22 — Convergence of Conditional Expectations
Topic: Conditional Expectation, Increasing Filtrations, UI Martingales, Kolmogorov 0–1, DCT for Conditional Expectations
1. Review from Monday
(1) Conditional expectations form a uniformly integrable family
Let $(\Omega, \mathcal{F}_0, P)$ with $E|X| < \infty$.
Then
\(\{ E_{\mathcal{F}}(X) : \mathcal{F} \subseteq \mathcal{F}_0 \}\)
is uniformly integrable (UI).
(2) UI martingales converge in $L^1$
If ${X_n, \mathcal{F}n}{n\ge1}$ is a UI martingale, then \(X_n \xrightarrow{\text{a.s.}} X, \qquad X_n \xrightarrow{L^1} X\) and equivalently, \(E|X_n - X| \to 0.\)
(3) Conditional expectations along filtrations
Given $X_n = E_{\mathcal{F}_n}(X)$, $n\ge1$.
2. Convergence of Conditional Expectations Along an Increasing Filtration
Theorem
Let $(\Omega, \mathcal{F}_0, P)$ with $E|X| < \infty$.
Suppose
\(\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \dots \subseteq \mathcal{F}_0,
\qquad \mathcal{F} = \sigma\left( \bigcup_{n\ge1} \mathcal{F}_n \right).\)
Then
\(E_{\mathcal{F}_n}(X) \xrightarrow[\text{a.s.}]{L^1} E_{\mathcal{F}}(X).\)
Proof Sketch
- $X_n := E_{\mathcal{F}_n}(X)$ is a martingale and UI
(by the earlier results). - By UI martingale convergence: \(X_n \to X_\infty \quad \text{a.s. and in } L^1.\)
-
Need to show $X_\infty = E_{\mathcal{F}}(X)$.
Check equality of integrals: \(E(X_n; A) = E(X; A), \qquad \forall A \in \mathcal{F}_n.\) Passing to limit: \(E(X_\infty; A) = E(X; A), \qquad \forall A \in \bigcup_n \mathcal{F}_n.\)
- Use Dynkin’s π–λ theorem to extend from the algebra
generated by $\cup_n \mathcal{F}_n$ to all of $\mathcal{F}$.
Conclude: \(X_\infty = E_{\mathcal{F}}(X).\)
3. Application (a): Conditional probabilities stabilize
Let $\mathcal{F} = \sigma{\mathcal{F}_n : n\ge1}$.
Then for any event $A\in\mathcal{F}$:
\(P_{\mathcal{F}_n}(A) \xrightarrow[\text{a.s.}]{L^1} \mathbf{1}_A.\)
Interpretation:
As the filtration gains more information, the conditional probability
of an event in the limit $\mathcal{F}$ collapses to either 0 or 1.
4. Application (b): New proof of Kolmogorov’s 0–1 Law
If $A$ is a tail event, then \(P(A) \in \{0,1\}.\)
Sketch
- Tail σ–field $\mathcal{T} = \bigcap_{n} \sigma(X_{n+1}, X_{n+2},\dots)$.
- For $A\in\mathcal{T}$, $P_{\mathcal{F}_n}(A)\to 1_A$ a.s.
- But $A$ is independent of $\mathcal{F}n$, hence
$P{\mathcal{F}_n}(A) = P(A)$. - Thus $P(A) = 1_A$ almost surely; hence $P(A)=0$ or $1$.
5. Dominated Convergence for Conditional Expectations
Let $Y_n \to Y$ a.s., and $|Y_n|\le X$ with $E|X|<\infty$.
Then
\(E_{\mathcal{G}}(Y_n) \to E_{\mathcal{G}}(Y) \quad \text{a.s. and in } L^1.\)
Key steps (from the handwritten derivation)
- Apply bounded convergence to
$E\big( |Y_n - Y| \,\big|\,\mathcal{G} \big)$. - Use that conditional expectation preserves inequalities.
- Extend to dominated case via absolute values and truncation.
6. Example (from last page)
Let $Y_n = Y \cdot \mathbf{1}_{{|Y| \le n}}$.
Then $Y_n \to Y$ a.s. and $|Y_n|\le |Y|$.
Thus
\(E_{\mathcal{G}}(Y_n) \to E_{\mathcal{G}}(Y).\)
Also shown:
If $\sum_k P(|Y_k|>1)=\infty$, pathological behaviors can occur
(example from the notes).
Summary
Lecture 23 develops:
- Conditional expectations along increasing filtrations.
- Martingale convergence using uniform integrability.
- The limit being $E_{\mathcal{F}}(X)$, proved with Dynkin’s π–λ theorem.
- Applications:
- Conditional probabilities converge to indicator functions.
- Kolmogorov’s 0–1 law (new proof).
- Dominated convergence theorem for conditional expectations.
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