Localization and Stopping Local Martingales

Stopping and Localization

Stopping a martingale freezes its value once a stopping time occurs. For a local martingale, this operation still preserves the local-martingale property, but it does not automatically produce a true martingale. The distinction matters: a localizing sequence gives a sequence of ordinary martingales on which standard arguments can be used, while passing from those stopped processes to the original process may require a separate convergence argument.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). A process \(H\) is in \(\mathcal{L}^2\) on \([0,T]\) when it is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). No stochastic integrals are needed here. A martingale \(M\) is an adapted, integrable process satisfying \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely whenever \(s\le t\). We use the standard right-continuous versions of processes and stopping-time conventions.

Definition: For a stopping time \(\tau\), the stopped process is \(M^\tau_t=M_{t\wedge\tau}\). A local martingale \(M\) has a nondecreasing sequence of stopping times \((\tau_n)\), with \(\tau_n\to\infty\) almost surely, such that each \(M^{\tau_n}\) is a martingale. Such a sequence is called a localizing sequence.

The sequence need not make the unstopped process a martingale: it says that each process stopped at \(\tau_n\) is a martingale. The key to stopping a local martingale is to combine its original localizer with deterministic time cutoffs. Those cutoffs ensure that the stopping operation applied to an ordinary martingale is bounded.

Stopping a Local Martingale

Theorem: If \(M\) is a local martingale with localizing sequence \((\tau_n)\), and \(\tau\) is any stopping time, then \(M^\tau\) is a local martingale. One localizing sequence for \(M^\tau\) is \(\rho_n=\tau_n\wedge n\).

We use the bounded optional sampling theorem: if \(N\) is a martingale and \(\eta\) is a bounded stopping time, then \(N^\eta\) is a martingale. This standard result was proved in the earlier material on bounded optional sampling.

Proof. The sequence \(\rho_n=\tau_n\wedge n\) is nondecreasing. Since both \(\tau_n\to\infty\) almost surely and \(n\to\infty\), we have \(\rho_n\to\infty\) almost surely. Fix \(n\). The process \(M^{\tau_n}\) is a martingale by the definition of \((\tau_n)\). The stopping time \(\eta_n=\tau\wedge n\) is bounded by \(n\), so bounded optional sampling implies that \((M^{\tau_n})^{\eta_n}\) is a martingale. At every time \(t\), its value is \(M_{t\wedge\tau_n\wedge\tau\wedge n}=M_{t\wedge\tau\wedge\rho_n}=(M^\tau)_{t\wedge\rho_n}\). Thus \((M^\tau)^{\rho_n}\) is a martingale for every \(n\), and \((\rho_n)\) tends to infinity almost surely. These are precisely the conditions for \(M^\tau\) to be a local martingale. \(\square\)

The deterministic cutoff is essential to this proof. It lets us apply bounded optional sampling to the martingale \(M^{\tau_n}\), without assuming that stopping it at the possibly unbounded time \(\tau\) preserves the martingale property. It also ensures the new localizing times \(\rho_n\) tend to infinity, even when \(\tau\) itself is finite.

Worked Example: Stopping a Two-Step Martingale

Let \(\xi_1,\xi_2\) be independent random variables, each taking the values \(1\) and \(-1\) with probability \(1/2\). Set \(M_0=0\), \(M_1=\xi_1\), and \(M_2=\xi_1+\xi_2\), with the filtration generated by the increments as they are revealed. Define \(\tau\) to be the first \(k\in\{1,2\}\) with \(M_k=1\), or \(2\) if there is no such \(k\). Find the stopped terminal value and check the martingale property.

There are four equally likely increment pairs. For \((\xi_1,\xi_2)=(1,1)\), \(\tau=1\), so \(M^\tau_2=1\). For \((1,-1)\), \(\tau=1\), so \(M^\tau_2=1\). For \((-1,1)\), \(\tau=2\), so \(M^\tau_2=-1+1=0\). For \((-1,-1)\), \(\tau=2\), so \(M^\tau_2=-1-1=-2\). Therefore

\mathbb{E}[M^\tau_2]=\frac{1+1+0-2}{4}=0.

At time \(0\), the conditional expectation of the terminal value is \(0=M^\tau_0\). At time \(1\), if \(\xi_1=1\), the process has already stopped and its value at both times \(1\) and \(2\) is \(1\). If \(\xi_1=-1\), its time-\(1\) value is \(-1\), and its two possible terminal values are \(0\) and \(-2\), each with conditional probability \(1/2\). Their conditional mean is \((0-2)/2=-1\). Thus \(\mathbb{E}[M^\tau_2\mid\mathcal{F}_1]=M^\tau_1\); the time-\(0\) and time-\(1\) conditional-expectation identities verify the stopped martingale property.

Check: the four terminal values sum to \(1+1+0-2=0\), agreeing with the computed expectation and the initial value. In particular, the pair \((-1,1)\) gives \(0\), not \(-2\); the pair \((-1,-1)\) gives \(-2\), not \(0\).

How Localization Supports an Argument

Localization is especially useful when a claim about a process can first be proved for its stopped versions. For a local martingale \(M\) and a localizing sequence \((\tau_n)\), each \(M^{\tau_n}\) is an ordinary martingale, so its integrability and conditional-expectation identities are available. If a conclusion is a pathwise identity and holds for every stopped process, it can often be transferred directly to \(M\).

Proposition: Let \(\rho_n\) be nondecreasing stopping times with \(\rho_n\to\infty\) almost surely. If two processes \(X\) and \(Y\) satisfy \(X^{\rho_n}=Y^{\rho_n}\) indistinguishably for every \(n\), then \(X=Y\) indistinguishably.

Proof. For each \(n\), indistinguishability gives an event of probability one on which \(X_{t\wedge\rho_n}=Y_{t\wedge\rho_n}\) for every \(t\ge0\). Intersect these countably many events with the probability-one event on which \(\rho_n\to\infty\). On the resulting event, fix any \(t\ge0\). Since \(\rho_n\to\infty\), some \(n\) has \(\rho_n\ge t\). For that \(n\), \(t\wedge\rho_n=t\), so \(X_t=Y_t\). This holds for every \(t\) on the same event, which proves indistinguishability. \(\square\)

Worked Example: Localizing a Stopped Brownian Motion

Take \(M_t=B_t\), which is a martingale, and let \(\tau\) be any stopping time, possibly unbounded. Use \(\rho_n=n\) to verify directly that \((B^\tau)^{\rho_n}\) is a martingale.

For each \(n\), the stopping time \(\eta_n=\tau\wedge n\) is bounded by \(n\). Brownian motion is an integrable martingale: its increments have mean zero and are independent of the past, so \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\) for \(s\le t\). Bounded optional sampling therefore shows that \(B^{\eta_n}\) is a martingale. At each \(t\),

(B^\tau)_{t\wedge n}=B_{t\wedge\tau\wedge n}=B^{\eta_n}_t.

Hence \((B^\tau)^{\rho_n}\) is a martingale for every \(n\), and the deterministic times \(\rho_n=n\) increase to infinity. This verifies the local-martingale property of \(B^\tau\). The argument does not assert that \(B^\tau\) is a true martingale for every unbounded \(\tau\); the theorem establishes the local property, and any stronger claim needs its own integrability justification.

Check: at a fixed time \(t\), choosing any integer \(n\ge t\) gives \((B^\tau)_{t\wedge n}=B_{t\wedge\tau}\), the value of the process being localized. The sequence of deterministic cutoffs therefore eventually leaves each fixed-time value unchanged.

Pathwise identities can pass from stopped processes to the original process by the proposition, but expectations require care. Almost-sure convergence alone does not preserve expectations or conditional expectations. The following elementary condition explains one safe way to pass a martingale identity to a limit.

Proposition: Suppose \(M^{\tau_n}\) is a martingale for every \(n\), \(\tau_n\to\infty\) almost surely, and for every fixed \(t\), \(M_{t\wedge\tau_n}\to M_t\) in \(L^1(\mathbb{P})\). Then \(M\) is a martingale.

Proof. Fix \(s\le t\). For every \(n\), the martingale property gives \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}\). Conditional expectation is a contraction in \(L^1\): \(\mathbb{E}\big[|\mathbb{E}[U\mid\mathcal{F}_s]|\big]\le\mathbb{E}[|U|]\) for integrable \(U\). Therefore \(L^1\) convergence at time \(t\) implies convergence of the left sides in \(L^1\) to \(\mathbb{E}[M_t\mid\mathcal{F}_s]\); \(L^1\) convergence at time \(s\) implies convergence of the right sides to \(M_s\). The two limits are equal almost surely. Since this holds for every \(s\le t\), and \(M_t\) is integrable by its \(L^1\) convergence, \(M\) is a martingale. \(\square\)

Worked Example: Why Almost-Sure Convergence Is Not Enough

Let \(U\) be uniform on \((0,1)\). Use a filtration that is trivial for \(t<1\) and reveals \(U\) at \(t\ge1\). For each positive integer \(n\), let \(A_n=\{U\le1/n\}\), \(X_n=n\mathbf{1}_{A_n}\), and define \(N^n_t=1\) for \(t<1\) and \(N^n_t=X_n\) for \(t\ge1\).

Since \(\mathbb{P}(A_n)=1/n\), \(\mathbb{E}[X_n]=n(1/n)=1\). Before time \(1\), \(\mathbb{E}[X_n\mid\mathcal{F}_t]=1=N^n_t\); at and after time \(1\), \(X_n\) is known, so its conditional expectation is \(X_n=N^n_t\). Thus \(N^n\) is a martingale. For every \(U>0\), \(U\le1/n\) holds for only finitely many \(n\), so \(X_n\to0\) almost surely. Nevertheless, \(\mathbb{E}[X_n]=1\) for every \(n\), whereas the expectation of the almost-sure limit is \(0\).

Check: the expectations are computed directly from the event probabilities: \(\mathbb{E}[n\mathbf{1}_{A_n}]=n\mathbb{P}(A_n)=n/n=1\). Thus almost-sure convergence does not justify passing the martingale conditional-expectation identity to a limit. The \(L^1\) assumption in the proposition supplies the needed control.

Common Pitfall: Confusing Local and True Martingales

A frequent error is to say that \(M^\tau\) is a martingale merely because \(M\) is a local martingale and \(\tau\) is bounded. The theorem proved here guarantees that \(M^\tau\) is a local martingale. To conclude that it is a true martingale, one must also establish integrability and the conditional-expectation identities for the unstopped process, or verify a sufficient convergence condition such as the \(L^1\) hypothesis above. Bounded optional sampling applies to an actual martingale; it cannot be applied directly to a local martingale without first controlling its localization.

For example, the invalid calculation is “\(M\) is a local martingale, so \(\mathbb{E}[M_{t\wedge\tau}\mid\mathcal{F}_s]=M_{s\wedge\tau}\) by optional sampling.” It assumes the very martingale property that localization has not established. The correction is to apply bounded optional sampling to \(M^{\tau_n}\), where it is a true martingale, and then pass to the limit only when a justified convergence condition is available. For pathwise equalities, localization and increasing stopping times can suffice; for expectations, almost-sure convergence by itself does not.

Check Your Understanding

For each question, distinguish which step uses an ordinary martingale and which conclusion is only local.

If \((\tau_n)\) localizes \(M\) and \(\tau\) is a stopping time, why is \(\rho_n=\tau_n\wedge n\) suitable for localizing \(M^\tau\)?
In the two-step example, what are the stopped terminal values for the pairs \((-1,1)\) and \((-1,-1)\), respectively?
Why does bounded optional sampling apply to \(M^{\tau_n}\) stopped at \(\tau\wedge n\), but not automatically to \(M\) stopped at \(\tau\)?
State the convergence assumption in the proposition that permits passage of conditional-expectation identities to the limit.
Why can a pathwise identity pass through a localizing sequence even when an expectation identity cannot?

Localization and Stopping Local Martingales

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