# American options and stopping times

The price of an American put option can be written as the following optimal stopping problem: $$V(0) = \mathop {\sup }\limits_{\tau \in \mathcal{T}} {\mathbb{E}^\mathbb{Q}}\left[ {{e^{ - r\tau }}\max [K - S(\tau ),0]} \right]$$, where $$\mathcal{T}$$ is the set of all stopping (exercise) times. Assume no-dividend case.

If I look at $${\mathbb{E}^\mathbb{Q}}\left[ {{e^{ - r\tau }}\max [K - S(\tau ),0]} \right]$$, then this is a price of a Black-Scholes European option maturing at $$\tau$$.

To solve the American option pricing problem using a common sense logic - why can't I just compute Black-Scholes price for each $$\tau$$-maturity option and then take a maximum price?

• The various options maturing at different $\tau$ are not independent of each other: if $S(\tau_1)$ is large then $S(\tau_2)$ for $\tau_2\approx \tau_1$ is probably also large and vice versa. You cannot treat the problem as choosing the best among independent assets (options). Instead the $S(\tau)$ belong to a single random trajectory of stock price. Mar 28 at 18:33
• Thanks @noob2, that I perfectly understand. The true solution is the "recursive" dynamic programming. But I don't see this in the general formula. Which element in the general formula have I overlooked that would bring in the dependency? Mar 28 at 18:46
• All I am saying is you left out the SDE $\frac{dS}{S}=r dt + \sigma dW$ that ties the $S(\cdot)$ values togther. Which is obvious, I know. Mar 28 at 23:08

You could but there are difficulties associated with this approach. The main one is that $$\tau$$ is stochastic, ie it is different for different paths of $$S$$, so the standard Black-Scholes formula does not apply. For example some $$\tau$$s you need to check are of the form $$\tau =\inf\{t : S(t) in which case you need to value a barrier option with barrier $$B$$, and other choices for $$\tau$$ that you need to check that are even more complicated