Question regarding the derivation of American Put option

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?

Hint

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

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

Reference

• Chapter 7, page 106. – user16651 Nov 25 '16 at 16:59
• 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? – JonathanDoeing Nov 26 '16 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.

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