Why Localization Needs a Limit Test
For a stochastic exponential, stopping often turns a local martingale into a martingale. That fact is useful, but it does not by itself imply that the unstopped process is a martingale: the expectations of the stopped variables may fail to pass to the limit. Uniform integrability is a condition that prevents this loss of expectation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For a finite horizon \(T\), let \(\theta_t\) be progressively measurable. If \(\mathbb{E}\int_0^T\theta_s^2\,ds<\infty\), then \(\theta\) belongs to \(\mathcal{L}^2\), and the Ito integral \(\int_0^t\theta_s\,dB_s\) is defined for \(0\le t\le T\). More generally, a locally square-integrable integrand is integrated by consistently defining its stopped Ito integrals for a sequence of stopping times on which the integrands belong to \(\mathcal{L}^2\).
The stochastic-exponential identity follows from Ito’s formula. For \(f(x,v)=\exp(x-\tfrac12v)\), the derivatives are \(f_x=f\), \(f_v=-\tfrac12f\), and \(f_{xx}=f\). Since \(d[M]_t=\theta_t^2dt\) and \(dM_t=\theta_t\,dB_t\), Ito’s formula gives \(dZ_t=Z_t\theta_t\,dB_t\). In particular, this is an Ito integral, with \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); it is not an ordinary chain-rule calculation.
The Localization Test
A sequence of stopping times \(\tau_n\) localizes a local martingale \(Z\) if \(\tau_n\) increases to the end of the time interval (or to infinity) and each stopped process \(Z_{t\wedge\tau_n}\) is a true martingale. For stochastic exponentials, one common way to justify the stopped martingale property is Novikov’s condition: if a progressively measurable integrand \(H\) satisfies \(\mathbb{E}\exp(\tfrac12\int_0^T H_s^2\,ds)<\infty\), then \(\mathcal{E}(\int H\,dB)\) is a true martingale on \([0,T]\). This is the standard Novikov criterion; its proof is part of the earlier material on exponential martingales.
Proof. Fix \(t\le T\). Because \(\tau_n\) increases to the end of the interval and \(Z\) has continuous paths, \(Z_{t\wedge\tau_n}\to Z_t\) almost surely. Uniform integrability and almost-sure convergence imply convergence in \(L^1\): \(\mathbb{E}|Z_{t\wedge\tau_n}-Z_t|\to0\). This implication can be seen by truncating at a common level \(K\): the tails have uniformly small expectations for large \(K\), while the bounded truncations converge in \(L^1\) by bounded convergence.
Now fix \(0\le s\le t\le T\). The stopped martingale identity is \(\mathbb{E}[Z_{t\wedge\tau_n}\mid\mathcal{F}_s]=Z_{s\wedge\tau_n}\). The \(L^1\) convergence just established holds at both \(s\) and \(t\). Conditional expectation is an \(L^1\) contraction, so the left sides converge in \(L^1\) to \(\mathbb{E}[Z_t\mid\mathcal{F}_s]\), and the right sides converge in \(L^1\) to \(Z_s\). Therefore \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) almost surely. Also \(Z_t\) is integrable because it is an \(L^1\) limit of integrable variables. This is exactly the martingale property. \(\square\)
Uniform integrability is a sufficient test for passing the stopped martingale identities to the limit. For a nonnegative local martingale, there is also a useful exact diagnostic: it is a supermartingale, and it is a martingale precisely when its expectation stays constant. The supermartingale assertion follows by applying conditional Fatou’s lemma to a localizing sequence of stopped martingales. If \(\mathbb{E}Z_t=\mathbb{E}Z_s\), then \(Z_s-\mathbb{E}[Z_t\mid\mathcal{F}_s]\ge0\) has expectation zero, so it vanishes almost surely. Conversely, a martingale has constant expectation.
Worked Localization Tests
Worked Example: A Deterministic Integrand Gives Uniform Integrability
Take \(T=1\) and \(\theta_s=2\). This deterministic integrand is progressively measurable and \(\mathbb{E}\int_0^1\theta_s^2\,ds=4<\infty\). Its stochastic exponential is \(Z_t=\exp(2B_t-2t)\). For \(s\le t\), independence of \(B_t-B_s\) from \(\mathcal{F}_s\), and the normal moment-generating function, give $$ \mathbb{E}[Z_t\mid\mathcal{F}_s] =Z_s\,\mathbb{E}\!\left[e^{2(B_t-B_s)-2(t-s)}\right] =Z_s e^{2(t-s)-2(t-s)} =Z_s. $$ Thus \(Z\) is a martingale. Its second moment is $$ \mathbb{E}[Z_t^2]=\mathbb{E}[e^{4B_t-4t}] =e^{8t-4t}=e^{4t}\le e^4,\qquad 0\le t\le1. $$ For any \(K>0\), \(Z_t\mathbf{1}_{\{Z_t>K\}}\le Z_t^2/K\), so the expectations of these tails are at most \(e^4/K\), uniformly over \(t\in[0,1]\). The family is therefore uniformly integrable. This bound also checks directly that large values cannot carry a fixed positive share of the expectation.
Worked Example: Stopping at a Level Produces a Bounded Martingale
Again let \(\theta_s=1\), so \(Z_t=\exp(B_t-\tfrac12t)\), and define \(\tau=\inf\{t\ge0:Z_t\ge2\}\wedge1\), where the infimum of an empty set is infinity. The stopped process is \(Z_{t\wedge\tau}\). By continuity, it never exceeds \(2\) on \([0,1]\), and it is positive. Its stopped integrand is \(H_s=\mathbf{1}_{\{s\le\tau\}}\), which is progressively measurable and satisfies \(\mathbb{E}\int_0^1H_s^2\,ds\le1\). Moreover, $$ \mathbb{E}\exp\left(\tfrac12\int_0^1H_s^2\,ds\right)\le e^{1/2}<\infty. $$ Novikov’s condition therefore shows that \(Z_{t\wedge\tau}\) is a true martingale. Since it is bounded by \(2\), the family of its values over \(0\le t\le1\) is uniformly integrable: for \(K\ge2\), every tail event \(\{Z_{t\wedge\tau}>K\}\) is empty. In particular, \(\mathbb{E}Z_{1\wedge\tau}=\mathbb{E}Z_0=1\). The stopping has made the process bounded, but this example alone says nothing about whether the unstopped process is a martingale in other problems; that requires a limit test.
Worked Example: Stopped Expectations Do Not Prevent Loss at the Limit
Set \(T=1\) and, for \(0\le s<1\), take the deterministic integrand \(\theta_s=(1-s)^{-1/2}\). On every interval \([0,t]\) with \(t<1\), it is progressively measurable and square-integrable, with $$ q(t)=\int_0^t\theta_s^2\,ds=-\log(1-t)<\infty. $$ Define \(Z_t=\exp(\int_0^t\theta_s\,dB_s-\tfrac12q(t))\) for \(t<1\). The integral on the entire interval \([0,1]\) is not in \(\mathcal{L}^2\), since \(\int_0^1\theta_s^2ds=\infty\); here the process is defined by its consistent stopped Ito integrals.
For each \(n\ge1\), put \(\tau_n=1-e^{-n}\), so \(q(\tau_n)=n\). Up to \(\tau_n\), the integrand is in \(\mathcal{L}^2\) and Novikov’s condition holds, because \(\exp(\tfrac12q(\tau_n))=e^{n/2}\) is finite. Thus \(Z_{t\wedge\tau_n}\) is a true martingale and \(\mathbb{E}Z_{\tau_n}=1\). At the terminal stopped time its second moment is $$ \mathbb{E}[Z_{\tau_n}^2] =\mathbb{E}\!\left[\exp\left(2\int_0^{\tau_n}\theta_s\,dB_s-n\right)\right] =e^{2n-n}=e^n. $$ This growing second moment does not, by itself, prove failure of uniform integrability; the decisive limit is as follows.
The Gaussian process \(Y_t=\int_0^t(1-s)^{-1/2}dB_s\), when indexed by its increasing variance \(u=q(t)\), has the law of Brownian motion at time \(u\). Brownian motion satisfies \(B_u/u\to0\) almost surely as \(u\to\infty\). Consequently \(Y_t-\tfrac12q(t)\to-\infty\) almost surely as \(t\uparrow1\), and \(Z_t\to0\). Define \(Z_1=0\); this gives a continuous extension. But for every \(t<1\), Novikov’s condition on \([0,t]\) gives \(\mathbb{E}Z_t=1\), whereas \(\mathbb{E}Z_1=0\). Thus \(Z\) is a nonnegative strict local martingale, meaning a local martingale that is not a true martingale on \([0,1]\).
The stopped terminal values \(Z_{\tau_n}\) converge almost surely to \(0\), while their expectations all equal \(1\). They cannot be uniformly integrable: if they were, almost-sure convergence would imply \(L^1\) convergence and hence \(\mathbb{E}Z_{\tau_n}\to\mathbb{E}0=0\), contradicting \(\mathbb{E}Z_{\tau_n}=1\). This is the precise failure that stopping alone cannot rule out.
Common Pitfall: Passing Expectations Through Almost-Sure Limits
A tempting but invalid calculation is: “Each stopped exponential has expectation \(1\); the stopping times increase to the terminal time; therefore the terminal exponential has expectation \(1\).” The last implication does not follow from almost-sure convergence. In the preceding example, the stopped values converge to zero but keep expectation \(1\). Uniform integrability, not just convergence of stopping times, supplies the missing justification.
For a nonnegative local martingale \(Z\) with \(Z_0=1\), conditional Fatou’s lemma gives \(\mathbb{E}Z_t\le1\). A strict inequality at even one time rules out the martingale property. In applications, one can use Novikov’s condition directly when it applies, or construct stopped martingales and test whether their values are uniformly integrable before passing to the limit. A local martingale is not automatically a martingale, and a collection of martingales need not retain its expectations under an almost-sure limit.
Check Your Understanding
Use the localization test and distinguish what stopping proves from what must still be checked.
- For a nonnegative local martingale, why does constant expectation imply the conditional-expectation identity?
- In the bounded-level example, which property of the stopped process gives uniform integrability?
- For \(\theta_s=(1-s)^{-1/2}\), compute \(\int_0^{\tau_n}\theta_s^2\,ds\) from the definition of \(\tau_n\).
- Why does almost-sure convergence of \(Z_{\tau_n}\) to zero contradict uniform integrability when every stopped expectation is one?
- State the uniform-integrability condition needed to pass the stopped martingale identities to the limit.