Feynman-Kac Formula and the Black-Scholes Solution¶

The preceding sections derived the Black-Scholes formula by transforming the PDE --- into the heat equation, into frequency space via Fourier, Mellin, or Laplace transforms, or by exploiting scaling structure. All of these are ultimately PDE techniques: they manipulate the equation directly. The Feynman-Kac formula takes a fundamentally different approach. It establishes that the solution of a parabolic PDE can be written as a probabilistic expectation, bypassing the PDE entirely in favor of stochastic calculus.

This shift from differential equations to expectations is not merely a technical alternative --- it is the conceptual foundation of risk-neutral pricing. This section presents the general Feynman-Kac theorem, applies it rigorously to the Black-Scholes PDE, and derives the closed-form option pricing formulas through detailed probabilistic calculations.

The General Feynman-Kac Theorem¶

1. Statement¶

Consider a stochastic process \(X_t\) satisfying the stochastic differential equation:

\[ dX_t = \mu(X_t, t)dt + \sigma(X_t, t)dW_t \]

Let \(u(x,t)\) be a function solving the parabolic PDE:

\[ \frac{\partial u}{\partial t} + \mu(x,t)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(x,t)\frac{\partial^2 u}{\partial x^2} - r(x,t)u + f(x,t) = 0 \]

with terminal condition:

\[ u(x,T) = \Phi(x) \]

Then the Feynman-Kac formula states:

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

2. Interpretation of Terms¶

1. Terminal condition term:

\[ e^{-\int_t^T r(X_\tau,\tau)d\tau}\Phi(X_T) \]

This represents the discounted terminal payoff at maturity \(T\).

2. Running cost/reward term:

\[ \int_t^T e^{-\int_t^s r(X_\tau,\tau)d\tau}f(X_s,s)ds \]

This represents accumulated discounted cash flows between \(t\) and \(T\).

3. For option pricing: - \(f \equiv 0\) (no intermediate cash flows for European options) - \(r\) is constant (risk-free rate) - \(\Phi(X_T)\) is the option payoff at maturity

The formula simplifies to:

\[ u(x,t) = \mathbb{E}\left[e^{-r(T-t)}\Phi(X_T) \mid X_t = x\right] \]

This is the risk-neutral valuation formula.

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. This is the probabilistic representation of the pricing semigroup \(\mathcal{P}_\tau = e^{\tau\mathcal{L}}\) introduced in the chapter overview.

This converts the PDE problem into a probabilistic expectation problem.

Rigorous Derivation of Feynman-Kac¶

We now prove the Feynman-Kac formula rigorously using Itô's lemma and martingale theory.

1. Step 1: Apply Itô's Lemma to u(X_t, t)¶

For a function \(u(X_t, t)\) where \(dX_t = \mu dt + \sigma dW_t\), Itô's lemma gives:

\[ du = \frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dX_t + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}(dX_t)^2 \]

Since \((dX_t)^2 = \sigma^2 dt\):

\[ du = \left[\frac{\partial u}{\partial t} + \mu\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}\right]dt + \sigma\frac{\partial u}{\partial x}dW_t \]

2. Step 2: Introduce Discounting¶

Define the discounted value:

\[ Y_t = e^{-\int_0^t r(X_s,s)ds}u(X_t,t) \]

Apply the product rule (Itô's lemma for products):

\[ dY_t = d\left(e^{-\int_0^t r ds}\right) \cdot u + e^{-\int_0^t r ds} \cdot du \]

The discount factor satisfies:

\[ d\left(e^{-\int_0^t r ds}\right) = -r(X_t,t)e^{-\int_0^t r ds}dt \]

Therefore:

\[ \begin{aligned} dY_t &= -re^{-\int_0^t r ds}u \cdot dt + e^{-\int_0^t r ds}\left[\frac{\partial u}{\partial t} + \mu\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}\right]dt \\ &\quad + e^{-\int_0^t r ds}\sigma\frac{\partial u}{\partial x}dW_t \end{aligned} \]

Combining:

\[ dY_t = e^{-\int_0^t r ds}\left[\frac{\partial u}{\partial t} + \mu\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} - ru\right]dt + e^{-\int_0^t r ds}\sigma\frac{\partial u}{\partial x}dW_t \]

3. Step 3: Use the PDE Condition¶

If \(u\) satisfies the Feynman-Kac PDE (with \(f=0\)):

\[ \frac{\partial u}{\partial t} + \mu\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} - ru = 0 \]

Then the drift term vanishes:

\[ \boxed{dY_t = e^{-\int_0^t r ds}\sigma\frac{\partial u}{\partial x}dW_t} \]

Key observation: \(Y_t\) is a martingale (has no \(dt\) term, only \(dW_t\) term).

4. Step 4: Take Conditional Expectation¶

Since \(Y_t\) is a martingale:

\[ \mathbb{E}[Y_T \mid \mathcal{F}_t] = Y_t \]

Expanding:

\[ \mathbb{E}\left[e^{-\int_0^T r ds}u(X_T,T) \mid \mathcal{F}_t\right] = e^{-\int_0^t r ds}u(X_t,t) \]

Using the terminal condition \(u(X_T,T) = \Phi(X_T)\):

\[ e^{-\int_0^t r ds}u(X_t,t) = \mathbb{E}\left[e^{-\int_0^T r ds}\Phi(X_T) \mid \mathcal{F}_t\right] \]

Multiply both sides by \(e^{\int_0^t r ds}\):

\[ u(X_t,t) = \mathbb{E}\left[e^{-\int_t^T r ds}\Phi(X_T) \mid \mathcal{F}_t\right] \]

This is the Feynman-Kac representation.

5. Application to Black-Scholes¶

For the Black-Scholes case with constant \(r\): - \(X_t = S_t\) - \(\mu(S_t, t) = rS_t\) - \(\sigma(S_t, t) = \sigma S_t\) - \(r(S_t, t) = r\)

The Feynman-Kac formula gives:

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

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

Conclusion: Any solution to the Black-Scholes PDE can be represented as a risk-neutral expectation.

Solving the SDE: Distribution of S_T¶

To evaluate the expectation in the Feynman-Kac formula, we need the distribution of \(S_T\). The result --- that \(\ln S_T\) is normally distributed --- is exactly the lognormal transition density whose Gaussian kernel we already met in the heat equation section. We rederive it here via Ito's lemma, which is the natural probabilistic route and confirms that the two approaches yield the same object.

1. The Stochastic Differential Equation¶

Under the risk-neutral measure:

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

2. Apply Itô's Lemma to ln S_t¶

Let \(f(S) = \ln S\). Then:

\[ \frac{\partial f}{\partial S} = \frac{1}{S}, \quad \frac{\partial^2 f}{\partial S^2} = -\frac{1}{S^2} \]

By Itô's lemma:

\[ d(\ln S_t) = \frac{1}{S_t}dS_t - \frac{1}{2}\frac{1}{S_t^2}(dS_t)^2 \]

Substituting \(dS_t = rS_t dt + \sigma S_t dW_t\) and \((dS_t)^2 = \sigma^2 S_t^2 dt\):

\[ d(\ln S_t) = \frac{1}{S_t}(rS_t dt + \sigma S_t dW_t) - \frac{1}{2}\frac{1}{S_t^2}\sigma^2 S_t^2 dt \]

\[ = rdt + \sigma dW_t - \frac{1}{2}\sigma^2 dt \]

\[ = \left(r - \frac{1}{2}\sigma^2\right)dt + \sigma dW_t^{\mathbb{Q}} \]

3. Integrate from t to T¶

\[ \ln S_T - \ln S_t = \left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma(W_T^{\mathbb{Q}} - W_t^{\mathbb{Q}}) \]

Since \(W_T^{\mathbb{Q}} - W_t^{\mathbb{Q}} \sim \mathcal{N}(0, T-t)\), we can write:

\[ W_T^{\mathbb{Q}} - W_t^{\mathbb{Q}} = \sqrt{T-t} \cdot Z \]

where \(Z \sim \mathcal{N}(0,1)\).

4. Explicit Solution¶

\[ \boxed{S_T = S_t \exp\left[\left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma\sqrt{T-t} \cdot Z\right]} \]

5. Distribution of S_T¶

Define \(\tau = T - t\). Then:

\[ \ln S_T \mid S_t \sim \mathcal{N}\left(\ln S_t + \left(r - \frac{1}{2}\sigma^2\right)\tau, \sigma^2\tau\right) \]

Log-normal distribution: \(S_T\) is log-normally distributed with mean \(\mathbb{E}[\ln S_T | S_t] = \ln S_t + (r - \frac{1}{2}\sigma^2)\tau\) and variance \(\text{Var}[\ln S_T | S_t] = \sigma^2\tau\). The corresponding transition density \(p(S_T \mid S_t)\) is the Gaussian kernel derived in the heat equation section --- expressed here in original \((S,t)\) coordinates rather than transformed \((x,\tau)\) coordinates.

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.

Combine the exponents:

\[ y - \frac{(y-m)^2}{2v^2} = -\frac{1}{2v^2}\left[(y-m)^2 - 2v^2 y\right] \]

Expand:

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

Complete the square:

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

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

Therefore:

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

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.

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. ✓

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.

Derivation: From our calculation of \(I_2\), we showed:

\[ \int_K^{\infty}p(S_T)dS_T = \mathbb{Q}(S_T > K) = \mathcal{N}(d_2) \]

3. Meaning of N(d_1)¶

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

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

Change of numeraire interpretation: Under the stock as numeraire, the stock price dynamics shift:

\[ \frac{dS_t}{S_t} = (r + \sigma^2)dt + \sigma dW_t^{\mathbb{Q}^S} \]

The drift changes by \(\sigma^2\) (via Girsanov's theorem), which explains the relationship:

\[ d_1 = d_2 + \sigma\sqrt{\tau} \]

Alternative interpretation: \(\mathcal{N}(d_1)\) is also the delta (hedge ratio) of the call option:

\[ \Delta_{\text{call}} = \frac{\partial C}{\partial S} = \mathcal{N}(d_1) \]

4. Decomposition of Call Value¶

The call formula can be written as:

\[ C = S \cdot \mathbb{Q}^S(S_T > K) - Ke^{-r\tau} \cdot \mathbb{Q}(S_T > K) \]

Interpretation: - First term: Expected value of stock received upon exercise (under stock measure) - Second term: Expected present value of strike paid upon exercise (under risk-neutral measure)

The difference between the two measures accounts for the convexity of the payoff.

Connection to Kolmogorov Equations¶

1. Backward Equation¶

The Black-Scholes PDE is the Kolmogorov backward equation for the process \(S_t\) under \(\mathbb{Q}\):

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

"Backward" because: - We start from terminal condition \(V(S,T) = \Phi(S)\) at time \(T\) - We solve backward in time to find \(V(S,t)\) for \(t < T\) - The PDE describes how the option value evolves as we move backward from maturity

2. Forward Equation (Fokker-Planck)¶

The transition density \(p(S_T, T \mid S_t, t)\) satisfies the Kolmogorov forward equation:

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

"Forward" because: - Given initial distribution at time \(t\) - We evolve forward in time to find distribution at time \(T\) - The PDE describes how the probability distribution evolves forward

3. Duality Relationship¶

The option value can be expressed as:

\[ V(S,t) = e^{-r\tau}\int_0^{\infty}\Phi(S_T)p(S_T, T \mid S, t)dS_T \]

Key insight: - The backward equation (PDE for \(V\)) describes the evolution of functionals - The forward equation (PDE for \(p\)) describes the evolution of densities - They are dual to each other through the Feynman-Kac representation

Why Feynman-Kac Works: Deep Intuition¶

The rigorous derivation in the previous sections 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: 1. Discounted asset prices are martingales 2. All derivatives can be priced consistently

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

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.

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 section. 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.