From Local Martingales to Supermartingales
Localization allows a process to behave like a martingale after suitable stopping, even when the unstopped process does not have the integrability needed to be a martingale. Nonnegativity supplies the key extra control: Fatou’s lemma turns the conditional martingale identities for the stopped processes into a one-sided conditional inequality for their limit.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). Let \(M=(M_t)_{t\ge0}\) be an adapted process with right-continuous paths and integrable \(M_0\). A sequence of stopping times \((\tau_n)\) is a localizing sequence if \(\tau_n\uparrow\infty\) almost surely and each stopped process \(M_{t\wedge\tau_n}\) is a martingale. The process \(M\) is a local martingale if it has such a sequence. A supermartingale is an adapted process with integrable values at every time and \(\mathbb{E}[M_t\mid\mathcal{F}_s]\le M_s\) almost surely whenever \(s\le t\).
The proof uses conditional Fatou’s lemma. In the form needed here, if \(X_n\ge0\) and \(\mathcal{G}\) is a sigma-field, then \(\mathbb{E}[\liminf_n X_n\mid\mathcal{G}]\le\liminf_n\mathbb{E}[X_n\mid\mathcal{G}]\). For completeness, set \(A_n=\inf_{k\ge n}X_k\). Then \(A_n\uparrow\liminf_kX_k\). Conditional monotone convergence gives \(\mathbb{E}[A_n\mid\mathcal{G}]\uparrow\mathbb{E}[\liminf_kX_k\mid\mathcal{G}]\). Since \(A_n\le X_k\) for every \(k\ge n\), we have \(\mathbb{E}[A_n\mid\mathcal{G}]\le\inf_{k\ge n}\mathbb{E}[X_k\mid\mathcal{G}]\). Taking limits proves the stated inequality.
Proof. Fix deterministic times \(0\le s\le t\), and choose a localizing sequence \((\tau_n)\). Because \(\tau_n\uparrow\infty\) almost surely, \(M_{t\wedge\tau_n}\to M_t\) and \(M_{s\wedge\tau_n}\to M_s\) almost surely. Nonnegativity ensures that every \(M_{t\wedge\tau_n}\) is nonnegative. Applying conditional Fatou’s lemma gives $$ \mathbb{E}[M_t\mid\mathcal{F}_s] \le \liminf_{n\to\infty}\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s] =\liminf_{n\to\infty}M_{s\wedge\tau_n} =M_s. $$ Here the middle equality is the martingale property of the stopped process \(M_{\cdot\wedge\tau_n}\). To verify integrability rather than assume it, apply ordinary Fatou’s lemma: $$ \mathbb{E}[M_t]\le\liminf_{n\to\infty}\mathbb{E}[M_{t\wedge\tau_n}] =\mathbb{E}[M_0]<\infty. $$ The last equality holds because each stopped process is a martingale. Thus \(M_t\) is integrable, and the conditional inequality proves that \(M\) is a supermartingale. Taking expectations in that inequality, or using the displayed Fatou bound, gives the expectation bound. \(\square\)
Proof. Taking expectations of \(\mathbb{E}[M_t\mid\mathcal{F}_s]\le M_s\) gives \(\mathbb{E}[M_t]\le\mathbb{E}[M_s]\), by the defining property of conditional expectation for integrable variables. If \(M_0=0\), the theorem gives \(0\le\mathbb{E}[M_t]\le0\). A nonnegative integrable random variable with expectation zero is zero almost surely, so \(M_t=0\) almost surely. \(\square\)
Examples Where the Bound Is an Equality
Worked Example: A Deterministic-Integrand Exponential
Let \(B_t\) be Brownian motion on the filtered probability space, and fix a constant \(\theta\in\mathbb{R}\). Consider \(Z_t=\exp(\theta B_t-\tfrac12\theta^2t)\). Verify directly that it is a martingale and compare its expectation with the theorem’s bound.
For each \(t\), the Gaussian moment-generating function gives \(\mathbb{E}[Z_t]=e^{-\theta^2t/2}\mathbb{E}[e^{\theta B_t}] =e^{-\theta^2t/2}e^{\theta^2t/2}=1\), so \(Z_t\) is integrable. If \(s\le t\), write \(B_t=B_s+(B_t-B_s)\). The increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Therefore $$ \mathbb{E}[Z_t\mid\mathcal{F}_s] =e^{\theta B_s-\theta^2t/2}\mathbb{E}[e^{\theta(B_t-B_s)}] =e^{\theta B_s-\theta^2t/2}e^{\theta^2(t-s)/2} =Z_s. $$ Thus the conditional expectation verifies the martingale property, not merely the local martingale property.
Check: substituting \(s=0\) and \(Z_0=1\) gives \(\mathbb{E}[Z_t]=1=\mathbb{E}[Z_0]\), so the expectation bound holds with equality.
Worked Example: A Nonnegative Stopped Brownian Martingale
Let \(B_0=0\), and define \(\tau=\inf\{t\ge0:|B_t|=1\}\) and \(N_t=1+B_{t\wedge\tau}\). Show that \(N\) is nonnegative and that its expectation remains one.
Before \(\tau\), continuity gives \(-1<B_t<1\); at \(\tau\), \(B_\tau\) is either \(-1\) or \(1\). Hence \(0\le N_t\le2\) for every \(t\). The process \(B_{t\wedge\tau}\) is a martingale: for fixed \(s\le t\), it is Brownian motion stopped at the bounded stopping times \(s\wedge\tau\le t\wedge\tau\), so bounded optional sampling for the martingale \(B\) gives \(\mathbb{E}[B_{t\wedge\tau}\mid\mathcal{F}_s]=B_{s\wedge\tau}\). Consequently \(\mathbb{E}[N_t\mid\mathcal{F}_s]=N_s\). The bound \(0\le N_t\le2\) ensures integrability at every time, and taking expectations yields \(\mathbb{E}[N_t]=\mathbb{E}[N_0]=1\).
Check: \(N_0=1+B_0=1\), and the conditional identity at \(s=0\) gives \(\mathbb{E}[N_t]=1\). The theorem guarantees only that the expectation cannot increase; this example shows that stopping can preserve it exactly.
When the Expectation Bound Is Strict
The theorem does not assert that a nonnegative local martingale is a martingale. A martingale must satisfy the conditional equality and be integrable; a local martingale is only required to become a martingale after stopping. The inequality in the theorem can be strict, as the next example demonstrates.
Worked Example: The Reciprocal of a Three-Dimensional Bessel Process
Let \(R_t\) be a three-dimensional Bessel process started at \(R_0=r>0\). It is positive and has stochastic differential \(dR_t=dB_t+\frac1{R_t}\,dt\), where \(B_t\) is Brownian motion. Consider \(M_t=1/R_t\). Show that it is a nonnegative local martingale and compute its expectation.
Apply Ito’s formula to \(f(x)=1/x\) on \(x>0\). Its derivatives are \(f'(x)=-x^{-2}\) and \(f''(x)=2x^{-3}\), so the second-order correction is \(\tfrac12 f''(R_t)(dR_t)^2=R_t^{-3}\,dt\), using \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). The drift contribution from \(f'(R_t)dR_t\) is \(-R_t^{-3}\,dt\), which cancels the correction. Thus $$ dM_t=-R_t^{-2}\,dB_t. $$ This Ito integral is understood locally: stop when \(R_t\) leaves \((1/n,n)\). On each such stopped interval the integrand is bounded, progressively measurable, and satisfies \(\mathbb{E}\int_0^T R_s^{-4}\mathbf{1}_{\{s\le\tau_n\}}\,ds\le n^4T<\infty\). The stopped integrals are therefore square-integrable martingales. Since \(R\) stays positive and has finite continuous paths on compact time intervals, these stopping times increase to infinity almost surely. Hence \(M\) is a nonnegative local martingale.
The standard radial transition density of \(R_t\), for \(y>0\), is $$ p_t(r,y)=\frac{y}{r}\left(\frac{e^{-(y-r)^2/(2t)}}{\sqrt{2\pi t}}-\frac{e^{-(y+r)^2/(2t)}}{\sqrt{2\pi t}}\right). $$ Using this density, with \(\Phi\) the standard normal cumulative distribution function, gives $$ \mathbb{E}[M_t] =\int_0^\infty \frac1y p_t(r,y)\,dy =\frac1r\left(\Phi\!\left(\frac r{\sqrt t}\right)-\left[1-\Phi\!\left(\frac r{\sqrt t}\right)\right]\right) =\frac{2\Phi(r/\sqrt t)-1}{r}. $$ For \(r=1\) and \(t=1\), this is \(2\Phi(1)-1\), which is strictly less than \(1=\mathbb{E}[M_0]\), since \(\Phi(1)<1\). It is positive since \(\Phi(1)>\tfrac12\). This explicitly verifies the theorem’s expectation bound and shows that it need not be an equality.
Check: as \(t\downarrow0\), \(r/\sqrt t\to\infty\), so \(\Phi(r/\sqrt t)\to1\) and the expectation formula tends to \(1/r=M_0\). For every \(t>0\), however, \(\Phi(r/\sqrt t)<1\), so \(\mathbb{E}[M_t]<1/r\).
A Common Limit-Passing Error
A tempting but invalid argument is: “Each stopped process is a martingale, and \(M_{t\wedge\tau_n}\to M_t\), so take limits in \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}\) to obtain \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\).” Almost-sure convergence alone does not justify interchanging a limit and conditional expectation. In the Bessel example, such an interchange would imply \(\mathbb{E}[M_t]=1/r\), contradicting the computed value \((2\Phi(r/\sqrt t)-1)/r\). The valid step is conditional Fatou’s lemma, which gives the one-sided inequality. Additional hypotheses may yield equality, but the theorem here does not supply them.
In short, localization supplies martingale identities before the limit, while nonnegativity permits Fatou’s lemma after the limit. The result is an integrable supermartingale and a nonincreasing expectation. Equality requires more than nonnegativity and localization alone.
Check Your Understanding
Use the definitions and the conditional Fatou argument to answer the following.
- In the theorem’s proof, where is nonnegativity used, and why is it important for the limit argument?
- For the exponential process \(Z_t\), compute \(\mathbb{E}[Z_t\mid\mathcal{F}_s]\) from the Gaussian increment and identify the step that uses independence.
- Why does \(0\le N_t\le2\) ensure that the stopped Brownian example is integrable at every time?
- For the reciprocal Bessel example, identify the Ito correction term and the drift term that cancels it.
- What additional limit-interchange claim would be needed to turn the theorem’s conditional inequality into equality?