Why Localization Does Not Preserve Expectations Automatically
A local martingale behaves like a martingale after suitable stopping, but this does not make it a martingale before stopping. The distinction matters whenever an argument takes expectations in a stochastic differential equation: an identity that is valid for every stopped process may fail for the unstopped process if the stopped variables do not have enough integrability to pass to a limit.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard three-dimensional Brownian motion \(\mathbf{B}_t=(B^1_t,B^2_t,B^3_t)\), with its usual augmented filtration. A progressively measurable process \(H\) belongs to \(\mathcal{L}^2\) on \([0,T]\) when \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); in that case its Ito integral is a square-integrable martingale. Integrals used only locally are defined by stopping so that their integrands belong to \(\mathcal{L}^2\) on every stopped finite interval.
A Positive Local Martingale from a Brownian Radius
Fix \(a>0\), let \(Y_t=(a,0,0)+\mathbf{B}_t\), and set \(R_t=|Y_t|\). The process \(R\) does not hit zero almost surely. One way to justify this fact is to apply Ito's formula to the harmonic function \(y\mapsto1/|y|\) between spheres of radii \(\varepsilon\) and \(L\), where \(0<\varepsilon<a<L\). The stopped function is bounded, so optional stopping gives the probability of reaching \(\varepsilon\) before \(L\) as \[ \frac{1/a-1/L}{1/\varepsilon-1/L}. \] Letting \(L\) increase to infinity gives probability \(\varepsilon/a\) of ever reaching radius \(\varepsilon\). A path that hits zero must reach every such radius, and \(\varepsilon/a\to0\) as \(\varepsilon\downarrow0\). Thus \(R_t>0\) for all finite \(t\), almost surely.
On this event, apply Ito's formula to \(g(y)=|y|\) away from zero. Its gradient is \(y/|y|\), and its Laplacian in three dimensions is \(2/|y|\). The resulting radial equation is \(dR_t=d\beta_t+R_t^{-1}\,dt\), where \(\beta_t=\sum_{j=1}^3\int_0^t(Y_s^j/R_s)\,dB_s^j\). Each integral is understood locally: after stopping away from zero, its progressively measurable integrand is bounded and hence in \(\mathcal{L}^2\). The quadratic variation of \(\beta\) is \(t\); Levy's characterization says that this continuous local martingale is a Brownian motion. This characterization is used here as a standard result.
Now define \(X_t=1/R_t\), so \(X_0=1/a\). For \(f(r)=1/r\), the derivatives are \(f'(r)=-r^{-2}\) and \(f''(r)=2r^{-3}\). Ito's formula, including its second-derivative correction, gives \[ dX_t=f'(R_t)d\beta_t+f'(R_t)R_t^{-1}dt+\tfrac12 f''(R_t)dt =-X_t^2\,d\beta_t-X_t^3dt+X_t^3dt =-X_t^2\,d\beta_t. \] The drift and the Ito correction cancel. The integral in this equation is not asserted to be globally square-integrable; its local meaning will be made precise by stopping.
Proof. For integers \(n\) large enough that \(1/n<a<n\), let \(\tau_n\) be the first time \(R\) leaves \((1/n,n)\). These stopping times increase to infinity almost surely: on every finite time interval, the continuous positive path \(R\) has a positive minimum and a finite maximum. Before \(\tau_n\), \(X\le n\), so the stopped stochastic-integral integrand \(X_s^2\mathbf{1}_{\{s\le\tau_n\}}\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T X_s^4\mathbf{1}_{\{s\le\tau_n\}}ds\le Tn^4\). The stopped equation is therefore an Ito integral with an \(\mathcal{L}^2\) integrand, and its conditional expectation is zero. This proves the local martingale assertion. The expectation formula follows from the radial density computation below; since \(a/\sqrt t\) is finite, \(0<\Phi(a/\sqrt t)<1\), and therefore \(2\Phi(a/\sqrt t)-1<1\). Its expectation differs from \(X_0\), so \(X\) cannot be a martingale. \(\square\)
Computing the Lost Expectation
Worked Example: The Expectation at a Fixed Time
Compute \(\mathbb{E}[X_t]\) for \(t>0\). The vector \(Y_t\) has Gaussian density \((2\pi t)^{-3/2}\exp(-|y-(a,0,0)|^2/(2t))\). Integrating over spheres of radius \(r\), using polar angle \(\theta\) measured from the first coordinate axis, gives the density of \(R_t\): \[ p_t(r)=\frac{r}{a\sqrt{2\pi t}}\left(\exp\left(-\frac{(r-a)^2}{2t}\right)-\exp\left(-\frac{(r+a)^2}{2t}\right)\right),\qquad r>0. \] Indeed, the angular integral is \(2\pi\int_{-1}^{1}\exp(ar u/t)\,du\); multiplying by the radial factor \(r^2\) and the Gaussian normalizing constant yields the displayed density.
Since \(X_t=1/R_t\), its expectation is the integral of \(p_t(r)/r\). Substituting the density gives \[ \mathbb{E}[X_t]=\frac{1}{a\sqrt{2\pi t}}\int_0^\infty\left(\exp\left(-\frac{(r-a)^2}{2t}\right)-\exp\left(-\frac{(r+a)^2}{2t}\right)\right)dr =\frac{\Phi(a/\sqrt t)-\Phi(-a/\sqrt t)}{a}. \] The symmetry identity \(\Phi(-z)=1-\Phi(z)\) gives \(\mathbb{E}[X_t]=(2\Phi(a/\sqrt t)-1)/a\). This is finite and strictly less than \(1/a\) for every \(t>0\).
Check: as \(t\downarrow0\), \(a/\sqrt t\to\infty\), so \(\Phi(a/\sqrt t)\to1\) and the expectation tends to \(1/a=X_0\). For any fixed positive \(t\), the argument of \(\Phi\) is finite, so the expectation is strictly below \(X_0\), consistent with the proposition.
Worked Example: Stopping Restores the Martingale Expectation
For the same process, check the expectation after stopping at \(\tau_n\). On \([0,\tau_n]\), \(X\) lies between \(1/n\) and \(n\). The stopped equation is \[ X_{t\wedge\tau_n}=\frac1a-\int_0^t \mathbf{1}_{\{s\le\tau_n\}}X_s^2\,d\beta_s. \] The integrand is progressively measurable and its squared integral has expectation at most \(Tn^4\) over any fixed interval \([0,T]\). The Ito integral is consequently a square-integrable martingale, so its expectation is zero. Hence, for every \(t\ge0\), \[ \mathbb{E}[X_{t\wedge\tau_n}]=\frac1a. \]
For fixed \(t\), \(\tau_n\to\infty\) almost surely, hence \(X_{t\wedge\tau_n}\to X_t\) almost surely. Yet the expectations do not converge to \(\mathbb{E}[X_t]\), which is strictly less than \(1/a\) when \(t>0\). Thus almost-sure convergence alone does not permit passing expectations through this limit.
Check: the stopped variables have expectation \(1/a\) at every \(n\), whereas the limit has expectation \((2\Phi(a/\sqrt t)-1)/a\). These are unequal for \(t>0\); the discrepancy is precisely why a limit theorem requiring additional integrability cannot be omitted.
What Goes Wrong with the Expectation Argument
A tempting but invalid calculation starts from \(dX_t=-X_t^2\,d\beta_t\) and concludes that \(\mathbb{E}[X_t]=X_0\) because “the stochastic integral has mean zero.” That conclusion requires more than a formal SDE. The standard zero-mean statement applies when the integrand is in \(\mathcal{L}^2\) on the time interval in question. The localized integrands \(X^2\mathbf{1}_{\{s\le\tau_n\}}\) meet that requirement; the unstopped integrand has not been shown to.
In this example it is not globally square-integrable on any interval \([0,T]\) with \(T>0\). For each \(s>0\), the density \(p_s(r)\) behaves like a positive constant times \(r^2\) as \(r\downarrow0\), as follows either by expanding the two exponentials in its formula or by their first-order difference. Consequently, the integral for \(\mathbb{E}[X_s^4]=\int_0^\infty r^{-4}p_s(r)\,dr\) diverges near zero like \(\int_0 r^{-2}\,dr\). In particular, \(\mathbb{E}\int_0^T X_s^4\,ds=\infty\). The global Ito-isometry argument for a square-integrable integrand is unavailable.
There is still a useful general conclusion for nonnegative local martingales: they are supermartingales, not necessarily martingales. This means they are integrable and satisfy \(\mathbb{E}[M_t\mid\mathcal{F}_s]\le M_s\), but equality need not hold.
Proof. Choose a localizing sequence \(\tau_n\) for which \(M_{u\wedge\tau_n}\) is a martingale, and fix \(s\le t\). Nonnegativity and Fatou's lemma give \(\mathbb{E}[M_t]\le\liminf_n\mathbb{E}[M_{t\wedge\tau_n}]=\mathbb{E}[M_0]\), so \(M_t\) is integrable. For the conditional inequality, \(M_{t\wedge\tau_n}\) converges almost surely to \(M_t\). Conditional Fatou's lemma and the martingale identity for the stopped processes imply \[ \mathbb{E}[M_t\mid\mathcal{F}_s]\le\liminf_n\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s] =\liminf_n M_{s\wedge\tau_n}=M_s \] almost surely. The last limit holds because \(\tau_n\to\infty\) almost surely. The inequality may be strict; the reciprocal-radius example shows that it can be strict even after taking unconditional expectations. \(\square\)
The stopped variables in the example cannot be uniformly integrable at a fixed \(t>0\). Uniform integrability means that \(\sup_n\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]\to0\) as \(K\to\infty\). The standard uniform-integrability convergence theorem says that almost-sure convergence together with uniform integrability implies convergence of expectations. Here \(X_{t\wedge\tau_n}\to X_t\) almost surely, but their expectations are all \(1/a\), while the limit has strictly smaller expectation. Therefore the stopped variables cannot be uniformly integrable. This is exactly the missing condition in the attempted passage from stopped martingales to the unstopped process.
A Practical Test for Expectation-Preservation Claims
When an argument says that an SDE solution has constant expectation because its stochastic-integral term has mean zero, check the following points before accepting the conclusion:
- Is the integrand progressively measurable and in \(\mathcal{L}^2\) on the interval, or is another condition justifying a true martingale explicitly available?
- If the argument first stops the process, what proves that the stopped variables are uniformly integrable, or otherwise justifies the limit of their expectations?
- Is the process known only to be a local martingale? If it is nonnegative, the general conclusion is the supermartingale inequality, not expectation preservation.
Common mistake. From the local equation \(dX_t=-X_t^2\,d\beta_t\), one might write \(\mathbb{E}[X_t]=1/a\) by assigning zero expectation to the stochastic integral. The correction is to verify the integrability hypothesis for that integral. Here it fails: the stopped integrals have zero mean, but the unstopped integrand is not in \(\mathcal{L}^2\); moreover, the exact expectation is \((2\Phi(a/\sqrt t)-1)/a<1/a\). Localization proves a local martingale identity, not the expectation identity that the invalid calculation claims.
Check Your Understanding
Use the definitions and calculations above to assess each claim.
- Why does \(dX_t=-X_t^2\,d\beta_t\) establish a local martingale without establishing a true martingale?
- For \(a=2\) and \(t=4\), write the exact value of \(\mathbb{E}[X_t]\) and compare it with \(X_0\).
- Which integrability bound ensures that the stopped integrand \(X_s^2\mathbf{1}_{\{s\le\tau_n\}}\) belongs to \(\mathcal{L}^2\) on \([0,T]\)?
- Why does almost-sure convergence of \(X_{t\wedge\tau_n}\) not justify convergence of their expectations in this example?
- State the conditional expectation inequality that applies to a nonnegative local martingale, and explain when it can be strict.