17 — Optional Stopping, Doob Inequalities, Maximal Inequalities
Setup
We work on a probability space
\((\Omega, \mathcal{F}_0, P), \qquad \mathcal{F} \subset \mathcal{F}_0.\)
Let $X \ge 0$ with $E(X) < \infty$.
Define a finite measure on $(\Omega,\mathcal{F})$ by
\(M(A) = E(X;A), \qquad A \in \mathcal{F}.\)
Absolute Continuity
If $P(A)=0$, then $M(A)=0$.
So $M \ll P$.
By Radon–Nikodym,
\(\frac{dM}{dP} = Y \in \mathcal{F}, \qquad
M(A) = E(Y;A) = \int_A Y\, dP, \quad A\in\mathcal{F}.\)
Hence \(E_{\mathcal{F}}(X) = Y.\)
If $X$ is not positive, write $X = X^+ - X^-$.
Then:
\(E_{\mathcal{F}}(X) = E_{\mathcal{F}}(X^+) - E_{\mathcal{F}}(X^-).\)
Section 4: Optional Stopping, $L^1$ Convergence
(Durrett §5.4, p. 212)
Let ${X_n, \mathcal{F}n}{n\ge0}$ be a submartingale.
Let $T$ be a stopping time with $T \le M < \infty$.
Theorem (Optional Stopping, Bounded Stopping Times)
\(\boxed{ E(X_0) \le E(X_T) \le E(X_M) } \tag{1}\)
Durrett Probability 4e - Theorem 5.4.1.
If $X_n$ is a submartingale and $N$ is a stopping time with $P(N\le k)=1$ then \(\mathbb{E} X_0 \le \mathbb{E} X_N \le \mathbb{E} X_k\)
Proof Idea
Use a “gambling system”: predictable sequences $H_k$.
A predictable sequence means $H_k \in \mathcal{F}_{k-1}$.
Let increments be $D_k = X_k - X_{k-1}$.
Define the stochastic integral \((H\cdot X)_n = \sum_{k=1}^n H_k D_k.\)
If $X$ is positive, then $(H\cdot X)_n$ is a submartingale.
Part (I): Show $\;E(X_T) \ge E(X_0)$
Choose
\(H_k = \mathbf{1}_{\{k \le T\}}.\)
Then \((H\cdot X)_n = X_{T\wedge n} - X_0.\)
At $n=M$: \((H\cdot X)_M = X_T - X_0.\)
Submartingale implies
\(E[X_T - X_0] \ge 0.\)
Part (II): Show $\;E(X_T) \le E(X_M)$
Set \(H_k = \mathbf{1}_{\{T < k\}}.\)
Then \((H\cdot X)_n = X_n - X_{T\wedge n}.\)
At $n=M$: \((H\cdot X)_M = X_M - X_T \ge 0.\)
Thus \(E(X_M) \ge E(X_T).\)
Corollary: Conditional Version
With the same setup, \(\boxed{ E(X_M \mid \mathcal{F}_T) \ge X_T \quad \text{a.s.} }\)
Proof:
Let $A\in\mathcal{F}_T$. Define a new stopping time
\(\tilde{T}=
\begin{cases}
T & \text{on } A, \\
M & \text{on } A^c.
\end{cases}\)
Since $\tilde{T}\le M$, apply the theorem:
\(E(X_M) \ge E(X_{\tilde T}).\)
Then restrict to $A$, obtaining:
\(E(X_M;A) \ge E(X_T;A).\)
Doob’s Inequality (Maximal Inequality)
“Kolmogorov maximality” is a special case.
Let ${X_n,\mathcal{F}_n}$ be a submartingale with $X_n \ge 0$.
Let $\lambda>0$.
Define
\(A = \{\max_{0\le m\le n} X_m \ge \lambda\}.\)
Then Doob’s inequality states \(\boxed{ \lambda P(A) \le E(X_n \mathbf{1}_A) \le E(X_n^+) }\)
Proof Sketch:
Define the stopping time
\(T=\inf\{m\le n : X_m \ge \lambda\}.\)
Then $X_T \mathbf{1}_A \ge \lambda \mathbf{1}_A$.
Using optional stopping,
\(E(X_n;A) \ge E(X_T;A) \ge \lambda P(A).\)
Kolmogorov’s Inequality
Let ${Z_k}$ be independent with
\(E(Z_k)=0, \qquad E(Z_k^2)<\infty.\)
Let
\(S_n=\sum_{k=1}^n Z_k.\)
Then \(P\!\left(\max_{1\le m\le n} |S_m| \ge x\right) \le \frac{E(S_n^2)}{x^2}.\)
Notes:
- ${S_n,\mathcal{F}_n}$ is a martingale.
- ${S_n^2}$ is a submartingale.
- Apply Doob’s inequality to $S_n^2$.
$L^p$ Maximal Inequality ($p>1$)
Let $p>1$, $q=\frac{p}{p-1}$.
Let ${X_n}$ be an $L^p$ submartingale.
Then \(\boxed{ \left\|\max_{1\le m\le n} X_m^+\right\|_p \;\le\; q \,\|X_n\|_p. }\)
Sketch:
Use the tail integral formula:
\(E\!\left[\max_{m\le n} X_m^+\right]^p
= \int_0^\infty p\lambda^{p-1} P(\max X_m > \lambda)\, d\lambda.\)
Apply Doob’s inequality to the probability inside.
Why $p=1$ Fails
For $p=1$, we only get a weaker inequality:
If ${X_n}$ is a submartingale, \(E \big[\max_{m\le n} X_m^+ \big] \le \frac{e}{e-1} \left( 1 + E\!\left[X_n^+ (\log^+ X_n^+)^{-1}\right] \right).\)
This ties into the $L^1$ version of the Dominated Convergence Theorem.
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