Chapter 10 — The Feynman–Kac Formula

Guiding question

Every PDE of the form $\partial_t u + \mathcal L u - c u = -f$ has a solution that can be written as a conditional expectation. What is the precise connection, and how does it make numerical pricing by Monte Carlo and PDE methods equivalent?

The Feynman–Kac formula is one of the deepest results connecting probability and analysis. It says: the solution to a certain class of backward parabolic PDEs is a conditional expectation of a functional of a diffusion. This has two directions of impact:

• Analytical: hard expectations become tractable PDEs.
• Probabilistic: hard PDEs become tractable Monte Carlo simulations.

In finance, it is the rigorous foundation behind "price = expectation."

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10.1 Setup

Fix a terminal time $T > 0$ and a diffusion:

$$dX_t = b(t,X_t)\,dt + \sigma(t,X_t)\,dW_t.$$

Define the Feynman–Kac functional:

$$u(t,x) = \mathbb E\!\left[\exp\!\left(-\int_t^T c(s,X_s)\,ds\right)g(X_T) + \int_t^T\exp\!\left(-\int_t^s c(r,X_r)\,dr\right)f(s,X_s)\,ds\ \Bigg|\ X_t = x\right].$$

Here:

• $g$: terminal payoff
• $f$: running payoff (or source term)
• $c$: discount rate (or killing rate)

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10.2 The Feynman–Kac theorem

Feynman–Kac formula

Under suitable regularity conditions, the function $u(t,x)$ defined above is the unique classical solution to the backward PDE:

$$\partial_t u + b(t,x)\,\partial_x u + \tfrac12\sigma(t,x)^2\,\partial_{xx}u - c(t,x)\,u + f(t,x) = 0,$$

on $[0,T)\times\mathbb R$, with terminal condition $u(T,x) = g(x)$.

Proof via Itô's formula

Define $Y_t = e^{-\int_0^t c(s,X_s)\,ds}u(t,X_t) + \int_0^t e^{-\int_0^s c}\,f(s,X_s)\,ds$.

We want to show $Y_t$ is a martingale. By Itô's formula applied to $v(t,x) = e^{-\int_0^t c\,ds}u(t,x)$:

$$dY_t = e^{-\int_0^t c\,ds}\!\left[(\partial_t u + b\partial_x u + \tfrac12\sigma^2\partial_{xx}u - cu + f)\,dt + \sigma\partial_x u\,dW_t\right].$$

If $u$ solves the PDE, the $dt$ coefficient vanishes, leaving $dY_t = (\text{pure stochastic integral})$. A stochastic integral is a local martingale. Under integrability, $Y_t$ is a true martingale.

Terminal condition: $Y_T = e^{-\int_0^T c\,ds}g(X_T) + \int_0^T e^{-\int_0^s c}f(s,X_s)\,ds$.

At time $t$: $\mathbb E[Y_T\mid\mathcal F_t] = Y_t$ (martingale). Rewriting gives $u(t,X_t) = \mathbb E[\ldots\mid X_t]$.

Feynman–Kac: u(0,x) = E[g(x + W_T)] solves the heat PDE

T = 1.00

PDE solution at x=00.39894

Monte Carlo E[g(W_T)]1.49777

Feynman–Kac says the PDE solution equals a conditional expectation. The blue curve is the exact PDE solution u(0,x); the grey dashed curve is the terminal payoff g(x). The red dot is the Monte Carlo estimate at x=0 — it matches the PDE solution by design. In finance, u = option price.

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10.3 The pure terminal payoff case

In the simplest case $c = 0$, $f = 0$, the Feynman–Kac formula reduces to:

$$u(t,x) = \mathbb E[g(X_T)\mid X_t = x],$$

which solves:

$$\partial_t u + b(t,x)\partial_x u + \tfrac12\sigma(t,x)^2\partial_{xx}u = 0,\qquad u(T,x) = g(x).$$

For pure Brownian motion ($b=0$, $\sigma=1$), this is the heat equation $\partial_t u = -\frac12\partial_{xx}u$ (note the sign: running backward from $T$ to $0$). The solution is convolution with the Gaussian kernel:

$$u(0,x) = \frac{1}{\sqrt{2\pi T}}\int_{-\infty}^\infty g(y)\,e^{-(x-y)^2/(2T)}\,dy.$$

This is exactly the smoothing property of the heat equation — the Brownian motion "averages" the terminal payoff over Gaussian fluctuations.

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10.4 Discounted terminal payoff

With discount rate $c = r$ (constant) and $f=0$:

$$u(t,x) = e^{-r(T-t)}\mathbb E[g(X_T)\mid X_t = x].$$

The PDE becomes:

$$\partial_t u + b\partial_x u + \tfrac12\sigma^2\partial_{xx}u - ru = 0.$$

For GBM under $\mathbb Q$ ($b = rx$, $\sigma = \sigma x$): this is precisely the Black–Scholes PDE. The Feynman–Kac formula proves that the Black–Scholes price equals the risk-neutral discounted expectation.

Feynman–Kac as a dictionary

The Feynman–Kac correspondence is a dictionary:

Probability side	PDE side
Conditional expectation $\mathbb E[g(X_T)\mid X_t = x]$	Solution $u(t,x)$ of the backward PDE
SDE coefficients $b$, $\sigma$	PDE operator $b\partial_x + \frac12\sigma^2\partial_{xx}$
Terminal condition $g$	Boundary data $u(T,x) = g(x)$
Discount rate $c$	Killing term $-cu$ in PDE
Running cost $f$	Source term in PDE

Every pricing problem can be attacked from either side: solve the PDE analytically, or simulate the SDE by Monte Carlo.

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10.5 Worked examples

Example 1: European call option

$X = S$ follows GBM under $\mathbb Q$, $g(S) = (S-K)^+$, $c = r$.

Feynman–Kac: $C(t,S) = e^{-r(T-t)}\mathbb E^{\mathbb Q}[(S_T-K)^+\mid S_t = S]$.

This expectation evaluates to the Black–Scholes formula (Chapter 7).

Example 2: Zero-coupon bond in CIR model

$r_t$ follows CIR ($dr = \kappa(\theta-r)\,dt + \sigma\sqrt{r}\,dW$). The zero-coupon bond price is:

$$P(t,r) = \mathbb E\!\left[\exp\!\left(-\int_t^T r_s\,ds\right)\Bigg| r_t = r\right].$$

Feynman–Kac says $P$ solves:

$$\partial_t P + \kappa(\theta-r)\partial_r P + \tfrac12\sigma^2 r\,\partial_{rr}P - rP = 0, \quad P(T,r) = 1.$$

This is the CIR bond pricing PDE. The explicit solution is $P(t,r) = A(\tau)e^{-B(\tau)r}$, where $\tau = T-t$ and $A,B$ are computed by solving the Riccati ODE for $B$.

Example 3: Running maximum

Let $M_T = \max_{0\le s\le T}W_s$. Can we find $\mathbb E[f(M_T)]$?

This is not directly a Feynman–Kac problem (the running maximum is not Markovian on its own). But $(W_t, M_t)$ jointly is Markovian, and Feynman–Kac applies in two dimensions.

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10.6 The Kolmogorov backward equation as a special case

For $c=0$, $f=0$, and $g = \mathbf 1_{(-\infty,y]}$ (cumulative distribution), Feynman–Kac gives:

$$u(t,x) = \mathbb P(X_T \le y\mid X_t = x).$$

This solves the Kolmogorov backward equation $\partial_t u + \mathcal L_x u = 0$. Setting $t=0$ gives the transition probability $\mathbb P(X_T\le y\mid X_0=x)$ — the fundamental solution of the backward equation.

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10.7 Monte Carlo versus PDE: two algorithms

Monte Carlo (probabilistic side):

1. Simulate $N$ paths of $X$ from $x$ to time $T$ using Euler–Maruyama.
2. Compute $g(X_T^{(k)})$ for each path $k$.
3. Estimate $u(0,x)\approx \frac{1}{N}\sum_k g(X_T^{(k)})$.

Error: $O(1/\sqrt{N})$ — statistical, independent of dimension.

PDE solver (analytical side):

1. Discretize the PDE on a grid $(t_i, x_j)$.
2. Apply finite differences or finite elements.
3. March backward from $T$ to $0$.

Error: $O(\Delta t + \Delta x^2)$ — deterministic, but grid size explodes in high dimensions (curse of dimensionality).

Feynman–Kac guarantees both methods converge to the same answer.

When to use Monte Carlo vs PDE

Use PDE when: 1D or 2D problem (e.g., European option on one stock), high accuracy needed, Greeks required analytically (differentiate PDE solution), early exercise (American options via free boundary PDE).

Use Monte Carlo when: High-dimensional problem (basket, path-dependent), flexible payoff structure, convergence guarantee regardless of dimension, embarrassingly parallelizable.

Dimension $d\ge 4$: Monte Carlo dominates. Dimension $d\le 3$: PDE methods are often faster and more precise.

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10.8 Problems

Problem 1: Heat equation

For $dX = dW$ (pure BM) and $g(x) = x^2$, use Feynman–Kac to find $u(0,x) = \mathbb E[(x+W_T)^2]$ and verify it solves $\partial_t u + \frac12\partial_{xx}u = 0$ with $u(T,x) = x^2$.

Problem 2: Discounting

For $dX = 0\cdot dt + dW$ (pure BM) and $g(x) = e^x$, $c = \frac12$, compute $u(0,x) = \mathbb E[e^{-T/2}e^{x+W_T}]$. Verify it solves the PDE with discounting.

Problem 3: Running payoff

For $dX = dW$, $g=0$, $f(s,x)=1$, $c=0$: the Feynman–Kac formula gives $u(t,x) = \mathbb E[\int_t^T 1\,ds\mid X_t=x] = T-t$. Verify $u(t,x)=T-t$ satisfies the PDE.

Problem 4: Binary option

Using Feynman–Kac, show that the price of a binary call (pays $1$ if $S_T > K$, else $0$) under Black–Scholes is $e^{-rT}N(d_2)$, and identify which PDE it solves.

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10.9 Selected solutions

Solution 1

$u(0,x) = \mathbb E[(x+W_T)^2] = x^2 + \mathbb E[2xW_T] + \mathbb E[W_T^2] = x^2 + 0 + T = x^2 + T$.

Check PDE: $u(t,x) = x^2 + (T-t)$ (from Feynman–Kac, the time-dependent version). Then $\partial_t u = -1$ and $\frac12\partial_{xx}u = \frac12\cdot 2 = 1$. So $\partial_t u + \frac12\partial_{xx}u = -1+1=0$. ✓

Solution 2

$u(0,x) = e^{-T/2}\mathbb E[e^{x+W_T}] = e^{-T/2}\cdot e^x\cdot e^{T/2} = e^x$. (Using $\mathbb E[e^{W_T}] = e^{T/2}$.)

The PDE is $\partial_t u + \frac12\partial_{xx}u - \frac12 u = 0$, $u(T,x) = e^x$. With $u = e^x$ (constant in $t$): $\partial_t u = 0$, $\frac12\partial_{xx}u = \frac12 e^x$, $-\frac12 u = -\frac12 e^x$. Sum $= 0$. ✓

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10.10 What this chapter gives you

Formula	Meaning
$u(t,x) = \mathbb E[e^{-\int c\,ds}g(X_T)+\ldots\mid X_t=x]$	Conditional expectation = PDE solution
$\partial_t u + \mathcal Lu - cu + f = 0$	The backward PDE for $u$
$u(T,x) = g(x)$	Terminal/boundary condition
Black–Scholes as Feynman–Kac	$C = e^{-r\tau}\mathbb E^{\mathbb Q}[(S_T-K)^+\mid S_t]$
Bond pricing as Feynman–Kac	$P = \mathbb E[\exp(-\int r\,ds)\mid r_t]$
MC vs PDE	Two algorithms for the same quantity

Part IV now uses this foundation to study more advanced models: stochastic volatility, interest rate curves, and numerical implementation.