Martingale Convergence¶
Martingales "want to converge." The conditional expectation property forces limiting behavior under mild conditions, distinguishing martingales from arbitrary sequences of integrable random variables. This section establishes when — and in what sense — a martingale admits a limit, and isolates a gap that will be filled in Uniform Integrability.
Almost Sure Convergence¶
The path to a.s. convergence runs through upcrossings. For \(a < b\), an upcrossing of \([a,b]\) by \(X_0, X_1, \ldots\) is a pair \(s < t\) with \(X_s \le a\) and \(X_t \ge b\). If a sequence oscillates infinitely, it upcrosses some rational interval infinitely often. Bounding the expected number of upcrossings therefore forces convergence.
Doob's upcrossing inequality
If \((X_n)_{n=0}^N\) is a submartingale and \(U_N^{[a,b]}\) is the number of upcrossings of \([a,b]\), then
Idea of proof. Define a predictable "betting strategy" \(H_n = 1\) when the process is inside an unfinished upcrossing, \(0\) otherwise. Each completed upcrossing contributes at least \(b-a\) to the gain \(G_N = \sum H_n (X_n - X_{n-1})\), and \(\mathbb{E}[G_N] \ge 0\) for submartingales. The inequality follows.
Martingale convergence theorem (Doob)
Let \((M_n)\) be a submartingale with \(\sup_n \mathbb{E}|M_n| < \infty\). Then there exists \(M_\infty \in L^1\) such that
Proof. For each rational \(a < b\), the upcrossing inequality gives \(\mathbb{E}[U_\infty^{[a,b]}] \le \sup_n \mathbb{E}[(M_n - a)^+]/(b-a) < \infty\), so \(U_\infty^{[a,b]} < \infty\) a.s. The event "\((M_n)\) oscillates" is the countable union over rational \(a < b\) of \(\{U_\infty^{[a,b]} = \infty\}\), hence null. So \(\liminf M_n = \limsup M_n =: M_\infty\) a.s. Finiteness follows from Markov's inequality applied to \(\sup_n |M_n|\). \(\square\)
The L¹ Gap¶
A.s. convergence does not imply \(L^1\) convergence. Mass can escape to rare high values:
Canonical counterexample
Let \(\xi_i\) be i.i.d. with \(\mathbb{P}(\xi_i = 2) = \mathbb{P}(\xi_i = 0) = 1/2\) and \(M_n = \prod_{i=1}^n \xi_i\). Then \((M_n)\) is a martingale, \(M_n \to 0\) a.s. (some \(\xi_i\) eventually vanishes), yet \(\mathbb{E}[M_n] = 1\) for all \(n\). The mass is carried by the event \(\{\xi_1 = \cdots = \xi_n = 2\}\) of probability \(2^{-n}\), on which \(M_n = 2^n\).
Recall (see § Uniform Integrability): \(\{X_\alpha\}\) is UI iff \(\lim_{K\to\infty} \sup_\alpha \mathbb{E}[|X_\alpha|\,\mathbf{1}_{\{|X_\alpha|>K\}}] = 0\); sufficient conditions include \(L^p\)-boundedness (\(p>1\)), being a family of conditional expectations of a single \(L^1\) variable, or domination by an integrable variable.
UI Martingales and Closure¶
UI convergence theorem
For a martingale \((M_n)\), the following are equivalent:
- \((M_n)\) is uniformly integrable.
- \(M_n \to M_\infty\) in \(L^1\) for some \(M_\infty \in L^1\).
- \(M_n \to M_\infty\) a.s. and in \(L^1\).
- \((M_n)\) is closed: \(M_n = \mathbb{E}[X \mid \mathcal{F}_n]\) for some \(X \in L^1\).
Key implications. \((4) \Rightarrow (1)\) because conditional expectations of a single \(L^1\) variable are UI. \((1) \Rightarrow (3)\) combines Doob's a.s. theorem with Vitali's theorem (UI + a.s. \(\Rightarrow L^1\)). \((3) \Rightarrow (4)\) follows because for \(A \in \mathcal{F}_n\), \(\mathbb{E}[M_\infty \mathbf{1}_A] = \lim_m \mathbb{E}[M_m \mathbf{1}_A] = \mathbb{E}[M_n \mathbf{1}_A]\).
L^p case, p > 1
If \(\sup_n \mathbb{E}|M_n|^p < \infty\) for some \(p > 1\), then \(M_n \to M_\infty\) a.s. and in \(L^p\). Reason: \(L^p\)-boundedness with \(p>1\) implies UI (hence a.s. and \(L^1\) convergence), and Doob's maximal inequality gives the \(L^p\) domination \(\mathbb{E}[\sup_n |M_n|^p] \le (p/(p-1))^p \sup_n \mathbb{E}|M_n|^p\), so \(L^p\) convergence follows by dominated convergence.
Backward Martingales¶
A backward martingale \((M_n, \mathcal{F}_n)_{n\ge 1}\) satisfies \(\mathcal{F}_n \supseteq \mathcal{F}_{n+1}\) and \(\mathbb{E}[M_n \mid \mathcal{F}_{n+1}] = M_{n+1}\): as \(n\) grows, information shrinks.
Backward convergence
Every \(L^1\)-bounded backward martingale converges a.s. and in \(L^1\) to \(M_\infty = \mathbb{E}[M_1 \mid \mathcal{F}_\infty]\), where \(\mathcal{F}_\infty = \bigcap_n \mathcal{F}_n\).
UI is automatic here: the sequence consists of conditional expectations of the single variable \(M_1\).
Strong law of large numbers
Let \(X_i\) be i.i.d. with \(\mathbb{E}|X_1| < \infty\) and mean \(\mu\). Set \(S_n = X_1 + \cdots + X_n\), \(\bar X_n = S_n/n\), and \(\mathcal{F}_n = \sigma(S_n, S_{n+1}, \ldots)\) (decreasing). By exchangeability, \(\mathbb{E}[X_k \mid \mathcal{F}_{n+1}]\) is the same for all \(k \le n+1\), so
making \((\bar X_n)\) a backward martingale. Convergence gives \(\bar X_n \to \mathbb{E}[X_1 \mid \mathcal{F}_\infty]\), and Kolmogorov's 0–1 law makes \(\mathcal{F}_\infty\) trivial, yielding \(\bar X_n \to \mu\) a.s. and in \(L^1\).
Continuous Time¶
For right-continuous martingales \((M_t)_{t \ge 0}\) with \(\sup_t \mathbb{E}|M_t| < \infty\), the same conclusions hold: \(M_\infty := \lim_{t\to\infty} M_t\) exists a.s. and is finite. If \((M_t)\) is UI, convergence is also in \(L^1\) and \(M_t = \mathbb{E}[M_\infty \mid \mathcal{F}_t]\). Doob's regularization theorem provides a càdlàg modification whenever the filtration satisfies the usual conditions.
Three behaviors to remember
- Lévy martingale \(M_t = \mathbb{E}[X \mid \mathcal{F}_t]\): UI, so \(M_t \to X\) a.s. and in \(L^1\).
- Exponential martingale \(Z_t = \exp(\theta W_t - \theta^2 t/2)\), \(\theta \ne 0\): \(\mathbb{E}[Z_t] = 1\) but \(Z_t \to 0\) a.s. Not UI.
- Simple random walk \(S_n\): \(\mathbb{E}[S_n^2] = n \to \infty\), not \(L^1\)-bounded, does not converge a.s.
Optional Sampling at Infinity¶
UI bridges optional sampling and convergence: for a UI martingale \((M_n)\) and any stopping time \(\tau\) (possibly infinite),
This extends the bounded optional sampling theorem — already established for bounded stopping times — to the unbounded case, using \(L^1\) convergence to pass to the limit \(\tau \wedge k \to \tau\).
The Convergence Hierarchy¶
| Hypothesis | a.s. | \(L^1\) | \(L^p\), \(p>1\) |
|---|---|---|---|
| \(L^1\)-bounded | yes | no (need UI) | — |
| UI | yes | yes | — |
| \(L^p\)-bounded, \(p>1\) | yes | yes | yes |
| Closed: \(M_n = \mathbb{E}[X\mid\mathcal{F}_n]\) | yes | yes | iff \(X \in L^p\) |
A.s. convergence comes from \(L^1\)-boundedness alone (upcrossings). \(L^1\) convergence requires UI, which is equivalent to closure. \(L^p\)-boundedness with \(p>1\) is strictly stronger and yields \(L^p\) convergence. Backward martingales always close automatically.
Exercises¶
Exercise 1. Let \((M_n)\) be a submartingale with \(\sup_n \mathbb{E}[M_n^+] < \infty\). Prove \(M_n\) converges a.s.
Solution to Exercise 1
For each rational \(a < b\), Doob's upcrossing inequality gives
so \(U_\infty^{[a,b]} < \infty\) a.s. The union over rational pairs is still a null set, so \(\liminf M_n = \limsup M_n\) a.s. For finiteness, note \(\mathbb{E}[M_n^-] = \mathbb{E}[M_n^+] - \mathbb{E}[M_n] \le \mathbb{E}[M_n^+] - \mathbb{E}[M_0]\) (submartingale means \(\mathbb{E}[M_n] \ge \mathbb{E}[M_0]\)), so \(\sup_n \mathbb{E}|M_n| < \infty\) and Markov's inequality forces the limit to be finite. \(\square\)
Exercise 2. Show directly that \(Z_t = e^{W_t - t/2}\) is not uniformly integrable.
Solution to Exercise 2
\(Z_t\) is a martingale, so \(\mathbb{E}[Z_t] = 1\). By the law of the iterated logarithm, \(W_t/t \to 0\) a.s., so \(\log Z_t = W_t - t/2 \to -\infty\) a.s., giving \(Z_t \to 0\) a.s. If \((Z_t)\) were UI, Vitali would force \(\mathbb{E}[Z_t] \to \mathbb{E}[0] = 0\), contradicting \(\mathbb{E}[Z_t] = 1\). \(\square\)
Exercise 3. Let \(X_i\) be i.i.d. with mean \(\mu\) and \(\mathbb{E}|X_1| < \infty\), \(\bar X_n = n^{-1}\sum_{k=1}^n X_k\), and \(\mathcal{F}_n = \sigma(S_n, S_{n+1}, \ldots)\). Show \((\bar X_n, \mathcal{F}_n)\) is a backward martingale and deduce the SLLN.
Solution to Exercise 3
The filtration is decreasing since functions of \((S_{n+1}, S_{n+2}, \ldots)\) are functions of \((S_n, S_{n+1}, \ldots)\). Given \(\mathcal{F}_{n+1}\), the summands \(X_1, \ldots, X_{n+1}\) are exchangeable by the i.i.d. assumption, so \(\mathbb{E}[X_k \mid \mathcal{F}_{n+1}]\) does not depend on \(k\). Summing yields \((n+1)\mathbb{E}[X_k \mid \mathcal{F}_{n+1}] = S_{n+1}\), so each equals \(\bar X_{n+1}\). Then
Backward convergence gives \(\bar X_n \to \mathbb{E}[X_1 \mid \mathcal{F}_\infty]\) a.s. and in \(L^1\). Since \(\mathcal{F}_\infty \subseteq \bigcap_n \sigma(X_n, X_{n+1}, \ldots)\), Kolmogorov's 0–1 law makes \(\mathcal{F}_\infty\) trivial and the limit equals \(\mu\). \(\square\)
Exercise 4. Show that \(L^p\)-boundedness with \(p > 1\) implies UI. Deduce that an \(L^p\)-bounded martingale converges in \(L^p\).
Solution to Exercise 4
For \(p > 1\) and \(C = \sup_\alpha \mathbb{E}|X_\alpha|^p\),
as \(K \to \infty\), uniformly in \(\alpha\); so the family is UI. For a martingale \((M_n)\), Doob's maximal inequality gives \(\mathbb{E}[\sup_n |M_n|^p] \le (p/(p-1))^p \,C < \infty\). Combined with a.s. convergence \(M_n \to M_\infty\) (from \(L^1\)-boundedness), dominated convergence yields \(\mathbb{E}|M_n - M_\infty|^p \to 0\). \(\square\)
Exercise 5. Let \((M_t)\) be a UI martingale with limit \(M_\infty\) and \(\tau\) any stopping time. Prove \(M_\tau = \mathbb{E}[M_\infty \mid \mathcal{F}_\tau]\) a.s.
Solution to Exercise 5
Let \(\tau_k = \tau \wedge k\), a bounded stopping time. Bounded optional sampling applied to the closed martingale \(M_n = \mathbb{E}[M_\infty \mid \mathcal{F}_n]\) gives \(M_{\tau_k} = \mathbb{E}[M_\infty \mid \mathcal{F}_{\tau_k}]\). For \(A \in \mathcal{F}_\tau \subseteq \mathcal{F}_{\tau_k}\),
Since \(\tau_k \to \tau\) a.s. and \(M\) converges a.s., \(M_{\tau_k} \to M_\tau\) a.s. UI of \((M_n)\) carries over to \((M_{\tau_k})\) (it is bounded by the Doob maximal \(L^1\) process), so \(M_{\tau_k} \to M_\tau\) in \(L^1\). Passing to the limit: \(\mathbb{E}[M_\tau \mathbf{1}_A] = \mathbb{E}[M_\infty \mathbf{1}_A]\) for all \(A \in \mathcal{F}_\tau\), giving the claim. \(\square\)