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An L1-bounded martingale need not converge in L1

Statement

Assume AC. On $([0, 1], B, λ)$ put $A_{n} = (0, 2^{- n}]$ for $n \geq 1$ , let $F_{0} = {\emptyset, [0, 1]}$ and $F_{n} = σ (A_{1}, \dots, A_{n})$ , and define $M_{0} = 1$ and $M_{n} = 2^{n} 1_{A_{n}}$ for $n \geq 1$ . Then $M$ is a nonnegative martingale with $E M_{n} = 1$ for every $n$ , hence $\sup_{n} E ∣ M_{n} ∣ = 1$ , but $M_{n} \to 0$ almost surely and not in $L^{1}$ .

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Martingale submartingale and supermartingale defines a martingale through conditional expectations.

[F2]

Conditional expectation given a sigma algebra supplies the event-integral characterization of conditional expectation.

[F3]

Expectation of a nonnegative or integrable random variable evaluates the displayed simple variables.

[F4]

The Axiom of Choice is assumed because the conditional-expectation and martingale interfaces used here require it.

Counterexample

1.1

The displayed process is nonnegative and $E M_{n} = 2^{n} λ (A_{n}) = 1$ for $n \geq 1$ , while $E M_{0} = 1$ .

2.1

For $n \geq 1$ , the atoms of $F_{n}$ are $A_{n}$ , the shells $A_{j} ∖ A_{j + 1}$ for $1 \leq j < n$ , and the outside atom $B = [0, 1] ∖ A_{1} = {0} \cup (1 / 2, 1]$ . The singleton ${0}$ is not a separate atom of this sigma-algebra. On $A_{n}$ , $A_{n + 1}$ occupies half the measure and $\frac{1}{λ (A_{n})} \int_{A_{n}} M_{n + 1} d λ = 2^{n} = M_{n} ∣_{A_{n}};$ on every shell and on $B$ , both $M_{n}$ and $M_{n + 1}$ vanish. For $n = 0$ the same calculation uses total mean one. Thus the event-integral characterization in F2 and the definition in F1 give $E [M_{n + 1} ∣ F_{n}] = M_{n}$ .

F1F2step 1.1

3.1

The sets $A_{n}$ decrease to the empty set, so $M_{n} (x)$ is eventually zero for every $x \in [0, 1]$ and $M_{n} \to 0$ pointwise. But $∥ M_{n} - 0 ∥_{1} = 1$ for every $n$ . Hence the martingale is $L^{1}$ -bounded without $L^{1}$ convergence. AC is used only as recorded in F4.

F3F4step 1.1step 2.1∎

Depends on

Used by

Almost-sure martingale convergence need not preserve expectation Counterexample

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, martingale-convergence warnings in §§2.5 and 2.8 (standard reference, not scraped)