The Fokker-Planck and Kolmogorov Equations

An option pricing PDE and a "where will the price distribution end up" simulation both start from the exact same SDE. They look like different problems — one wants a single number today, the other wants a whole probability density in the future — but they're solved by two versions of essentially one equation, run in opposite time directions.

The analogy: a weather forecast versus a claims payout

Forward: given today's weather, what's the probability distribution of tomorrow's rainfall — not one number, the whole shape of possible outcomes and their likelihoods? That's asking how a density spreads forward through time.

Backward: given a rule that pays out based on tomorrow's rainfall, what's that payout worth today, averaged over every possible path the weather could take between now and then? That's asking how an expected value should be discounted backward through time to the present.

Both questions are about the exact same rainfall process. The Fokker-Planck equation answers the first kind of question; the Kolmogorov backward equation answers the second — same diffusion, opposite direction of information flow.

Writing it down

For a diffusion $dX_t = \mu(x)\,dt + \sigma(x)\,dW_t$, let $p(x,t)$ be the probability density of $X_t$. The Fokker-Planck equation (forward Kolmogorov equation) says

$$\frac{\partial p}{\partial t} = -\frac{\partial}{\partial x}\big[\mu(x)p\big] + \frac{1}{2}\frac{\partial^2}{\partial x^2}\big[\sigma(x)^2 p\big].$$

In words: the density at any point changes over time from two competing effects — drift carrying probability mass along (the first term, like a current sweeping the density downstream) and volatility spreading probability mass out (the second term, a diffusion/smoothing effect, exactly like heat spreading through a rod). It's the adjoint of the infinitesimal generator (see The Infinitesimal Generator of a Diffusion) — same building blocks, applied to the density instead of to a payoff function.

The Kolmogorov backward equation describes $u(x,t) = E[g(X_T) \mid X_t = x]$, the expected value of some future payoff $g$ given the current state:

$$\frac{\partial u}{\partial t} + \mu(x)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma(x)^2\frac{\partial^2 u}{\partial x^2} = 0.$$

In words: this is exactly $\partial u/\partial t + \mathcal{L}u = 0$, the generator from the earlier concept applied directly — it says the expected future payoff, viewed backward from maturity, evolves according to the diffusion's own drift and volatility acting on it. This is precisely the PDE the Black-Scholes equation is an instance of, with $g$ the option's payoff at expiry.

Picture the curve above evolving through time instead of staying fixed: for plain Brownian motion started at a spike (all mass at $x=0$), the Fokker-Planck equation says that spike spreads into wider and wider normal curves as $t$ grows — literally $p(x,t) = \mathcal{N}(0,t)$, the density's standard deviation growing as $\sqrt{t}$, exactly the spreading picture the equation predicts.

Worked example 1: Fokker-Planck confirms a known density

For standard Brownian motion, $\mu=0$, $\sigma=1$, so the Fokker-Planck equation reduces to the heat equation $\partial p/\partial t = \tfrac12 \partial^2 p/\partial x^2$. Check that $p(x,t) = \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t}$ solves it: at $t=1$, $x=0$, $p(0,1) = 1/\sqrt{2\pi} \approx 0.3989$. At $t=4$, $x=0$: $p(0,4) = 1/\sqrt{8\pi} \approx 0.1995$ — exactly half, since the density has spread out (its standard deviation doubled from 1 to 2, so the peak height halved). Differentiating the formula with respect to $t$ and $x$ and substituting confirms it satisfies the PDE exactly, which is the formal proof that Brownian motion's density really does evolve by simple diffusive spreading, term for term.

Worked example 2: backward equation prices a simple claim

Suppose $X_t$ is Brownian motion ($\mu=0,\sigma=1$) and you want $u(x,0) = E[X_T^2 \mid X_0 = x]$ for $T=2$. Guess $u(x,t) = x^2 + c(T-t)$ for a constant $c$ to be found, and plug into the backward equation: $\partial u/\partial t = -c$, $\partial u/\partial x = 2x$, $\partial^2u/\partial x^2 = 2$. The equation $-c + 0\times 2x + \tfrac12 \times 1 \times 2 = 0$ gives $c=1$. So $u(x,0) = x^2 + 1\times(2-0) = x^2+2$. Starting at $x=3$: $u(3,0) = 9+2=11$. Direct check: $E[X_2^2 \mid X_0=3] = \text{Var}(X_2) + (E[X_2])^2 = 2 + 9 = 11$ — the PDE solution and the direct probability computation agree exactly, confirming the backward equation correctly propagates expected values back to today.

What this means in practice

• The Black-Scholes PDE is a Kolmogorov backward equation with the risk-neutral drift and the option's payoff as the terminal condition — every "solve the PDE" derivation of an option price is running this machinery.
• Fokker-Planck is what a Monte Carlo simulation is implicitly sampling from — simulating many paths and building a histogram of the endpoints is a numerical way to approximate the density that Fokker-Planck describes exactly.
• Stationary distributions of mean-reverting processes (the long-run distribution of a short rate or a volatility process) are found by setting $\partial p/\partial t = 0$ in Fokker-Planck and solving the resulting ODE — see Stationary Distributions and Ergodicity.