Martingale Convergence Theorem proof doubt

convergence-divergencemartingalesprobability theoryproof-explanation

Martingale convergence theorem:
Let $(X_n)_{n\in\mathbb{N}}$ be a submartingale with $\sup\{E(X^+_n):n\geqslant 0\}<\infty$. Then there exists an $\mathscr{F}_\infty$-measurable random variable $X_\infty$ with $E(|X_\infty|)<\infty$ and $X_n\to X_\infty$ as $n\to\infty$ almost surely.

Proof: Let $a<b$. Since $E[(X_n-a)^+]\leqslant |a|+E(X_n^+)$,applying the following inequality $E(U_n^{a,b})\leqslant\frac{E[(X_n-a)^+]-E[(X_0-a)^+]}{b-a}$ (where $U_n^{a,b}$ is the number of upcrossings over $[a,b]$ until time $n$) we get:

$E(U_n^{a,b})\leqslant\frac{|a|+E[X_n^+]}{b-a}$.

Manifestly, the monotone limit $U^{a,b}:=\lim_{n\to\infty}U_n^{a,b}$ exists. By assumption, we have $E(U^{a,b})=\lim_{n\to\infty}E(U_n^{a,b})<\infty$. In particular, $P(U^{a,b}<\infty)=1$. Define the $\mathscr{F}_\infty$-measurable events

$C^{a,b}=\{\liminf_{n\to\infty}X_n<a\}\cap\{\limsup_{n\to\infty}X_n>b\}\subset\{U^{a,b}=\infty\} $

and for $a<b$

$C=\bigcup_\limits{{a,b\in\mathbb{Q}}}C^{a,b}.$

Then $P(C^{a,b})=0$ and thus also $P(C)=0$. However, by construction, $(X_n)_{n\in\mathbb{N}}$ is convergent on $C^c$. Hence there exists the almost sure limit $X_\infty=\lim_{n\to\infty X_n}$.

By Fatou's lemma:

$E(X^+_\infty)\leqslant\sup\{E(X_n^+):n\geqslant 0\}<\infty$

On the other hand (since X is submartingale), again by Fatou's lemma,

$$E(X^-_\infty)\leqslant\liminf_{n\to\infty}E(X^-_n)=\liminf_{n\to\infty}E(X^+_n)-E(X_n)\leqslant \sup\{E(X_n^+):n\geqslant 0\}-E(X_0)<\infty.$$

$\blacksquare$

I am not understanding the author claim $P(C^{a,b})=0$. If I have for example a symmetric discrete random walk so that $X_n=1$ or $X_n=0$ for all $n\in\mathbb{N}$. $\liminf X_n=0$ and $\liminf X_n=1$ if I chose $a=\frac{1}{3}$ and $b=\frac{1}{2}$. Both limits are obviously higher and lower than a and b respectively.

Questions:

1) How does the author conclude $P(C^{a,b})=0$?

2) Does the author creates $C^{a,b}$ to prove $X_n$ converges almost surely?

Best Answer

  1. If $\omega\in C^{a,b}$, then the sequence $\left(X_n(\omega)\right)_{n\geqslant 1}$ goes under $a$ an infinite amount of times, and over $b$ an infinite amount of times. Therefore, $\omega\in \{U^{a,b}=+\infty\}$ and the latter set has probability zero hence all the sets $C^{a,b}$, $a<b$, $a,b\in\mathbb Q$ have probability zero.

  2. Indeed, $\omega\in C$ if and only if $\left(X_n(\omega)\right)_{n\geqslant 1}$ is not convergent so proving that $C$ has probability zero gives the almost sure convergence.

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