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20 — L² Martingales, BCII+, Polya Scheme, Extensions

Setup: L² Martingales

Let

  • $X_0 = 0$
  • ${X_n, \mathcal{F}n}{n\ge 0}$ be a martingale
  • $E(X_n^2) < \infty$

Define increments
\(D_k = X_k - X_{k-1}\) and the predictable increasing process \(A_n = \sum_{k=1}^n E(D_k^2 \mid \mathcal{F}_{k-1}), \qquad A_0 = 0.\)

Then

  • $A_n$ is $\mathcal{F}_{n-1}$-measurable,
  • $A_n \uparrow A_\infty$,
  • $X_n^2 = M_n + A_n$ is the Doob decomposition.

Doob’s $L^2$ inequality: \(E(A_\infty) = E\!\left( \sup_{n\ge0} |X_n|^2 \right) \le 4 E(A_\infty).\)

Convergence of Series Using Martingales

Let \(X_n = \sum_{k=1}^n (1_{B_k} - P_k),\) where $B_k \in \mathcal{F}k$ and $P_k = P(B_k \mid \mathcal{F}{k-1})$.

Then $X_n$ is a martingale with increments
\(D_k = 1_{B_k} - P_k, \qquad E(D_k \mid \mathcal{F}_{k-1}) = 0.\)

Variance: \(E(D_k^2 \mid \mathcal{F}_{k-1}) = P_k(1 - P_k).\)

Thus \(A_n = \sum_{k=1}^n P_k(1-P_k).\)

Extended Borel–Cantelli II (BCII+)

If $B_k \in \mathcal{F}_k$, then
\(\{B_k \text{ i.o.}\} \quad=\quad \left\{ \sum_{k=1}^\infty P_k = \infty \right\} \quad \text{a.s.}\)

More precisely:

If

\(\sum_{k=1}^\infty P_k(1-P_k) < \infty,\) then
\(\sum_{k=1}^\infty (1_{B_k} - P_k) \quad\text{converges a.s.}\)

If

\(A_\infty = \sum_{k=1}^\infty P_k(1-P_k) = \infty,\) then using SLLN for martingales, \(\frac{X_n}{A_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0,\) hence \(\frac{\sum_{k=1}^n 1_{B_k}}{\sum_{k=1}^n P_k} \xrightarrow[n\to\infty]{\text{a.s.}} 1.\)

Where Is This Used? Polya Scheme

Classical Polya Urn

Initial:

  • $r$ red,
  • $g$ green.

Each draw:

  1. Draw one ball.
  2. Observe its color.
  3. Replace it and add $c$ balls of the same color.

Let

  • $G_n =#$ green after $n$ steps
  • $R_n =#$ red after $n$ steps
  • Total: $G_n + R_n = g + r + nc$

The fraction of greens: \(X_n = \frac{G_n}{G_n + R_n}.\)

Conditioning on $\mathcal{F}_{n}$: \(P(B_{n+1}\text{ is green}) = X_n.\)

Thus ${X_n}$ is a martingale.

As $n\to\infty$, \(X_n \xrightarrow{\text{a.s.}} X,\) where \(X \sim \mathrm{Beta}\!\left(\frac{g}{c},\,\frac{r}{c}\right).\)

Another Example: Friedman’s Urn

Same philosophy but with slightly altered reinforcement (add 1 to opposite color).
Still:

  • Define indicators $D_n = 1_{B_n}$.
  • Show $X_n = G_n/(G_n + R_n)$ is a martingale.
  • Use BCII+ to show almost sure convergence.

Doob $L^2$ Maximal Inequality (Alternate Version)

For an $L^2$ martingale ${X_n}$: \(E\!\left( \sup_{n\ge 0} |X_n| \right) \le 3\, E\!\left( \sqrt{A_\infty} \right).\)

Corollary

If $E(\sqrt{A_\infty}) < \infty$, then:

  1. $X_n \to X$ a.s. (MGCT)
  2. $E X_n - X \to 0$ (DCT + UI)

Extension of Wald’s First Equation

Let ${\xi_k}$ be iid with:

  • $E(\xi_k)=0$,
  • $E(\xi_k^2) < \infty$.

Let the stopping time $T$ satisfy:

If $\mathbb{E}(T)<\infty$

then \(E(S_T) = 0, \qquad S_T = \sum_{k=1}^T \xi_k.\)

New version:

Even if $E(T)=\infty$ but
\(E\sqrt{T} < \infty,\) then still \(E(S_T)=0.\)

Reason:

  • $S_{T\wedge n}$ is uniformly integrable because
    \(A_{T\wedge n} = \sigma^2 (T\wedge n),\) so \(E\sqrt{A_\infty} = \sigma\, E\sqrt{T} < \infty.\)

Thus \(E(S_T) = \lim_{n\to\infty} E(S_{T\wedge n}) = 0.\)

Summary

  • Extended BCII shows when $\sum P_k = \infty$ implies $B_k$ infinitely often.
  • Polya urn process uses martingale fraction-of-green to derive Beta-limit.
  • Doob $L^2$ inequality gives strong control of martingale maxima.
  • Wald’s equation extends to cases with infinite expectation of $T$, provided square-root integrability.
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