19 — L² Martingales, Doob Decomposition, L² Inequality, and Kronecker Lemma
L² Martingales
Let
- ${X_n, \mathcal{F}n}{n\ge 0}$ be a martingale,
- $X_0 = 0$,
- $E(X_n^2) < \infty$ for all $n\ge 1$.
A submartingale is convex if $x \mapsto x^2$ is convex.
We apply Doob’s decomposition to $X_n^2$.
Doob Decomposition
Because $x \mapsto x^2$ is convex,
\(E(X_n^2 \mid \mathcal{F}_{n-1}) \ge X_{n-1}^2,\)
so ${X_n^2}$ is a submartingale.
Hence there exist unique predictable $A_n$ and martingale $M_n$ such that
\(X_n^2 = M_n + A_n,\qquad A_0 = 0,\qquad A_n \uparrow,\quad A_n \in \mathcal{F}_{n-1}.\)
The predictable quadratic variation is \(A_n = \sum_{k=1}^n E\big[(X_k - X_{k-1})^2 \mid \mathcal{F}_{k-1}\big].\)
Taking expectations:
\(E(A_n) = E(X_n^2).\)
Since $A_n \uparrow$, define
\(A_\infty = \lim_{n\to\infty} A_n \quad\text{a.s.}\)
By MCT,
\(E(A_\infty) = \sup_n E(X_n^2).\)
Durrett Probability 4e - Theorem 5.2.10 Doob’s Decomposition.
Any submartingale $X_n$, $n\ge 0$, can be written in a unique way as $X_n=M_n+A_n$, where $M_n$ is a martingale and $A_n$ is a predictable increasing sequence with $A_0=0$.
Doob’s $L^2$ Inequality
For martingales, \(E\left( \sup_{1\le k\le n} X_k^2 \right) \le 4\, E(X_n^2) = 4\,E(A_n).\)
This is Doob’s $L^2$ maximal inequality (special case of the general $L^p$ inequality for $p>1$).
Durrett Probablity 4e - Theorem 5.4.3. $L^p$ Maximum Inequality
If $X_n$ is a submartingale then or $1<p<\infty$, \(\mathbb{E}(\bar{X}^p_n)\le (\frac{p}{1-p})^p \mathbb{E}(X^+_n)^p\) Consequently, if $Y_n$ is a martingale and $Y_n^*=\max_{0\le m\le n}\vert Y_n\vert^p$, \(\mathbb{E}(Y^*_n)\le (\frac{p}{1-p})^p \mathbb{E}[\vert Y_n\vert^p]\)
Convergence Theorem (No independence assumed)
Let \(X_n = \sum_{k=1}^n D_k,\) where $D_k = X_k - X_{k-1}$.
Then $X_n$ converges a.s. to a finite r.v. on the event ${A_\infty < \infty}$.
Proof Sketch
Define the stopping time \(T_a = \inf\{n : A_{n+1} > a\}.\)
Then \(E\left(\sup_{m\le n} X_{m\wedge T_a}^2\right) \le 4E(A_{n\wedge T_a}) \le 4a.\)
Thus ${X_{n\wedge T_a}}$ is bounded in $L^2$, hence Cauchy, hence convergent a.s.
Letting $a\to\infty$ yields convergence on ${A_\infty<\infty}$.
Kronecker Lemma (Martingale Difference Version)
Let $f(x)\ge 1$ be increasing with
\(\int_1^\infty \frac{dx}{f(x)^2} < \infty.\)
Let ${D_k}$ be martingale differences and define \(Y_n = \frac{1}{f(A_n)}\sum_{k=1}^n D_k.\)
Then on ${A_\infty = \infty}$, \(Y_n \to 0 \quad\text{a.s.}\)
Durrett Probability 4e - Theorem 2.5.5. Kronecker’s Lemma.
If $a_n\uparrow\infty$ and $\sum_{n=1}^\infty x_n/a_n$ converges then \(a^{-1}_n\sum_{m=1}^n x_m\to 0.\)
Sketch
Expand using the martingale transform idea: \(E\big[Y_{n+1} - Y_n \mid \mathcal{F}_n\big] = 0.\)
Compute quadratic variation and use MCT plus the integrability condition \(\sum \frac{E(D_k^2 \mid \mathcal{F}_{k-1})}{f(A_k)^2} < \infty.\)
Example (Application)
Suppose
- $E(D_k)=0$,
- $\sum_k E(D_k^2)=\infty$,
- $\sum_k E(D_k^2)/f(A_k)^2 < \infty$.
Then
\(\frac{\sum_{k=1}^n D_k}{f(A_n)} \to 0 \quad\text{a.s.}\)
Back to Borel–Cantelli II (martingale version)
If $B_n \in \mathcal{F}n$, let $P_n = P(B_n \mid \mathcal{F}{n-1})$.
Then
\(\sum_n P_n = \infty
\quad\Longleftrightarrow\quad
B_n \text{ i.o.}\)
This mirrors earlier results from Lecture 16–17 but in the martingale framework.
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