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15 — Extended Borel–Cantelli II, Polya’s Urn, Martingales

Setup from last time

We have a martingale \(X_0 = 0,\qquad \{X_n,\mathcal F_n\}_{n\ge 0} \ \text{MG},\) with bounded increments \(|D_k| = |X_k - X_{k-1}| \le M < \infty \quad\text{a.s.},\ k\ge 1.\)

Then \(\Omega = C \cup D,\qquad C\cap D = \varnothing,\) where \(C = \left\{ \lim_{n\to\infty} X_n = x,\ |x|<\infty \right\},\) \(D = \left\{ \lim_{n\to\infty} X_n = +\infty,\ \lim_{n\to\infty} X_n = -\infty \right\}.\)

Extended Borel–Cantelli II

Let

  • $A_n \in \mathcal F_n,\ n\ge 0,$
  • $\mathcal F_n \subset \mathcal F_{n+1}$.

Claim \(\{ A_n \ \text{i.o.} \} = \left\{ \sum_{n=1}^\infty \mathbf 1_{A_n} = \infty \right\} = \left\{ \sum_{n=1}^\infty P_{\mathcal F_{n-1}}(A_n) = \infty \right\}.\)

Proof idea

Define the martingale \(Z_n = \sum_{k=1}^n\Big( \mathbf 1_{A_k} - P_{\mathcal F_{k-1}}(A_k) \Big),\qquad Z_0=0,\) which is a sum of increasing predictable processes.
Also let \(Y_n = \sum_{k=1}^n \mathbf 1_{A_k},\) which is increasing, hence a submartingale.

Because

  • ${X_n}$ is a MG,
  • $ D_n \le 1$,
  • bounded increments satisfy the submartingale convergence theorem,

on the set $C$ we have: \(\sum_{n=1}^\infty \mathbf 1_{A_n} = \infty \quad\text{iff}\quad \sum_{n=1}^\infty P_{\mathcal F_{n-1}}(A_n) = \infty.\)

On $D$ (the divergent sets), both diverge automatically.

Polya’s Urn Scheme

(Durrett §5.3.2 p.205)

Initial urn contains

  • $g$ green marbles,
  • $r$ red marbles.

At each draw:

  1. Pick a marble uniformly at random.
  2. Replace it and add $c$ more of the same color.

Let

  • $G_n =$ number of greens after $n$ selections,
  • $R_n =$ number of reds,
  • total $G_n + R_n = g + r + nc.$

Define the fraction of greens: \(X_0 = \frac{g}{g+r},\qquad X_n = \frac{G_n}{G_n + R_n},\quad 0\le X_n \le 1.\)

Probability next draw is green: \(P(\text{green at }n+1 \mid \mathcal F_n) = X_n.\)

Claim: ${X_n, \mathcal F_n}$ is a martingale.

Check: \(X_{n+1} = \begin{cases} \dfrac{G_n + c}{g+r+(n+1)c} & \text{with prob } \dfrac{G_n}{g+r+nc}, \dfrac{G_n}{g+r+(n+1)c} & \text{with prob } \dfrac{R_n}{g+r+nc}. \end{cases}\)

Compute: $$ E[X_{n+1}\mid \mathcal F_n] = \frac{G_n}{g+r+nc} \cdot \frac{G_n + c}{g+r+(n+1)c}

  • \frac{R_n}{g+r+nc} \cdot \frac{G_n}{g+r+(n+1)c} = X_n. $$

Martingale convergence theorem

Since $0\le X_n\le 1$, MGCT gives: \(X_n \xrightarrow[n\to\infty]{a.s.} X,\qquad E|X_n - X|\to 0.\)

The limit $X$ is random, not deterministic.

Distribution of the limit

\(X \sim \operatorname{Beta}\left(\frac{g}{c},\, \frac{r}{c}\right), \qquad 0\le x\le 1.\)

Density: \(f_X(x) = C_{g+r,c}\, x^{g/c -1} (1-x)^{r/c -1}.\)

Radon–Nikodym via Martingales (Example)

Let $\mu$ and $\gamma$ be two probability measures on $([0,1],\mathcal B)$.
Let \(\rho = \frac{\mu+\gamma}{2}.\)

Partition $[0,1]$ dyadically: \(I_{n,k} = \left( \frac{k-1}{2^n},\frac{k}{2^n} \right),\qquad k=1,\dots, 2^n.\)

Define \(X_n(t) = \frac{\mu(I_{n,k})}{\rho(I_{n,k})},\qquad t\in I_{n,k}.\)

Then:

  • $\mathcal F_n = \sigma(I_{n,k})$ is an increasing filtration,
  • $0\le X_n \le 2$,
  • ${X_n,\mathcal F_n}$ is a martingale.

Check the martingale property

For each interval $I_{n,k}$: \(\int_{I_{n,k}} X_{n+1}\, d\rho = \mu(I_{n,k}) = \int_{I_{n,k}} X_n\, d\rho.\)

Thus $E[X_{n+1}\mid \mathcal F_n] = X_n$.

By MGCT,
\(X_n \to X \quad \rho\text{-a.s.,}\qquad 0\le X\le 2.\)

Reconstruction of measure

For any Borel set $A$, \(\mu(A) = \int_A X\, d\rho.\)

First show this for $A\in \mathcal F_n$, then extend to
$\bigcup_n \mathcal F_n$, then to all Borel sets via Dynkin’s π–λ theorem.

Summary

  • Extended Borel–Cantelli II:
    $\sum 1{A_n}=\infty$ iff $\sum P{\mathcal F_{n-1}}(A_n)=\infty.$
  • Polya’s urn: the color fraction is a martingale;
    $X_n\to X\sim\mathrm{Beta}(g/c,r/c).$
  • Dyadic martingale construction shows the Radon–Nikodym derivative exists.
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