From Deterministic Times to Stopping Times
The Markov property at a fixed time says that, conditional on the present state, the future evolution of a diffusion does not depend on its earlier history. The strong Markov property asserts the same conclusion when the time is a random stopping time: a time whose occurrence can be determined from observations made up to that time. For a time-homogeneous diffusion, this is the basis for restarting the process at random hitting and exit times.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Consider the one-dimensional SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with deterministic initial state \(X_0=x\). The first integral is an ordinary time integral; the second is an Ito integral. Whenever a stochastic integral is used below, its integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The coefficients are time-homogeneous and globally Lipschitz: for some \(L<\infty\), \(|\mu(x)-\mu(y)|+|\sigma(x)-\sigma(y)|\le L|x-y|\) for all \(x,y\). In particular, they have linear growth, since \(|\mu(x)|+|\sigma(x)|\le|\mu(0)|+|\sigma(0)|+L|x|\).
A stopping time \(\tau\) satisfies \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\); \(\mathcal{F}_\tau\) is the information available at \(\tau\). The assertion is a conditional expectation identity, not a claim that \(X_{\tau+s}\) is independent of \(\mathcal{F}_\tau\): its conditional law depends on the random state \(X_\tau\).
Stability and the Restart Argument
We use the standard existence theorem for globally Lipschitz SDEs: for every square-integrable initial value measurable at the starting time, there is a strong solution with finite second moments on bounded time intervals. Here “strong” means the solution is adapted to the given filtration and driven by its given Brownian motion. The Lipschitz conditions also give pathwise uniqueness: solutions with the same initial value and Brownian motion agree almost surely. A useful stability estimate can be obtained directly from the integral equation, without requiring a fourth-moment estimate.
Indeed, let \(X^x,X^y\) be solutions started from deterministic \(x,y\) and driven by the same Brownian motion. For \(t\le T\), subtract their integral equations. The inequality \(|a+b+c|^2\le3(|a|^2+|b|^2+|c|^2)\), Cauchy–Schwarz for the time integral, and the Ito isometry give
The stochastic-integral term is justified by the Ito isometry; the integrand difference is square-integrable because of the finite second moments and the Lipschitz bound. Gronwall’s inequality therefore yields \(\mathbb{E}|X_t^x-X_t^y|^2\le3|x-y|^2\exp(3L^2(T+1)t)\). In particular, for bounded continuous \(f\), \(P_s f(x)\) is continuous in \(x\): if \(x_n\to x\), this estimate gives \(X_s^{x_n}\to X_s^x\) in probability under the common driving Brownian motion, and boundedness and continuity of \(f\) imply convergence of the expectations.
Proof. First take a stopping time \(\rho\) with countably many deterministic values \(t_k\). On the event \(\{\rho=t_k\}\), the future Brownian increments \(B_{t_k+u}-B_{t_k}\), \(u\ge0\), are independent of \(\mathcal{F}_{t_k}\) and form a Brownian motion. Starting the SDE at \(X_{t_k}\) and driving it by these increments gives the post-\(t_k\) process. Pathwise uniqueness identifies that solution with \((X_{t_k+u})_{u\ge0}\). Consequently, for bounded continuous \(f\), \(\mathbb{E}[f(X_{\rho+s})\mid\mathcal{F}_\rho]=P_s f(X_\rho)\). This conditional identity follows by applying the deterministic-time identity on each event \(\{\rho=t_k\}\); the events form a countable partition. The deterministic-time identity itself follows from independent Brownian increments and the same restart and uniqueness argument.
Now let \(\tau\) be any almost surely finite stopping time and set \(\tau_n=2^{-n}\lceil2^n\tau\rceil\). Each \(\tau_n\) is a countably valued stopping time, \(\tau\le\tau_n\), and \(\tau_n\downarrow\tau\). Also \(\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}\). For any \(A\in\mathcal{F}_\tau\), the countably valued case gives, for bounded continuous \(f\), \[ \mathbb{E}[\mathbf{1}_A f(X_{\tau_n+s})] =\mathbb{E}[\mathbf{1}_A P_s f(X_{\tau_n})]. \] Continuity of sample paths implies \(X_{\tau_n+s}\to X_{\tau+s}\) and \(X_{\tau_n}\to X_\tau\) almost surely. Bounded convergence applies to the left side; continuity of \(P_s f\), proved above, and bounded convergence apply to the right side. Passing to the limit proves the conditional identity for every bounded continuous \(f\).
To extend the result carefully, take regular conditional distributions of \(X_{\tau+s}\) given \(\mathcal{F}_\tau\) and of the restart kernel \(P_s(X_\tau,\cdot)\); these exist for real-valued random variables. For rational \(a<b\) and positive integers \(m\), the functions \(\phi_{a,b,m}(z)=\max(0,1-m\,\operatorname{dist}(z,[a,b]))\) are bounded and continuous. There are countably many such functions. Intersect the probability-one events on which the conditional identity holds for each one, and also for the constant function \(1\). On this common event, dominated convergence as \(m\to\infty\) gives equality of the two conditional probability measures on every rational closed interval \([a,b]\). These intervals, together with the empty set and \(\mathbb{R}\), form a measure-determining class: they generate the Borel sets, so the probability measures agree on all Borel sets. Their integrals consequently agree for every bounded Borel \(f\), which is the stated conditional identity. \(\square\)
Worked Consequences
Worked Example: Brownian Motion with Drift
Let \(X_t=x+\tfrac12t+2B_t\). Find the conditional expectation of \(\cos(X_{\tau+1})\) given \(\mathcal{F}_\tau\), for an almost surely finite stopping time \(\tau\).
The post-\(\tau\) increment has the same law as \(\tfrac12+2Z\), where \(Z\sim N(0,1)\), independent of \(\mathcal{F}_\tau\). Thus the strong Markov formula gives \(\mathbb{E}[\cos(X_{\tau+1})\mid\mathcal{F}_\tau]=\mathbb{E}[\cos(X_\tau+\tfrac12+2Z)\mid X_\tau]\). Using \(\mathbb{E}[e^{iuZ}]=e^{-u^2/2}\) at \(u=2\), this is \(e^{-2}\cos(X_\tau+\tfrac12)\).
Check: at a deterministic time \(\tau=0\), the same computation gives \(\mathbb{E}[\cos(x+\tfrac12+2B_1)]=e^{-2}\cos(x+\tfrac12)\). The damping factor is \(e^{-(2)^2/2}=e^{-2}\), as required by the normal characteristic function.
Worked Example: Restarting an Ornstein–Uhlenbeck Process
Consider \(dX_t=(2-X_t)\,dt+dB_t\). Find the conditional mean and variance of \(X_{\tau+\log 2}\) given \(\mathcal{F}_\tau\).
For the integrating factor, apply Ito’s formula to \(g(t,x)=e^t x\): \(g_t=e^t x\), \(g_x=e^t\), and \(g_{xx}=0\), so the correction \(\tfrac12g_{xx}\,dt\) is zero. The product calculation gives \(d(e^tX_t)=2e^t\,dt+e^t\,dB_t\). Restarting at \(\tau\), integration over an interval of length \(s\) yields \(X_{\tau+s}=e^{-s}X_\tau+2(1-e^{-s})+\int_0^s e^{-(s-u)}\,d\widetilde B_u\), where \(\widetilde B_u=B_{\tau+u}-B_\tau\). This Ito integrand is progressively measurable and square-integrable, since \(\int_0^s e^{-2(s-u)}\,du<\infty\). Its integral has conditional mean zero and conditional variance equal to that integral, by the Ito isometry and the Brownian strong Markov property.
Set \(s=\log 2\), so \(e^{-s}=\tfrac12\). The conditional mean is \(\tfrac12X_\tau+2(1-\tfrac12)=\tfrac12X_\tau+1\). The conditional variance is \(\int_0^{\log 2}e^{-2(\log 2-u)}du=(1-e^{-2\log 2})/2=(1-\tfrac14)/2=\tfrac38\).
Check: the stochastic integral has conditional mean zero, so taking the conditional expectation of the displayed restart formula gives exactly \(\tfrac12X_\tau+1\). Its variance is \(\tfrac38\), independently of \(X_\tau\), as the Ito isometry gives.
Worked Example: A Geometric Brownian Motion at a Stopping Time
Let \(dS_t=\tfrac14S_t\,dt+\tfrac12S_t\,dB_t\), with \(S_0>0\). Compute \(\mathbb{E}[S_{\tau+1}\mid\mathcal{F}_\tau]\).
To verify the explicit solution, set \(G(t,z)=S_0\exp((\tfrac14-\tfrac18)t+\tfrac12z)\). Its derivatives satisfy \(G_t=\tfrac18G\), \(G_z=\tfrac12G\), and \(G_{zz}=\tfrac14G\). Ito’s formula, including the correction \(\tfrac12G_{zz}\,dt=\tfrac18G\,dt\), gives drift \((\tfrac18+\tfrac18)G\,dt=\tfrac14G\,dt\) and diffusion \(\tfrac12G\,dB_t\), verifying the SDE. Restarting at \(\tau\) gives \(S_{\tau+1}=S_\tau\exp(\tfrac18+\tfrac12Z)\) conditionally, where \(Z\sim N(0,1)\) is independent of \(\mathcal{F}_\tau\). Therefore \(\mathbb{E}[S_{\tau+1}\mid\mathcal{F}_\tau]=S_\tau e^{1/8}\mathbb{E}[e^{Z/2}]=S_\tau e^{1/8}e^{(1/2)^2/2}=S_\tau e^{1/4}\).
Check: at a deterministic starting state \(S_\tau=s_0\), the expected multiplicative factor over one time unit is \(e^{1/4}\), matching the drift coefficient in the SDE. The calculation uses the normal moment-generating function at \(1/2\), namely \(e^{1/8}\).
Why the Hypotheses Matter
Time homogeneity matters: the kernel \(P_s\) depends only on the elapsed time \(s\) and current state. If coefficients also depend on calendar time, the state \(X_\tau\) alone need not determine the future law; augmenting the state with time is the appropriate treatment. Global Lipschitz conditions provide a strong solution, pathwise uniqueness, finite second moments, and the stability needed for continuity of \(P_s f\). They are sufficient conditions, not necessary ones.
Common mistake: using the absolute Brownian value instead of its future increment. A wrong calculation for \(X_t=x+\tfrac12t+2B_t\) is to write \(X_{\tau+1}=X_\tau+\tfrac12+2B_{\tau+1}\) and then treat \(B_{\tau+1}\) as a fresh standard normal independent of \(\mathcal{F}_\tau\). The correct subtraction is \(X_{\tau+1}-X_\tau=\tfrac12+2(B_{\tau+1}-B_\tau)\). It is the increment \(B_{\tau+1}-B_\tau\), not \(B_{\tau+1}\), that has the fresh Brownian law independently of the information at \(\tau\). Omitting the subtraction double-counts the past Brownian value.
The proof’s countable-class step is also essential. An almost-sure identity for each bounded continuous test function separately does not automatically supply one common probability-one event for an uncountable collection of functions. A countable measure-determining class gives a single event on which the conditional laws agree, after which all bounded Borel tests follow.
Check Your Understanding
Use the assumptions and proof above to answer the following questions.
- Why does the stability estimate imply continuity of \(P_s f(x)\) when \(f\) is bounded and continuous?
- For \(\tau_n=2^{-n}\lceil2^n\tau\rceil\), why is every event in \(\mathcal{F}_\tau\) also in \(\mathcal{F}_{\tau_n}\)?
- In the Ornstein–Uhlenbeck example, recompute the conditional variance at \(s=\log 2\) from the Ito-isometry integral.
- Why is pathwise uniqueness needed when identifying the process after a deterministic or countably valued restart time?
- Explain why proving the conditional identity separately for every bounded continuous function is not, by itself, sufficient to assert equality of conditional laws on one common probability-one event.