Mathematics has two great languages for change. The analyst's is the partial differential equation: a local law, obeyed point by point, solved by calculus. The probabilist's is the random path: a particle jittering through time, its future summarised by an expectation. The Feynman–Kac formula says they are the same language: the solution of a parabolic PDE can be written as an expectation over random paths — and any such expectation solves a PDE. Cross the bridge one way and a probability problem becomes calculus; cross it the other and a differential equation can be solved by rolling dice.

For us the stakes are concrete. We have priced an option two ways: by solving the Black–Scholes PDE, and by taking a risk-neutral expectation. They had better agree — and Feynman–Kac is the theorem that guarantees they do. In general form: let X follow the diffusion dX = a(t, X)\,dt + b(t, X)\,d\tilde W, and suppose u(t, x) solves the terminal-value problem

Then u is an expectation of the terminal data g, carried along the diffusion and discounted at r:

The Black–Scholes PDE is exactly this with a = rS and b = \sigma S. But before proving it, learn to read it — every symbol in the PDE names an ingredient of the random path.

The dictionary: read the PDE as a recipe for a path

Feynman–Kac is best memorised as a dictionary between the two languages. Lay the Black–Scholes PDE and the risk-neutral SDE side by side,

and translate term by term:

• The coefficient of u_S is the SDE's drift a. Here it is rS — the risk-neutral drift, because the PDE came from a no-arbitrage argument. Whatever multiplies the first spatial derivative tells the particle which way to lean.
• The coefficient of u_{SS} is half the squared diffusion, \tfrac12 b^2. Here \tfrac12\sigma^2 S^2, so b = \sigma S. Second derivatives are the fingerprint of randomness — they enter through the (dX)^2 = b^2\,dt term of Itô's lemma, which is why the \tfrac12 is there.
• The zeroth-order term -ru is the discount factor e^{-r(T-t)} in front of the expectation. Probabilists call it killing: imagine the particle dying at rate r, and the expectation only counting payoffs from survivors.
• The terminal condition u(T, x) = g(x) is the payoff — the function you evaluate at the path's endpoint X_T.

The dictionary works in both directions. Handed a PDE you cannot solve on paper, read off a, b, r, g, simulate the SDE, and average. Handed an expectation you cannot compute, write down its PDE and hit it with a century of numerical analysis. One theorem, two solvers.

The bridge, line by line

Why is the theorem true? The strategy is the one that keeps recurring in this subject: form the right martingale, then take its expectation. We apply Itô's lemma to the discounted process Y_t = e^{-rt}\,u(t, X_t) and watch the PDE make all the drift vanish. Every step below is mechanical; the miracle is in step 3, where the equation and the probability lock together.

Step 1 — Itô on the discounted process. With Y_t = e^{-rt}u(t, X_t), the product/chain rule of Itô gives a discount term plus the Itô expansion of u:

Step 2 — group the dt terms. The bracket multiplying dt is

Step 3 — invoke the PDE. But that bracket is exactly the left-hand side of the PDE u was assumed to solve — and it equals zero. The entire drift annihilates:

Step 4 — recognise a martingale. What remains is a pure d\tilde W differential — no dt drift left — so (under the usual integrability conditions) Y_t = e^{-rt}u(t, X_t) is a martingale. The PDE was precisely the condition needed to kill the drift. This is the heart of the whole subject: a parabolic PDE is nothing but the statement that a certain process is a martingale.

Step 5 — take the conditional expectation. A martingale forecasts itself: Y_t = \mathbb{E}[Y_T \mid \mathcal{F}_t]. Writing both ends out,

Step 6 — apply the terminal condition and rearrange. At T, u(T, X_T) = g(X_T). Multiply by e^{rt} and use the Markov property (X from X_t = x forgets its past):

The solution of the PDE is a discounted expectation along the diffusion — that is the bridge. Specialised to a = rS, b = \sigma S, and the call payoff g(S) = (S - K)^+, it reproduces the risk-neutral price exactly.

Worked example 1: the heat equation is Brownian motion

Strip the theorem to its simplest case and it becomes something you can verify with bare hands. Set the drift a = 0, the diffusion b = 1 and the rate r = 0. The diffusion is then plain Brownian motion, dX = d W, and the PDE collapses to the (backward) heat equation:

In words: the temperature at x is the average of the terminal data over where a Brownian particle started at x ends up. Since W_T given W_t = x is Gaussian with mean x and variance \tau = T - t, the expectation is a convolution with a Gaussian bell — which is exactly the classical heat kernel. Brownian motion is the heat kernel wearing a probabilist's hat.

Check it on an explicit payoff. Take g(x) = x^2. Writing W_T = x + Z with Z \sim N(0, \tau):

Now differentiate the answer and confirm it really solves the PDE: u_t = -1, u_x = 2x, u_{xx} = 2, so u_t + \tfrac12 u_{xx} = -1 + 1 = 0 — satisfied — and at t = T the extra term dies, leaving u(T, x) = x^2 = g(x). The expectation solved the equation without a single separation of variables. Notice what the formula is saying: far from expiry the surface x^2 + \tau floats above the payoff by the accumulated variance; as t \to T it settles down onto it.

The figure below shows the same phenomenon for the option-like payoff g(x) = x^+ = \max(x, 0). The dashed hockey-stick is the terminal data; the solid curve is u(t,x) = \mathbb{E}[\,g(x + W_\tau)\,], which works out to x\,\Phi(x/\sqrt{\tau}) + \sqrt{\tau}\,\varphi(x/\sqrt{\tau}). Slide the time-to-maturity \tau: more time means more Brownian spread, so the expectation smooths the kink away — heat diffusing through a sharp corner. At \tau = 0 the two curves coincide, exactly as the terminal condition demands.

Worked example 2: two toy contracts, priced both ways

Back in the Black–Scholes world (a = rS, b = \sigma S), it is worth confirming the bridge holds in cases you can verify on both sides. Take first the trivial payoff g(S) = 1 — a contract paying one unit of cash at T no matter what. The expectation side is immediate:

Now check the PDE side. With u = e^{-r(T - t)} independent of S, the spatial derivatives vanish, and u_t = r\,e^{-r(T - t)} = r u. The PDE reads u_t + rS u_S + \tfrac12\sigma^2 S^2 u_{SS} - r u = r u + 0 + 0 - r u = 0 — satisfied, with terminal value u(T, S) = 1. The discount bond prices itself.

Second contract: the stock itself, g(S) = S (equivalently, a forward). The expectation side uses the fact that under the risk-neutral measure the stock grows at rate r on average, \mathbb{E}[S_T \mid S_t = S] = S\,e^{r(T-t)}:

The discounting and the risk-neutral growth cancel exactly — today's price of receiving the stock at T is just today's stock price. PDE side: try u = S. Then u_t = 0, u_S = 1, u_{SS} = 0, and 0 + rS\cdot 1 + 0 - rS = 0 — satisfied, with u(T, S) = S = g(S). Both toy contracts agree on both sides of the bridge, for structurally different reasons: the bond because the -ru term eats u_t, the forward because the drift term eats it. The same machinery, applied to (S - K)^+, yields the full Black–Scholes formula — the integral is just longer.

Pricing by dice: the Monte-Carlo consequence

Read the theorem right-to-left one more time and a numerical method falls out for free. The price is an expectation; the law of large numbers says an expectation is (almost surely) the limit of a sample average. So: simulate N independent risk-neutral paths of the SDE, evaluate the payoff on each terminal value, average, and discount:

That is a PDE being solved by a random-number generator. The error shrinks like 1/\sqrt{N} — slow, but blissfully independent of dimension, which is why simulation wins the moment the payoff depends on many underlyings. Run it yourself: the snippet prices a call by both routes and prints them side by side. Bump N and watch the dice converge on the calculus.