Solving Black-Scholes: From Heat to Formula

Introduction: The Final Stretch

In Part 2, we performed a coordinate transformation to turn the Black-Scholes PDE into the 1D Heat Equation:

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

Now, we must solve it. Because this is the standard diffusion equation, we can use the fundamental solution (Green’s Function). The value of the option $F(x, \tau)$ is simply the convolution of the initial condition (payoff) with the Gaussian heat kernel.

1. Setting Up the Integral

First, recall our transformed variables:

$x = \ln(S) + (r - \frac{1}{2}\sigma^2)\tau$
Payoff at $\tau=0$ (expiration): $F(z, 0) = \max(e^z - K, 0)$

The solution is given by the integral:

\[F(x, \tau) = \int_{-\infty}^{\infty} F(z, 0) \frac{1}{\sqrt{2\pi \sigma^2 \tau}} e^{-\frac{(x-z)^2}{2\sigma^2 \tau}} dz\]

Since the payoff $\max(e^z - K, 0)$ is only non-zero when $e^z > K$ (i.e., $z > \ln K$), we change the lower limit of integration to $\ln K$:

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

We split this into two separate integrals, which we will solve individually:

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

2. Solving Term 2: The Strike Term

Let’s start with the second term.

\[\text{Term 2} = K \int_{\ln K}^{\infty} \frac{1}{\sigma\sqrt{2\pi\tau}} e^{-\frac{(z-x)^2}{2\sigma^2 \tau}} dz\]

We define a standard normal variable $Z$:

\[Z = \frac{z - x}{\sigma\sqrt{\tau}} \quad \text{where} \quad z \sim N[x, \sigma^2\tau]\]

We change the limit of integration. Since $z \ge \ln K$:

\[x + \sigma\sqrt{\tau}Z \ge \ln K \implies Z \ge \frac{\ln K - x}{\sigma\sqrt{\tau}}\]

Substituting these in:

\[\text{Term 2} = K \int_{\frac{\ln K - x}{\sigma\sqrt{\tau}}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{Z^2}{2}} dZ\]

Because the Normal distribution is symmetric, the area from a point $a$ to $\infty$ is the same as the area from $-\infty$ to $-a$. We define $-d_2$ as the lower limit:

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

Thus, the integral becomes the Cumulative Distribution Function, $\Phi(d_2)$:

\[\text{Term 2} = K \cdot \Phi(d_2)\]

3. Solving Term 1: Completing the Square

Now for the first term. This is trickier because of the extra $e^z$ factor.

\[\text{Term 1} = \int_{\ln K}^{\infty} e^z \frac{1}{\sigma\sqrt{2\pi \tau}} e^{-\frac{(z-x)^2}{2\sigma^2 \tau}} dz\]

To solve this, we must combine the exponentials ($e^z$ and the Gaussian). We look at the exponent terms:

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

We perform the algebra to complete the square:

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

Essentially, the $e^z$ term shifts the mean of our distribution by exactly $\sigma^2\tau$.

\[z \sim N[x + \sigma^2\tau, \sigma^2\tau]\]

When we factor out the constants and solve the integral (just like we did for Term 2), we get a new limit $d_1$:

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

And the final result for Term 1:

\[\text{Term 1} = e^{x + \frac{\sigma^2\tau}{2}} \Phi(d_1)\]

4. The Grand Unification

We now combine the terms to get the value of the transformed option $F$:

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

Finally, we must transform back to the real world variables. Recall our definitions:

$V = e^{-r\tau}F$
$x = \ln S + (r - \frac{1}{2}\sigma^2)\tau$

Substitute $F$ into the first equation:

\[V = e^{-r\tau} \left[ e^{x + \frac{\sigma^2 \tau}{2}} \Phi(d_1) - K \Phi(d_2) \right]\]

Now, look at the first term closely. Substitute $x$:

\[e^{-r\tau} e^{\ln S + (r - \frac{1}{2}\sigma^2)\tau + \frac{\sigma^2 \tau}{2}}\]

The terms in the exponent simplify beautifully:

$-r\tau$ cancels $+r\tau$
$-\frac{1}{2}\sigma^2 \tau$ cancels $+\frac{1}{2}\sigma^2 \tau$

We are left with $e^{\ln S}$, which is just $S$.

Conclusion: The Black-Scholes Formula

Replacing $\tau$ with $T-t$, we arrive at the final pricing formula:

\[C(S, t) = S \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2)\]

Where:

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

We have successfully journeyed from the thermodynamics of heat diffusion to the pricing of financial derivatives, proving that the language of math is truly universal.