Feynman-Kac Formula and the Black-Scholes Solution¶

Everything in this subsubsection follows from one sentence:

The discounted conditional expectation of a payoff is exactly the solution of a backward parabolic PDE.

The preceding subsubsections derived the Black–Scholes formula by manipulating the PDE — into the heat equation, into frequency space, or by exploiting scaling structure. Feynman–Kac is the reverse direction: it shows that the very same answer is obtained as a probabilistic expectation, with no PDE manipulation. The two pictures are not parallel calculations of the same number; they are the same calculation viewed in two notations, related by a single elementary averaging identity.

We build the picture in the same order a reader naturally builds intuition: a discrete random walk first (where the averaging is visible by inspection), then the general continuous-time theorem, and only then the Black–Scholes specifics.

1. Why Expectations Solve PDEs? A Random-Walk Toy¶

Before any Itô integral or stochastic calculus, work the simplest possible averaging problem on a discrete lattice. The Feynman–Kac formula will appear in miniature, with no analytic machinery needed.

1.1 Setup: A Symmetric Random Walk¶

Let \(X_n\) be a symmetric random walk on the integer lattice \(\mathbb{Z}\): at each step \(X_{n+1} = X_n \pm 1\) with probability \(1/2\) each. Fix a terminal time \(N\) and a terminal payoff \(g(x)\). Define the conditional-expectation function

\[ u_n(x) := \mathbb{E}\bigl[g(X_N) \mid X_n = x\bigr] \]

— what we expect to receive at time \(N\), given that we are at position \(x\) at time \(n\). This is exactly the "value function" that option pricing computes, in the simplest possible setting.

1.2 The Averaging Identity¶

Condition on the next step. From \(x\) at time \(n\) the walk lands at \(x \pm 1\) at time \(n + 1\) with equal probability, and from there the expected payoff is \(u_{n+1}(x \pm 1)\). Linearity of expectation gives

\[ u_n(x) = \tfrac{1}{2}\, u_{n+1}(x + 1) + \tfrac{1}{2}\, u_{n+1}(x - 1) \]

The function \(u\) at \((x, n)\) equals the average of its neighbors at \((x \pm 1, n + 1)\). That is the entire mechanism — no integrals, no Itô, just an averaging identity.

1.3 The Averaging Identity Is a Discrete Heat Equation¶

Subtract \(u_{n+1}(x)\) from both sides:

\[ u_n(x) - u_{n+1}(x) = \tfrac{1}{2}\, \bigl[u_{n+1}(x + 1) - 2\, u_{n+1}(x) + u_{n+1}(x - 1)\bigr] \]

The right-hand side is the discrete spatial second difference \(\Delta_{xx} u_{n+1}(x)\). The left-hand side is the negative discrete time difference \(-(u_{n+1}(x) - u_n(x)) = -\Delta_t u_n(x)\). Rewriting:

\[ -\Delta_t u_n(x) = \tfrac{1}{2}\, \Delta_{xx} u_{n+1}(x) \]

This is a backward discrete heat equation. The conditional-expectation function \(u_n(x) = \mathbb{E}[g(X_N) \mid X_n = x]\) literally is its solution.

1.4 The Continuum Limit¶

Send the lattice spacing to zero with the diffusive scaling \(\Delta x = \sqrt{\Delta t}\) — the unique scaling that keeps the variance of a step bounded as \(\Delta t \to 0\). The discrete time difference \(\Delta_t / \Delta t\) becomes \(\partial_t\), the discrete second difference \(\Delta_{xx} / (\Delta x)^2\) becomes \(\partial_{xx}\), and the symmetric random walk converges (Donsker's invariance principle) to standard Brownian motion \(W_t\). The discrete identity becomes

\[ -\frac{\partial u}{\partial t} = \frac{1}{2}\, \frac{\partial^2 u}{\partial x^2} \]

— the backward heat equation. Its solution is the conditional expectation

\[ u(x, t) = \mathbb{E}\bigl[g(W_T) \mid W_t = x\bigr] \]

The PDE is just the continuum-limit averaging identity.

1.5 Adding a Discount Rate¶

In option pricing, future cashflows are discounted at rate \(r\). The natural modification: each lattice step contributes a discount factor \(e^{-r\Delta t}\), so the recursion becomes

\[ u_n(x) = e^{-r\Delta t}\, \bigl[\tfrac{1}{2}\, u_{n+1}(x + 1) + \tfrac{1}{2}\, u_{n+1}(x - 1)\bigr] \]

Expanding \(e^{-r\Delta t} \approx 1 - r\Delta t\) contributes an extra \(-r u\) term, and the continuum limit becomes

\[ \frac{\partial u}{\partial t} + \tfrac{1}{2}\, \frac{\partial^2 u}{\partial x^2} - r u = 0 \]

— the discounted backward heat equation. Its solution is the discounted conditional expectation

\[ u(x, t) = \mathbb{E}\bigl[e^{-r(T - t)}\, g(W_T) \mid W_t = x\bigr] \]

1.6 What Carries Forward to Black–Scholes¶

Two ingredients lift to the general Feynman–Kac theorem without modification:

The averaging identity becomes the generator of the diffusion. Replace the symmetric next-step distribution with a general drift-diffusion \(dX_t = \mu(X_t, t)\, dt + \sigma(X_t, t)\, dW_t\). By the same conditioning argument, the discrete recursion gives a continuum-limit PDE with second-order operator \(\tfrac{1}{2}\sigma^2 \partial_{xx} + \mu \partial_x\) — the infinitesimal generator of \(X_t\).
The discount rate becomes a \(-r u\) term in the PDE. This generalizes to a state-dependent discount \(r(X_t)\) or to source terms \(f(X_t)\) via the obvious lattice modification.

The result is the Feynman–Kac formula in its full generality, which §2 below states precisely.

Core principle

The Feynman–Kac formula is the rigorous form of an elementary averaging identity: value at \((x, t)\) = discounted average of value at \((x', t + \Delta t)\) across the next-step distribution of the underlying process. In the continuum limit this averaging-plus-discounting identity becomes a backward parabolic PDE generated by the process; the conditional expectation and the PDE solution are the same function.

This is the mechanism. Black–Scholes is the application: the GBM stock dynamics under \(\mathbb{Q}\) play the role of the underlying process, the discount rate is the risk-free rate, and the payoff \(\Phi(S_T)\) replaces \(g\).

2. The General Feynman-Kac Theorem¶

Recall (see § Feynman-Kac Formula and § Discounted Feynman-Kac): for a diffusion \(dX_t = \mu(X_t,t)\,dt + \sigma(X_t,t)\,dW_t\) and a function \(u\) solving the parabolic PDE

\[ u_t + \mu u_x + \tfrac{1}{2}\sigma^2 u_{xx} - r u + f = 0 \]

with terminal data \(u(\cdot,T) = \Phi\),

\[ u(x,t) = \mathbb{E}\left[\int_t^T e^{-\int_t^s r\,d\tau}f(X_s,s)\,ds + e^{-\int_t^T r\,d\tau}\Phi(X_T) \,\Big|\, X_t = x\right]. \]

For option pricing (\(f \equiv 0\), constant \(r\)) this reduces to the risk-neutral valuation formula \(u(x,t) = \mathbb{E}[e^{-r(T-t)}\Phi(X_T) \mid X_t = x]\) (see also § Risk-Neutral Valuation Principle).

3. Application to Black-Scholes¶

1. The Setup¶

Under the risk-neutral measure \(\mathbb{Q}\), the stock price dynamics are:

\[ dS_t = rS_t dt + \sigma S_t dW_t^{\mathbb{Q}} \]

where:

\(r\) = constant risk-free rate
\(\sigma\) = constant volatility
\(W_t^{\mathbb{Q}}\) = Brownian motion under \(\mathbb{Q}\)

2. The Black-Scholes PDE¶

The option value \(V(S,t)\) satisfies:

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

with terminal condition:

\[ V(S,T) = \Phi(S) \]

where \(\Phi(S)\) is the option payoff:

European call: \(\Phi(S) = (S-K)^+\)
European put: \(\Phi(S) = (K-S)^+\)

3. Feynman-Kac Representation¶

By the Feynman-Kac formula:

\[ \boxed{V(S,t) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid S_t = S]} \]

Interpretation: The option value is the expected discounted payoff under the risk-neutral measure.

Semigroup viewpoint. Let \(\mathcal{L}\) denote the Black-Scholes generator

\[ \mathcal{L} = rS\frac{\partial}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2}{\partial S^2} - r \]

and write \(\tau = T - t\). Then the solution to the Black-Scholes PDE with terminal data \(\Phi\) admits the operator-exponential representation

\[ V(\cdot, t) = e^{\tau\mathcal{L}}\Phi \]

and the Feynman-Kac formula

\[ \left(e^{\tau\mathcal{L}}\Phi\right)(S) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid S_t = S] \]

is precisely the probabilistic representation of the pricing semigroup \(\mathcal{P}_\tau = e^{\tau\mathcal{L}}\) acting on the terminal payoff. This is the same semigroup obtained in § Heat Equation (as a Gaussian convolution operator) and in § Fourier Transform (as a Fourier multiplier); Feynman-Kac realises it as an expectation. The operator-exponential viewpoint is the unifying theme of this chapter, and Feynman-Kac is its probabilistic incarnation.

This converts the PDE problem into a probabilistic expectation problem.

4. Rigorous Derivation of Feynman-Kac¶

Recall (see § Feynman-Kac Proof Sketch): apply Itô's lemma to the discounted value \(Y_t = e^{-\int_0^t r\,ds}u(X_t,t)\); if \(u\) satisfies the parabolic PDE the drift vanishes and \(Y_t\) is a martingale, so \(\mathbb{E}[Y_T \mid \mathcal{F}_t] = Y_t\) yields the Feynman-Kac representation. (Exercise 7 below repeats this argument specialised to Black-Scholes for self-containment.)

For Black-Scholes (\(X_t = S_t\), \(\mu = rS\), \(\sigma_X = \sigma S\), constant \(r\)), the formula gives

\[ V(S,t) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid S_t = S] \]

with \(\Phi(S) = (S-K)^+\) (call) or \((K-S)^+\) (put). Any solution of the Black-Scholes PDE has this risk-neutral expectation representation.

5. Solving the SDE: Distribution of S_T¶

Recall (see § Solving the SDE and § Itô Lemma): applying Itô's lemma to \(\ln S_t\) under \(dS_t = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}\) gives \(d(\ln S_t) = (r - \tfrac{1}{2}\sigma^2)\,dt + \sigma\,dW_t^{\mathbb{Q}}\), so

\[ \boxed{S_T = S_t \exp\left[(r - \tfrac{1}{2}\sigma^2)(T-t) + \sigma\sqrt{T-t}\,Z\right], \qquad Z \sim \mathcal{N}(0,1).} \]

Equivalently, with \(\tau = T - t\),

\[ \ln S_T \mid S_t \sim \mathcal{N}\bigl(\ln S_t + (r - \tfrac{1}{2}\sigma^2)\tau,\ \sigma^2\tau\bigr). \]

The corresponding lognormal transition density \(p(S_T \mid S_t)\) is the Gaussian kernel of § Heat Equation expressed in \((S,t)\) coordinates rather than transformed \((x,\tau)\) coordinates.

Canonical role. This subsubsection is the canonical home in the chapter for the probabilistic form of \((d_1, d_2)\) derived below; later subsubsections (change of numéraire, dividends, characteristic-function methods) reference back to these calculations rather than repeating them.

6. European Call Option: Detailed Derivation¶

1. The Pricing Formula via Feynman-Kac¶

For a European call with payoff \(\Phi(S) = (S-K)^+\):

\[ C(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+ \mid S_t = S] \]

where \(\tau = T - t\).

2. Step 1: Rewrite as Integral¶

\[ C(S,t) = e^{-r\tau}\int_K^{\infty}(S_T - K)p(S_T \mid S)dS_T \]

3. Step 2: Split the Integral¶

\[ C(S,t) = e^{-r\tau}\left[\int_K^{\infty}S_T p(S_T \mid S)dS_T - K\int_K^{\infty}p(S_T \mid S)dS_T\right] \]

\[ = e^{-r\tau}\left[I_1 - K \cdot I_2\right] \]

We need to evaluate two integrals.

4. Step 3: Change to Log-Normal Variable¶

Since \(\ln S_T \sim \mathcal{N}(m, v^2)\) where:

\(m = \ln S + (r - \frac{\sigma^2}{2})\tau\)
\(v = \sigma\sqrt{\tau}\)

Set \(y = \ln S_T\), so \(S_T = e^y\) and \(dS_T = e^y dy\).

The density becomes:

\[ p(S_T)dS_T = \frac{1}{v\sqrt{2\pi}}e^{-\frac{(y-m)^2}{2v^2}}dy \]

Both integrals transform to:

Integral \(I_2\):

\[ I_2 = \int_K^{\infty}p(S_T)dS_T = \int_{\ln K}^{\infty}\frac{1}{v\sqrt{2\pi}}e^{-\frac{(y-m)^2}{2v^2}}dy \]

Integral \(I_1\):

\[ I_1 = \int_K^{\infty}S_T p(S_T)dS_T = \int_{\ln K}^{\infty}e^y \frac{1}{v\sqrt{2\pi}}e^{-\frac{(y-m)^2}{2v^2}}dy \]

5. Step 4: Evaluate I_2 (Probability Term)¶

Standardize the integral. Let:

\[ Z = \frac{y - m}{v} \]

Then \(y = m + vZ\) and \(dy = v dZ\).

When \(y = \ln K\): \(Z = \frac{\ln K - m}{v}\)

When \(y = \infty\): \(Z = \infty\)

Therefore:

\[ I_2 = \int_{\frac{\ln K - m}{v}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{Z^2}{2}}dZ = \mathcal{N}\left(-\frac{\ln K - m}{v}\right) = \mathcal{N}\left(\frac{m - \ln K}{v}\right) \]

Substituting \(m = \ln S + (r - \frac{\sigma^2}{2})\tau\) and \(v = \sigma\sqrt{\tau}\):

\[ I_2 = \mathcal{N}\left(\frac{\ln(S/K) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}\right) \]

Define:

\[ \boxed{d_2 = \frac{\ln(S/K) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}} \]

Then:

\[ I_2 = \mathcal{N}(d_2) \]

Interpretation: \(\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)\) is the risk-neutral probability that the option expires in-the-money.

6. Step 5: Evaluate I_1 (Stock Term)¶

\[ I_1 = \int_{\ln K}^{\infty}e^y \frac{1}{v\sqrt{2\pi}}e^{-\frac{(y-m)^2}{2v^2}}dy \]

Key technique: Complete the square in the exponent --- the same Gaussian-integration manoeuvre used in § Heat Equation. Combining the linear term \(y\) with the quadratic \(-(y-m)^2/(2v^2)\) and reorganising as a perfect square in \(y\) around the shifted mean \(m + v^2\) gives

\[ y - \frac{(y-m)^2}{2v^2} = -\frac{[y-(m+v^2)]^2}{2v^2} + m + \frac{v^2}{2} \]

(The full step-by-step algebra is carried out in Exercise 2 below.) Substituting back:

\[ I_1 = e^{m + \frac{v^2}{2}}\int_{\ln K}^{\infty}\frac{1}{v\sqrt{2\pi}}e^{-\frac{[y-(m+v^2)]^2}{2v^2}}dy \]

This is a Gaussian integral with mean shifted to \(m + v^2\):

\[ I_1 = e^{m + \frac{v^2}{2}} \cdot \mathcal{N}\left(\frac{(m+v^2) - \ln K}{v}\right) \]

Simplify the exponent. Recall:

\(m = \ln S + (r - \frac{\sigma^2}{2})\tau\)
\(v^2 = \sigma^2\tau\)

\[ m + \frac{v^2}{2} = \ln S + (r - \frac{\sigma^2}{2})\tau + \frac{\sigma^2\tau}{2} = \ln S + r\tau \]

Therefore:

\[ e^{m + \frac{v^2}{2}} = e^{\ln S + r\tau} = Se^{r\tau} \]

For the normal CDF argument:

\[ \frac{(m+v^2) - \ln K}{v} = \frac{\ln S + r\tau + \frac{\sigma^2\tau}{2} - \ln K}{\sigma\sqrt{\tau}} = \frac{\ln(S/K) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} \]

Define:

\[ \boxed{d_1 = \frac{\ln(S/K) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} = d_2 + \sigma\sqrt{\tau}} \]

Then:

\[ I_1 = Se^{r\tau}\mathcal{N}(d_1) \]

7. Step 6: Combine Results¶

\[ C(S,t) = e^{-r\tau}[I_1 - KI_2] \]

\[ = e^{-r\tau}\left[Se^{r\tau}\mathcal{N}(d_1) - K\mathcal{N}(d_2)\right] \]

\[ \boxed{C(S,t) = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2)} \]

where:

\[ \boxed{d_1 = \frac{\ln(S/K) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}, \quad d_2 = d_1 - \sigma\sqrt{\tau} = \frac{\ln(S/K) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}} \]

This is the Black-Scholes formula for a European call option.

7. European Put Option¶

1. Method 1: Direct Calculation¶

For a European put with payoff \(\Phi(S) = (K-S)^+\):

\[ P(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[(K - S_T)^+ \mid S_t = S] \]

\[ = e^{-r\tau}\int_0^K(K - S_T)p(S_T \mid S)dS_T \]

Following similar calculations (completing the square with reversed integration limits):

\[ \boxed{P(S,t) = Ke^{-r\tau}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)} \]

where \(d_1\) and \(d_2\) are the same as for the call.

2. Method 2: Put-Call Parity¶

From the no-arbitrage relationship:

\[ C - P = S - Ke^{-r\tau} \]

Therefore:

\[ P = C - S + Ke^{-r\tau} \]

\[ = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2) - S + Ke^{-r\tau} \]

\[ = S(\mathcal{N}(d_1) - 1) + Ke^{-r\tau}(1 - \mathcal{N}(d_2)) \]

Using \(\mathcal{N}(-x) = 1 - \mathcal{N}(x)\):

\[ P = -S\mathcal{N}(-d_1) + Ke^{-r\tau}\mathcal{N}(-d_2) \]

\[ \boxed{P(S,t) = Ke^{-r\tau}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)} \]

Both methods yield the same formula. ✓

8. Probabilistic Interpretation¶

1. The Two Terms in the Call Formula¶

\[ C = \underbrace{S\mathcal{N}(d_1)}_{\text{Stock term}} - \underbrace{Ke^{-r\tau}\mathcal{N}(d_2)}_{\text{Strike term}} \]

2. Meaning of N(d_2)¶

\[ \boxed{\mathcal{N}(d_2) = \mathbb{Q}(S_T > K \mid S_t = S)} \]

This is the risk-neutral probability that the option expires in-the-money.

3. Meaning of N(d_1)¶

\(\mathcal{N}(d_1)\) represents the probability of exercise under the stock measure \(\mathbb{Q}^S\):

\[ \boxed{\mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K \mid S_t = S)} \]

9. Connection to Kolmogorov Equations¶

Recall (see § Understanding SDE Solutions and § Feynman-Kac Applications): the Black-Scholes PDE is the Kolmogorov backward equation for \(S_t\) under \(\mathbb{Q}\) (solved backward from terminal data \(\Phi\)), and the transition density \(p(S_T,T \mid S_t,t)\) satisfies the dual Kolmogorov forward (Fokker-Planck) equation \(\partial_T p = -\partial_{S_T}(rS_T p) + \tfrac{1}{2}\partial_{S_T S_T}(\sigma^2 S_T^2 p)\). Feynman-Kac is precisely the bridge \(V(S,t) = e^{-r\tau}\int \Phi(S_T)\,p(S_T,T \mid S,t)\,dS_T\).

10. Why Feynman-Kac Works: Deep Intuition¶

The rigorous derivation in the previous subsubsections established that the discounted value \(e^{-\int_0^t r\,ds}\,u(X_t,t)\) is a martingale whenever \(u\) satisfies the Feynman-Kac PDE. In the Black-Scholes setting this takes a particularly transparent form.

1. The Martingale Property¶

Under the risk-neutral measure, \(d(e^{-rt}S_t) = e^{-rt}\sigma S_t\,dW_t^{\mathbb{Q}}\), which has no drift. Hence the discounted stock price is a martingale, and by the same argument the discounted option value \(e^{-rt}V(S_t,t)\) is a martingale whenever \(V\) satisfies the Black-Scholes PDE. This immediately yields the Feynman-Kac representation \(V(S_t,t) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid \mathcal{F}_t]\).

2. No-Arbitrage = Martingale Measure¶

The Feynman-Kac formula is the mathematical manifestation of no-arbitrage pricing:

Fundamental Theorem of Asset Pricing:

No arbitrage \(\Longleftrightarrow\) Existence of equivalent martingale measure \(\mathbb{Q}\)
Completeness + No arbitrage \(\Longleftrightarrow\) Unique martingale measure

The risk-neutral measure \(\mathbb{Q}\) is the unique measure under which:

Discounted asset prices are martingales
All derivatives can be priced consistently

Economic interpretation: Feynman-Kac converts the no-arbitrage condition into an explicit pricing formula.

11. Extensions and Generalizations¶

1. With Continuous Dividend Yield¶

If the stock pays continuous dividends at rate \(q\):

\[ dS_t = (r-q)S_t dt + \sigma S_t dW_t^{\mathbb{Q}} \]

The Black-Scholes formula becomes:

\[ \boxed{C(S,t) = Se^{-q\tau}\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2)} \]

where:

\[ d_1 = \frac{\ln(S/K) + (r - q + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}, \quad d_2 = d_1 - \sigma\sqrt{\tau} \]

Application: Foreign exchange options (treat foreign interest rate as dividend yield).

2. With Time-Dependent Parameters¶

For time-varying \(r(t)\), \(\sigma(t)\), \(q(t)\):

\[ V(S,t) = \mathbb{E}^{\mathbb{Q}}\left[\exp\left(-\int_t^T r(s)ds\right)\Phi(S_T) \mid S_t = S\right] \]

where:

\[ S_T = S_t\exp\left[\int_t^T\left(r(s)-q(s)-\frac{\sigma^2(s)}{2}\right)ds + \int_t^T\sigma(s)dW_s^{\mathbb{Q}}\right] \]

The log-return is normally distributed with:

Mean: \(\int_t^T(r(s) - q(s) - \frac{\sigma^2(s)}{2})ds\)
Variance: \(\int_t^T\sigma^2(s)ds\)

Modified \(d_1\) and \(d_2\): Replace \(r\tau\) with \(\int_t^T r(s)ds\) and \(\sigma^2\tau\) with \(\int_t^T\sigma^2(s)ds\).

3. Multi-Dimensional Case¶

For basket options with \(n\) assets \(S_1, \ldots, S_n\):

\[ V(\mathbf{S},t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[\Phi(\mathbf{S}_T) \mid \mathbf{S}_t = \mathbf{S}] \]

where \(\mathbf{S}_T = (S_1^T, \ldots, S_n^T)\) follows a multivariate log-normal distribution with correlation matrix \(\rho\).

Example payoffs:

Basket call: \(\Phi = (\sum_{i=1}^n w_i S_i^T - K)^+\)
Exchange option: \(\Phi = (S_1^T - S_2^T)^+\)
Spread option: \(\Phi = (S_1^T - S_2^T - K)^+\)

Evaluation: Typically requires numerical integration or Monte Carlo simulation.

12. Summary¶

The Feynman-Kac formula establishes the fundamental connection:

\[ \text{PDE Solution} = \text{Probabilistic Expectation} \]

For Black-Scholes:

\[ V(S,t) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid S_t = S] \]

Derivation steps:

Apply Ito's lemma to discounted \(u(X_t,t)\)
Show it becomes a martingale if \(u\) satisfies the PDE
Take expectation to get probabilistic representation
Solve SDE to find distribution of \(S_T\) (log-normal)
Evaluate expectation by integrating against density
Complete the square to obtain \(\mathcal{N}(d_1)\) and \(\mathcal{N}(d_2)\) terms

Black-Scholes formulas:

Call: \(C = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2)\)

Put: \(P = Ke^{-r\tau}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)\)

where:

\[ d_1 = \frac{\ln(S/K) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}, \quad d_2 = d_1 - \sigma\sqrt{\tau} \]

Interpretation:

\(\mathcal{N}(d_2)\) = Risk-neutral probability of exercise
\(\mathcal{N}(d_1)\) = Delta = Stock-measure probability of exercise

Why Feynman-Kac is powerful:

Converts PDE to probability: expectations are often easier to compute than solving PDEs, and carry intuitive economic meaning
Enables Monte Carlo simulation: particularly useful for high-dimensional and path-dependent problems
Generalizes naturally: extends to time-dependent parameters, multiple assets, and exotic payoffs
Theoretical foundation: links stochastic calculus with classical PDE theory and the Fundamental Theorem of Asset Pricing

The formula demonstrates that PDE, probability, and stochastic process are three equivalent perspectives on the same mathematical object --- a trinity that underlies modern quantitative finance and enables both theoretical insight and practical computation.

The three core methods are one object¶

The expectation \(\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) \mid S_t = S]\) is a convolution of the payoff with the transition density of \(\ln S_T\) --- exactly the Green's function integral derived in the heat equation subsubsection. The heat equation convolution and the Feynman-Kac expectation are the same integral; the Green's function and the transition density are the same function. Conversely, writing the expectation as \(\mathbb{E}^{\mathbb{Q}}[e^{i\omega \ln S_T}]\) in frequency space yields the characteristic function \(\phi_T(\omega)\) that drives Fourier methods. The three core approaches of this chapter are therefore not independent techniques. They compute the same object --- the pricing semigroup \(\mathcal{P}_\tau = e^{\tau\mathcal{L}}\) applied to the payoff --- in three coordinate systems:

Method	Coordinate system	Key object
Heat equation	Spatial \((x, \tau)\)	Green's function (Gaussian kernel)
Feynman-Kac	Probabilistic	Transition density / expectation
Fourier	Spectral \((\omega)\)	Characteristic function

Exercises¶

Exercise 1. Verify the Feynman-Kac formula for the trivial case \(\Phi(S) = S\) (the stock itself). Show that \(V(S,t) = S\) satisfies both the Black-Scholes PDE and the expectation \(e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[S_T \mid S_t = S]\).

Solution to Exercise 1

We need to verify that \(V(S,t) = S\) satisfies both the Black-Scholes PDE and the Feynman-Kac representation.

PDE verification. With \(V = S\): \(\frac{\partial V}{\partial t} = 0\), \(\frac{\partial V}{\partial S} = 1\), \(\frac{\partial^2 V}{\partial S^2} = 0\). Substituting:

\[ 0 + rS \cdot 1 + \frac{1}{2}\sigma^2 S^2 \cdot 0 - rS = rS - rS = 0 \]

The PDE is satisfied.

Feynman-Kac verification. We need \(e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[S_T \mid S_t = S] = S\).

Under \(\mathbb{Q}\), \(S_T = S \exp\left((r - \frac{1}{2}\sigma^2)(T-t) + \sigma(W_T - W_t)\right)\). Taking the expectation:

\[ \mathbb{E}^{\mathbb{Q}}[S_T \mid S_t = S] = S \exp\left((r - \frac{1}{2}\sigma^2)(T-t)\right) \cdot \mathbb{E}\left[\exp(\sigma(W_T - W_t))\right] \]

Since \(W_T - W_t \sim \mathcal{N}(0, T-t)\):

\[ \mathbb{E}[\exp(\sigma(W_T - W_t))] = \exp\left(\frac{1}{2}\sigma^2(T-t)\right) \]

Therefore:

\[ \mathbb{E}^{\mathbb{Q}}[S_T \mid S_t = S] = S \exp\left((r - \frac{1}{2}\sigma^2 + \frac{1}{2}\sigma^2)(T-t)\right) = Se^{r(T-t)} \]

Discounting: \(e^{-r(T-t)} \cdot Se^{r(T-t)} = S\). Both representations agree.

Exercise 2. Carry out the "completing the square" step in the evaluation of \(I_1\) in full detail. Starting from the combined exponent \(y - \frac{(y-m)^2}{2v^2}\), show every algebraic step leading to the factorization into \(e^{m + v^2/2}\) times a Gaussian integral with shifted mean \(m + v^2\).

Solution to Exercise 2

Starting from the combined exponent \(y - \frac{(y-m)^2}{2v^2}\), we proceed step by step.

Step 1: Write as a single fraction.

\[ y - \frac{(y-m)^2}{2v^2} = \frac{2v^2 y - (y-m)^2}{2v^2} \]

Step 2: Expand the numerator.

\[ 2v^2 y - (y-m)^2 = 2v^2 y - y^2 + 2ym - m^2 \]

\[ = -y^2 + 2y(m + v^2) - m^2 \]

Step 3: Complete the square in \(y\).

\[ -y^2 + 2y(m+v^2) - m^2 = -\left[y^2 - 2y(m+v^2)\right] - m^2 \]

\[ = -\left[(y - (m+v^2))^2 - (m+v^2)^2\right] - m^2 \]

\[ = -(y - (m+v^2))^2 + (m+v^2)^2 - m^2 \]

Step 4: Simplify the constant.

\[ (m+v^2)^2 - m^2 = m^2 + 2mv^2 + v^4 - m^2 = 2mv^2 + v^4 \]

Step 5: Substitute back.

\[ y - \frac{(y-m)^2}{2v^2} = \frac{-(y-(m+v^2))^2 + 2mv^2 + v^4}{2v^2} \]

\[ = -\frac{(y-(m+v^2))^2}{2v^2} + m + \frac{v^2}{2} \]

Step 6: Factor out the constant. Therefore:

\[ \exp\left(y - \frac{(y-m)^2}{2v^2}\right) = \exp\left(m + \frac{v^2}{2}\right) \cdot \exp\left(-\frac{(y-(m+v^2))^2}{2v^2}\right) \]

The second factor is the kernel of a Gaussian with mean \(m + v^2\) and variance \(v^2\), confirming that \(I_1 = e^{m+v^2/2} \cdot \mathcal{N}\left(\frac{(m+v^2) - \ln K}{v}\right)\).

Exercise 3. Use the Feynman-Kac representation to derive the price of a power option with payoff \(\Phi(S_T) = S_T^2\) at maturity. Compute \(e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[S_T^2 \mid S_t = S]\) using the log-normal distribution of \(S_T\).

Solution to Exercise 3

The payoff is \(\Phi(S_T) = S_T^2\). By Feynman-Kac:

\[ V(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T^2 \mid S_t = S] \]

where \(\tau = T - t\).

Under \(\mathbb{Q}\), \(S_T = S\exp\left((r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z\right)\) with \(Z \sim \mathcal{N}(0,1)\). Therefore:

\[ S_T^2 = S^2 \exp\left(2(r - \frac{1}{2}\sigma^2)\tau + 2\sigma\sqrt{\tau}Z\right) \]

Taking the expectation:

\[ \mathbb{E}[S_T^2] = S^2 \exp\left(2(r - \frac{1}{2}\sigma^2)\tau\right) \cdot \mathbb{E}[\exp(2\sigma\sqrt{\tau}Z)] \]

Using \(\mathbb{E}[e^{aZ}] = e^{a^2/2}\) for \(Z \sim \mathcal{N}(0,1)\):

\[ \mathbb{E}[\exp(2\sigma\sqrt{\tau}Z)] = \exp(2\sigma^2\tau) \]

Therefore:

\[ \mathbb{E}[S_T^2] = S^2 \exp\left(2r\tau - \sigma^2\tau + 2\sigma^2\tau\right) = S^2 \exp\left((2r + \sigma^2)\tau\right) \]

Discounting:

\[ V(S,t) = e^{-r\tau} \cdot S^2 e^{(2r+\sigma^2)\tau} = S^2 e^{(r+\sigma^2)\tau} \]

One can verify this satisfies the Black-Scholes PDE: with \(V = S^2 e^{(r+\sigma^2)\tau}\), computing \(V_t = -(r+\sigma^2)V\), \(V_S = 2Se^{(r+\sigma^2)\tau}\), \(V_{SS} = 2e^{(r+\sigma^2)\tau}\), and substituting into \(V_t + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS} - rV = 0\) confirms the identity.

Exercise 4. The Kolmogorov backward equation for Black-Scholes governs \(V(S,t)\), while the forward equation governs the transition density \(p(S_T, T \mid S, t)\). Starting from the Black-Scholes backward equation, write down the corresponding forward (Fokker-Planck) equation and verify that the log-normal density \(p(S_T, T \mid S, t)\) satisfies it.

Solution to Exercise 4

Backward equation (Black-Scholes PDE for \(V(S,t)\)):

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

Deriving the forward equation. The transition density \(p(S_T, T \mid S, t)\) satisfies the Kolmogorov forward (Fokker-Planck) equation with respect to \((S_T, T)\):

\[ \frac{\partial p}{\partial T} = -\frac{\partial}{\partial S_T}(rS_T \, p) + \frac{1}{2}\frac{\partial^2}{\partial S_T^2}(\sigma^2 S_T^2 \, p) \]

Verification. The log-normal density is:

\[ p(S_T, T \mid S, t) = \frac{1}{S_T \sigma\sqrt{2\pi\tau}}\exp\left(-\frac{(\ln(S_T/S) - (r - \frac{\sigma^2}{2})\tau)^2}{2\sigma^2\tau}\right) \]

Let \(y = \ln S_T\), \(m = \ln S + (r - \frac{\sigma^2}{2})\tau\), \(v^2 = \sigma^2\tau\). The density in \(y\)-space is Gaussian: \(q(y) = \frac{1}{\sqrt{2\pi v^2}}e^{-(y-m)^2/(2v^2)}\).

Taking derivatives of \(q\) with respect to \(T\) (noting \(m\) and \(v^2\) both depend on \(T\)), one can verify by direct computation that the forward equation is satisfied. The key steps involve:

\(\frac{\partial m}{\partial T} = r - \frac{\sigma^2}{2}\) and \(\frac{\partial v^2}{\partial T} = \sigma^2\)
Converting derivatives with respect to \(S_T\) to derivatives with respect to \(y\) using \(\frac{\partial}{\partial S_T} = \frac{1}{S_T}\frac{\partial}{\partial y}\)
Expanding the forward equation terms and verifying cancellation

Exercise 5. Using the Feynman-Kac formula with time-dependent volatility \(\sigma(t)\), show that the Black-Scholes call price takes the same functional form as the constant-volatility case, but with \(\sigma^2 T\) replaced by \(\int_t^T \sigma^2(s) \, ds\). Define the effective volatility \(\bar{\sigma}\) and express \(d_1\) and \(d_2\) in terms of \(\bar{\sigma}\).

Solution to Exercise 5

With time-dependent volatility \(\sigma(t)\), the SDE under \(\mathbb{Q}\) is:

\[ dS_s = rS_s \, ds + \sigma(s) S_s \, dW_s^{\mathbb{Q}} \]

By Ito's lemma applied to \(\ln S\):

\[ \ln S_T = \ln S_t + \int_t^T \left(r - \frac{\sigma^2(s)}{2}\right)ds + \int_t^T \sigma(s) \, dW_s \]

The stochastic integral \(\int_t^T \sigma(s) \, dW_s\) is Gaussian with mean 0 and variance \(\int_t^T \sigma^2(s) \, ds\).

Define the effective (total) variance:

\[ \Sigma^2 = \int_t^T \sigma^2(s) \, ds \]

and the effective volatility:

\[ \bar{\sigma} = \sqrt{\frac{1}{\tau}\int_t^T \sigma^2(s) \, ds} \]

so that \(\Sigma^2 = \bar{\sigma}^2 \tau\).

The distribution of \(\ln S_T\) is:

\[ \ln S_T \mid S_t \sim \mathcal{N}\left(\ln S_t + r\tau - \frac{\Sigma^2}{2}, \, \Sigma^2\right) \]

This is identical to the constant-volatility case with \(\sigma^2\tau\) replaced by \(\Sigma^2 = \int_t^T \sigma^2(s) \, ds\). The call price therefore takes the same functional form:

\[ C = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2) \]

with:

\[ d_1 = \frac{\ln(S/K) + r\tau + \frac{1}{2}\bar{\sigma}^2\tau}{\bar{\sigma}\sqrt{\tau}} = \frac{\ln(S/K) + r\tau + \frac{1}{2}\Sigma^2}{\Sigma} \]

\[ d_2 = d_1 - \bar{\sigma}\sqrt{\tau} = d_1 - \Sigma \]

Exercise 6. Consider a European option with payoff \(\Phi(S_T) = \ln(S_T)\) (a log contract). Use the Feynman-Kac representation to compute its price \(V(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[\ln S_T \mid S_t = S]\). Show that the result is \(V = e^{-r\tau}[\ln S + (r - \frac{1}{2}\sigma^2)\tau]\), and verify that this satisfies the Black-Scholes PDE.

Solution to Exercise 6

The payoff is \(\Phi(S_T) = \ln(S_T)\). By Feynman-Kac:

\[ V(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[\ln S_T \mid S_t = S] \]

Under \(\mathbb{Q}\):

\[ \ln S_T = \ln S + (r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z \]

Taking the expectation (\(\mathbb{E}[Z] = 0\)):

\[ \mathbb{E}^{\mathbb{Q}}[\ln S_T \mid S_t = S] = \ln S + (r - \frac{1}{2}\sigma^2)\tau \]

Therefore:

\[ V(S,t) = e^{-r\tau}\left[\ln S + (r - \frac{1}{2}\sigma^2)\tau\right] \]

PDE verification. Let \(\tau = T - t\) and compute derivatives:

\[ \frac{\partial V}{\partial t} = re^{-r\tau}\left[\ln S + (r - \frac{1}{2}\sigma^2)\tau\right] + e^{-r\tau}(r - \frac{1}{2}\sigma^2)(-1) \]

Since \(\frac{\partial \tau}{\partial t} = -1\):

\[ \frac{\partial V}{\partial t} = re^{-r\tau}\left[\ln S + (r-\frac{1}{2}\sigma^2)\tau\right] - e^{-r\tau}(r - \frac{1}{2}\sigma^2) \]

\[ \frac{\partial V}{\partial S} = \frac{e^{-r\tau}}{S}, \quad \frac{\partial^2 V}{\partial S^2} = -\frac{e^{-r\tau}}{S^2} \]

Substituting into the BS PDE:

\[ re^{-r\tau}[\ln S + (r-\frac{1}{2}\sigma^2)\tau] - e^{-r\tau}(r-\frac{1}{2}\sigma^2) + rS \cdot \frac{e^{-r\tau}}{S} + \frac{\sigma^2 S^2}{2}\cdot\left(-\frac{e^{-r\tau}}{S^2}\right) - re^{-r\tau}[\ln S + (r-\frac{1}{2}\sigma^2)\tau] \]

Factoring out \(e^{-r\tau}\):

\[ -(r-\frac{1}{2}\sigma^2) + r - \frac{1}{2}\sigma^2 = -r + \frac{1}{2}\sigma^2 + r - \frac{1}{2}\sigma^2 = 0 \]

The PDE is satisfied.

Exercise 7. The discounted option value \(e^{-rt}V(S_t, t)\) is a martingale under \(\mathbb{Q}\). Use Ito's lemma to compute \(d(e^{-rt}V)\) and show that the drift vanishes if and only if \(V\) satisfies the Black-Scholes PDE. This provides an alternative proof of the Feynman-Kac connection.

Solution to Exercise 7

Let \(Y_t = e^{-rt}V(S_t, t)\). Apply the product rule (Ito's lemma for products):

\[ dY_t = -re^{-rt}V \, dt + e^{-rt} \, dV \]

Apply Ito's lemma to \(V(S_t, t)\):

\[ dV = \frac{\partial V}{\partial t} \, dt + \frac{\partial V}{\partial S} \, dS_t + \frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS_t)^2 \]

Under \(\mathbb{Q}\), \(dS_t = rS_t \, dt + \sigma S_t \, dW_t\) and \((dS_t)^2 = \sigma^2 S_t^2 \, dt\). Substituting:

\[ dV = \left[\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2}\right]dt + \sigma S\frac{\partial V}{\partial S} \, dW_t \]

Therefore:

\[ dY_t = e^{-rt}\left[\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV\right]dt + e^{-rt}\sigma S\frac{\partial V}{\partial S} \, dW_t \]

For \(Y_t\) to be a martingale, the \(dt\) coefficient (drift) must vanish:

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

This is exactly the Black-Scholes PDE. Conversely, if \(V\) satisfies the PDE, then:

\[ dY_t = e^{-rt}\sigma S_t\frac{\partial V}{\partial S}(S_t, t) \, dW_t \]

which has no drift, so \(Y_t = e^{-rt}V(S_t,t)\) is a local martingale (and a martingale under standard integrability conditions). This provides an alternative derivation of the Feynman-Kac representation.