Black-Scholes Formula via Heat Equation¶

The Black-Scholes PDE can be transformed into the heat equation, one of the most studied partial differential equations in mathematical physics. This transformation allows us to leverage the well-known fundamental solution (Green's function) to derive the Black-Scholes formula analytically.

This approach reveals the deep connection between option pricing and diffusion processes, demonstrating how financial derivatives can be priced using classical PDE techniques.

The Black-Scholes PDE¶

1. Standard Form¶

For a European option value $V(S,t)$:

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

Terminal condition (European call):

\[ V(S,T) = \max(S-K, 0) \]

Challenge: The PDE has: - Variable coefficients ($S^2$ and $S$ terms) - A zero-order term ($-rV$) - Terminal condition (not initial)

Strategy: Transform into the heat equation, which has constant coefficients and is thoroughly understood.

Change of Variables¶

We apply three transformations to reduce the BS PDE to canonical heat equation form.

1. Transformation 1: Time to Maturity¶

\[ \boxed{\tau = T - t} \]

Meaning: Time remaining until option expiration.

Example: If current time is $t = 0.5$ years and expiration is $T = 1$ year:

\[ \tau = 1 - 0.5 = 0.5 \text{ years (6 months remaining)} \]

Effect: Converts terminal condition into initial condition and reverses time direction:

\[ \frac{\partial}{\partial t} = -\frac{\partial}{\partial \tau} \]

2. Transformation 2: Expected Log-Price¶

\[ \boxed{x = \log S + \left(r - \frac{1}{2}\sigma^2\right)\tau} \]

Meaning: This is the expected log-price at maturity under the risk-neutral measure.

Derivation: Under geometric Brownian motion:

\[ S_T = S_t \exp\left(\left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma W_\tau\right) \]

Taking logarithm and expectation:

\[ \mathbb{E}^{\mathbb{Q}}[\log S_T | S_t] = \log S_t + \left(r - \frac{1}{2}\sigma^2\right)\tau = x \]

Example: For $S = 100$, $r = 5\%$, $\sigma = 20\%$, $\tau = 0.5$:

\[ \begin{aligned} x &= \log 100 + \left(0.05 - \frac{1}{2}(0.2)^2\right) \cdot 0.5 \\ &= 4.605 + (0.05 - 0.02) \cdot 0.5 \\ &= 4.605 + 0.015 = 4.620 \end{aligned} \]

Effect: Centers the process by removing drift, converting first-order terms to second-order.

3. Transformation 3: Forward Option Value¶

\[ \boxed{F(x,\tau) = V(S,t) e^{r\tau}} \]

Meaning: Option value compounded forward to expiration at the risk-free rate.

Intuition: Under risk-neutral pricing:

\[ V(S,t) = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[\text{Payoff}] \]

so

\[ F(x,\tau) = \mathbb{E}^{\mathbb{Q}}[\text{Payoff}] \]

is the undiscounted expected payoff.

Example: If current option value is $V = 10$, $r = 5\%$, $\tau = 0.5$:

\[ F = 10 \cdot e^{0.05 \times 0.5} = 10 \cdot e^{0.025} \approx 10.25 \]

Effect: Eliminates the $-rV$ term from the PDE.

Transformation of Derivatives¶

1. Time Derivative¶

Using chain rule:

\[ \frac{\partial V}{\partial t} = \frac{\partial V}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial V}{\partial \tau}\frac{\partial \tau}{\partial t} \]

Since: - $\frac{\partial x}{\partial t} = -\left(r - \frac{1}{2}\sigma^2\right)$ (from definition of $x$) - $\frac{\partial \tau}{\partial t} = -1$

We get:

\[ \frac{\partial V}{\partial t} = -\left(r - \frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial x} - \frac{\partial V}{\partial \tau} \]

2. First Spatial Derivative¶

\[ \frac{\partial V}{\partial S} = \frac{\partial V}{\partial x}\frac{\partial x}{\partial S} = \frac{\partial V}{\partial x} \cdot \frac{1}{S} \]

3. Second Spatial Derivative¶

\[ \begin{aligned} \frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}\left(\frac{1}{S}\frac{\partial V}{\partial x}\right) \\ &= \frac{\partial}{\partial x}\left(\frac{1}{S}\frac{\partial V}{\partial x}\right)\frac{\partial x}{\partial S} \\ &= \frac{1}{S}\left[\frac{\partial^2 V}{\partial x^2}\frac{1}{S} + \frac{\partial V}{\partial x}\frac{\partial}{\partial x}\left(\frac{1}{S}\right)\right] \end{aligned} \]

Key observation: Since $S = e^{x - (r-\frac{1}{2}\sigma^2)\tau}$:

\[ \frac{\partial}{\partial x}\left(\frac{1}{S}\right) = -\frac{1}{S} \]

Therefore:

\[ \frac{\partial^2 V}{\partial S^2} = \frac{1}{S^2}\left[\frac{\partial^2 V}{\partial x^2} - \frac{\partial V}{\partial x}\right] \]

Derivation of the Heat Equation¶

1. Step 1: Substitute into BS PDE¶

Starting with:

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

Substitute transformed derivatives:

\[ \begin{aligned} &-\left(r - \frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial x} - \frac{\partial V}{\partial \tau} \\ &\quad + \frac{1}{2}\sigma^2\left[\frac{\partial^2 V}{\partial x^2} - \frac{\partial V}{\partial x}\right] \\ &\quad + r\frac{\partial V}{\partial x} - rV = 0 \end{aligned} \]

2. Step 2: Collect Terms¶

Grouping by derivative order:

Second-order: $\frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial x^2}$
First-order: $-\left(r - \frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial x} - \frac{1}{2}\sigma^2\frac{\partial V}{\partial x} + r\frac{\partial V}{\partial x} = 0$ (cancels!)
Zero-order: $-\frac{\partial V}{\partial \tau} - rV = 0$

Result:

\[ -\frac{\partial V}{\partial \tau} + \frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial x^2} - rV = 0 \]

3. Step 3: Eliminate -rV Term¶

Since $V = Fe^{-r\tau}$, compute derivatives:

\[ \frac{\partial V}{\partial \tau} = \frac{\partial F}{\partial \tau}e^{-r\tau} - rFe^{-r\tau} \]

\[ \frac{\partial^2 V}{\partial x^2} = \frac{\partial^2 F}{\partial x^2}e^{-r\tau} \]

Substitute into the PDE:

\[ -\left[\frac{\partial F}{\partial \tau}e^{-r\tau} - rFe^{-r\tau}\right] + \frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2}e^{-r\tau} - rFe^{-r\tau} = 0 \]

Divide by $e^{-r\tau}$:

\[ -\frac{\partial F}{\partial \tau} + rF + \frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2} - rF = 0 \]

Simplify:

\[ \boxed{\frac{\partial F}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2}} \]

This is the heat equation with thermal diffusivity $\kappa = \frac{1}{2}\sigma^2$.

Initial Condition¶

The terminal condition $V(S,T) = (S-K)^+$ becomes an initial condition for $F$.

At $\tau = 0$ (maturity): - $x = \log S$ (since $\tau = 0$ in the definition of $x$) - $F(x,0) = V(S,T)e^{r \cdot 0} = V(S,T)$

For a European call:

\[ \boxed{F(x,0) = \psi(x) = \max(e^x - K, 0) = (e^x - K)^+} \]

This is a piecewise function:

\[ \psi(x) = \begin{cases} e^x - K & \text{if } x > \log K \\ 0 & \text{if } x \leq \log K \end{cases} \]

Summary: We've transformed the terminal-value problem (BS PDE backward in time) into an initial-value problem (heat equation forward in time).

Green's Function (Fundamental Solution)¶

1. Definition¶

The fundamental solution or Green's function of the heat equation is the solution to:

\[ \frac{\partial G}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 G}{\partial x^2} \]

with initial condition:

\[ G(x,0; z) = \delta(x-z) \]

where $\delta$ is the Dirac delta function (unit point source at $z$).

Physical interpretation: $G(x,\tau; z)$ represents the temperature distribution at position $x$ and time $\tau$ resulting from an instantaneous unit heat source placed at position $z$ at time $0$.

2. Explicit Form¶

The fundamental solution is:

\[ \boxed{G(x,\tau; z) = \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right)} \]

This is a Gaussian kernel with: - Mean: $z$ (centered at source location) - Variance: $\sigma^2\tau$ (spreads with time) - Normalization: $\int_{-\infty}^\infty G(x,\tau; z) dx = 1$

3. Verification¶

Property 1 (Satisfies heat equation):

Direct calculation shows:

\[ \frac{\partial G}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 G}{\partial x^2} \]

using standard Gaussian differentiation formulas.

Property 2 (Initial condition):

\[ \lim_{\tau \to 0^+} G(x,\tau; z) = \delta(x-z) \]

As $\tau \to 0$, the Gaussian concentrates at $z$.

Property 3 (Probability conservation):

\[ \int_{-\infty}^\infty G(x,\tau; z) dx = 1 \quad \text{for all } \tau > 0 \]

4. Intuition¶

At $\tau = 0$: All heat concentrated at point $z$
As $\tau$ increases: Heat diffuses according to Gaussian spread
As $\tau \to \infty$: Heat spreads everywhere uniformly

Solution via Superposition¶

1. Duhamel's Principle¶

For any initial condition $\psi(x)$, the solution to the heat equation is obtained by superposition of fundamental solutions:

\[ \boxed{F(x,\tau) = \int_{-\infty}^\infty \psi(z) G(x,\tau; z) dz} \]

Explicitly:

\[ F(x,\tau) = \int_{-\infty}^\infty \psi(z) \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz \]

Interpretation: - Decompose initial condition $\psi(x)$ into point sources: $\psi(z)\delta(z)$ - Each point source at $z$ evolves according to $G(x,\tau; z)$ - The total solution is the integral (continuous sum) of all contributions

Probabilistic view: If $X \sim \mathcal{N}(x, \sigma^2\tau)$, then:

\[ F(x,\tau) = \mathbb{E}[\psi(X)] = \int_{-\infty}^\infty \psi(z) \cdot \text{pdf}(z; x, \sigma^2\tau) dz \]

This connects the PDE solution to expectation under Brownian motion.

Application to European Call¶

1. Setup¶

For a call option with $\psi(x) = (e^x - K)^+$:

\[ F(x,\tau) = \int_{-\infty}^\infty (e^z - K)^+ \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz \]

Since $(e^z - K)^+ = 0$ for $z \leq \log K$:

\[ F(x,\tau) = \int_{\log K}^\infty (e^z - K) \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz \]

2. Split into Two Integrals¶

\[ F(x,\tau) = \underbrace{\int_{\log K}^\infty e^z \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz}_{I_1} - K\underbrace{\int_{\log K}^\infty \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz}_{I_2} \]

We evaluate each integral separately.

Evaluation of I_2 (Strike Term)¶

\[ I_2 = \int_{\log K}^\infty \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz \]

This is a Gaussian tail probability. Define standard normal variable:

\[ Z = \frac{z - x}{\sigma\sqrt{\tau}} \sim \mathcal{N}(0,1) \]

Then:

\[ \begin{aligned} I_2 &= \mathbb{P}(z \geq \log K) \\ &= \mathbb{P}\left(\frac{z-x}{\sigma\sqrt{\tau}} \geq \frac{\log K - x}{\sigma\sqrt{\tau}}\right) \\ &= \mathbb{P}\left(Z \geq \frac{\log K - x}{\sigma\sqrt{\tau}}\right) \\ &= \mathbb{P}\left(Z \leq \frac{x - \log K}{\sigma\sqrt{\tau}}\right) \\ &= \mathcal{N}\left(\frac{x - \log K}{\sigma\sqrt{\tau}}\right) \end{aligned} \]

Define:

\[ \boxed{d_2 = \frac{x - \log K}{\sigma\sqrt{\tau}}} \]

Therefore:

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

Evaluation of I_1 (Stock Term)¶

\[ I_1 = \int_{\log K}^\infty e^z \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right) dz \]

Key technique: Complete the square in the exponent.

1. Step 1: Combine Exponents¶

\[ z - \frac{(x-z)^2}{2\sigma^2\tau} = -\frac{1}{2\sigma^2\tau}\left[(x-z)^2 - 2\sigma^2\tau z\right] \]

Expand:

\[ (x-z)^2 - 2\sigma^2\tau z = z^2 - 2zx + x^2 - 2\sigma^2\tau z = z^2 - 2z(x + \sigma^2\tau) + x^2 \]

Complete the square:

\[ \begin{aligned} &= [z - (x + \sigma^2\tau)]^2 - (x + \sigma^2\tau)^2 + x^2 \\ &= [z - (x + \sigma^2\tau)]^2 - 2x\sigma^2\tau - \sigma^4\tau^2 \end{aligned} \]

Therefore:

\[ z - \frac{(x-z)^2}{2\sigma^2\tau} = -\frac{[z - (x+\sigma^2\tau)]^2}{2\sigma^2\tau} + x + \frac{\sigma^2\tau}{2} \]

2. Step 2: Factor Out Constant¶

\[ I_1 = e^{x + \frac{\sigma^2\tau}{2}} \int_{\log K}^\infty \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{[z - (x+\sigma^2\tau)]^2}{2\sigma^2\tau}\right) dz \]

The integral is a Gaussian tail probability with shifted mean $x + \sigma^2\tau$:

\[ \int_{\log K}^\infty \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{[z - (x+\sigma^2\tau)]^2}{2\sigma^2\tau}\right) dz = \mathbb{P}(W \geq \log K) \]

where $W \sim \mathcal{N}(x+\sigma^2\tau, \sigma^2\tau)$.

3. Step 3: Standardize¶

\[ \begin{aligned} \mathbb{P}(W \geq \log K) &= \mathbb{P}\left(\frac{W - (x+\sigma^2\tau)}{\sigma\sqrt{\tau}} \geq \frac{\log K - (x+\sigma^2\tau)}{\sigma\sqrt{\tau}}\right) \\ &= \mathbb{P}\left(Z \geq \frac{\log K - x - \sigma^2\tau}{\sigma\sqrt{\tau}}\right) \\ &= \mathcal{N}\left(\frac{x + \sigma^2\tau - \log K}{\sigma\sqrt{\tau}}\right) \end{aligned} \]

Define:

\[ \boxed{d_1 = \frac{x + \sigma^2\tau - \log K}{\sigma\sqrt{\tau}} = d_2 + \sigma\sqrt{\tau}} \]

Therefore:

\[ I_1 = e^{x + \frac{\sigma^2\tau}{2}}\mathcal{N}(d_1) \]

Synthesis: Black-Scholes Formula¶

1. Forward Value¶

Combining $I_1$ and $I_2$:

\[ F(x,\tau) = e^{x + \frac{\sigma^2\tau}{2}}\mathcal{N}(d_1) - K\mathcal{N}(d_2) \]

2. Transform Back to Original Variables¶

Recall: - $x = \log S + (r - \frac{1}{2}\sigma^2)\tau$ - $e^x = Se^{(r - \frac{1}{2}\sigma^2)\tau}$

Therefore:

\[ e^{x + \frac{\sigma^2\tau}{2}} = Se^{(r - \frac{1}{2}\sigma^2)\tau + \frac{\sigma^2\tau}{2}} = Se^{r\tau} \]

Substituting:

\[ F(x,\tau) = Se^{r\tau}\mathcal{N}(d_1) - K\mathcal{N}(d_2) \]

3. Discount Back¶

Since $V = Fe^{-r\tau}$:

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

where $\tau = T - t$ and:

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

\[ \boxed{d_2 = d_1 - \sigma\sqrt{\tau} = \frac{\log(S/K) + (r - \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}} \]

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

Interpretation¶

1. Probabilistic Meaning¶

\[ C = S_0\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2) \]

Term 1: $S_0\mathcal{N}(d_1)$ - Expected stock value conditional on exercise under the stock measure - $\mathcal{N}(d_1)$ is the probability of exercise under a measure where the stock is the numeraire

Term 2: $Ke^{-rT}\mathcal{N}(d_2)$ - Expected discounted strike payment - $\mathcal{N}(d_2)$ is the probability of exercise under the risk-neutral measure

2. Connection to Diffusion¶

The heat equation derivation reveals: - Option pricing ≡ Diffusion of payoff backward in time - Green's function = Transition density of log-price under Brownian motion - Black-Scholes formula = Weighted average of terminal payoffs over Gaussian distribution

Physical analogy: If the option payoff were a temperature distribution at maturity, the current value is how that temperature has "diffused backward in time."

Summary¶

The heat equation approach provides a complete analytical derivation of the Black-Scholes formula:

1. Key Steps¶

Transform: BS PDE → Heat equation via change of variables $(S,t) \to (x,\tau)$ and $V \to F$
Recognize: Heat equation has known Green's function (Gaussian kernel)
Apply superposition: Solution is convolution of initial condition with Green's function
Evaluate integrals: Use completing the square for Gaussian integrals
Transform back: Return to original variables $(S,t)$ and discount

2. Advantages¶

Explicit solution: Closed-form formula obtained analytically
Clear probabilistic interpretation: Connection to Brownian motion apparent
Classical PDE theory: Connects to well-studied heat/diffusion equations
Generalizable: Extends to other parabolic PDEs and payoffs

3. Limitations¶

European options only: Requires fixed terminal condition
Smooth payoffs work best: Discontinuous payoffs require distribution theory
Limited flexibility: Less adaptable than numerical methods for exotics

4. Theoretical Significance¶

This derivation reveals the fundamental triad in quantitative finance:

Stochastic Processes (Brownian motion) ↕ PDE Theory (Heat equation) ↕ Probability (Gaussian distributions)

The heat equation method demonstrates that these three perspectives are mathematically equivalent, each providing complementary insights into derivative pricing.

5. What the kernel really is¶

The Gaussian kernel $G(x,\tau;z)$ used throughout this derivation is exactly the transition density of Brownian motion with drift $(r - \frac{1}{2}\sigma^2)$ and diffusion coefficient $\sigma$. The convolution integral

\[ F(x,\tau) = \int_{-\infty}^{\infty} \psi(z)\, G(x,\tau;z)\, dz \]

is therefore an expectation: the expected payoff under the risk-neutral measure. This is not a coincidence. The Feynman-Kac theorem (developed later in this chapter) establishes that solutions of parabolic PDEs are always expressible as such expectations. Put differently: the heat equation approach and the probabilistic approach compute exactly the same object --- the convolution is the expectation, and the Green's function is the transition density. Recognizing this equivalence is essential, because it means every PDE method in this chapter has a probabilistic counterpart, and vice versa. In the operator language of the introduction, the convolution with $G$ is the kernel representation of the pricing semigroup $\mathcal{P}_\tau = e^{\tau\mathcal{L}}$.

Exercises¶

Exercise 1. Verify that the Green's function $G(x,\tau;z) = \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right)$ satisfies the heat equation $\frac{\partial G}{\partial \tau} = \frac{1}{2}\sigma^2 \frac{\partial^2 G}{\partial x^2}$ by computing both sides explicitly and showing they are equal.

Solution to Exercise 1

The Green's function is $G(x,\tau;z) = \frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right)$.

Compute the left side $\frac{\partial G}{\partial \tau}$. Let $u = \frac{(x-z)^2}{2\sigma^2\tau}$. Then:

\[ G = (2\pi\sigma^2\tau)^{-1/2} e^{-u} \]

\[ \frac{\partial G}{\partial \tau} = -\frac{1}{2\tau}G + G \cdot \frac{(x-z)^2}{2\sigma^2\tau^2} = G\left[-\frac{1}{2\tau} + \frac{(x-z)^2}{2\sigma^2\tau^2}\right] \]

Compute the right side $\frac{1}{2}\sigma^2\frac{\partial^2 G}{\partial x^2}$. First:

\[ \frac{\partial G}{\partial x} = G \cdot \left(-\frac{x-z}{\sigma^2\tau}\right) \]

\[ \frac{\partial^2 G}{\partial x^2} = G \cdot \frac{(x-z)^2}{\sigma^4\tau^2} + G \cdot \left(-\frac{1}{\sigma^2\tau}\right) = G\left[\frac{(x-z)^2}{\sigma^4\tau^2} - \frac{1}{\sigma^2\tau}\right] \]

Therefore:

\[ \frac{1}{2}\sigma^2\frac{\partial^2 G}{\partial x^2} = G\left[\frac{(x-z)^2}{2\sigma^2\tau^2} - \frac{1}{2\tau}\right] \]

This equals $\frac{\partial G}{\partial \tau}$, confirming that $G$ satisfies the heat equation.

Exercise 2. Derive the Black-Scholes put formula using the heat equation approach. Start from the initial condition $\psi(x) = (K - e^x)^+$ and evaluate the two resulting Gaussian integrals. Verify that your answer matches the put formula $P = Ke^{-r\tau}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)$.

Solution to Exercise 2

For the European put, the initial condition is $\psi(x) = (K - e^x)^+ = K - e^x$ for $x < \ln K$ and $0$ for $x \geq \ln K$.

The superposition integral is:

\[ F(x,\tau) = \int_{-\infty}^{\ln K}(K - e^z)\frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right)dz \]

Split into two integrals:

\[ F = K\underbrace{\int_{-\infty}^{\ln K}\frac{1}{\sqrt{2\pi\sigma^2\tau}}e^{-(x-z)^2/(2\sigma^2\tau)}dz}_{J_2} - \underbrace{\int_{-\infty}^{\ln K}e^z \frac{1}{\sqrt{2\pi\sigma^2\tau}}e^{-(x-z)^2/(2\sigma^2\tau)}dz}_{J_1} \]

Integral $J_2$: With the substitution $Z = (z-x)/(\sigma\sqrt{\tau})$:

\[ J_2 = \mathbb{P}\left(Z \leq \frac{\ln K - x}{\sigma\sqrt{\tau}}\right) = \mathcal{N}\left(\frac{\ln K - x}{\sigma\sqrt{\tau}}\right) = \mathcal{N}(-d_2) \]

Integral $J_1$: By the same completing-the-square technique as for the call:

\[ J_1 = e^{x + \sigma^2\tau/2}\mathcal{N}\left(\frac{\ln K - x - \sigma^2\tau}{\sigma\sqrt{\tau}}\right) = e^{x+\sigma^2\tau/2}\mathcal{N}(-d_1) \]

Therefore:

\[ F(x,\tau) = K\mathcal{N}(-d_2) - e^{x+\sigma^2\tau/2}\mathcal{N}(-d_1) \]

Transforming back with $e^{x+\sigma^2\tau/2} = Se^{r\tau}$ and $V = Fe^{-r\tau}$:

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

This matches the standard put formula.

Exercise 3. In the transformation from the BS PDE to the heat equation, the first-order term cancels. Show this cancellation in detail: substitute the transformed derivatives into the PDE and verify that all $\frac{\partial V}{\partial x}$ terms cancel exactly.

Solution to Exercise 3

The transformed derivatives yield the following first-order terms in the PDE:

\[ -\left(r - \frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial x} + r\frac{\partial V}{\partial x} - \frac{1}{2}\sigma^2\frac{\partial V}{\partial x} \]

The first term comes from the time derivative transformation ($\frac{\partial V}{\partial t}$ contribution). The second term comes from the $rS\frac{\partial V}{\partial S} = r\frac{\partial V}{\partial x}$ term. The third term comes from the second-order term: $\frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} = \frac{1}{2}\sigma^2\left(\frac{\partial^2 V}{\partial x^2} - \frac{\partial V}{\partial x}\right)$, contributing $-\frac{1}{2}\sigma^2\frac{\partial V}{\partial x}$.

Summing all first-order coefficients:

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

The cancellation is exact. This happens because the transformation $x = \ln S + (r - \frac{1}{2}\sigma^2)\tau$ was specifically designed to incorporate the risk-neutral drift, thereby eliminating the convection (first-order) term from the PDE.

Exercise 4. The three transformations $(\tau, x, F)$ reduce the BS PDE to the heat equation with diffusivity $\kappa = \frac{1}{2}\sigma^2$. If we instead define $y = x / (\sigma\sqrt{2})$ and $F(y,\tau)$, show that we obtain the standard heat equation $\frac{\partial F}{\partial \tau} = \frac{\partial^2 F}{\partial y^2}$ with unit diffusivity. Express $d_1$ and $d_2$ in terms of $y$ and $\tau$.

Solution to Exercise 4

Define $y = x/(\sigma\sqrt{2})$. Then $x = \sigma\sqrt{2}\,y$ and:

\[ \frac{\partial F}{\partial y} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial y} = \sigma\sqrt{2}\frac{\partial F}{\partial x} \]

\[ \frac{\partial^2 F}{\partial y^2} = 2\sigma^2\frac{\partial^2 F}{\partial x^2} \]

Substituting into $\frac{\partial F}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2}$:

\[ \frac{\partial F}{\partial \tau} = \frac{1}{2}\sigma^2 \cdot \frac{1}{2\sigma^2}\frac{\partial^2 F}{\partial y^2} = \frac{1}{2} \cdot \frac{1}{2}\frac{\partial^2 F}{\partial y^2} \]

This gives $\frac{\partial F}{\partial \tau} = \frac{1}{4}\frac{\partial^2 F}{\partial y^2}$, not unit diffusivity. To obtain the standard form $\frac{\partial F}{\partial \tau} = \frac{\partial^2 F}{\partial y^2}$, introduce a time rescaling $\tilde{\tau} = \frac{1}{2}\sigma^2\tau$:

\[ \frac{\partial F}{\partial \tilde{\tau}} = \frac{1}{\frac{1}{2}\sigma^2}\frac{\partial F}{\partial \tau} = \frac{1}{\frac{1}{2}\sigma^2}\cdot\frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2} = \frac{\partial^2 F}{\partial x^2} \]

So with $y = x$ and $\tilde{\tau} = \frac{1}{2}\sigma^2\tau$, we obtain unit diffusivity. In these variables:

\[ d_1 = \frac{x + \sigma^2\tau - \ln K}{\sigma\sqrt{\tau}} = \frac{x + 2\tilde{\tau} - \ln K}{\sqrt{2\tilde{\tau}}} \]

\[ d_2 = \frac{x - \ln K}{\sigma\sqrt{\tau}} = \frac{x - \ln K}{\sqrt{2\tilde{\tau}}} \]

Exercise 5. Use the superposition integral to price a digital call with payoff $\psi(x) = \mathbf{1}_{\{e^x > K\}} = \mathbf{1}_{\{x > \ln K\}}$. Evaluate the resulting Gaussian integral and transform back to original variables to obtain $D_0 = e^{-rT}\mathcal{N}(d_2)$.

Solution to Exercise 5

The digital call payoff is $\psi(x) = \mathbf{1}_{\{x > \ln K\}}$. The superposition integral gives:

\[ F(x,\tau) = \int_{\ln K}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2\tau}}\exp\left(-\frac{(x-z)^2}{2\sigma^2\tau}\right)dz \]

This is a standard Gaussian tail probability. With $Z = (z - x)/(\sigma\sqrt{\tau})$:

\[ F(x,\tau) = \mathbb{P}\left(Z \geq \frac{\ln K - x}{\sigma\sqrt{\tau}}\right) = \mathcal{N}\left(\frac{x - \ln K}{\sigma\sqrt{\tau}}\right) = \mathcal{N}(d_2) \]

where $d_2 = \frac{x - \ln K}{\sigma\sqrt{\tau}}$.

Transforming back to original variables: $x = \ln S + (r - \frac{1}{2}\sigma^2)\tau$, so:

\[ d_2 = \frac{\ln S + (r - \frac{1}{2}\sigma^2)\tau - \ln K}{\sigma\sqrt{\tau}} = \frac{\ln(S/K) + (r - \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}} \]

Since $V = Fe^{-r\tau}$ and $F = \mathcal{N}(d_2)$:

\[ D_0 = e^{-rT}\mathcal{N}(d_2) \]

This is the price of a digital (cash-or-nothing) call paying $1 if $S_T > K$.

Exercise 6. The heat equation approach requires the initial condition $\psi(x) = (e^x - K)^+$ to be integrable against the Green's function. Discuss what happens if the payoff grows faster than $e^{|x|}$ as $|x| \to \infty$. Give an example of a payoff for which the superposition integral diverges, and explain what this means financially.

Solution to Exercise 6

The superposition integral is $F(x,\tau) = \int_{-\infty}^{\infty}\psi(z)G(x,\tau;z)dz$ where $G$ is the Gaussian kernel with variance $\sigma^2\tau$.

Growth condition. For the integral to converge, we need:

\[ \int_{-\infty}^{\infty}|\psi(z)| \cdot \frac{1}{\sqrt{2\pi\sigma^2\tau}}e^{-(x-z)^2/(2\sigma^2\tau)}dz < \infty \]

The Gaussian kernel decays as $e^{-z^2/(2\sigma^2\tau)}$ for large $|z|$, so $\psi(z)$ can grow at most as $e^{cz^2}$ for some $c < 1/(2\sigma^2\tau)$ and the integral still converges. In particular, any payoff growing as $e^{\alpha|z|}$ (polynomial exponential growth) is integrable.

Divergent example. Consider the payoff $\psi(z) = e^{z^2}$ (super-exponential growth). The integral becomes:

\[ \int_{-\infty}^{\infty}e^{z^2}\frac{1}{\sqrt{2\pi\sigma^2\tau}}e^{-(x-z)^2/(2\sigma^2\tau)}dz \]

The exponent grows as $z^2 - z^2/(2\sigma^2\tau) = z^2(1 - 1/(2\sigma^2\tau))$. For $\tau > 1/(2\sigma^2)$, the coefficient is positive and the integral diverges.

Financial interpretation. A payoff that grows faster than exponentially in $\ln S$ (i.e., super-polynomially in $S$) cannot be priced using the standard Green's function approach because the expected payoff under the risk-neutral measure is infinite. Such payoffs violate the integrability conditions needed for the risk-neutral pricing formula to hold. Financially, no finite amount of capital can replicate such a payoff, so it has no well-defined arbitrage-free price. The standard call payoff $(e^x - K)^+$ grows only as $e^x$ (linearly in $S$), which is within the allowable growth rate.