The Feynman-Kac Theorem

Feynman-Kac is the theorem that says two apparently different objects are the same: the solution of a certain partial differential equation, and the expected value of a payoff along a random path. It is the reason Black-Scholes can be derived either as a PDE (via delta-hedging) or as a discounted risk-neutral expectation, Feynman-Kac guarantees the two routes give the identical price. It is the formal dictionary between the analyst's world of PDEs and the probabilist's world of expectations, and it underlies every Monte-Carlo-versus-PDE choice a quant makes.

The statement

Let $X_t$ be a diffusion $dX_t = \mu(X_t,t)\,dt + \sigma(X_t,t)\,dW_t$, and suppose $u(x,t)$ solves the backward parabolic PDE

$\frac{\partial u}{\partial t} + \mu(x,t)\frac{\partial u}{\partial x} + \tfrac12\sigma^2(x,t)\frac{\partial^2 u}{\partial x^2} - r(x,t)\,u = 0,$

with terminal condition $u(x,T) = \phi(x)$. Then $u$ has the probabilistic representation

$\boxed{\,u(x,t) = \mathbb{E}\!\left[\,e^{-\int_t^T r(X_s,s)\,ds}\,\phi(X_T)\;\Big|\;X_t = x\right].\,}$

The PDE and the expectation are two faces of the same function. The differential operator $\mathcal{L} = \mu\,\partial_x + \tfrac12\sigma^2\,\partial_{xx}$ is the generator of the diffusion; the $-ru$ term is discounting; the terminal condition is the payoff.

Derivation: discount, then martingalize

The proof is a one-line application of Itô's lemma plus the observation that a driftless Itô process is a martingale. Define the discounted process

$M_t = e^{-\int_0^t r(X_s,s)\,ds}\,u(X_t, t).$

Apply Itô, treating the discount factor and $u(X_t,t)$ as a product (the discount factor has finite variation, so no extra covariation term):

$dM_t = e^{-\int_0^t r\,ds}\Big[\underbrace{\big(u_t + \mu u_x + \tfrac12\sigma^2 u_{xx} - r\,u\big)}_{=\,0\ \text{by the PDE}}\,dt + \sigma u_x\,dW_t\Big].$

The entire $dt$ drift is precisely the left side of the PDE, which is zero. So $dM_t = e^{-\int r}\sigma u_x\,dW_t$, a driftless process, hence (under integrability) a martingale. The martingale property $M_t = \mathbb{E}[M_T \mid \mathcal{F}_t]$ then gives, using $u(X_T,T) = \phi(X_T)$,

$e^{-\int_0^t r\,ds}\,u(X_t,t) = \mathbb{E}\!\left[e^{-\int_0^T r\,ds}\,\phi(X_T)\,\Big|\,\mathcal{F}_t\right].$

Divide by the discount to time $t$ and set $X_t = x$ to recover the boxed formula. The PDE was exactly the condition that makes the discounted value a martingale, which is the same condition as no-arbitrage pricing.

Why the Black-Scholes PDE equals a discounted expectation

Specialize to the Black-Scholes setting: under the risk-neutral measure $\mathbb{Q}$, the stock has generator with drift $rS$ and diffusion $\sigma S$, and constant discount $r$. Feynman-Kac with $\mu = rS$, $\sigma_X = \sigma S$, $\phi(S) = (S-K)^+$ says the solution of

$V_t + rS\,V_S + \tfrac12\sigma^2 S^2 V_{SS} - rV = 0, \qquad V(S,T) = (S-K)^+$

is exactly

$V(S,t) = e^{-r(T-t)}\,\mathbb{E}^{\mathbb{Q}}\big[(S_T - K)^+ \mid S_t = S\big].$

The delta-hedging derivation produces the PDE; Feynman-Kac converts it to the expectation, which the lognormal integral evaluates to $SN(d_1) - Ke^{-r(T-t)}N(d_2)$. The two "derivations" of Black-Scholes are not independent results, Feynman-Kac is the theorem that makes them provably identical.

Both directions

The dictionary runs both ways, and quants exploit each:

• PDE $\to$ expectation: when a diffusion is easy to simulate but the PDE is high-dimensional or the payoff is path-dependent, price by Monte Carlo of the expectation. This is why Feynman-Kac is the theoretical license for Monte Carlo pricing.
• Expectation $\to$ PDE: when the state space is low-dimensional (one or two factors) and you want the whole solution surface, boundary behaviour, or early-exercise (American) features, solve the PDE by finite differences. The curse of dimensionality decides: PDEs win in 1–3 factors, Monte Carlo wins in high dimension.

Worked example

Price a claim paying $\phi(S_T) = S_T^2$ (a "power" payoff) under $\mathbb{Q}$ with $dS = rS\,dt + \sigma S\,dW$. Rather than solve the PDE, use the expectation side directly. Since $S_T = S_0\exp((r-\tfrac12\sigma^2)T + \sigma W_T)$,

$V_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T^2] = e^{-rT}S_0^2\,e^{2(r-\frac12\sigma^2)T}\,\mathbb{E}[e^{2\sigma W_T}] = e^{-rT}S_0^2\,e^{(2r-\sigma^2)T}\,e^{2\sigma^2 T} = S_0^2\,e^{(r+\sigma^2)T}.$

You can verify this solves the Black-Scholes PDE with terminal condition $S^2$, the two agree, which is Feynman-Kac in action. Notice the price grows with $\sigma^2$: a convex payoff is long volatility, exactly as the PDE's $\tfrac12\sigma^2 S^2 V_{SS}$ term (with $V_{SS} = 2 > 0$) predicts.

What breaks in practice

• Regularity and growth conditions. The clean theorem needs $\mu,\sigma$ Lipschitz and $\phi$ of controlled growth; kinked payoffs like $(S-K)^+$ are handled in the viscosity/weak sense, and blow-up payoffs can make the expectation infinite even when the PDE looks fine.
• Early exercise breaks the equivalence. American options are a free-boundary PDE (variational inequality), and the naive Feynman-Kac expectation must be replaced by an optimal-stopping representation, $u = \sup_\tau \mathbb{E}[e^{-r\tau}\phi(X_\tau)]$, the source of all the difficulty in American/Bermudan pricing.
• Jumps. With jumps the "PDE" becomes a partial integro-differential equation (PIDE) and the generator gains a non-local term; standard Feynman-Kac is for pure diffusions.
• Dimensionality. The theorem is exact in any dimension, but solving the PDE numerically suffers the curse of dimensionality above ~3–4 factors, forcing the Monte Carlo (expectation) route.

In interviews

State the correspondence: the solution of the discounted backward PDE $u_t + \mu u_x + \tfrac12\sigma^2 u_{xx} - ru = 0$ with terminal $\phi$ equals the discounted expectation $\mathbb{E}[e^{-\int r}\phi(X_T)]$. The derivation they want is the martingale argument: apply Itô to the discounted $u(X_t,t)$, note the drift is the PDE and vanishes, so the process is a martingale, then take expectations. The payoff line: this is why the Black-Scholes PDE and the risk-neutral pricing integral are the same object, and why you can price either by finite-difference PDE or by Monte Carlo, the choice being dimensionality. A good closer is the American caveat: early exercise turns it into optimal stopping, which Feynman-Kac in its basic form does not cover.