advanced32 min studyLesson 677 in Mathematics

Kolmogorov backward and forward equations

Distinguish evolution of observables from evolution of laws and qualify every density claim.

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What you’ll understand

  • Derive the backward terminal-value equation for expected observables.
  • Write the distributional forward equation and its density form.
  • Check drift signs and the half-variance factor in a Gaussian example.

The same diffusion produces two complementary PDE descriptions. The backward equation evolves an observable as a function of its starting point; the forward equation evolves the law of the endpoint. Their drift signs differ because the operators are adjoints. We first write weak identities that require no density, then identify extra smoothness and nondegeneracy assumptions under which classical density equations are legitimate. Keeping these levels separate prevents deterministic or degenerate diffusions from being assigned nonexistent smooth densities.

Backward evolution acts on starting states

For a nonexplosive Markov diffusion and a terminal observable gg, define u(t,x)u(t,x) as the expected value of g(XT)g(X_T) when the process starts from xx at time tt. If this function is once differentiable in time and twice in space with the needed integrability, applying Itô's formula along the process shows that its drift must vanish. It therefore solves the backward equation with terminal, rather than initial, data. Time-homogeneous elapsed-time notation reverses the time derivative and yields the semigroup equation instead.

A clean sufficient classical setting is smooth bounded coefficients with bounded derivatives and a uniformly positive definite covariance, together with smooth compactly supported terminal data. More general settings exist, but global Lipschitz existence of an SDE alone does not automatically supply all these PDE derivatives. Unbounded observables can also be treated when suitable moment and growth controls are separately established. An explicit polynomial example below verifies those controls through a Gaussian solution rather than invoking a bounded-data theorem outside its assumptions.

u(t,x)=Et,x[g(XT)],∂tu+Lu=0,u(T,x)=g(x)u(t,x)=\mathbb E_{t,x}[g(X_T)],\qquad \partial_tu+Lu=0,\qquad u(T,x)=g(x)
The spatial operator acts on the starting-state variable, and the time direction runs toward a fixed terminal horizon.↔ Scroll wide equations horizontally.

Forward evolution acts on measures

Let μt\mu_t be the distribution of the process. The martingale test identity implies that its pairing with a compactly supported smooth test changes by the pairing with LL applied to that test. This is a distributional forward equation for a family of measures. It remains meaningful when the law has atoms. For example, a zero-diffusion equation with a deterministic initial state keeps a point mass, so there is no ordinary Lebesgue density to differentiate.

When a sufficiently smooth density exists and integration by parts has no remaining boundary terms, the adjoint operator gives the Fokker–Planck equation. In one dimension the drift becomes a negative first derivative of the drift-density product, and the diffusion becomes one half of a second derivative of the variance-density product. Coefficients cannot be pulled outside these derivatives when they depend on position. On a bounded domain, killing, reflection, or another boundary rule also changes the corresponding flux or boundary condition.

ddt∫φ dμt=∫Lφ dμt,∂tρ=−∂x(bρ)+12∂xx(σ2ρ)\frac{d}{dt}\int\varphi\,d\mu_t=\int L\varphi\,d\mu_t,\qquad \partial_t\rho=-\partial_x(b\rho)+\tfrac12\partial_{xx}(\sigma^2\rho)
The measure identity is fundamental; the density PDE is its adjoint form when a density and sufficient regularity exist.↔ Scroll wide equations horizontally.

The initial delta is not a smooth initial density

For constant drift and positive constant volatility, transition laws from a fixed point are Gaussian at every positive elapsed time. Their variance is volatility squared times elapsed time, not half that quantity. The half appears in the differential operator. At zero elapsed time the law is a point mass, so its density formula is singular there. The correct initial condition is convergence of measures to the point mass, or convergence of expectations of bounded continuous tests.

The backward and forward equations are linked by duality: evolving a test and pairing with an initial law equals pairing that test with the evolved law. This equality explains their different variables and signs without choosing one as more physical. A PDE calculation should identify which variable is being differentiated, whether time is elapsed or terminal, and whether the claimed solution is classical or distributional. These three labels prevent most apparent contradictions between correct versions of the same transition formula.

A Gaussian law and a backward quadratic

Let dXs=ds+2dWsdX_s=ds+2dW_s. Starting from Xt=xX_t=x, compute the endpoint law at a later time T and the backward expected value of g(XT)=XT2g(X_T)=X_T^2. Verify the backward PDE directly.

  1. Write h=T−th=T-t. The endpoint is x+h+2(WT−Wt)x+h+2(W_T-W_t), hence has Gaussian mean x+hx+h and variance 4h4h. Its expected square is u(t,x)=(x+h)2+4hu(t,x)=(x+h)^2+4h.

  2. Differentiate with respect to the starting-time variable: ut=−2(x+h)−4u_t=-2(x+h)-4. The spatial derivatives are ux=2(x+h)u_x=2(x+h) and uxx=2u_{xx}=2.

  3. The generator is L=∂x+2∂xxL=\partial_x+2\partial_{xx}, giving Lu=2(x+h)+4Lu=2(x+h)+4. Therefore ut+Lu=0u_t+Lu=0, and at t=Tt=T the formula reduces to x2x^2. Gaussian finite moments justify this unbounded terminal observable.

The result

The terminal variance is four times elapsed time, while the generator's second-derivative coefficient is two. The explicit derivatives check both the time sign and the factor one half.

Explore the model

Interactive lab

Interactive lab · heat

A periodic forward mode

Pause/Reset, then set both mode to 2 and diffusivity to 1. Predict amplitude exp(−2) after one Step at time 0.5. Reset and set both mode to 1 and diffusivity to 1; predict exp(−0.5). These are exact periodic heat modes with covariance coefficient twice the diffusivity, not full Gaussian densities on the line.

A periodic forward modeThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA diffusion damps higher spatial frequencies faster. The signed sine mode is a PDE test component, not a nonnegative probability density by itself.

Check your understanding

Which forward statement remains meaningful without assuming a density exists?

Try it yourself

For dXt=−2dt+6 dWtdX_t=-2dt+\sqrt6\,dW_t with X0=3X_0=3, find the law at time two and write the constant-coefficient forward density PDE for positive times. State the initial condition as a measure.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Multiply elapsed time by the drift for the mean shift and by volatility squared for the variance. The forward operator is the adjoint.

Compare with the solution

The law at time two is N(−1,12)N(-1,12), because its mean is 3−4=−13-4=-1 and variance is 6⋅2=126\cdot2=12. The forward equation is ∂tρ=2∂xρ+3∂xxρ\partial_t\rho=2\partial_x\rho+3\partial_{xx}\rho. Its initial law is δ3\delta_3, interpreted weakly through integrals against bounded continuous tests. The positive drift-derivative coefficient in the forward PDE is consistent with a negative trajectory drift, because the adjoint reverses the first-derivative sign.

Take it with you

  • Backward equations evolve observables in starting-state variables.
  • Forward equations evolve measures and need not begin with densities.
  • Classical density formulas and boundary flux statements require additional hypotheses.

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