Two Time Variables, Two Equations
For a time-homogeneous diffusion, the same generator describes evolution regardless of when the process starts. With time-dependent coefficients, the generator changes with the clock. This makes it especially important to distinguish the starting time \(s\) from the ending time \(t\): the backward equation differentiates in \(s\), while the forward equation differentiates in \(t\).
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Consider the one-dimensional SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\). Here \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. Assume that \(\mu,\sigma\) are continuous in time, globally Lipschitz in space uniformly on each finite time interval, and satisfy a uniform linear growth bound there: for each \(T<\infty\), some \(C_T<\infty\) gives \(|\mu(t,x)|+|\sigma(t,x)|\le C_T(1+|x|)\) for \(0\le t\le T\). The standard Lipschitz existence theorem then gives a unique strong solution from each deterministic starting pair \((s,x)\); that theorem is used here as a standard result.
All stochastic integrals below are Ito integrals. Their integrands are progressively measurable and belong to \(\mathcal{L}^2\) on the time interval in question, meaning \(\mathbb{E}\int H_r^2\,dr<\infty\). The stated growth assumptions and finite-time moment bounds ensure this for the SDE integrands; when applying Ito's formula to a test function, its derivatives and the coefficients are taken to satisfy the corresponding integrability conditions.
The Two-Time Kolmogorov Equations
The equations below are stated for \(s<t\). The backward equation is for \(u(s,x)=P_{s,t}g(x)\), with \(t\) held fixed. It requires sufficient smoothness of \(u\) for a classical equation; otherwise its corresponding weak formulation is used. The forward equation is first understood weakly, and it becomes a pointwise PDE when the transition density has the required derivatives. Smoothness of coefficients alone is not being used here as a claim that every transition law has a classical density.
Proof. First fix the terminal time \(t\), and take \(s<t\). The Markov property gives \(u(s,x)=\mathbb{E}[u(s+h,X_{s+h})\mid X_s=x]\) for \(0<h<t-s\). Apply Ito's formula to \(F(r,z)=u(r,z)\) along \(X_r\) on \([s,s+h]\). Its derivatives are \(F_r=u_r\), \(F_z=u_x\), and \(F_{zz}=u_{xx}\), so the formula, including the quadratic-variation correction, is \(du(r,X_r)=[u_r(r,X_r)+\mu(r,X_r)u_x(r,X_r)+\tfrac12\sigma^2(r,X_r)u_{xx}(r,X_r)]\,dr+u_x(r,X_r)\sigma(r,X_r)\,dB_r\). The stochastic integral has expectation zero because its integrand is in \(\mathcal{L}^2\). Taking conditional expectations and using the Markov identity gives \(0=\mathbb{E}[\int_s^{s+h}(u_r+\mathcal{L}_r u)(r,X_r)\,dr\mid X_s=x]\). Divide by \(h\), then let \(h\downarrow0\). Continuity of the integrand and of the solution yields \(u_s(s,x)+\mathcal{L}_s u(s,x)=0\). At \(s=t\), the definition of the transition operator gives \(u(t,x)=g(x)\).
For the forward statement, apply Ito's formula instead to \(F(r,z)=\varphi(z)\). Here \(F_r=0\), \(F_z=\varphi'\), and \(F_{zz}=\varphi''\), giving \(d\varphi(X_r)=[\mu(r,X_r)\varphi'(X_r)+\tfrac12\sigma^2(r,X_r)\varphi''(X_r)]\,dr+\varphi'(X_r)\sigma(r,X_r)\,dB_r\). The stochastic integral has expectation zero under the stated square-integrability condition. Taking expectations and integrating from \(s\) to \(t\) proves \(\mathbb{E}[\varphi(X_t)]-\mathbb{E}[\varphi(X_s)]=\int_s^t\mathbb{E}[(\mathcal{L}_r\varphi)(X_r)]\,dr\). If the law has density \(p(s,x;r,y)\), this identity says that its time derivative, tested against \(\varphi\), is the density paired with \(\mathcal{L}_t\varphi\). Integration by parts transfers the derivatives from \(\varphi\) to the density and coefficients, giving \(\mathcal{L}_t^*p\). Compact support of \(\varphi\) removes boundary terms in this local weak calculation. At \(t=s\), the process starts at \(x\), so its law is the point mass \(\delta_x\). This proves the weak statement and the asserted classical equation whenever the density has the required regularity. \(\square\)
The two equations describe the same transition family from opposite sides. In the backward equation, the terminal time is fixed and the initial time moves. In the forward equation, the initial pair \((s,x)\) is fixed and the terminal law evolves. In particular, the forward operator is the adjoint in the spatial variable; it is not obtained by simply changing the sign of the backward generator.
Worked Calculations
Worked Example: A Gaussian Transition with Time-Varying Drift and Volatility
Take \(\mu(r,x)=r\), \(\sigma(r,x)=1+r\), and start at \(X_s=x\). The coefficients do not depend on the state, so integrating the SDE gives \(X_t=x+\int_s^t r\,dr+\int_s^t(1+r)\,dB_r\). The first integral is \((t^2-s^2)/2\). The second is a centered normal variable with variance \(\int_s^t(1+r)^2\,dr=((1+t)^3-(1+s)^3)/3\). Thus define \(m(s,x;t)=x+(t^2-s^2)/2\) and \(v(s;t)=((1+t)^3-(1+s)^3)/3\); for \(s<t\), \(v>0\), and the transition density is \(p(s,x;t,y)=(2\pi v)^{-1/2}\exp(-(y-m)^2/(2v))\).
Check the forward equation directly. Write \(z=y-m\). Since \(m_t=t\), \(v_t=(1+t)^2\), and \(p_y=-zp/v\), differentiation gives \(p_t=p[-v_t/(2v)+zt/v+z^2v_t/(2v^2)]\). On the other hand, the forward operator is \(-\partial_y(tp)+\tfrac12(1+t)^2\partial_{yy}p\). Using \(p_{yy}=(z^2/v^2-1/v)p\), it equals \(p[tz/v+\tfrac12v_t(z^2/v^2-1/v)]\), the same expression. The initial condition is recovered as \(t\downarrow s\): \(m\to x\), \(v\to0\), and the Gaussian density converges to \(\delta_x\) as a distribution.
Check: the mean of the displayed transition is \(x+(t^2-s^2)/2\), exactly the starting value plus the integrated drift. Its variance is \(((1+t)^3-(1+s)^3)/3\), exactly the integrated squared diffusion coefficient.
Worked Example: A Backward Solution for a Quadratic Payoff
Use the same coefficients and terminal payoff \(g(x)=x^2\). For \(s\le t\), set \(M(s,t)=(t^2-s^2)/2\) and \(V(s,t)=((1+t)^3-(1+s)^3)/3\). Since \(X_t\) is normal with mean \(x+M\) and variance \(V\), its second moment gives \(u(s,x)=P_{s,t}g(x)=(x+M)^2+V\).
Here \(u_x=2(x+M)\), \(u_{xx}=2\), and \(M_s=-s\), \(V_s=-(1+s)^2\), so \(u_s=-2s(x+M)-(1+s)^2\). The generator at time \(s\) gives \(\mathcal{L}_s u=s\,2(x+M)+\tfrac12(1+s)^2\,2=2s(x+M)+(1+s)^2\). Therefore \(u_s+\mathcal{L}_s u=0\). At \(s=t\), \(M=V=0\), so \(u(t,x)=x^2\), as required.
Check: the transition calculation gives \(\mathbb{E}[X_t^2\mid X_s=x]=(x+M)^2+V\), the same function as the PDE calculation. The variance term \(V\) is necessary: omitting it would give the square of the mean rather than the second moment.
Worked Example: A Time-Dependent Ornstein–Uhlenbeck Transition
Consider \(dX_r=-X_r/(1+r)\,dr+dB_r\!/(1+r)\), with \(X_s=x\). Apply the deterministic product rule to \(Y_r=(1+r)X_r\). Since \(1+r\) has finite variation, its quadratic covariation with \(X\) is zero, and \(dY_r=X_r\,dr+(1+r)dX_r=X_r\,dr-X_r\,dr+dB_r=dB_r\). Consequently \(X_t=((1+s)x+B_t-B_s)/(1+t)\). Its transition is normal with mean \(m=(1+s)x/(1+t)\) and variance \(v=(t-s)/(1+t)^2\).
For \(s=0,t=1\), substitution gives \(m=x/2\) and \(v=1/4\), hence \(p(0,x;1,y)=\sqrt{2/\pi}\exp(-2(y-x/2)^2)\). Indeed, the normal density factor is \(1/\sqrt{2\pi(1/4)}=2/\sqrt{2\pi}=\sqrt{2/\pi}\), and its exponent is \(-(y-x/2)^2/(2(1/4))=-2(y-x/2)^2\).
Check: the explicit representation gives \(\mathbb{E}[X_1\mid X_0=x]=x/2\) and \(\operatorname{Var}(X_1\mid X_0=x)=\operatorname{Var}(B_1)/4=1/4\), matching the density. This example also shows that a time-dependent drift can change the conditional mean by a time-dependent factor, even when the diffusion is Gaussian.
Keeping the Variables Straight
| Equation | Fixed variables | Variable being evolved | Spatial operator |
|---|---|---|---|
| Backward | Terminal time \(t\), payoff \(g\) | Initial time \(s\) and state \(x\) | \(\mathcal{L}_s\) acts on the function |
| Forward | Initial time \(s\) and state \(x\) | Terminal time \(t\) and state \(y\) | \(\mathcal{L}_t^*\) acts on the density |
Common mistake. Writing \(\partial_t u=\mathcal{L}_t u\) for \(u(s,x)=\mathbb{E}[g(X_t)\mid X_s=x]\) confuses the terminal-time evolution of a density with the backward equation for a payoff. The backward equation differentiates in the initial time and has a plus sign: \(\partial_s u+\mathcal{L}_s u=0\). For a quick check, in the first example the mean payoff \(g(y)=y\) is \(u(s,x)=x+(t^2-s^2)/2\). Its initial-time derivative is \(-s\), while \(\mathcal{L}_s u=s\); their sum is zero. Using a terminal-time derivative in its place would not give this identity.
These equations matter whenever coefficients change with calendar time, including diffusions with time-varying drift or volatility. The transition operator \(P_{s,t}\) depends on both times, so it is not generally a one-parameter semigroup depending only on \(t-s\). The generator’s time label must therefore be retained: \(\mathcal{L}_s\) in the backward equation and \(\mathcal{L}_t^*\) in the forward equation.
Check Your Understanding
Use the definitions and equations above to answer each question. In density questions, assume \(s<t\).
- For \(\mu(r,x)=c(r)\) and \(\sigma(r,x)=a(r)\), write the mean and variance of \(X_t\) given \(X_s=x\) as time integrals.
- For a smooth \(u(s,x)\), expand \(\partial_su+\mathcal{L}_su=0\) using \(\mu\) and \(\sigma\).
- Write the forward equation in one dimension and identify which variable its derivatives act on.
- For the time-dependent Ornstein–Uhlenbeck example, compute the transition mean and variance at \(s=1,t=2\).
- Explain why the initial condition for a transition density is a point mass rather than an ordinary smooth density at \(t=s\).