Why Localize a Martingale?
A martingale must be integrable at every fixed time and satisfy a conditional-expectation identity. Some processes do not meet these requirements globally, even though their behavior becomes martingale-like when observed up to suitable stopping times. Localization makes that idea precise: stop the process at a sequence of increasingly distant stopping times, and require each stopped process to be a martingale. The stopping times move farther away so that, on any fixed finite time interval, they eventually do not affect the process.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). A process \(M=(M_t)_{t\ge0}\) is adapted if \(M_t\) is \(\mathcal{F}_t\)-measurable for every \(t\). We consider adapted processes with right-continuous paths and left limits. For a stopping time \(\tau\), the stopped process is \(M^\tau_t=M_{t\wedge\tau}\). A martingale is an adapted process with \(\mathbb{E}|M_t|<\infty\) for every \(t\), and with \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely whenever \(0\le s\le t\).
The condition \(\tau_n\to\infty\) is pathwise: outside a single null set, for every fixed \(T<\infty\), eventually \(\tau_n>T\). Thus \(M^{\tau_n}\) eventually agrees with \(M\) throughout \([0,T]\) on each such path. The definition does not require \(M_t\) itself to be integrable. That distinction is essential: the stopped processes may be integrable even when the original process is not.
Martingales Are Local Martingales
Proof. Each \(\tau_n=n\) is a stopping time, the sequence is nondecreasing, and \(\tau_n\to\infty\). For fixed \(n\), \(M^{\tau_n}\) is a martingale by the bounded optional sampling theorem: if \(s\le t\), the stopping times \(s\wedge n\) and \(t\wedge n\) are bounded, and optional sampling gives \(\mathbb{E}[M_{t\wedge n}\mid\mathcal{F}_{s\wedge n}]=M_{s\wedge n}\). This implies the martingale identity for the stopped process with respect to \((\mathcal{F}_t)\). Hence \((n)\) is a localizing sequence. \(\square\)
Here the only stopping result being used is the bounded optional sampling theorem: for a martingale and bounded stopping times \(\sigma\le\tau\), \(\mathbb{E}[M_\tau\mid\mathcal{F}_\sigma]=M_\sigma\). The substantive direction is the converse question: when does being a local martingale imply being a true martingale? Boundedness is one sufficient condition, as the next result shows.
Proof. Let \((\tau_n)\) localize \(M\), and fix \(0\le s\le t\). For every \(n\), \(M^{\tau_n}\) is a martingale, so for every \(A\in\mathcal{F}_s\), \(\mathbb{E}[\mathbf{1}_A M_{t\wedge\tau_n}]=\mathbb{E}[\mathbf{1}_A M_{s\wedge\tau_n}]\). Since \(\tau_n\to\infty\) almost surely, \(M_{t\wedge\tau_n}\to M_t\) and \(M_{s\wedge\tau_n}\to M_s\) almost surely. Both sequences are bounded in absolute value by \(C\). The dominated convergence theorem therefore gives \(\mathbb{E}[\mathbf{1}_A M_t]=\mathbb{E}[\mathbf{1}_A M_s]\). Also \(\mathbb{E}|M_t|\le C\). Since the equality holds for every \(A\in\mathcal{F}_s\), it is exactly the defining conditional-expectation identity \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). Thus \(M\) is a martingale. \(\square\)
Worked Examples: Localization in Practice
Worked Example: A Bounded Martingale with a Simple Localizer
Let \(A\) be an event with \(\mathbb{P}(A)=\tfrac13\). Use a filtration that reveals whether \(A\) occurred at time \(1\): take \(\mathcal{F}_t\) trivial for \(0\le t<1\), and \(\mathcal{F}_t=\sigma(A)\) for \(t\ge1\). Define \(M_t=\tfrac13\) for \(t<1\), and \(M_t=\mathbf{1}_A\) for \(t\ge1\).
For \(s\le t<1\), both values are \(\tfrac13\). For \(1\le s\le t\), both values equal \(\mathbf{1}_A\). If \(s<1\le t\), then \(\mathbb{E}[M_t\mid\mathcal{F}_s]=\mathbb{E}[\mathbf{1}_A]=\tfrac13=M_s\), because \(\mathcal{F}_s\) is trivial. These cases verify the martingale identity for every \(s\le t\). The process is integrable and bounded by \(1\).
For every \(n\ge1\), the stopped process \(M^{n}\) equals \(M\): after time \(1\), \(M\) is already constant in time, so stopping it at \(n\ge1\) changes nothing. Thus \(\tau_n=n\) is a localizing sequence. Check: at every time \(t\), \(\mathbb{E}[M_t]=\tfrac13\): before time \(1\) this is the constant value, and after time \(1\) it follows from \(\mathbb{E}[\mathbf{1}_A]=\tfrac13\).
Worked Example: Localizing Brownian Motion by Exit Times
Let \(B_t\) be a standard Brownian motion on the filtered probability space, so \(B_0=0\), its paths are continuous, and its increments are independent of the past with \(B_t-B_s\sim N(0,t-s)\) for \(s<t\). For \(n\ge1\), set \(\tau_n=\inf\{t\ge0:|B_t|\ge n\}\), with the infimum of the empty set defined as \(\infty\).
Each \(\tau_n\) is a stopping time by the hitting-time property for continuous adapted processes. The thresholds increase, so \(\tau_n\le\tau_{n+1}\). For any fixed \(T<\infty\), continuity makes the path bounded on \([0,T]\); once \(n>\sup_{0\le u\le T}|B_u|\), one has \(\tau_n>T\). Therefore \(\tau_n\to\infty\) almost surely.
Before \(\tau_n\), the path stays inside \((-n,n)\); at a finite exit it reaches \(n\) or \(-n\) by continuity. Hence \(|B_{t\wedge\tau_n}|\le n\). To verify the martingale property, fix \(s\le t\) and choose an integer \(k\ge t\). The stopping time \(\tau_n\wedge k\) is bounded, so the bounded stopping theorem for Brownian motion says \(B^{\tau_n\wedge k}\) is a martingale. At both times \(s\) and \(t\), stopping at \(\tau_n\wedge k\) gives the same value as stopping at \(\tau_n\), since \(s\le t\le k\). Thus \(B^{\tau_n}\) is a martingale. This verifies all parts of the local-martingale definition for \(B\). Check: each stopped process is bounded by its own threshold \(n\), while the thresholds tend to infinity; no single bound on the unstopped Brownian path over all times is asserted.
A Local Martingale That Is Not a True Martingale
The boundedness theorem does not say every local martingale is a martingale. Here is a direct example in which the obstruction is failure of integrability. Let \(B\) be a standard Brownian motion, and let \(Y\) be independent of the entire Brownian path with density \(y^{-2}\) on \(y\ge1\). This is a probability density because \(\int_1^\infty y^{-2}\,dy=1\), while \(\mathbb{E}[Y]=\int_1^\infty y^{-1}\,dy=\infty\). Use the filtration containing \(Y\) at time \(0\) and the Brownian history up to time \(t\) thereafter. Independence ensures that \(B\) remains Brownian relative to this filtration.
Worked Example: A Strict Local Martingale from a Heavy-Tailed Variable
Define \(M_t=YB_t\). For each integer \(n\ge1\), define \(\tau_n=0\) on \(\{Y>n\}\) and \(\tau_n=\infty\) on \(\{Y\le n\}\). This is a stopping time because \(\{Y>n\}\in\mathcal{F}_0\); for every \(t\ge0\), \(\{\tau_n\le t\}=\{Y>n\}\). The sequence is nondecreasing: for fixed \(Y\), its values are \(0\) until \(n\ge Y\), and are then \(\infty\). Since \(Y\) is finite almost surely, \(\tau_n\to\infty\) almost surely.
Because \(B_0=0\), the stopped process is \(M_{t\wedge\tau_n}=Y\mathbf{1}_{\{Y\le n\}}B_t\). Its coefficient is \(\mathcal{F}_0\)-measurable and bounded by \(n\), so it is integrable at each time. For \(s\le t\), independence of the Brownian increment from \(\mathcal{F}_s\) gives \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=Y\mathbf{1}_{\{Y\le n\}}\mathbb{E}[B_t\mid\mathcal{F}_s]=Y\mathbf{1}_{\{Y\le n\}}B_s=M_{s\wedge\tau_n}\). Thus each stopped process is a martingale.
For \(t>0\), however, \(M_t\) is not integrable. Independence and nonnegativity give \(\mathbb{E}|YB_t|=\mathbb{E}[Y]\mathbb{E}|B_t|=\infty\), since \(\mathbb{E}[Y]=\infty\) and \(B_t\) is nonzero with positive probability. Therefore \(M\) is a local martingale but not a true martingale. Check: the stopped coefficient has finite mean, \(\mathbb{E}[Y\mathbf{1}_{\{Y\le n\}}]=\int_1^n y\cdot y^{-2}\,dy=\log n\) for \(n\ge1\) (with value \(0\) at \(n=1\)); the unstopped coefficient has infinite mean. The localizers remove precisely the large values responsible for nonintegrability.
Common Pitfalls and What Localization Does Not Say
A local martingale is not automatically a martingale. The strict-local example satisfies the stopped martingale identities but fails the integrability requirement at every positive time. It is not legitimate to write \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for the original process when \(\mathbb{E}|M_t|=\infty\).
Stopping times must move out to infinity. A single stopping time that makes a process harmless is not enough. In the Brownian example, stopping at a fixed level gives one bounded martingale, but the local-martingale definition requires a nondecreasing sequence whose limit is infinity almost surely. The increasing exit thresholds provide that sequence.
Bounded stopped processes do not imply a globally bounded process. In the Brownian example, the bound for \(B^{\tau_n}\) is \(n\), which depends on \(n\). The bounded-local-martingale theorem applies only when one constant \(C\) bounds \(M_t\) for all times, not when each localization has its own growing bound.
| Property | True martingale | Local martingale |
|---|---|---|
| Integrable at each fixed time | Required | Not required for the original process |
| Conditional-expectation identity | For the original process | For each process stopped at a localizer |
| Stopping sequence tending to infinity | Not part of the definition | Required |
| Relationship | Every martingale is local | May fail to be a true martingale |
Check Your Understanding
For each question, distinguish the requirements on the original process from those on its stopped versions.
- State the three defining properties of a localizing sequence.
- Why does a uniform bound \(|M_t|\le C\) allow the stopped martingale identities to pass to \(M\)? Identify the convergence theorem used.
- For the Brownian exit times \(\tau_n\), explain why continuity implies \(\tau_n\to\infty\) almost surely.
- In the heavy-tailed example, which integrability condition fails for \(M_t\) when \(t>0\), and why does it hold for \(M_{t\wedge\tau_n}\)?
- Does the fact that every \(M^{\tau_n}\) is bounded by a finite constant depending on \(n\) imply that \(M\) is a true martingale? Explain.