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23 — Conditioning, UI vs DCT, Reverse Martingales

1. Setup

Let ${Z_n}$ be IID.
Since they are independent, conditioning on $\sigma{Z_{n_k}}$ behaves differently from the usual filtration conditioning.

Define \(X_n = \begin{cases} Z_{n_k}, & n \in \{n_k : k \ge 1\}, \\ 0, & n \notin \{n_k : k \ge 1\}. \end{cases}\)

Then:

  • The subsequence ${Z_{n_k}}$ is IID.
  • $\mathbb E(Z_{n}) = \mathbb E(Z_1)$.
  • ${X_n}$ is uniformly integrable.
  • But after conditioning, UI may be lost and DCT may fail.

2. Loss of DCT under conditioning

Your notes (p.1–2) show:

\[P\left(X_n \xrightarrow[n\to\infty]{a.s.} 0\right)=1, \qquad P_Y\left(X_n \xrightarrow[n\to\infty]{a.s.} 0\right)=1.\]

Even though the convergence survives, the dominated convergence theorem does not.

If $Y \ge |W_n|$ and $\mathbb E|Y|<\infty$, then \(E_{\mathcal F}(Y) \;\ge\; E_{\mathcal F}(|W_n|) \;\ge\; | E_{\mathcal F}(W_n) | \quad \text{a.s.}\)

This shows that conditional expectations interact differently with domination.

3. DCT under conditional expectation

Suppose:

  • $Y_n \to Y$ a.s.,
  • $ Y_n \le Z$ for all $n$,
  • $\mathbb E(Z) < \infty$,
  • and $\mathcal F_n \uparrow \mathcal F$.

Then: \(E_{\mathcal F_n}(Y_n) \xrightarrow[n\to\infty]{a.s.} E_{\mathcal F}(Y).\)

Proof sketch (from p.2)

Write $W_n = Y_n - Y$.
Then:

  • $ W_n \to 0$,
  • $ W_n \le 2Z$, and $\mathbb E(2Z) < \infty$.

Therefore, \(|E_{\mathcal F_n}(W_n)| \le E_{\mathcal F_n}(|W_n|) \le \limsup_{n\to\infty} E_{\mathcal F_n}(|W_n|).\)

To finish, use the tail supremum trick:

\[E_{\mathcal F_n}(|W_n|) \le E_{\mathcal F_n}\Big( \sup_{k \ge 0} |W_{N+k}| \Big) \xrightarrow[N\to\infty]{a.s.} 0.\]

Thus the conditional dominated convergence theorem holds in this setting.

4. Reverse (Backwards) Martingales

(Durrett §5.6, p.225)

A reverse martingale is indexed by negative integers:

\[\{\mathcal F_{-n}\}_{n\ge 0}, \qquad \mathcal F_{-(n+1)} \subset \mathcal F_{-n}.\]

A sequence ${X_{-n}}$ is a reverse martingale if: \(E(X_{-n} \mid \mathcal F_{-(n+1)}) = X_{-(n+1)}.\)

Key facts (from p.2–3)

  1. The filtration is decreasing: \(\mathcal F_0 \supset \mathcal F_{-1} \supset \mathcal F_{-2} \supset \cdots\)

  2. Reverse martingale convergence theorem (RMCT): \(X_{-n} \xrightarrow[n\to\infty]{a.s.,\, L^1} X_{-\infty}.\)

  3. Moreover, \(X_{-\infty} = E(X_0 \mid \mathcal F_{-\infty}), \qquad \mathcal F_{-\infty} = \bigcap_{n=1}^\infty \mathcal F_{-n}.\)

Proof sketch (from your notes)

For $A \in \mathcal F_{-n}$:

\[E(X_0; A) = E(X_{-n}; A) \to E(X_{-\infty}; A).\]

Thus $X_{-\infty} = E(X_0 \mid \mathcal F_{-\infty})$.

Reverse filtrations “shrink” and the variables become more constant as conditioning increases.

5. Application: Proving the SLLN

This is where the reverse martingale theorem is used.

  • For IID $X_i$, define
    \(\mathcal F_n = \sigma(X_n, X_{n+1}, \ldots)\) which is a reverse filtration.
  • Then \(\frac{S_n}{n} = E(X_1 \mid \mathcal F_n)\) is a reverse martingale.

RMCT tells us: \(\frac{S_n}{n} \xrightarrow[n\to\infty]{a.s.} E(X_1 \mid \mathcal F_\infty) = E(X_1)\) giving the Strong Law of Large Numbers.

6. Summary of Lecture 24

  • UI can be destroyed by conditioning.
  • DCT can be restored under appropriate conditional expectations.
  • Reverse martingales behave like forward martingales under a reversed filtration.
  • Reverse martingale convergence theorem leads directly to the SLLN.
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