Complete Black-Scholes Proofs: From First Principles
1 Part I: Mathematical Foundations and Definitions
This section establishes all the mathematical machinery needed for the Black-Scholes proofs. We assume knowledge of measure theory and basic stochastic calculus but will define all financial and probabilistic concepts explicitly.
1.0.1 Financial Market Framework
The explicit solution to geometric Brownian motion is: \[S_t = S_0 \exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right)\] This can be verified using Itô’s lemma on \(f(t,x) = \ln x\).
1.0.2 Portfolio and Trading Strategy Concepts
1.0.3 Martingale Theory for Finance
1.0.4 Options and Derivatives
1.0.5 Fundamental Theorems of Asset Pricing
1.1 Part II: The Black-Scholes Model Setup
1.1.1 Model Specification
We consider a financial market on a probability space \((\Omega, \mathcal{F}, \mathbb{P})\) with filtration \(\{\mathcal{F}_t\}_{t \geq 0}\) and two traded assets:
1.1.2 Key Properties of the Model
Let \(Y_t = \ln S_t\). By Itô’s lemma with \(f(x) = \ln x\):
\[\begin{align} dY_t &= f'(S_t) dS_t + \frac{1}{2}f''(S_t)(dS_t)^2 \\ &= \frac{1}{S_t} dS_t + \frac{1}{2}\left(-\frac{1}{S_t^2}\right)(dS_t)^2 \\ &= \frac{1}{S_t}(\mu S_t \, dt + \sigma S_t \, dW_t) - \frac{1}{2S_t^2}(\sigma S_t)^2 dt \\ &= \mu \, dt + \sigma \, dW_t - \frac{\sigma^2}{2} dt \\ &= \left(\mu - \frac{\sigma^2}{2}\right) dt + \sigma \, dW_t \end{align}\]Integrating from \(0\) to \(t\): \[Y_t = Y_0 + \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\]
Since \(Y_t = \ln S_t\) and \(Y_0 = \ln S_0\): \[\ln S_t = \ln S_0 + \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\]
Therefore: \[S_t = S_0 \exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right)\]
1.2 Part III: Proof via Martingale Approach
1.2.1 Step 1: Construction of Risk-Neutral Measure
Step 1: Define the market price of risk: \[\theta = \frac{\mu - r}{\sigma}\]
This is the constant that will appear in Girsanov’s theorem.
Step 2: Define the Radon-Nikodym density process: \[\begin{align} Z_t &= \exp\left(-\theta W_t - \frac{1}{2}\theta^2 t\right) \\ &= \exp\left(-\frac{\mu - r}{\sigma} W_t - \frac{1}{2}\left(\frac{\mu - r}{\sigma}\right)^2 t\right) \end{align}\]
Step 3: Verify that \(Z_t\) is a martingale and \(\mathbb{E}[Z_T] = 1\).
Since \(\theta\) is constant, we can compute: \[\mathbb{E}[Z_t] = \mathbb{E}\left[\exp\left(-\theta W_t - \frac{1}{2}\theta^2 t\right)\right]\]
Since \(W_t \sim N(0,t)\) under \(\mathbb{P}\): \[\begin{align} \mathbb{E}[Z_t] &= \int_{-\infty}^{\infty} \exp\left(-\theta w - \frac{1}{2}\theta^2 t\right) \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{w^2}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{w^2 + 2\theta tw}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{(w + \theta t)^2 - \theta^2 t^2}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \exp\left(\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{(w + \theta t)^2}{2t}\right) dw \\ &= 1 \end{align}\]
The last integral equals 1 since it’s the integral of a normal density.
Step 4: Define \(\mathbb{Q}\) by \(\frac{d\mathbb{Q}}{d\mathbb{P}} = Z_T\).
Step 5: Apply Girsanov’s theorem. Under \(\mathbb{Q}\), the process: \[\tilde{W}_t = W_t + \int_0^t \theta \, ds = W_t + \theta t\] is a \(\mathbb{Q}\)-Brownian motion.
Step 6: Transform the stock price dynamics. Under \(\mathbb{Q}\): \[\begin{align} dS_t &= \mu S_t \, dt + \sigma S_t \, dW_t \\ &= \mu S_t \, dt + \sigma S_t (d\tilde{W}_t - \theta \, dt) \\ &= \mu S_t \, dt + \sigma S_t \, d\tilde{W}_t - \sigma S_t \theta \, dt \\ &= (\mu - \sigma\theta) S_t \, dt + \sigma S_t \, d\tilde{W}_t \\ &= \left(\mu - \sigma \cdot \frac{\mu - r}{\sigma}\right) S_t \, dt + \sigma S_t \, d\tilde{W}_t \\ &= r S_t \, dt + \sigma S_t \, d\tilde{W}_t \end{align}\]
Step 7: Verify the martingale property. The discounted stock price is: \[e^{-rt}S_t\]
Its differential is: \[\begin{align} d(e^{-rt}S_t) &= -re^{-rt}S_t \, dt + e^{-rt} dS_t \\ &= -re^{-rt}S_t \, dt + e^{-rt}(rS_t \, dt + \sigma S_t \, d\tilde{W}_t) \\ &= e^{-rt}\sigma S_t \, d\tilde{W}_t \end{align}\]
Since this has no \(dt\) term, \(e^{-rt}S_t\) is a \(\mathbb{Q}\)-martingale.
This fundamental question gets to the heart of mathematical finance. The answer reveals why the risk-neutral measure is not just a mathematical convenience, but an essential concept for correct option pricing.
1.2.1.0.1 The Fundamental Problem with the Physical Measure \(\mathbb{P}\)
Under the physical measure \(\mathbb{P}\), the stock follows: \[dS_t = \mu S_t dt + \sigma S_t dW_t\]
If we naively tried to price the option directly under \(\mathbb{P}\): \[V(0, S_0) = e^{-rT} \mathbb{E}_\mathbb{P}[(S_T - K)^+]\]
This formula is fundamentally incorrect! Here’s why:
1.2.1.0.1.1 Issue 1: The Discount Rate Problem
The expectation under \(\mathbb{P}\) gives us the expected payoff in the “real world” where the stock has expected return \(\mu\). But what discount rate should we use?
- If \(\mu > r\): The stock is expected to grow faster than the risk-free rate
- We might think we should discount at rate \(\mu\), not \(r\)
- But the option’s risk profile is completely different from the stock’s risk profile
- The appropriate discount rate for the option depends on its specific risk characteristics
The key insight: We don’t know what discount rate to use for the option under \(\mathbb{P}\) because the option’s risk premium is unclear and depends on complex risk preferences.
1.2.1.0.2 Why the Risk-Neutral Measure \(\mathbb{Q}\) Solves This Elegantly
The brilliant insight is to change to a probability measure where all assets have the same expected return equal to the risk-free rate.
1.2.1.0.2.1 Under \(\mathbb{Q}\):
- Stock expected return: \(r\)
- Bond expected return: \(r\)
- Any derivative’s expected return: \(r\)
This means we can discount everything at the risk-free rate \(r\)!
\[V(0, S_0) = e^{-rT} \mathbb{E}_\mathbb{Q}[(S_T - K)^+]\]
Now the formula is correct because: 1. The expected return of the option under \(\mathbb{Q}\) is exactly \(r\) 2. So we discount at rate \(r\) 3. The mathematics works out perfectly 4. No risk premiums need to be determined
1.2.1.0.3 The Economic Intuition
Think of the two measures this way:
Physical measure \(\mathbb{P}\): “What will actually happen in the real world?” - Stocks have risk premiums reflecting investor risk aversion - Different assets have different expected returns - Discount rates are asset-specific and difficult to determine - Requires knowledge of risk preferences and market prices of risk
Risk-neutral measure \(\mathbb{Q}\): “What would happen in a hypothetical world where all investors are risk-neutral?” - All assets earn the risk-free rate in expectation - No risk premiums exist - Everything can be discounted at the same rate \(r\) - Completely bypasses the need to know risk preferences
1.2.1.0.4 The Arbitrage Connection
The risk-neutral measure exists precisely because there are no arbitrage opportunities. Here’s the fundamental logic:
- No arbitrage ⟺ Risk-neutral measure exists (First Fundamental Theorem of Asset Pricing)
- If we can replicate the option with a portfolio of stock and bonds, then the option price must equal the portfolio value (no arbitrage principle)
- The replicating portfolio approach automatically gives us the risk-neutral valuation
- The \(\mathbb{Q}\) measure is the unique measure that makes this work
1.2.1.0.5 A Concrete Example
Consider a simple one-period binomial model:
Setup: - Stock price: \(S_0 = 100\) - Up move: \(S_1 = 120\) with probability \(p = 0.6\) under \(\mathbb{P}\) - Down move: \(S_1 = 80\) with probability \(1-p = 0.4\) under \(\mathbb{P}\) - Risk-free rate: \(r = 5\%\)
Under \(\mathbb{P}\): Expected stock return is \(0.6 \times 20\% + 0.4 \times (-20\%) = 4\%\)
For a call option with \(K = 100\): - Payoff if stock goes up: \(\max(120-100, 0) = 20\) - Payoff if stock goes down: \(\max(80-100, 0) = 0\)
1.2.1.0.5.1 Wrong Approach (using \(\mathbb{P}\)):
\[V_0 = \frac{0.6 \times 20 + 0.4 \times 0}{1.05} = \frac{12}{1.05} = 11.43\]
This is incorrect because we’re using the wrong probabilities for discounting at the risk-free rate.
1.2.1.0.5.2 Correct Approach (using \(\mathbb{Q}\)):
First, find the risk-neutral probabilities. Under \(\mathbb{Q}\), the stock must have expected return equal to the risk-free rate \(r = 5\%\):
\[q \times 120 + (1-q) \times 80 = 100 \times 1.05 = 105\] \[40q + 80 = 105\] \[q = 0.625\]
Now we can price correctly: \[V_0 = \frac{0.625 \times 20 + 0.375 \times 0}{1.05} = \frac{12.5}{1.05} = 11.90\]
The risk-neutral approach gives the unique arbitrage-free price!
1.2.1.0.5.3 Verification by Replication:
We can verify this is correct by constructing a replicating portfolio: - Buy \(\Delta\) shares of stock - Invest \(B\) in bonds
Portfolio value in up state: \(120\Delta + 1.05B = 20\) Portfolio value in down state: \(80\Delta + 1.05B = 0\)
Solving: \(\Delta = 0.5\), \(B = -38.10\)
Initial portfolio cost: \(100 \times 0.5 - 38.10 = 11.90\) ✓
1.2.1.0.6 Why This Matters for Black-Scholes
In the Black-Scholes framework:
- Physical measure: Stock has drift \(\mu\), but option pricing would require determining the option’s risk premium
- Risk-neutral measure: Stock has drift \(r\), and option pricing becomes a pure expectation calculation
- Girsanov’s theorem: Provides the mathematical machinery to change from \(\mathbb{P}\) to \(\mathbb{Q}\)
- Hedging connection: The \(\mathbb{Q}\) measure emerges naturally from the delta-hedging argument
1.2.1.0.7 Summary: Why We Need \(\mathbb{Q}\)
The risk-neutral measure is essential because it:
- Eliminates risk premium confusion - all assets earn rate \(r\) in expectation
- Provides the correct discount rate - always the risk-free rate \(r\)
- Gives the unique arbitrage-free approach - guaranteed by fundamental theorems
- Connects to replication strategies - matches the hedging-based derivation perfectly
- Makes pricing computationally tractable - turns complex pricing into expectation calculations
- Bypasses investor preferences - no need to know risk aversion parameters
- Ensures market completeness - works for any derivative in a complete market
The key insight: The risk-neutral measure is not about what will actually happen in reality - it’s about finding the unique arbitrage-free price in a mathematically elegant and practically implementable way. It transforms the complex problem of determining risk-adjusted discount rates into the simpler problem of computing expectations under an artificial but mathematically convenient probability measure.
1.2.2 Step 2: Stock Price Under Risk-Neutral Measure
Under \(\mathbb{Q}\), the stock follows: \[dS_t = rS_t \, dt + \sigma S_t \, d\tilde{W}_t\]
Using the same technique as in the original measure, let \(Y_t = \ln S_t\): \[\begin{align} dY_t &= \frac{1}{S_t} dS_t - \frac{1}{2S_t^2}(dS_t)^2 \\ &= \frac{1}{S_t}(rS_t \, dt + \sigma S_t \, d\tilde{W}_t) - \frac{1}{2S_t^2}(\sigma S_t)^2 dt \\ &= r \, dt + \sigma \, d\tilde{W}_t - \frac{\sigma^2}{2} dt \\ &= \left(r - \frac{\sigma^2}{2}\right) dt + \sigma \, d\tilde{W}_t \end{align}\]
Integrating: \[Y_t = Y_0 + \left(r - \frac{\sigma^2}{2}\right)t + \sigma \tilde{W}_t\]
Therefore: \[S_t = S_0 \exp\left(\left(r - \frac{\sigma^2}{2}\right)t + \sigma \tilde{W}_t\right)\]
1.2.3 Step 3: Risk-Neutral Valuation
This follows directly from the risk-neutral valuation principle. In an arbitrage-free complete market, the option price equals the discounted expected payoff under the risk-neutral measure.
1.2.4 Step 4: Computing the Expectation
Step 1: Express the stock price at maturity. Under \(\mathbb{Q}\), using the strong Markov property: \[S_T = S_t \exp\left(\left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma (\tilde{W}_T - \tilde{W}_t)\right)\]
Since \(\tilde{W}_T - \tilde{W}_t \sim N(0, T-t)\) under \(\mathbb{Q}\), let \(Z \sim N(0,1)\). Then: \[S_T = S_t \exp\left(\left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma\sqrt{T-t} \cdot Z\right)\]
Step 2: Set up the expectation. \[\begin{align} V(t, S_t) &= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[(S_T - K)^+ | \mathcal{F}_t] \\ &= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}\left[\left(S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z} - K\right)^+\right] \end{align}\]
Step 3: Find the exercise region. The option is exercised when \(S_T > K\), i.e., when: \[S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z} > K\]
Taking logarithms: \[\begin{align} &(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z > \ln(K/S_t) \\ &Z > \frac{\ln(K/S_t) - (r-\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}} = -d_2 \end{align}\]
Step 4: Evaluate the integral. \[\begin{align} V(t, S_t) &= e^{-r(T-t)} \int_{-d_2}^{\infty} \left(S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} z} - K\right) \\ &\quad \times \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &= e^{-r(T-t)} S_t e^{(r-\sigma^2/2)(T-t)} \int_{-d_2}^{\infty} e^{\sigma\sqrt{T-t} z} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &\quad - e^{-r(T-t)} K \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \end{align}\]
Step 5: Evaluate the first integral. For the first integral, complete the square in the exponent: \[\sigma\sqrt{T-t} z - \frac{z^2}{2} = -\frac{1}{2}(z - \sigma\sqrt{T-t})^2 + \frac{\sigma^2(T-t)}{2}\]
Therefore: \[\begin{align} &\int_{-d_2}^{\infty} e^{\sigma\sqrt{T-t} z} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &= e^{\sigma^2(T-t)/2} \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-(z-\sigma\sqrt{T-t})^2/2} dz \\ &= e^{\sigma^2(T-t)/2} \int_{-d_2-\sigma\sqrt{T-t}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2} du \\ &= e^{\sigma^2(T-t)/2} \Phi(d_2 + \sigma\sqrt{T-t}) \\ &= e^{\sigma^2(T-t)/2} \Phi(d_1) \end{align}\]
where we used the substitution \(u = z - \sigma\sqrt{T-t}\) and the fact that: \[d_2 + \sigma\sqrt{T-t} = d_1\]
Step 6: Evaluate the second integral. \[\int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz = \Phi(d_2)\]
Step 7: Combine the results. \[\begin{align} V(t, S_t) &= e^{-r(T-t)} S_t e^{(r-\sigma^2/2)(T-t)} e^{\sigma^2(T-t)/2} \Phi(d_1) - e^{-r(T-t)} K \Phi(d_2) \\ &= S_t e^{-r(T-t)} e^{(r-\sigma^2/2)(T-t)} e^{\sigma^2(T-t)/2} \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \\ &= S_t e^{-r(T-t) + r(T-t) - \sigma^2(T-t)/2 + \sigma^2(T-t)/2} \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \\ &= S_t \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \end{align}\]
1.3 Part IV: Proof via PDE Approach
1.3.1 Step 1: Delta Hedging Strategy
1.3.2 Step 2: Application of Itô’s Lemma to Option Value
Since \(dS_t = \mu S_t \, dt + \sigma S_t \, dW_t\), we have: \((dS_t)^2 = (\mu S_t \, dt + \sigma S_t \, dW_t)^2 = \sigma^2 S_t^2 (dW_t)^2 = \sigma^2 S_t^2 dt\)
where we used the fact that \((dW_t)^2 = dt\), \(dt \cdot dW_t = 0\), and \((dt)^2 = 0\).
Applying Itô’s lemma: \[\begin{align} dV &= \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS_t + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} (dS_t)^2 \\ &= \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} (\mu S_t \, dt + \sigma S_t \, dW_t) + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2 S_t^2 dt \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt + \sigma S_t \frac{\partial V}{\partial S} dW_t \end{align}\]
1.3.3 Step 3: Portfolio Dynamics
We have: \[\begin{align} d\Pi_t &= dV - \Delta_t dS_t \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt + \sigma S_t \frac{\partial V}{\partial S} dW_t \\ &\quad - \Delta_t (\mu S_t dt + \sigma S_t dW_t) \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} - \Delta_t \mu S_t\right] dt \\ &\quad + \sigma S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) dW_t \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt \\ &\quad + \sigma S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) dW_t \end{align}\]
1.3.4 Step 4: Eliminating Risk
1.3.5 Step 5: No-Arbitrage Condition
If a risk-free portfolio earned more than the risk-free rate, we could borrow at rate \(r\), invest in the portfolio, and earn a risk-free profit (arbitrage). If it earned less, we could short the portfolio, lend at rate \(r\), and again earn risk-free profit.
1.3.6 Step 6: Deriving the Black-Scholes PDE
Equating the two expressions for \(d\Pi_t\): \(\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} = r\left(V - \frac{\partial V}{\partial S} S_t\right)\)
Rearranging: \(\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} = rV - rS_t \frac{\partial V}{\partial S}\)
Therefore: \(\frac{\partial V}{\partial t} + rS_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} - rV = 0\)
The terminal condition comes from the option payoff at maturity.
1.3.7 Step 7: Solving the PDE
To solve this PDE, we use a change of variables to transform it into the heat equation.
We need to compute the partial derivatives of \(V\) in terms of \(u\). We have: \(V(t,S) = K e^{-r(T-t)} u(T-t, \ln(S/K))\)
Let \(\tau = T-t\) and \(x = \ln(S/K)\). Then:
\[\begin{align} \frac{\partial V}{\partial t} &= K e^{-r\tau} \left[r u - \frac{\partial u}{\partial \tau}\right] \\ \frac{\partial V}{\partial S} &= K e^{-r\tau} \frac{1}{S} \frac{\partial u}{\partial x} \\ \frac{\partial^2 V}{\partial S^2} &= K e^{-r\tau} \frac{1}{S^2} \left[\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x}\right] \end{align}\]Substituting into the Black-Scholes PDE and simplifying (after dividing by \(K e^{-r\tau}\)): \(-\frac{\partial u}{\partial \tau} + \left(r - \frac{\sigma^2}{2}\right) \frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2 \frac{\partial^2 u}{\partial x^2} = 0\)
Therefore: \(\frac{\partial u}{\partial \tau} = \frac{1}{2}\sigma^2 \frac{\partial^2 u}{\partial x^2} + \left(r - \frac{\sigma^2}{2}\right) \frac{\partial u}{\partial x}\)
The initial condition at \(\tau = 0\) (i.e., \(t = T\)) is: \(u(0, x) = \frac{(Ke^x - K)^+}{K} = (e^x - 1)^+\)
1.3.8 Step 8: Computing the Solution
We have: \(V(t,S) = K e^{-r(T-t)} u(T-t, \ln(S/K))\)
The transformation from the \(u\) solution to the \(V\) solution follows the same integration techniques as in the martingale approach. The key insight is that both methods yield identical results, demonstrating the equivalence of risk-neutral valuation and PDE approaches.
1.4 Verification and Conclusion
1.4.1 Economic Interpretation
1.4.2 Key Insights
1.4.3 Greeks and Risk Management
1.4.4 Extensions and Limitations
1.4.5 Computational Implementation
This complete treatment of the Black-Scholes model provides both theoretical rigor and practical applicability, showing how mathematical finance bridges pure mathematics and real-world financial applications.
1.5 Introduction
The Feynman-Kac theorem provides a fundamental connection between partial differential equations and stochastic processes. While our previous Black-Scholes derivations used either pure martingale methods or pure PDE methods, the Feynman-Kac theorem elegantly unifies both approaches and provides the most direct path from the Black-Scholes PDE to the risk-neutral valuation formula.
The Feynman-Kac theorem shows that the solution to certain PDEs can be represented as expectations of stochastic processes. This is exactly what we need in finance: we derive a PDE from no-arbitrage arguments, then use Feynman-Kac to get the risk-neutral pricing formula.
1.6 Part I: The Feynman-Kac Theorem
1.6.1 Statement of the Theorem
The proof uses Itô’s lemma and the martingale representation theorem.
Step 1: Define the discounted process Let \(Y_s = e^{-\int_t^s r(u,X_u) du} u(s,X_s)\) for \(s \in [t,T]\).
Step 2: Apply Itô’s lemma to \(Y_s\) We need to compute \(dY_s\). Using the product rule and Itô’s lemma:
\[\begin{align} dY_s &= d\left(e^{-\int_t^s r(u,X_u) du}\right) u(s,X_s) + e^{-\int_t^s r(u,X_u) du} du(s,X_s) \\ &\quad + d\left(e^{-\int_t^s r(u,X_u) du}\right) du(s,X_s) \end{align}\]Step 3: Compute each term
For the discount factor: \[d\left(e^{-\int_t^s r(u,X_u) du}\right) = -r(s,X_s) e^{-\int_t^s r(u,X_u) du} ds\]
For \(u(s,X_s)\), by Itô’s lemma: \[\begin{align} du(s,X_s) &= \frac{\partial u}{\partial s} ds + \frac{\partial u}{\partial x} dX_s + \frac{1}{2}\frac{\partial^2 u}{\partial x^2} (dX_s)^2 \\ &= \left[\frac{\partial u}{\partial s} + b(s,X_s)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(s,X_s)\frac{\partial^2 u}{\partial x^2}\right] ds \\ &\quad + \sigma(s,X_s)\frac{\partial u}{\partial x} dW_s \end{align}\]
Step 4: Substitute the PDE From our PDE, we know: \[\frac{\partial u}{\partial s} + b(s,X_s)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(s,X_s)\frac{\partial^2 u}{\partial x^2} = r(s,X_s)u - f(s,X_s)\]
Step 5: Combine terms \[\begin{align} dY_s &= e^{-\int_t^s r(u,X_u) du} \left[-r(s,X_s)u(s,X_s) + r(s,X_s)u(s,X_s) - f(s,X_s)\right] ds \\ &\quad + e^{-\int_t^s r(u,X_u) du} \sigma(s,X_s)\frac{\partial u}{\partial x} dW_s \\ &= -e^{-\int_t^s r(u,X_u) du} f(s,X_s) ds + e^{-\int_t^s r(u,X_u) du} \sigma(s,X_s)\frac{\partial u}{\partial x} dW_s \end{align}\]
Step 6: Integrate from \(t\) to \(T\) \[Y_T - Y_t = -\int_t^T e^{-\int_t^s r(u,X_u) du} f(s,X_s) ds + \int_t^T e^{-\int_t^s r(u,X_u) du} \sigma(s,X_s)\frac{\partial u}{\partial x} dW_s\]
Step 7: Take expectation The stochastic integral has zero expectation (under appropriate integrability conditions): \[\mathbb{E}[Y_T | X_t = x] = Y_t - \mathbb{E}\left[\int_t^T e^{-\int_t^s r(u,X_u) du} f(s,X_s) ds \,\Big|\, X_t = x\right]\]
Step 8: Substitute definitions \[\begin{align} \mathbb{E}\left[e^{-\int_t^T r(u,X_u) du} u(T,X_T) \,\Big|\, X_t = x\right] &= u(t,x) \\ &\quad - \mathbb{E}\left[\int_t^T e^{-\int_t^s r(u,X_u) du} f(s,X_s) ds \,\Big|\, X_t = x\right] \end{align}\]
Step 9: Use terminal condition \(u(T,x) = g(x)\) \[u(t,x) = \mathbb{E}\left[e^{-\int_t^T r(u,X_u) du} g(X_T) + \int_t^T e^{-\int_t^s r(u,X_u) du} f(s,X_s) ds \,\Big|\, X_t = x\right]\]
This completes the proof.
1.6.2 Key Insights of Feynman-Kac
The Feynman-Kac theorem reveals a fundamental duality:
PDE Side: Deterministic partial differential equation with boundary conditions SDE Side: Stochastic differential equation with expectation of terminal payoff
This duality is the mathematical foundation of: - Risk-neutral pricing in finance - Monte Carlo methods for PDE solving - Connection between heat equations and Brownian motion - Quantum mechanics (Feynman path integrals)
1.7 Part II: Application to Black-Scholes
1.7.1 Black-Scholes PDE Setup
1.7.2 Transforming to Standard Feynman-Kac Form
Step 1: Express derivatives in terms of \(u(t,x)\).
Since \(V(t,S) = u(t,\ln S)\), by the chain rule: \[\frac{\partial V}{\partial S} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial S} = \frac{\partial u}{\partial x} \frac{1}{S}\]
For the second derivative: \[\begin{align} \frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}\left(\frac{1}{S}\frac{\partial u}{\partial x}\right) \\ &= -\frac{1}{S^2}\frac{\partial u}{\partial x} + \frac{1}{S}\frac{\partial}{\partial S}\left(\frac{\partial u}{\partial x}\right) \\ &= -\frac{1}{S^2}\frac{\partial u}{\partial x} + \frac{1}{S}\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial S} \\ &= -\frac{1}{S^2}\frac{\partial u}{\partial x} + \frac{1}{S^2}\frac{\partial^2 u}{\partial x^2} \\ &= \frac{1}{S^2}\left(\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x}\right) \end{align}\]
Step 2: Substitute into the Black-Scholes PDE. \[\begin{align} &\frac{\partial u}{\partial t} + rS \cdot \frac{1}{S}\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2 S^2 \cdot \frac{1}{S^2}\left(\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x}\right) - ru = 0 \\ &\frac{\partial u}{\partial t} + r\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\left(\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x}\right) - ru = 0 \\ &\frac{\partial u}{\partial t} + \left(r - \frac{\sigma^2}{2}\right)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} - ru = 0 \end{align}\]
Step 3: Rearrange to standard form. \[\frac{\partial u}{\partial t} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} + \left(r - \frac{\sigma^2}{2}\right)\frac{\partial u}{\partial x} - ru = 0\]
This is now in the standard Feynman-Kac form with: - \(b(t,x) = r - \frac{\sigma^2}{2}\) (drift coefficient) - \(\sigma(t,x) = \sigma\) (diffusion coefficient)
- \(r(t,x) = r\) (discount rate) - \(f(t,x) = 0\) (source term)
1.7.3 Applying Feynman-Kac to Black-Scholes
Step 1: Identify the parameters for Feynman-Kac. From our transformed PDE: - \(b(t,x) = r - \frac{\sigma^2}{2}\) - \(\sigma(t,x) = \sigma\)
- \(r(t,x) = r\) - \(f(t,x) = 0\) - \(g(x) = (e^x - K)^+\)
Step 2: Write the associated SDE. \[dX_s = \left(r - \frac{\sigma^2}{2}\right) ds + \sigma dW_s\]
Step 3: Apply the Feynman-Kac formula. Since \(f(t,x) = 0\), the formula simplifies to: \[u(t,x) = \mathbb{E}\left[e^{-r(T-t)} g(X_T) \,\Big|\, X_t = x\right]\]
Step 4: Solve the SDE. The solution to the linear SDE is: \[X_T = X_t + \left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma(W_T - W_t)\]
Step 5: Transform back to stock price. Since \(X = \ln S\), we have: \[\begin{align} \ln S_T &= \ln S_t + \left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma(W_T - W_t) \\ S_T &= S_t \exp\left(\left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma(W_T - W_t)\right) \end{align}\]
Step 6: Final formula. \[V(t,S_t) = e^{-r(T-t)} \mathbb{E}[(S_T - K)^+ | S_t]\]
This is exactly the risk-neutral valuation formula!
1.7.4 The Key Insight: Automatic Risk-Neutral Measure
The beauty of the Feynman-Kac approach is that it automatically produces the risk-neutral measure without explicitly constructing it via Girsanov’s theorem.
What happens: 1. We start with the Black-Scholes PDE (derived from no-arbitrage hedging) 2. Feynman-Kac gives us the stochastic representation 3. The resulting SDE for \(X_t = \ln S_t\) has drift \(r - \sigma^2/2\) 4. This corresponds to the stock having drift \(r\) under the measure induced by the Feynman-Kac expectation 5. This measure is exactly the risk-neutral measure \(\mathbb{Q}\)!
The connection: - PDE approach: Hedging → Black-Scholes PDE → Feynman-Kac → Risk-neutral expectation - Martingale approach: No-arbitrage → Risk-neutral measure → Expectation formula
Both paths lead to the same destination, but Feynman-Kac provides the bridge.
1.8 Part III: Complete Assumptions and Regularity Conditions
1.8.1 Mathematical Assumptions for Feynman-Kac
Assumption 1-3 (Coefficient Properties): ✓ All coefficients are constants, so they trivially satisfy: - Lipschitz continuity: \(|b(t,x) - b(t,y)| \leq L|x-y|\) (with \(L=0\)) - Linear growth: \(|b(t,x)| \leq C(1+|x|)\) (with \(C = |r-\sigma^2/2|\)) - Non-degeneracy: \(\sigma > 0\)
Assumption 4 (Domain): ✓ The log-price \(x = \ln S\) can take any real value, making \(\mathbb{R}\) the natural domain.
Assumption 5 (Terminal Condition): ✓ \((e^x - K)^+ \leq e^{|x|} + K\) has exponential growth, which satisfies polynomial growth conditions for the theorem.
Assumption 7 (SDE Solution): ✓ The SDE \(dX_s = \mu ds + \sigma dW_s\) with constant coefficients has the explicit solution: \[X_t = X_0 + \mu t + \sigma W_t\] This clearly has a unique strong solution.
Assumption 8 (Finite Moments): ✓ Since \(X_t\) is Gaussian, it has finite moments of all orders.
Assumption 9 (PDE Solution): ✓ The Black-Scholes PDE is a well-studied parabolic PDE with known classical solutions.
1.8.2 Financial Market Assumptions
1.8.3 Comparison with Alternative Approaches
Martingale Approach: - Strengths: Direct probabilistic interpretation, connects to fundamental theorems - Weaknesses: Requires explicit construction of risk-neutral measure via Girsanov
PDE Approach:
- Strengths: Uses familiar hedging intuition, connects to numerical methods - Weaknesses: Requires solving PDE, less direct connection to probability
Feynman-Kac Approach: - Strengths: Unifies PDE and probability, automatic risk-neutral measure, generalizes easily - Weaknesses: Requires understanding of both PDE and SDE theory, more abstract
When to Use Each: - Martingale: When risk-neutral measure is natural starting point - PDE: When hedging interpretation is most important
- Feynman-Kac: When connecting different mathematical frameworks or generalizing to complex payoffs
1.9 Part IV: Extensions and Applications
1.9.1 Generalizations of Feynman-Kac in Finance
1.9.2 Computational Advantages
The Feynman-Kac representation enables two complementary numerical approaches:
Monte Carlo Simulation: - Simulate the SDE paths - Compute sample average of discounted payoffs - Converges by Law of Large Numbers
PDE Finite Differences: - Discretize the PDE on a grid - Solve numerically using implicit/explicit schemes - Converges to PDE solution
Both methods solve the same problem via Feynman-Kac duality!
1.10 Conclusion
The Feynman-Kac approach reveals that: - PDEs and stochastic processes are dual representations of the same underlying mathematics - The risk-neutral measure emerges naturally from the PDE structure - Option pricing connects heat equations, Brownian motion, and financial hedging - Monte Carlo and finite difference methods are fundamentally solving the same problem
This unity suggests that: - Different mathematical tools often lead to the same financial insights - The choice of method depends on computational needs and theoretical focus - Understanding multiple approaches deepens comprehension of the underlying economics
The Feynman-Kac theorem thus provides not just another derivation method, but a unifying perspective that bridges deterministic and stochastic approaches to mathematical finance.
This derivation completes the trilogy of Black-Scholes approaches, showing how partial differential equations, stochastic processes, and martingale theory all contribute to our understanding of option pricing.