From Local Martingales to Martingales
Localization lets us study a process through stopped versions that are martingales. But the defining limit of those stopped processes does not, by itself, preserve expectations. To pass from a local martingale to a true martingale, we need a condition that controls the stopped values as the localization is removed. Boundedness, domination by an integrable random variable, and uniform integrability are three useful ways to supply that control.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying an adapted standard Brownian motion \(B_t\). The results below concern general adapted processes; no stochastic integral is needed. A process \(M_t\) is a true martingale if each \(M_t\) is integrable and, for every \(0\le s\le t\), \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely. It is a local martingale if there are increasing stopping times \(\tau_n\uparrow\infty\) almost surely such that each stopped process \(M_{t\wedge\tau_n}\) is a true martingale.
The Localization Criterion
Proof. Fix \(0\le s\le t\). Since \(\tau_n\uparrow\infty\) almost surely, for almost every outcome \(t\wedge\tau_n=t\) for all sufficiently large \(n\). Thus \(M_{t\wedge\tau_n}\to M_t\) almost surely, and \(M_{s\wedge\tau_n}\to M_s\) almost surely as well. Uniform integrability and almost-sure convergence imply convergence in \(L^1\), by the standard uniform-integrability convergence theorem. In particular, \(M_t\) and \(M_s\) are integrable. Each stopped process is a martingale, so \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}\). Conditional expectation is a contraction in \(L^1\); therefore the left sides converge in \(L^1\) to \(\mathbb{E}[M_t\mid\mathcal{F}_s]\), while the right sides converge in \(L^1\) to \(M_s\). Hence \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely. Since \(s,t\) were arbitrary, \(M\) is a true martingale. If \(M\) is class D, then each \(M_{t\wedge\tau_n}\) belongs to the class-D family because \(t\wedge\tau_n\) is a bounded stopping time. That family is uniformly integrable, so the first assertion applies. \(\square\)
The theorem identifies the exact issue: the localized martingales have the right conditional-expectation identities, but those identities pass to the limit only with adequate integrability control. A convenient sufficient condition is domination on each finite time interval.
Proof. Fix \(T\). For every \(t\le T\), \(|M_{t\wedge\tau_n}|\le Y_T\). Dominated convergence gives \(M_{t\wedge\tau_n}\to M_t\) in \(L^1\), and likewise at any \(s\le t\le T\). The stopped martingale identity can therefore be passed to the limit in \(L^1\), exactly as in the theorem’s proof, to give \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). Since \(T\) is arbitrary, the identity holds for all finite \(s\le t\). A deterministic bound is itself an integrable dominating random variable. \(\square\)
Worked Applications
Worked Example: A Brownian Motion Stopped at Two Levels
Let \(\tau=\inf\{t\ge0:|B_t|\ge1\}\) and \(M_t=B_{t\wedge\tau}\). Show that \(M\) is a true martingale and find its expectation.
Brownian motion is a martingale. Stopping it at \(\tau\wedge n\), a bounded stopping time, gives a martingale; hence \(M\) is a local martingale. By continuity of Brownian paths, \(|B_{t\wedge\tau}|\le1\) for every \(t\). The bounded-local-martingale corollary shows that \(M\) is a true martingale. Since \(M_0=0\), its expectation is \(\mathbb{E}[M_t]=\mathbb{E}[M_0]=0\) for every \(t\).
Check: the bound gives \(\mathbb{E}[|M_t|]\le1\), so each value is integrable as required. Also, the symmetry of Brownian motion and the symmetric stopping interval imply that \(M_t\) and \(-M_t\) have the same distribution, consistent with the computed mean zero.
Worked Example: Domination on a Fixed Time Interval
On \([0,1]\), use an integrable envelope to verify the martingale criterion for \(M_t=B_t\).
Set \(Y=\sup_{0\le u\le1}|B_u|\). The Doob \(L^2\) maximal inequality for the martingale \(B\), applied to the nonnegative submartingale \(|B|\), gives \(\mathbb{E}[Y^2]\le4\mathbb{E}[|B_1|^2]=4\), since \(\mathbb{E}[|B_1|^2]=1\). Cauchy–Schwarz then gives \(\mathbb{E}[Y]\le\sqrt{\mathbb{E}[Y^2]}\le2\). Thus \(Y\) is integrable and \(|B_t|\le Y\) for all \(t\in[0,1]\). The domination criterion applies on this interval.
Check: for \(0\le s\le t\le1\), the result gives \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\), as expected from the Brownian martingale property. The envelope proves integrability uniformly over the interval; it does not assert that \(Y\le2\) almost surely. The number \(2\) is an upper bound for \(\mathbb{E}[Y]\).
Worked Example: A Bounded Conditional-Expectation Martingale
Let \(A\in\mathcal{F}\) have probability \(\mathbb{P}(A)=\tfrac13\), and define \(X=2\mathbf{1}_A-\mathbf{1}_{A^c}\). For the martingale \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\), verify boundedness and compute its initial expectation when \(\mathcal{F}_0\) is trivial.
On \(A\), \(X=2\); on \(A^c\), \(X=-1\). Therefore \(\mathbb{E}[X]=2\cdot\tfrac13+(-1)\cdot\tfrac23=0\), and \(|X|\le2\). Conditional expectation preserves this bound: \(|M_t|=|\mathbb{E}[X\mid\mathcal{F}_t]|\le\mathbb{E}[|X|\mid\mathcal{F}_t]\le2\) almost surely. The conditional-expectation identity gives a martingale, and the bound also shows directly that its values are integrable and uniformly bounded. With \(\mathcal{F}_0\) trivial, \(M_0=\mathbb{E}[X]=0\), so \(\mathbb{E}[M_t]=0\) for every \(t\).
Check: if \(\mathcal{F}_t\) reveals whether \(A\) occurred, then \(M_t=2\) on \(A\) and \(M_t=-1\) on \(A^c\); its expectation is \(2\cdot\tfrac13-1\cdot\tfrac23=0\). If \(A\) has not been revealed, the conditional value remains \(0\). Both cases respect the bound \(2\).
Why Uniform Integrability Matters
A localizing sequence gives \(M_{t\wedge\tau_n}\to M_t\) almost surely for fixed \(t\), but almost-sure convergence alone does not justify passing expectations through the limit. Uniform integrability is a sufficient condition for that passage. The same point applies to conditional expectations: the identity for each stopped process is useful only when the two sides converge in a mode strong enough to preserve the identity.
A useful way to see the danger is to let \(U\) be uniform on \((0,1)\) and set \(X_n=n\mathbf{1}_{\{U\le1/n\}}\). For every positive integer \(n\), \(\mathbb{E}[X_n]=n\cdot(1/n)=1\). For every \(U>0\), eventually \(U>1/n\), so \(X_n\to0\) almost surely and \(\mathbb{E}[\lim_n X_n]=0\). Thus the expectations do not converge to the expectation of the limit. This sequence is not offered as a localizing sequence; it demonstrates concretely why almost-sure convergence alone cannot support the expectation step in a localization argument.
Common mistake. A tempting but invalid calculation is: “\(\mathbb{E}[M_{t\wedge\tau_n}]=\mathbb{E}[M_0]\) for every \(n\), and \(M_{t\wedge\tau_n}\to M_t\) almost surely, so \(\mathbb{E}[M_t]=\mathbb{E}[M_0]\).” The conclusion does not follow from those premises alone. The correction is to verify an additional condition, such as uniform integrability of the stopped values or domination by an integrable envelope, and then justify convergence in \(L^1\). A local martingale is not automatically a true martingale.
| Criterion | What to verify | Why it helps |
|---|---|---|
| Boundedness | A deterministic bound for \(M_t\) on each finite time interval | Supplies an integrable envelope for localization |
| Domination | \(|M_u|\le Y_T\) for \(u\le T\), with \(\mathbb{E}[Y_T]<\infty\) | Allows dominated convergence for stopped values |
| Uniform integrability | The stopped values, or all bounded-stop values, form a uniformly integrable family | Turns almost-sure convergence into \(L^1\) convergence |
Check Your Understanding
For each question, identify the integrability condition needed to justify the conclusion.
- Why does a deterministic bound on every finite time interval imply that a local martingale is a true martingale?
- In the localization criterion, where is uniform integrability used to pass from stopped martingales to the limiting process?
- For the stopped Brownian motion in the first example, what bound holds at the exit time as well as before it?
- Why does almost-sure convergence of \(X_n\) not imply \(\mathbb{E}[X_n]\to\mathbb{E}[\lim_n X_n]\)? Use the displayed uniform-variable calculation.
- What additional property of a local martingale makes it class D, and how does that property yield the true-martingale conclusion?