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14 — Doob’s Decomposition and Uniqueness

Doob’s Decomposition Theorem (Summary)

Let ${X_n, \mathcal{F}n}{n\ge 0}$ be a submartingale.
Then there exists a predictable process ${A_n}_{n\ge 0}$ with:

  • $A_0 = 0$
  • $A_n \le A_{n+1}$ a.s.
  • ${M_n}_{n\ge 0}$ defined by $M_n = X_n - A_n$ is a martingale

such that
\(X_n = A_n + M_n,\qquad n\ge 0,\) and this decomposition is unique.

Example: Decomposing a Submartingale Using Increments

Define
\(D_k = X_k - X_{k-1},\qquad k\ge 1.\)

Then
\(X_n = X_0 + \sum_{k=1}^n D_k.\)

Submartingale condition implies
\(E(D_k \mid \mathcal{F}_{k-1}) \ge 0.\)

Define the martingale part: \(M_n = X_0 + \sum_{k=1}^n (D_k - E(D_k \mid \mathcal{F}_{k-1})).\)

Define the predictable increasing part: \(A_n = \sum_{k=1}^n E(D_k \mid \mathcal{F}_{k-1}).\)

Then
\(X_n = M_n + A_n.\)

Proof of Uniqueness

Suppose there are two decompositions: \(X_n = M_n + A_n = \tilde{M}_n + \tilde{A}_n.\)

Given $A_0 = \tilde{A}_0 = 0$, we get
\(M_0 = \tilde{M}_0.\)

Step 1: Compare consecutive differences

We want to show
\(M_n - M_{n-1} = \tilde{M}_n - \tilde{M}_{n-1}\quad \text{a.s.}\)

Then telescoping implies $M_n = \tilde{M}_n$ and hence $A_n = \tilde{A}_n$.

Define the difference \(D_n := M_n - \tilde{M}_n.\)

  • ${D_n, \mathcal{F}_n}$ is a martingale (page 2 of the PDF).
  • Also, $A_n - \tilde{A}n$ is predictable (in $\mathcal{F}{n-1}$).

Since
$$X_n - X_{n-1} = (M_n - \tilde{M}n) - (M{n-1} - \tilde{M}_{n-1})

  • (A_n - \tilde{A}n) - (A{n-1} - \tilde{A}_{n-1}),$$

and the increment of the predictable processes matches the increment of the martingale differences, we obtain:

\[M_n - M_{n-1} = \tilde{M}_n - \tilde{M}_{n-1}\quad\text{a.s.}\]

Thus
\(M_n = \tilde{M}_n,\qquad A_n = \tilde{A}_n,\quad n\ge 0.\)

Uniqueness proven.

Example (Durrett §5.3 p. 204)

Let ${D_k}_{k\ge 1}$ be iid with:

  • $E(D_k) = 0$
  • $ D_k \le M$ almost surely

Define
\(X_n = \sum_{k=1}^n D_k.\)

Then the sample space splits into:

\[\Omega = C \cup D,\qquad C\cap D = \varnothing\ \text{a.s.}\]

where

  • $C = {\lim_{n\to\infty} X_n \text{ exists and is finite}}$
  • $D = {\lim_{n\to\infty} X_n = +\infty \text{ or } -\infty}$

Because increments are bounded, page 2 shows the four possible long-term behaviors:

  1. $X_n \equiv 0$ (not happening unless trivial)
  2. $X_n \to +\infty$
  3. $X_n \to -\infty$
  4. $\liminf X_n < 0 < \limsup X_n$ (oscillation)

But with iid mean-zero bounded increments, we get (page 2):

\[P(D)=1,\qquad P(C)=0.\]

Example: Hitting Below a Level (Page 3)

Assume

  • negative part $D_k^- \le M$,
  • positive part $D_k^+ \le M$.

Define the hitting time
\(T_k = \inf\{n: X_n \le -k\}.\)

Then:

  • ${X_{n\wedge T_k}}$ is a martingale
  • $X_{n\wedge T_k} \ge -k - M$

By the Martingale Convergence Theorem (MGCT),
\(X_{n\wedge T_k} \to X_{\infty,k}\quad\text{a.s.}\)

Taking $k\to\infty$ gives the decomposition of the limit behavior into:

  • $C = {\lim X_n \text{ exists finitely}}$
  • $D = {\lim X_n = -\infty}$

Final statement on page 3:

In this chapter (Doob decomposition): \(A_n \in \mathcal{F}_{n-1},\quad \{A_n \text{ i.o.}\} = \left\{\sum_{n=1}^\infty P_{\!n-1}(A_n) = \infty\right\}.\)

This is the conditional version of Borel–Cantelli II.

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