I encountered the following derivation of the Black Scholes formula for call price. It may very well be an established method but I had never seen it before so I called it an alternative derivation.

I immediately start with the slightly transformed version of the Black Scholes PDE. $z$ denoting the log spot price and $C$ denoting the call price we have

$$\frac{\partial C}{\partial t} + \left(r-\frac{1}{2}\sigma^2\right)\frac{\partial C}{\partial z} + \frac{1}{2}\sigma^2\frac{\partial^2 C}{\partial z^2} = rC$$ with the boundary condition $C(T,z) = \max(e^z-K,0)$ where $K$ is strike and $T$ is expiry.

The derivation then assumes the following form for $C$

$$C(t,z) = e^zP(t,z) - Ke^{-r(T-t)}Q(t,z)$$

and substitutes the partial derivatives into the original PDE to obtain

$$e^z\left(\frac{\partial P}{\partial t} + \left(r+\frac{1}{2}\sigma^2\right)\frac{\partial P}{\partial z} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial z^2}\right) - Ke^{-r(T-t)}\left(\frac{\partial Q}{\partial t} + \left(r-\frac{1}{2}\sigma^2\right)\frac{\partial Q}{\partial z} + \frac{1}{2}\sigma^2\frac{\partial^2 Q}{\partial z^2}\right)= 0$$

The bit that I don't get is that according to the derivation this somehow implies the following.

$$\frac{\partial P}{\partial t} + \left(r+\frac{1}{2}\sigma^2\right)\frac{\partial P}{\partial z} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial z^2} = 0$$ $$\frac{\partial Q}{\partial t} + \left(r-\frac{1}{2}\sigma^2\right)\frac{\partial Q}{\partial z} + \frac{1}{2}\sigma^2\frac{\partial^2 Q}{\partial z^2} = 0$$

Why would the individual terms be zero?

  • $\begingroup$ Well this an ansatz approach when you postulate a form of the solution to make the equation easier to solve. There is a unique solution to the problem given boundary condition so it is fine. But you have not derived anything here because to solve your ansatz you need to solve twice almost the same equation. $\endgroup$
    – Ezy
    Nov 20, 2018 at 10:56

1 Answer 1


The option pricing formula must satisfy the PDE you have derived for all values of $K$. The only way this can be the case is if the two parts that you separate are both equal to zero. Suppose the joint PDE (before you separate it into two) is satisfied for some value of $K$. But suppose that the $Q$-part in parentheses on the second line is not zero. Change the value of $K$, and your PDE will no longer be satisfied. So the second line of your PDE must be zero. Since the second line must be zero, the first line must also be zero.

The solution to the $P$ differential equation is related to the price of an asset-or-nothing call, and the solution of the $Q$ differential equation is related to the price of a cash-or-nothing option.

Edited to add - this is actually not right, there are alternate $P$ and $Q$ such that the PDEs are not satisfied separately. Working on a fix . . .

  • $\begingroup$ Not sure if I fully understand your point. The functions $P$ and $Q$ depend on $K$. Consider the case $K = 0$. Then we obtain the individual PDE describing $P$ but that is then only true for $K = 0$. I don't see how that generalizes to arbitrary $K$. $\endgroup$
    – Calculon
    Nov 22, 2018 at 9:45
  • 1
    $\begingroup$ Yes, I realised my solution was broken earlier today, but was having trouble fixing it. Now I know why - the PDEs do not have to separate. There are not unique $P$ and $Q$ which are solutions to the combined PDE. For example, take the $P$ to be the Black-Scholes-Merton formula divided by $e^{z}$, and take $Q$ to be zero. That's a solution. You could also take $Q$ to be anything at all, that isn't a solution to the separate $Q$ PDE, and reverse-engineer $P$ such that the combination gives you the BSM formula. In that case, neither $P$ nor $Q$ will satisfy the individual PDEs. $\endgroup$
    – RLK
    Nov 22, 2018 at 12:44
  • 1
    $\begingroup$ So I think the best we could do would be to show that there is a particular $P$ and $Q$ such that the PDEs separate. When you write "The bit that I don't get is that according to the derivation this somehow implies the following.", you are correct not to get it, because the first PDE does not imply the second - out of the infinity many combinations of $P$ and $Q$ that satisfy the first, there is one combination such that the PDEs separate. $\endgroup$
    – RLK
    Nov 22, 2018 at 12:45
  • 1
    $\begingroup$ I'll see if I can think of a way to prove that. Otherwise, I think it has to be another conjecture, to be verified later. You've already got one conjecture, that the solution is of a particular form, maybe another won't hurt. Sorry for the broken solution. I'm used to interest rate models, where we often get a clean separation using the method I tried to apply here :( $\endgroup$
    – RLK
    Nov 22, 2018 at 12:47
  • $\begingroup$ Thank you very much for your reply. I appreciate it. $\endgroup$
    – Calculon
    Nov 22, 2018 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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