Time-Inhomogeneous Diffusions as Augmented Markov Processes

Why Add the Clock to the State?

A diffusion with coefficients that depend on calendar time can still be Markov: once the present time and present position are known, the future law does not require the earlier path. But its transition law generally depends on the starting time as well as the elapsed time. Adjoining the clock records that extra information in the state. The resulting pair has transitions that depend only on the elapsed time, so it is a time-homogeneous Markov process.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). On each finite time interval, consider the SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\). Assume \(\mu\) and \(\sigma\) are measurable in time, globally Lipschitz in the state variable uniformly in time, and satisfy a uniform linear growth bound. These standard existence and uniqueness conditions give a unique strong solution from each deterministic starting time \(s\) and state \(x\). The stochastic integrand \(\sigma(t,X_t)\) is progressively measurable and belongs to \(\mathcal{L}^2\) on finite intervals: \(\mathbb{E}\int_s^T \sigma(t,X_t)^2\,dt<\infty\). The drift is \(\mu\); the diffusion coefficient is \(\sigma\).

Definition: For \(s\le t\), let \(P_{s,t}(x,A)=\mathbb{P}(X_t^{s,x}\in A)\) be the transition kernel of the diffusion started from \(x\) at time \(s\), where \(A\) is a Borel subset of \(\mathbb{R}\). The augmented state is \(Y_t=(t,X_t)\), taking values in \(\mathbb{R}_+\times\mathbb{R}\). Its elapsed-time-\(h\) transition kernel is \(\widetilde P_h((s,x),C)=\int_{\mathbb{R}}\mathbf{1}_C(s+h,y)\,P_{s,s+h}(x,dy)\), for \(h\ge0\) and Borel \(C\subseteq\mathbb{R}_+\times\mathbb{R}\).

The indicator makes the deterministic clock update explicit: the first coordinate is \(s+h\), while the second coordinate is distributed according to the original diffusion's transition law. This definition applies also at \(h=0\); \(P_{s,s}(x,\cdot)\) is the point mass at \(x\), so \(\widetilde P_0((s,x),\cdot)\) is the point mass at \((s,x)\).

The Augmented Markov Property

Theorem: Under the stated assumptions, \(Y_t=(t,X_t)\) is a time-homogeneous Markov process with transition kernel \(\widetilde P_h\). More precisely, for every bounded Borel function \(g\), every \(s\le t\), and \(h\ge0\), \(\mathbb{E}[g(t+h,X_{t+h})\mid\mathcal{F}_t]=\int_{\mathbb{R}}g(t+h,y)\,P_{t,t+h}(X_t,dy)\) almost surely. The kernels also satisfy \(\widetilde P_{h+k}=\widetilde P_k\widetilde P_h\) for \(h,k\ge0\), where kernel composition means integrating the second kernel against the first.

Proof. At deterministic time \(t\), the Brownian increments after \(t\) are independent of \(\mathcal{F}_t\) and form a Brownian motion when viewed from time \(t\). Restart the SDE at \(t\) from \(X_t\), using these increments and the coefficients evaluated at the actual calendar times \(r\ge t\). Pathwise uniqueness makes this restarted solution agree with the original solution after \(t\). Consequently, conditional on \(\mathcal{F}_t\), its value at \(t+h\) has law \(P_{t,t+h}(X_t,\cdot)\). Integrating the bounded function \(y\mapsto g(t+h,y)\) gives the displayed conditional expectation. This is the Markov property, and the transition formula uses only the current augmented state \((t,X_t)\) and elapsed time \(h\), which is time homogeneity.

For the semigroup law, start from \((s,x)\). After elapsed time \(k\), the clock equals \(s+k\), and the position has law \(P_{s,s+k}(x,\cdot)\). From each resulting position \(z\), the next \(h\) units of time use \(P_{s+k,s+k+h}(z,\cdot)\). The Markov property just proved therefore gives the composition \(P_{s,s+k}P_{s+k,s+k+h}=P_{s,s+k+h}\). The clock after both steps is \(s+k+h\), exactly the clock after elapsed time \(k+h\). Thus the two augmented kernels agree on every Borel set \(C\), proving \(\widetilde P_{h+k}=\widetilde P_k\widetilde P_h\). \(\square\)

The composition order here means: first run the kernel for \(k\), then the kernel for \(h\). The time coordinate prevents confusion about which calendar-time transition law to use. In particular, the formula does not assert that \(P_{s,s+h}\) is independent of \(s\); it usually is not.

Worked Kernels

Worked Example: A Drift That Increases with Calendar Time

Consider \(dX_t=t\,dt+dB_t\), started from \(X_s=x\). Find its transition law after elapsed time \(h\), and its augmented kernel from \((2,1)\) after two time units.

Integrating the drift from \(s\) to \(s+h\) gives \(X_{s+h}=x+\int_s^{s+h}r\,dr+(B_{s+h}-B_s)=x+sh+\tfrac12h^2+(B_{s+h}-B_s)\). The Brownian increment is independent of \(\mathcal{F}_s\) and has law \(N(0,h)\) for \(h>0\). Hence \(P_{s,s+h}(x,\cdot)\) is the normal law with mean \(x+sh+\tfrac12h^2\) and variance \(h\).

For \(s=2,x=1,h=2\), the mean is \(1+2\cdot2+\tfrac12(2^2)=7\), and the variance is \(2\). Thus the augmented kernel puts the clock at \(4\) and assigns the position the law \(N(7,2)\): \(\widetilde P_2((2,1),C)=\mathbb{P}((4,Z)\in C)\) for \(Z\sim N(7,2)\).

Check: the drift accumulated over \([2,4]\) is \(\int_2^4r\,dr=(16-4)/2=6\), so the position's mean is \(1+6=7\), as computed. The variance is the variance of \(B_4-B_2\), namely \(4-2=2\).

Worked Example: A Time-Dependent Linear Drift

On the interval \([1,2]\), consider \(dX_t=-(\log 2)X_t\,dt+dB_t\), with \(X_1=x\). Find the transition law at time \(2\), then express the corresponding augmented transition.

Write \(c=\log 2\). Multiplying by the deterministic integrating factor \(e^{ct}\) gives the solution \(X_t=e^{-c(t-1)}x+\int_1^t e^{-c(t-r)}\,dB_r\). The integrand is deterministic and square-integrable on \([1,2]\), so the Ito integral is centered normal with variance equal to its squared-integrand integral. At \(t=2\), the mean is \(e^{-c}x=x/2\), and the variance is \(\int_1^2e^{-2c(2-r)}\,dr=(1-e^{-2c})/(2c)=3/(8\log 2)\). The transition law is therefore \(N(x/2,\,3/(8\log 2))\), and \(\widetilde P_1((1,x),C)=\mathbb{P}((2,Z)\in C)\) for \(Z\) with this law.

Check: \(e^{-c}=e^{-\log 2}=1/2\) and \(e^{-2c}=e^{-2\log 2}=1/4\), so the variance numerator is \(1-1/4=3/4\); dividing by \(2\log 2\) gives \(3/(8\log 2)\). The mean and variance also follow directly from the displayed solution: the stochastic integral has mean zero and variance equal to the integral of the squared integrand.

The integrating-factor calculation uses only the deterministic product rule for a smooth factor and an Ito process. Equivalently, Ito's formula applied to \(f(t,z)=e^{ct}z\) has \(f_t=ce^{ct}z\), \(f_z=e^{ct}\), and \(f_{zz}=0\); thus its \(\tfrac12 f_{zz}\,dt\) correction is zero. Substitution into the SDE yields \(d(e^{ct}X_t)=e^{ct}\,dB_t\), which integrates to the solution above.

Why the Clock Coordinate Matters

Worked Example: The Position Alone Is Not Time-Homogeneous

Let \(X_t=B_t+t^2\), so \(dX_t=2t\,dt+dB_t\). From time \(s\) and position \(x\), the position after elapsed time \(h>0\) has law \(N(x+2sh+h^2,h)\). Show that \(X_t\) cannot have a time-homogeneous transition kernel, and give the augmented kernel.

Take the same position \(x\) and elapsed time \(h=1\). Starting at calendar time \(s=0\), the next position has law \(N(x+1,1)\). Starting at calendar time \(s=1\), it has law \(N(x+3,1)\). These laws differ: their means differ by \(2\). A time-homogeneous kernel for \(X\) would assign the same transition law from \(x\) after one time unit, regardless of the starting calendar time. Hence no such kernel describes these transitions.

The augmented transition records the time and is \(\widetilde P_h((s,x),C)=\mathbb{P}((s+h,Z)\in C)\), where \(Z\sim N(x+2sh+h^2,h)\). For \(s=0,h=1\), the clock becomes \(1\) and the mean is \(x+1\). For \(s=1,h=1\), the clock becomes \(2\) and the mean is \(x+3\). The distinct transition laws are now associated with distinct augmented starting states.

Check: the drift accumulated on \([s,s+1]\) is \(\int_s^{s+1}2r\,dr=(s+1)^2-s^2=2s+1\). It equals \(1\) at \(s=0\) and \(3\) at \(s=1\), matching both means after adding the starting position \(x\). In each case the Brownian increment has variance \(1\).

Kernel Interpretation and Common Pitfalls

The augmented kernel is a genuine transition kernel on the product state space: for each starting state it is a probability measure in \(C\), and for each Borel set \(C\) it is measurable in the starting state under the usual transition-kernel measurability of the diffusion. Its time coordinate is deterministic, but that does not make the position transition independent of the clock. Rather, it makes the clock dependence part of the current state, so the Markov transition rule can be indexed by elapsed time alone.

A common mistake is to call \(X_t\) time-homogeneous merely because it is Markov. The Markov property says that the conditional future law depends on the current state and current time; time homogeneity is the stronger assertion that transitions depend on elapsed time but not on calendar time. The example \(B_t+t^2\) is Markov but fails that stronger property.

A second mistake is to write a single kernel \(P_h(x,\cdot)\) for a time-inhomogeneous diffusion without verifying that it does not depend on the starting time. The correct object before augmentation is \(P_{s,t}(x,\cdot)\). After augmentation, \(\widetilde P_h((s,x),\cdot)\) is indexed by elapsed time \(h\), and its starting state includes \(s\). This distinction is useful whenever one wants to use time-homogeneous Markov-process tools without discarding genuinely time-dependent coefficients.

Check Your Understanding

For each question, distinguish calendar time from elapsed time and include degenerate cases where relevant.

For \(dX_t=a(t)\,dt+dB_t\) with deterministic integrable \(a\), express the transition law from \(x\) at time \(s\) after elapsed time \(h\).
In the definition of \(\widetilde P_h\), what does the first coordinate equal when \(h=0\), and what is the position law in that case?
Explain why the Markov property alone does not imply that the transition law of \(X\) is time-homogeneous.
State the semigroup composition identity for \(\widetilde P_k\) followed by \(\widetilde P_h\), including the resulting clock coordinate.
For \(X_t=B_t+t^2\), compute the mean shift over an interval of length \(h\) beginning at time \(s\), and identify which term depends on calendar time.