A Strict Local Martingale and the Meaning of Boundedness

Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.

It is tempting to think that a continuous local martingale that is uniformly integrable or bounded in \(L^p\) for \(p > 1\) is automatically a martingale. The distinction is between bounds at deterministic times and bounds for the stopped variables used in localisation. A bound on the former family need not justify passing to the limit in the latter.

The example is the reciprocal distance of a three-dimensional Brownian motion from the origin. We first analyse radial Brownian motion in dimension \(d\geq2\). Let \(B_t=(B_t^1,\ldots,B_t^d)\) start at \(x\in\mathbb R^d\setminus\{0\}\). Itō’s formula gives

\[|B_t|^2=|x|^2+2\sum_{i=1}^d\int_0^tB_s^i\,\mathrm dB_s^i+dt.\]

Each stochastic integral is square-integrable on finite horizons, since \(\mathbb{E}\int_0^t(B_s^i)^2\,\mathrm ds=x_i^2t+t^2/2<\infty\). Define

\[\beta_t=\sum_{i=1}^d\int_0^t\frac{B_s^i}{|B_s|}\,\mathrm dB_s^i,\]

with the fraction set to zero at the origin. The integrands are bounded. For every fixed \(s>0\), the Gaussian vector \(B_s\) has a density and hence \(\mathbb P(B_s=0)=0\); Fubini gives zero Lebesgue time at the origin almost surely. Thus \(\langle\beta\rangle_t=\int_0^t\mathbb1_{\{B_s\ne0\}}\,\mathrm ds=t\), and Lévy’s characterization makes \(\beta\) Brownian motion. In particular,

\[|B_t|^2=|x|^2+2\int_0^t|B_s|\,\mathrm d\beta_s+dt.\]

To justify functions singular at zero, first stop inside an annulus. For \(0<\varepsilon<|x|<R\), let \(T_\varepsilon=\inf\{t\geq0:|B_t|\leq\varepsilon\}\) and \(S_R=\inf\{t\geq0:|B_t|\geq R\}\). Define

\[f(r)=\begin{cases}\log r,&d=2,\\r^{2-d},&d\geq3,\end{cases} \qquad F(y)=f(|y|),\quad y\in\mathbb R^d\setminus\{0\}.\]

The derivatives are

\[\begin{aligned} \partial_iF(y)&=\begin{cases}y_i|y|^{-2},&d=2,\\(2-d)y_i|y|^{-d},&d\geq3,\end{cases}\\ \partial_{ii}F(y)&=\begin{cases}|y|^{-2}-2y_i^2|y|^{-4},&d=2,\\(2-d)(|y|^{-d}-dy_i^2|y|^{-d-2}),&d\geq3. \end{cases} \end{aligned}\]

Summing over \(i\) gives \(\Delta F=0\). For \(\tau=T_\varepsilon\wedge S_R\), Itō’s formula therefore reads

\[F(B_{t\wedge\tau})=F(x)+\sum_{i=1}^d\int_0^{t\wedge\tau}\partial_iF(B_s)\,\mathrm dB_s^i.\]

This is a bounded martingale because \(F\) is bounded on the closed annulus. The exit time is finite almost surely: \(\mathbb P(S_R>n)\leq\mathbb P(|B_n^1|<R)\to0\), by the Gaussian density bound. Optional stopping and bounded convergence give

\[f(|x|)=f(\varepsilon)\mathbb P(T_\varepsilon<S_R)+f(R)\mathbb P(S_R<T_\varepsilon), \qquad \mathbb P(T_\varepsilon<S_R)=\frac{f(R)-f(|x|)}{f(R)-f(\varepsilon)}.\]

For fixed \(R>|x|\), this probability tends to zero as \(\varepsilon\downarrow0\). A path that hits the origin before \(S_R\) must first hit every smaller sphere. Hence the probability of hitting the origin before \(S_R\) is zero. Take a countable increasing sequence of radii tending to infinity. Continuity bounds a path on every finite time interval, so any finite hit of the origin would occur before exit from one of those balls. Thus \(B_t\ne0\) for every \(t\geq0\), almost surely.

We may now apply Itō’s formula to the radius, first stopped in annuli and then without stopping. Since \(\partial_i|y|=y_i/|y|\) and \(\partial_{ii}|y|=(|y|^2-y_i^2)/|y|^3\),

\[|B_t|=|x|+\beta_t+\frac{d-1}{2}\int_0^t\frac{\mathrm ds}{|B_s|}.\]

For \(d\geq3\), the positive harmonic function \(F:\mathbb R^d\setminus\{0\}\to(0,\infty)\), \(F(y)=|y|^{2-d}\), similarly gives

\[|B_t|^{2-d}=|x|^{2-d}+\sum_{i=1}^d\int_0^t\frac{(2-d)B_s^i}{|B_s|^d}\,\mathrm dB_s^i.\]

The right side defines a continuous local martingale, localized by exits from annuli. It is nonnegative, so it is a supermartingale and \(\mathbb{E}|B_t|^{2-d}\leq|x|^{2-d}\). The nonnegative-supermartingale convergence theorem gives an almost-sure finite limit. The radius is unbounded almost surely, since every \(S_R\) is finite. Therefore that limit must be zero, and the radius itself tends to infinity.

Now take \(d=3\) and write \(Z_t=|B_t|^{-1}\). We have proved that \(Z\) is a continuous local martingale and \(Z_t\to0\) almost surely. It is nevertheless bounded in \(L^2\) at deterministic times. Set \(\delta=|x|/2\). For \(t>0\), split its second moment as

\[\mathbb{E}Z_t^2=\left(\int_{|y|<\delta}+\int_{|y|\geq\delta}\right) \frac{|y|^{-2}}{(2\pi t)^{3/2}}\exp\left(-\frac{|y-x|^2}{2t}\right)\,\mathrm dy.\]

The second integral is at most \(\delta^{-2}\). In the first, \(|y-x|\geq|x|/2\), while polar coordinates give \(\int_{|y|<\delta}|y|^{-2}\,\mathrm dy=4\pi\delta\). Consequently

\[\mathbb{E}Z_t^2\leq\delta^{-2}+4\pi\delta\,(2\pi t)^{-3/2}e^{-|x|^2/(8t)}.\]

The last function of \(t\) is continuous and tends to zero both as \(t\downarrow0\) and as \(t\to\infty\), so its supremum is finite. Including \(Z_0=|x|^{-1}\), this proves \(\sup_{t\geq0}\mathbb{E}Z_t^2<\infty\).

If \(Z\) were a true martingale, its expectations would all equal \(|x|^{-1}>0\). The deterministic-time \(L^2\) bound makes this family uniformly integrable, and its almost-sure limit is zero. Hence \(\mathbb{E}Z_t\to0\), a contradiction. Thus \(Z\) is a strict local martingale. The example separates uniform integrability at deterministic times from the class-D control over stopped values needed to remove localization. ◻