When we derive the boundary conditions for the American put options, if we let $S_f(t)$ be the optimal exercise boundary, for $S \gt S_f(t)$ we get $$\frac{1}{2}\sigma^2S^2\frac{\partial^2P}{\partial S^2}+\frac{\partial P}{\partial t}-rP+rS\frac{\partial P}{\partial S} \le 0$$

Why is it $\le$ and not $=$ like the derivation for the European put option?


2 Answers 2



The slop of option , delta, at $S=S_f(t)$ is $-1$, indeed $$\frac{\partial P}{\partial S}=-1$$ thus we can say that the return from the portfolio can not be greater than the return from a bank deposit, therefore

$$\frac{1}{2}\sigma^2S^2\frac{\partial^2P}{\partial S^2}+\frac{\partial P}{\partial t}-rP+rS\frac{\partial P}{\partial S} \le 0$$


  • $\begingroup$ Chapter 7, page 106. $\endgroup$
    – user16651
    Commented Nov 25, 2016 at 16:59
  • $\begingroup$ Thank you for your response, I know the return can't be greater than the return of a bank deposit. But when doing other derivations it's equal to the return of a bank deposit. What's the difference? $\endgroup$ Commented Nov 26, 2016 at 12:10

Since we are specifying $S > S_f(t)$ you are correct, and equality holds. The author was probably trying to generalize to an equation where no restriction on being above the exercise boundary holds.

Mathematically, this is clumsy and useless but it often motivates the early exercise updates when we are using finite difference schemes (like trees or grids) to price the option.

Recall that in these schemes we discretize underling prices to a grid $S_n$ and backwardate from time $t^{m}$ to $t^{m+1}$ by applying a matrix operator to grid prices $P^m$.

If we use an explicit scheme like a tree this consists of computing $$ \tilde{P}^{m+1} = C \cdot P^m $$ for some matrix $C$ and then handling early exercise by setting $$ {P}^{m+1} = \max\left(\tilde{P}^{m+1} , X\right) $$ where $X$ are exercise values.

For nodes on the grid with no neighbors above exercise value, equality holds exactly. But right near exercise value the finite difference formula for $\frac{\partial^2 P}{\partial S^2}$ term gets an updated term inside, destroying the equality.

$$ \frac{\partial^2 P_n}{\partial S^2} \approx \frac{P_{n+1} - 2P_n + {\color{red} {{X}_{n-1}}}}{\Delta S^2} $$


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.