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Lp-bounded martingale convergence

Statement

Assume AC. Let $p > 1$ . If $M$ is a martingale and $C : = \sup_{n} E ∣ M_{n} ∣^{p} < \infty$ , then some $M_{\infty} \in L^{p}$ satisfies $M_{n} \to M_{\infty}$ almost surely and in $L^{p}$ . Moreover $M_{n} = E [M_{\infty} ∣ F_{n}] a.s., {∥ \sup_{n} ∣ M_{n} ∣ ∥}_{p} \leq \frac{p}{p - 1} \sup_{n} ∥ M_{n} ∥_{p} .$

Facts & Assumptions

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

[F1]

Doob submartingale convergence theorem gives almost-sure convergence from a uniform positive-part bound.

[F2]

Doob Lp maximal inequality gives the finite-horizon maximal estimate.

[F3]

Monotone convergence for the integral and Dominated convergence pass respectively to the infinite maximum and to the $L^{p}$ limit.

[F4]

Conditional lp contraction and Multistep martingale characterization identify the terminal conditional expectations.

[F5]

The Axiom of Choice is inherited from the martingale and conditional-expectation interfaces.

Proof

1.1F1

Since the underlying measure is a probability measure, Hölder gives $\sup_{n} E ∣ M_{n} ∣ \leq C^{1 / p}$ . In particular $\sup_{n} E (M_{n})^{+} \leq C^{1 / p}$ . The martingale $M$ is also a submartingale, so F1 applies directly to $M$ and gives $M_{n} \to M_{\infty}$ almost surely for a finite integrable $M_{\infty}$ .

1.2F2F3

For each $N$ , F2 gives ${∥ \max_{k \leq N} ∣ M_{k} ∣ ∥}_{p} \leq q \sup_{n} ∥ M_{n} ∥_{p}, q = \frac{p}{p - 1} .$ The maxima increase to $M^{*} = \sup_{n} ∣ M_{n} ∣$ , so F3 yields the displayed infinite-horizon bound and $M^{*} \in L^{p}$ . In particular $∣ M_{\infty} ∣ \leq M^{*}$ almost surely, hence $M_{\infty} \in L^{p}$ .

2.1F3step 1.1step 1.2

We have $∣ M_{n} - M_{\infty} ∣^{p} \leq (2 M^{*})^{p}$ and pointwise convergence to zero. Dominated convergence gives $∥ M_{n} - M_{\infty} ∥_{p} \to 0$ .

3.1F4F5∎

Fix $n$ and take $m \geq n$ . F4 gives $M_{n} = E [M_{m} ∣ F_{n}]$ . Conditional $L^{p}$ contraction and step 2.1 imply $∥ E [M_{m} - M_{\infty} ∣ F_{n}] ∥_{p} \leq ∥ M_{m} - M_{\infty} ∥_{p} ⟶ 0.$ The left conditional expectations therefore converge to zero while $M_{n}$ is fixed, proving $M_{n} = E [M_{\infty} ∣ F_{n}]$ almost surely. AC has only the inherited role in F5.

Depends on

Used by

An Lp-bounded martingale with an Lp terminal value Example

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, Theorem 2.25 and §2.9, pp. 16, 23–24 (standard reference, not scraped)