I know the derivation of the Black-Scholes differential equation and I understand (most of) the solution of the diffusion equation. What I am missing is the transformation from the Black-Scholes differential equation to the diffusion equation (with all the conditions) and back to the original problem.

All the transformations I have seen so far are not very clear or technically demanding (at least by my standards).

My question:
Could you provide me references for a very easily understood, step-by-step solution?

up vote 30 down vote accepted

One starts with the Black-Scholes equation $$\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}+ rS\frac{\partial C}{\partial S}-rC=0,\qquad\qquad\qquad\qquad\qquad(1)$$ supplemented with the terminal and boundary conditions (in the case of a European call) $$C(S,T)=\max(S-K,0),\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2)$$ $$C(0,t)=0,\qquad C(S,t)\sim S\ \mbox{ as } S\to\infty.\qquad\qquad\qquad\qquad\qquad\qquad$$ The option value $C(S,t)$ is defined over the domain $0< S < \infty$, $0\leq t\leq T$.

Step 1. The equation can be rewritten in the equivalent form $$\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2\left(S\frac{\partial }{\partial S}\right)^2C+\left(r-\frac{1}{2}\sigma^2\right)S\frac{\partial C}{\partial S}-rC=0.$$ The change of independent variables $$S=e^y,\qquad t=T-\tau$$ results in $$S\frac{\partial }{\partial S}\to\frac{\partial}{\partial y},\qquad \frac{\partial}{\partial t}\to - \frac{\partial}{\partial \tau},$$ so one gets the constant coefficient equation $$\frac{\partial C}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 C}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial C}{\partial y}+rC=0.\qquad\qquad\qquad(3)$$

Step 2. If we replace $C(y,\tau)$ in equation (3) with $u=e^{r\tau}C$, we will obtain that $$\frac{\partial u}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial u}{\partial y}=0.$$

Step 3. Finally, the substitution $x=y+(r-\sigma^2/2)\tau$ allows us to eliminate the first order term and to reduce the preceding equation to the form $$\frac{\partial u}{\partial \tau}=\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}$$ which is the standard heat equation. The function $u(x,\tau)$ is defined for $-\infty < x < \infty$, $0\leq\tau\leq T$. The terminal condition (2) turns into the initial condition $$u(x,0)=u_0(x)=\max(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0),$$ where $a=2r/\sigma^2$. The solution of the heat equation is given by the well-known formula $$u(x,\tau)=\frac{1}{\sigma\sqrt{2\pi \tau}}\int_{-\infty}^{\infty} u_0(s)\exp\left(-\frac{(x-s)^2}{2\sigma^2 \tau}\right)ds.$$

Now, if we evaluate the integral with our specific function $u_0$ and return to the old variables $(x,\tau,u)\to(S,t,C)$, we will arrive at the usual Black–Merton-Scholes formula for the value of a European call. The details of the calculation can be found e.g. in The Mathematics of Financial Derivatives by Wilmott, Howison, and Dewynne (see Section 5.4).

  • Sorry to resurrect an old question, but I'm having some difficulties doing the substitution in step 3. Can anyone give a little extra detail? – Mattheus Jan 15 '14 at 18:57
  • I managed to understand, through the help of this page planetmath.org/AnalyticSolutionOfBlackScholesPDE. I might edit the answer above to elaborate later. – Mattheus Jan 15 '14 at 20:17
  • for step 3, there are two standard ways to do it. One is as above, the other is multiplying by an exponential. I go into both in detail in my book "concepts" – Mark Joshi May 22 '15 at 7:01

I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform: \begin{align*} \mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy, \end{align*} where $\mathcal{F}[f] = \mathcal{F}[f](\omega)$.

We'll use the following well-known facts: \begin{align*} \mathcal{F}\left[ \frac{1}{s \sqrt{2\pi}} \exp\left( -\frac{1}{2}\left(\frac{y - m}{s}\right)^2\right) \right] & = \exp(-i \omega m - s^2\omega^2/2), \\ \mathcal{F}\left[\frac{ \partial^n f}{\partial y^n}\right] & = (i \omega)^n \mathcal{F}[f], \\ \mathcal{F}[cf] & = c \mathcal{F}[f], \\ \mathcal{F}[f \ast g] & = \mathcal{F}[f]\mathcal{F}[g], \end{align*} where the convolution $(f \ast g)(y) = \int_{-\infty}^\infty f(z)g(y-z)dz$.

Using @olaker's notation, start with the constant coefficient PDE ((3) above) $$ \frac{\partial C}{\partial \tau} = \frac{\sigma^2}{2}\frac{\partial^2 C}{\partial y^2} + \left(r-\frac{\sigma^2}{2}\right)\frac{\partial C}{\partial y} - rC. $$ Step 1. Take the Fourier transform of each term term above and solve the resulting (very simple) ODE: \begin{align*} \frac{\partial \hat{C}}{\partial \tau} & = -\frac{\sigma^2 \omega^2}{2}\hat{C} + i \omega\left(r-\frac{\sigma^2}{2}\right)\hat{C} - r\hat{C}, \\ \hat{C} & = \hat{C}_0 e^{-r\tau} \exp \left(-\frac{\sigma^2 \omega^2}{2}\tau + i \omega\left(r-\frac{\sigma^2}{2}\right)\tau\right). \end{align*}

Step 2. Letting $m = \left(\frac{\sigma^2}{2} - r\right)\tau$ and $s = \sigma \sqrt{\tau}$ from the Fourier transform notation, note $$ \exp \left(-\frac{\sigma^2 \omega^2}{2}\tau + i \omega\left(r-\frac{\sigma^2}{2}\right)\tau\right) = \mathcal{F}\left[ \frac{1}{\sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right], $$ so \begin{align*} \hat{C} & = \hat{C}_0 e^{-r\tau}\mathcal{F}\left[ \frac{1}{\sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right] \\ & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau}\mathcal{F}\left[C_0 \ast \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right]. \end{align*}

Step 3. Take inverse transform: \begin{align*} C(y, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_{-\infty}^\infty C_0(z) \exp\left( -\frac{1}{2}\left(\frac{y - z - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) dz \\ C(y, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_{-\infty}^\infty C_0(z) \exp\left( -\frac{1}{2}\left(\frac{z - \left(y + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) dz \end{align*}

Step 4. Finally, change variables back $y \to S$, where we had $y = \log S$. Before we do this, note $S$ is really the ``initial'' stock price in the usual sense, i.e. $S = S_0$, but to be consistent we'll stick with $S$ as the initial (known) stock price. We'll also transform the $z$ variable, suggestively calling it $S_T$ by $S_T = e^z$. \begin{align*} C(S, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_0^\infty C_0(S_T) \frac{1}{S_T} \exp\left( -\frac{1}{2}\left(\frac{\log S_T - \left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) dS_T. \end{align*} Just note \begin{align*} f(S_T) := \frac{1}{S_T \sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{\log S_T - \left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) \end{align*} is the probability density function for a log $\mathcal{N}\left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau, \sigma^2 \tau\right)$ random variable, and under Black-Scholes, this is indeed the distribution of $S_T$ under $\mathbb{Q}$. Hence \begin{align*} C(S,\tau) = e^{-r\tau}\mathbb{E}_{\mathbb{Q}}[C_0(S_T)|\mathcal{F}_t]. \end{align*}

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.