Discounted price process of an american put (perpetual) has a $dt$ part in it, which is negative if the price at time $t$ is less than the optimal exercise price. This is the only thing that drags the discounted process down as the time goes on and makes the whole process a supermartingale. So when you don't exercise the option at it's stopping time, it has a tendency to go down. However, I do not seem to be understanding the intuition behind here, as the process goes down only when the price is less than the optimal exercise price and shouldn't having a lower stock price make the option more valuable? So what am I getting wrong?


3 Answers 3


I would not say there is no link to what you say but here would be my view.

Intuitive explanation

If you wait for a delay $h$ before exercising, you lose your exercise right between $t$ and $t+h$, this leads to a loss in value.

Supermartingale property proof

(to apply it in your case : $\phi_t=e^{-rt}(L-S_t)^+$)

If we denote $\phi$ the obstacle, and $\text{Am}(\phi)$ the American perpetual option on pay-off $\phi$, assuming there is a optimal strategy $\tau^\star(t)$ to exercise the option knowing you buy the option at time $t$. Allowed strategies are stopping time (meaning you can take your decision only according to what you know at that time) bigger or equal than $t$.

You get :

$$\text{Am}(\phi)_t=\mathbb{E}(\phi_{\tau^{\star}(t)}|\mathcal{F}_t)=\sup_{\tau\geq t}\mathbb{E}(\phi_\tau|\mathcal{F}_t)$$

Setting $\tau=\tau^\star(t+h)$ on the right hand side leads you to : $$\text{Am}(\phi)_t\geq \mathbb{E}(\phi_{\tau^\star(t+h)}|\mathcal{F}_t)$$

using tower property of conditionnal expectation : $$\mathbb{E}(\phi_{\tau^\star(t+h)}|\mathcal{F}_t)=\mathbb{E}(\mathbb{E}(\phi_{\tau^\star(t+h)}|\mathcal{F}_{t+h})|\mathcal{F}_t)$$

using the first equality in $t+h$ rather in $t$ : $$\mathbb{E}(\phi_{\tau^\star(t+h)}|\mathcal{F}_{t+h})=\text{Am}(\phi)_{t+h}$$

pluggin this into previous inequality leads you to :

$$\text{Am}(\phi)_t\geq \mathbb{E}(\text{Am}(\phi)_{t+h}|\mathcal{F}_t)$$

  • $\begingroup$ Thank you, a very interesting proof. Can you please explain this part $\text{Am}(\phi)_t=\sup_{\tau\geq t}\mathbb{E}(\phi_\tau|\mathcal{F}_t)$ ? $\endgroup$
    – iNarek94
    Commented Apr 28, 2016 at 8:56
  • $\begingroup$ This kind of means you can look ahead, I always thought the equality took place for $\tau\leq t$. $\endgroup$
    – iNarek94
    Commented Apr 28, 2016 at 9:05
  • $\begingroup$ +1 for this nice answer! @iNarek94 $\text{Am}(\phi)_t=\mathbb{E}(\phi_{\tau^{\star}(t)}|\mathcal{F}_t)=\sup_{\tau \in [t,T]} \mathbb{E}(\phi_\tau|\mathcal{F}_t)$ simply means that, amongst all possible stopping times with values $\tau \in [t,T]$ ($T$ denoting the option expiry), the optimal stopping strategy $\tau^*(t)$ given the information at time $t$ is the one which maximises your expected gain (hence the supremum). This has nothing to do with seeing in the future. Or rather, it is not different from when you look at a future expectation pricing a European option for instance. $\endgroup$
    – Quantuple
    Commented Apr 28, 2016 at 9:24
  • $\begingroup$ And when not exercising on an optimal time, you loose value because of the interest you could earn on the intrinsic value and didn't? $\endgroup$
    – iNarek94
    Commented Apr 28, 2016 at 10:20
  • $\begingroup$ @iNarek94 No that's the whole idea: you aways exercice at an optimal time if you are rational! And this optimal time is the one which maximises your expected profit (amongst all possible early exercise times). $\endgroup$
    – Quantuple
    Commented Apr 28, 2016 at 10:30

Just to add an intuitive argument to @MJ73550's already very nice answer:

When holding an American option - or any option callable by the holder for that matter -, the question you ask yourself before exercising it is whether the proceeds from early exercise (i.e. exercise now to get the option's intrinsic value) are greater than what you could expect to earn if you were to exercise your right later (i.e. continuation value).

At any point in time, the value of your option is thus always the maximum between what you would receive in the 2 above scenarios, since you would like to exercise when it is optimal for you.

Without loss of generality, assume you can only exercise at fixed dates separated by an interval $h$ (as it would typically be the case for a Bermudan option). Then at any time $t$ you have, for an option expiring at $T$:

$ \text{Am}(t,T) = \max( (S_t - K)^+, \mathbb{E}[\text{Am}(t+h, T)] ) \geq \mathbb{E}[\text{Am}(t+h, T)]$


  • $\text{Am}(t,T)$ - current option value at $t$
  • $(S_t - K)^+$ - intrinsic value = proceeds if you were to exercise at $t$
  • $\mathbb{E}[\text{Am}(t+h, T)]$ - what you can expect to earn if you wait until $t+h$ to make your decision.

whence the supermartingale idea, or as @MJ73550's answer illustrates: the best early exercise strategy over $[t,T]$ is always at least as good as the best early exercise strategy over $[t+h,T]$ since the former interval includes the latter.

Some remarks:

  • the above holds for any $h>0$, particularly $h \rightarrow 0^+$;
  • at $t=T$, the continuation and exercise value are the same, since there is no choice left;
  • this process of comparing the continuation and intrinsic values is exactly what you do when using trees or Least-Squares Monte Carlo to price callable options.
  • $\begingroup$ +1, I realized I didn't get the idea behind pricing an american option, specially when we're talking about a set of stopping times and choosing a maximum over it. Thought it was looking ahead, turned out to be not. Thanks! $\endgroup$
    – iNarek94
    Commented Apr 28, 2016 at 10:19

Ok, so I have been thinking about it, and may have found the solution, but please correct me if I'm wrong. I guess the discounted process goes down, because when the holder of the option doesn't exercise it, as long as the price $S(t)$ is less than the optimal exercise price $L^*$ he's loosing cash from not investing into money market?


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