Questions: Kolmogorov Forward and Backward Equations

Question 1

The Kolmogorov backward equation acts on the initial point (x,t), while the forward equation acts on the terminal point (y,T). Which equation would you use to find the stationary (equilibrium) density of a diffusion?

• A. The backward equation, since stationarity is a condition on initial states
• B. The forward equation (Fokker-Planck), since setting ∂p/∂T = 0 gives the equation for the time-independent equilibrium density
• C. Neither — stationary densities require ergodic theory, not Kolmogorov equations
• D. Both equations simultaneously, since stationarity constrains both initial and final conditions

Answer: (B)

The stationary density π(y) satisfies the time-independent Fokker-Planck equation: 0 = -(d/dy)[μ(y)π(y)] + (1/2)(d²/dy²)[σ²(y)π(y)]. This is an ODE that can often be solved explicitly — for example, the OU process dX = -θX dt + σ dW has stationary density proportional to exp(-θx²/σ²), which is Gaussian N(0, σ²/2θ). The backward equation is the wrong tool here because the stationary density describes the long-run distribution of the terminal state, not a function of the initial state.

Question 2

The Fokker-Planck (forward) equation for the process dX = dW (pure Brownian motion, no drift) is:

• A. ∂p/∂T = (1/2)∂²p/∂y², the heat equation
• B. ∂p/∂T = ∂p/∂y, the transport equation
• C. ∂p/∂T = -∂²p/∂y², the backward heat equation
• D. ∂p/∂T = p, exponential growth

Answer: (A)

With μ = 0 and σ = 1, the Fokker-Planck equation becomes ∂p/∂T = (1/2)∂²p/∂y². This is the classical heat equation. The fundamental solution (starting from p(y,0) = δ(y-x)) is the Gaussian kernel p(y,T) = (2πT)^{-1/2}exp(-(y-x)²/(2T)), which is exactly the transition density of Brownian motion. The connection between Brownian motion and the heat equation — heat diffuses like probability diffuses — was one of the key insights of early 20th century mathematical physics.

Question 3

Explain the physical/probabilistic interpretation of the two terms in the Fokker-Planck equation ∂p/∂T = -(∂/∂y)[μp] + (1/2)(∂²/∂y²)[σ²p].

Model answer: The first term -(∂/∂y)[μp] is the transport (advection) term: it describes how the drift μ(y) shifts probability mass in the direction of the drift. If all particles have the same drift, probability is carried along like fluid in a flow. The second term (1/2)(∂²/∂y²)[σ²p] is the diffusion term: it describes how noise spreads probability mass, smearing concentrated distributions into broader ones. The balance between transport and diffusion determines the evolving shape of the density. In steady state (∂p/∂T = 0), the inward transport from mean reversion exactly balances the outward diffusion from noise.

This interpretation parallels the convection-diffusion equation in physics. The transport term preserves the total mass (probability) while moving it; the diffusion term also preserves total mass while spreading it. Their interplay determines whether the process converges to a stationary distribution (when drift dominates at large |y|) or diffuses to infinity (when drift is too weak).