_0\mathrm{.}$ The inequality can be strict, so neither equality nor $L^1$ convergence follows from nonnegativity alone. AC has only the role in F3.

Every positive-length level-$n$ cell is the union of its two level-$(n+1)$ children, while the singleton $\{0\}$ is itself a generator at every level. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. Dyadic half-open intervals form a countable base for the relative topology of $[0,1]$: every open interval is the countable union of dyadic cells whose closures lie inside it, with endpoints handled by the stated convention. Thus F1 gives $\sigma ({\bigcup }_n{\mathcal{F}}_n)=\mathcal{B}$.

F2 gives $M_n\to \mathbb{E}[X\mid \mathcal{B}]$ almost surely and in $L^1$. Since $X$ is $\mathcal{B}$-measurable, F3 identifies the latter conditional expectation with $X$. The singleton endpoint convention changes no $L^1$ or almost-sure statement. AC is used exactly through F4.

Deleting the first generator gives ${\mathcal{G}}_{n+1}\subseteq {\mathcal{G}}_n$, and F1 gives ${\bigcap }_n{\mathcal{G}}_n=\mathcal{T}$. Each $Y_n$ is integrable and ${\mathcal{G}}_n$-measurable by the conditional-expectation interface in F5. The tower identity in F5 gives $\mathbb{E}[Y_n\mid {\mathcal{G}}_{n+1}]=Y_{n+1}$ almost surely, which is the reverse-martingale identity. F2 now applies directly to $X$, proving both modes of convergence.

If the $Z_n$ are independent and $A\in \mathcal{T}$, take $X=1_A$. F3 gives $\mathbb{P}(A)\in \{0,1\}$, so $1_A$ and therefore $\mathbb{E}[1_A\mid \mathcal{T}]$ is almost surely constant. This extra conclusion is asserted only under independence. AC has exactly the inherited role in F4.

F1 makes $(M_n)$ a martingale, and F2 gives ${\sup }_n\mathrm{\parallel }{M}_n{\mathrm{\parallel }}_p\le \mathrm{\parallel }X{\mathrm{\parallel }}_p<\mathrm{\infty }$. F3 therefore supplies $M_{\mathrm{\infty }}\in L^p$ with both asserted modes of convergence. As a pointwise limit of variables measurable for ${\mathcal{F}}_{\mathrm{\infty }}$, it has an ${\mathcal{F}}_{\mathrm{\infty }}$-measurable version.

When ${\mathcal{F}}_{\mathrm{\infty }}=\sigma ({\bigcup }_n{\mathcal{F}}_n)$, F4 gives almost-sure and $L^1$ convergence of the same sequence to $\mathbb{E}[X\mid {\mathcal{F}}_{\mathrm{\infty }}]$. Limits in probability are unique, so it equals the $L^p$ limit $M_{\mathrm{\infty }}$ almost surely. AC is used exactly through F5.

The displayed process is nonnegative and $\mathbb{E}{M}_n=2^n\lambda (A_n)=1$ for $n\ge 1$, while $\mathbb{E}{M}_0=1$.

For $n\ge 1$, the atoms of ${\mathcal{F}}_n$ are $A_n$, the shells $A_j\setminus A_{j+1}$ for $1\le j<n$, and the outside atom $B=[0,1]\setminus A_1=\{0\}\cup (1\mathrm{/}2,1]$. The singleton $\{0\}$ is not a separate atom of this sigma-algebra. On $A_n$, $A_{n+1}$ occupies half the measure and $\frac{1}{\lambda (A_n)}{\int }_{A_n}{M}_{n+1}d\lambda =2^n=M_n{\mathrm{\mid }}_{A_n};$ on every shell and on $B$, both $M_n$ and $M_{n+1}$ vanish. For $n=0$ the same calculation uses total mean one. Thus the event-integral characterization in F2 and the definition in F1 give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$.

The sets $A_n$ decrease to the empty set, so $M_n(x)$ is eventually zero for every $x\in [0,1]$ and $M_n\to 0$ pointwise. But $\mathrm{\parallel }{M}_n-0{\mathrm{\parallel }}_1=1$ for every $n$. Hence the martingale is $L^1$-bounded without $L^1$ convergence. AC is used only as recorded in F4.

By F1, $M_n\to 0$ almost surely while $\mathbb{E}{M}_n=1$ for every $n$. Therefore ${\lim }_n\mathbb{E}{M}_n=1\mathrm{\ne }0=\mathbb{E}[{\lim }_n{M}_n]\mathrm{.}$ This is the required explicit failure.

If $(M_n)$ were uniformly integrable, F2 would force $L^1$ convergence to its almost-sure limit. That would imply $\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }\to 0$, contradicting the computed value one. Thus the failure is exactly outside the closed/UI regime. AC has only the inherited role in F3.

Fix $n$. On $[0,1]$ put $A_k=(0,2^{-k}]$ for $1\le k\le n$, $A_0=[0,1]$, and ${\mathcal{F}}_k=\sigma (A_1,\dots ,A_k)$. For $X=2^n{1}_{A_n}$, direct averaging on $A_k$ gives $M_k:=\mathbb{E}[X\mid {\mathcal{F}}_k]=2^k{1}_{A_k}(0\le k\le n)$ up to the null endpoint. F1 verifies the martingale, and $\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }=2^n{2}^{-n}=1$.

On the shell $A_j\setminus A_{j+1}$ for $0\le j<n$, the maximum is $2^j$ and the shell has measure $2^{-(j+1)}$. On $A_n$ the maximum is $2^n$ and the measure is $2^{-n}$. Therefore $\mathbb{E}{\max }_{k\le n}\mathrm{\mid }{M}_k\mathrm{\mid }={\sum }_{j=0}^{n-1}{2}^j{2}^{-(j+1)}+2^n{2}^{-n}=\frac{n}{2}+1.$

If a horizon-independent $C$ existed, step 1.1, step 2.1 would give $n\mathrm{/}2+1\le C$ for every $n$, impossible. Thus the $p>1$ restriction cannot be extended to this strong $L^1$ form. AC has exactly the inherited role in F3.

Let ${\xi }_k=S_k-S_{k-1}\in \{-1,1\}$ and use the natural filtration. Symmetry gives $\mathbb{E}{\xi }_k=0$, and independence from the past plus F1 yields $\mathbb{E}[S_k\mid {\mathcal{F}}_{k-1}]=S_{k-1}$. Thus $S$ is a martingale and $\mathrm{\mid }{S}_k-S_{k-1}\mathrm{\mid }=1$.

For the stated $n\ge 1$, apply F2 with $c_k=1$. Since ${\sum }_{k=1}^n{c}_k^2=n>0$, it gives exactly $2e^{-t^2\mathrm{/}(2n)}$. At $t=x\sqrt{n}$ the exponent is $-x^2\mathrm{/}2$; the prefactor and lack of lattice correction show this is a robust bound, not the exact binomial tail. AC is used only through F3.