Why Local Martingales Need a Test
A stochastic exponential is positive, but positivity alone does not ensure that its expectation stays constant. This distinction matters when an exponential is intended to serve as a probability density: a density must have expectation one, whereas a strict local martingale can lose expectation as time passes. The purpose here is to identify the loss and to see how stopping and uniform integrability detect it.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). For \(T>0\), the Ito integral \(\int_0^T H_s\,dB_s\) is defined for progressively measurable \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For locally square-integrable integrands, the integral and its stochastic exponential are defined by localization: stop the integrand so that each stopped integrand belongs to \(\mathcal{L}^2\), then let the stopping times increase. Expectations in this tutorial are under \(\mathbb{P}\).
The distinction is that a local martingale is a martingale after suitable stopping, but the unstopped process need not preserve expectations. In particular, \(Z_t\) is not automatically a valid density process. The measure-change applications of this fact require a true martingale; positivity by itself is insufficient.
Positive Local Martingales Are Supermartingales
Proof. By the definition of local martingale, choose stopping times \(\tau_n\) increasing to infinity almost surely such that each stopped process \(Z_{t\wedge\tau_n}\) is a martingale. For fixed \(s\le t\), conditional Fatou's lemma applies to the nonnegative random variables \(Z_{t\wedge\tau_n}\). Since \(Z_{t\wedge\tau_n}\to Z_t\) and \(Z_{s\wedge\tau_n}\to Z_s\) almost surely, it gives $$ \mathbb{E}[Z_t\mid\mathcal{F}_s] \le \liminf_{n\to\infty}\mathbb{E}[Z_{t\wedge\tau_n}\mid\mathcal{F}_s] =\liminf_{n\to\infty}Z_{s\wedge\tau_n} =Z_s. $$ Taking expectations with \(s=0\) shows \(\mathbb{E}[Z_t]\le1\), so \(Z_t\) is integrable. Thus \(Z\) is a supermartingale. If \(Z\) is a martingale, its expectations are constant. Conversely, if they are constant, the supermartingale inequality shows that the nonnegative random variable \(Z_s-\mathbb{E}[Z_t\mid\mathcal{F}_s]\) has expectation zero. It must therefore equal zero almost surely, which is the martingale conditional-expectation identity. \(\square\)
For a nonnegative local martingale starting at one, an expectation strictly below one at even one time proves that the process is not a true martingale. Conversely, on a fixed interval \([0,T]\), checking \(\mathbb{E}[Z_T]=1\) suffices: the supermartingale inequalities give \(1\ge\mathbb{E}[Z_s]\ge\mathbb{E}[Z_T]=1\) for every \(s\le T\).
Worked Example: A Bounded Integrand Preserves Unit Expectation
Let \(\tau\) be the first time \(B_t\) reaches either \(1\) or \(-1\), and take \(\theta_s=\mathbf{1}_{\{s\le\tau\}}\) on \([0,1]\). This integrand is progressively measurable and bounded by \(1\), so it belongs to \(\mathcal{L}^2\). We show directly that its stochastic exponential has expectation one.
Let \(Z_t=\exp(-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds)\). Stop \(Z\) when it first reaches \(n\), at time \(\rho_n\). Ito's formula applied to \(f(z)=z^2\), with \(f'(z)=2z\) and \(f''(z)=2\), gives the correction term \(\tfrac12 f''(Z_s)(Z_s\theta_s)^2\,ds=Z_s^2\theta_s^2\,ds\). Thus $$ d(Z_{t\wedge\rho_n}^2) =-2\mathbf{1}_{\{s\le\rho_n\}}Z_s^2\theta_s\,dB_s +\mathbf{1}_{\{s\le\rho_n\}}Z_s^2\theta_s^2\,ds. $$ The stopped stochastic integral has expectation zero: its integrand is bounded on the stopped interval, hence square-integrable. Taking expectations, using \(\theta_s^2\le1\), and applying Gronwall's inequality yields \(\mathbb{E}[Z_{t\wedge\rho_n}^2]\le e^t\) for \(0\le t\le1\). In particular, the stopped variables at each fixed \(t\) are uniformly integrable, since a uniform second-moment bound implies uniform integrability.
As \(n\to\infty\), \(Z_{t\wedge\rho_n}\to Z_t\) almost surely. Uniform integrability gives convergence in \(L^1\). Each stopped process has expectation \(1\), so \(\mathbb{E}[Z_t]=1\). The preceding theorem now implies that \(Z\) is a true martingale on \([0,1]\). Check: at \(t=0\), the exponential equals \(1\) and its expectation is \(1\), consistent with the result; at every \(t\in[0,1]\), the argument gives the same expectation, not merely an upper bound.
Localization and Uniform Integrability
Stopping converts a local martingale into martingales, but taking limits of their expectations is not automatic. Almost-sure convergence alone does not preserve expectations. Uniform integrability supplies the missing control: a family \(\mathcal{X}\) of integrable random variables is uniformly integrable if \(\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\to0\) as \(K\to\infty\).
Proof. Almost surely, \(Z_{T\wedge\tau_n}\to Z_T\). If the family is uniformly integrable, this convergence is in \(L^1\), so $$ \mathbb{E}[Z_T]=\lim_{n\to\infty}\mathbb{E}[Z_{T\wedge\tau_n}]=1. $$ The theorem then gives the martingale property throughout \([0,T]\). For the converse, optional sampling for a martingale at bounded stopping times gives \(Z_{T\wedge\tau_n}=\mathbb{E}[Z_T\mid\mathcal{F}_{T\wedge\tau_n}]\). Conditional expectations of one integrable random variable form a uniformly integrable family. Indeed, write \(Y=Z_T\) and \(X=\mathbb{E}[Y\mid\mathcal{G}]\) for any sigma-field \(\mathcal{G}\). For \(K>0\), split \(Y\) into its parts on \(\{|Y|\le\sqrt K\}\) and \(\{|Y|>\sqrt K\}\). The conditional expectation of the first part has absolute value at most \(\sqrt K\), so on \(\{|X|>2\sqrt K\}\) the absolute conditional expectation of the second part exceeds \(\sqrt K\). Consequently, $$ \mathbb{E}\big[|X|\mathbf{1}_{\{|X|>2\sqrt K\}}\big] \le 2\mathbb{E}\big[|Y|\mathbf{1}_{\{|Y|>\sqrt K\}}\big]. $$ The right side tends to zero and does not depend on \(\mathcal{G}\). This proves uniform integrability for the stopped family. \(\square\)
Thus the practical test is not simply whether stopped exponentials have expectation one—they do by construction. The key question is whether their terminal values retain enough integrability to pass that expectation to the limit.
Worked Example: A Strict Local Martingale from Three-Dimensional Brownian Motion
Let \(\mathbf{B}_t\) be three-dimensional Brownian motion, and start at \(x=(a,0,0)\) with \(a>0\). Put \(R_t=|x+\mathbf{B}_t|\). The standard three-dimensional Brownian fact that a point is almost surely not hit implies \(R_t>0\) for all \(t\ge0\). Its radial process satisfies \(dR_t=d\beta_t+\frac{1}{R_t}\,dt\), where \(\beta_t\) is a one-dimensional Brownian motion. This radial SDE and the non-hitting fact are standard; their proofs are beyond this tutorial.
Set \(Z_t=a/R_t\). Apply Ito's formula to \(f(r)=a/r\), for which \(f'(r)=-a/r^2\) and \(f''(r)=2a/r^3\). The second-order correction is \(\tfrac12 f''(R_t)(dR_t)^2=(a/R_t^3)\,dt\), since \(d\beta_t\,d\beta_t=dt\). Substitution of the drift and diffusion gives $$ dZ_t=-\frac{a}{R_t^2}\,d\beta_t -\frac{a}{R_t^3}\,dt+\frac{a}{R_t^3}\,dt =-\frac{a}{R_t^2}\,d\beta_t =-Z_t\frac{1}{R_t}\,d\beta_t. $$ Therefore \(Z\) is the stochastic exponential with integrand \(\theta_t=1/R_t\) under the convention \(dZ_t=-Z_t\theta_t\,d\beta_t\). In particular, the sign is positive in \(\theta_t=1/R_t\). The integrand is locally square-integrable: on each compact time interval each positive continuous path \(R\) has a positive minimum, so \(\int_0^T R_s^{-2}\,ds<\infty\) pathwise. Its expectation need not be finite, so the integral is understood by localization.
We compute the expectation to test whether \(Z\) is a true martingale. The radius of a three-dimensional Gaussian vector started at distance \(a\) has density $$ p_t(r)=\frac{r}{a}\left(\varphi_t(r-a)-\varphi_t(r+a)\right),\qquad r>0, $$ where \(\varphi_t(u)=(2\pi t)^{-1/2}e^{-u^2/(2t)}\). This density follows by integrating the Gaussian density over the sphere of radius \(r\). Hence, for \(t>0\), $$ \mathbb{E}[Z_t] =\int_0^\infty \frac{a}{r}p_t(r)\,dr =\int_0^\infty\big(\varphi_t(r-a)-\varphi_t(r+a)\big)\,dr =2\Phi\left(\frac{a}{\sqrt t}\right)-1<1. $$ The last inequality holds because \(a/\sqrt t\) is finite and the standard normal CDF satisfies \(\Phi(u)<1\) for finite \(u\). For the exact choice \(a=1,t=1\), the expectation is \(2\Phi(1)-1\), strictly less than \(1\); at time zero, \(Z_0=a/R_0=1\). Thus \(Z\) is a strict local martingale.
Check by localization: stop when \(R\) first reaches \(1/n\), for integers \(n\) large enough that \(1/n<a\). Before this stopping time, \(\theta=1/R\le n\), so the stopped integrand is bounded and the bounded-integrand argument gives stopped expectation \(1\) on every finite horizon. Yet the unstopped expectation at \(t>0\) is \(2\Phi(a/\sqrt t)-1<1\). Therefore those stopped terminal values cannot be uniformly integrable: if they were, the proposition would force \(\mathbb{E}[Z_t]=1\).
Common Pitfall: Fatou's Lemma Has a Direction
A tempting but incorrect argument is to claim that almost-sure convergence of \(Z_{t\wedge\tau_n}\) to \(Z_t\) implies \(\mathbb{E}[Z_{t\wedge\tau_n}]\le\mathbb{E}[Z_t]\). Fatou's lemma for nonnegative variables actually gives \(\mathbb{E}[Z_t]\le\liminf_n\mathbb{E}[Z_{t\wedge\tau_n}]\), the opposite direction. The supermartingale inequality must instead be obtained by applying conditional Fatou's lemma to the conditional expectations, as in the theorem's proof.
A second pitfall is to infer that a localizing sequence of unit-expectation martingales has a unit-expectation limit. The strict example shows why this fails: expectations can decrease in the limit when uniform integrability is absent. Conditions such as Novikov's condition provide useful sufficient tests for stochastic exponentials, but they are a separate criterion; the fundamental issue here is whether localization can be removed without losing expectation.
| Property | What it guarantees | What it does not guarantee alone |
|---|---|---|
| Positive local martingale | Supermartingale; expectation at most its initial value | Constant expectation |
| Stopped localizing processes are martingales | Each stopped expectation is preserved | Expectation is preserved in the limit |
| Uniform integrability of localized terminal values | Convergence in \(L^1\) and preservation of expectation | A particular sufficient exponential-integrability condition |
Check Your Understanding
Use the supermartingale and uniform-integrability tests, keeping track of whether each statement concerns the stopped or unstopped process.
- Why does \(\mathbb{E}[Z_T]=1\) imply that a nonnegative local martingale with \(Z_0=1\) is a martingale at every time \(s\le T\)?
- Which direction does Fatou's lemma give when \(Z_{t\wedge\tau_n}\to Z_t\) almost surely and all variables are nonnegative?
- Why does a uniform second-moment bound imply uniform integrability?
- For the radial example, what is \(\mathbb{E}[a/R_t]\), and why is it strictly less than one when \(a,t>0\)?
- What conclusion about the stopped terminal values follows if their almost-sure limit has expectation strictly below one?