# American option pricing formulation

Assuming the usual setup of:

• $$\left(\Omega, \mathcal{S}, \mathbb{P}\right)$$ our probability space endowed with a filtration $$\mathbb{F}=\left(\mathcal{F}_t\right)_{t\in[0,T]}$$,
• $$T>0$$ denoting the option maturity,
• an $$\mathbb{F}$$-adapted process $$Z=\left(Z_t\right)_{t\in[0,T]}$$ modeling the discounted value of the option payoff at time $$t$$;

Why do we define the problem of pricing an American option as: $${\text{ess}\sup}_{\tau\in\mathrm{T}_{[0, T]}} \mathbb{E}\left[Z_{\tau}|\mathcal{F}_0\right]$$ and not as: $${\text{ess}\sup}_{s\in[0, T]} \mathbb{E}\left[Z_{s}|\mathcal{F}_0\right]?$$ In the above $$\mathrm{T}_A$$ is the set of all stopping times (with respect to our filtration $$\mathbb{F}$$) with values in the set $$A$$.

Non-mathematical common sense suggests that the option holder is basically only interested in a moment $$s$$ when to exercise the option optimally, so why should he be interested in optimizing over all stopping times?

My further doubts stem from the fact that every $$s\in[0,T]$$ is obviously also a stopping time, therefore we have an inclusion of the second formulation in the first and it would appear reasonable to state that:

$${\text{ess}\sup}_{s\in[0, T]} \mathbb{E}\left[Z_{s}|\mathcal{F}_0\right] \leq {\text{ess}\sup}_{\tau\in\mathrm{T}_{[0, T]}} \mathbb{E}\left[Z_{\tau}|\mathcal{F}_0\right].$$

On the other hand, I am aware that finding the optimal stopping time provides the optimal moment of exercise, since our stopping times take values in $$[0,T]$$.

Therefore I have this uncomfortable feeling, that the first formulation might provide a bigger optimal value (because we are optimizing over a broader family of arguments) than the second, whereas I would imagine that both formulations should amount to the same result.

What sort of fallacies am I committing in following the presented line of thought?

1. On the one hand, indeed we have $${\text{ess}\sup}_{s\in[0,T]}\mathbb{E}\left[Z_s|\mathcal{F}_0\right] \leq {\text{ess}\sup}_{\tau\in\mathrm{T}_{[0,T]}}\mathbb{E}\left[Z_\tau|\mathcal{F}_0\right].$$
2. On the other, the optimal stopping time $$\tau^\star$$ obtained from solving $${\text{ess}\sup}_{\tau\in\mathrm{T}_{[0,T]}}\mathbb{E}\left[Z_\tau|\mathcal{F}_0\right]$$ provides us with an optimal time $$t^*$$ for the option exercise, as it takes values in $$[0,T]$$.
Speaking more formally, for any arbitrary stopping time $$\tau\in\mathrm{T}_{[0, T]}$$ we define the moment of exercise as: $$\sup\{t\in[0, T]:\left\{\tau \leq t\} = \emptyset\right\}$$ i.e. the last moment $$t\in[0,T]$$ such that the event $$\{\tau\leq t\}$$ is empty and $$\mathbb{P}\left(\{\tau\leq t\}\right) = 0$$ holds.
With that in mind, we can write $${\text{ess}\sup}_{\tau\in\mathrm{T}_{[0,T]}}\mathbb{E}\left[Z_\tau|\mathcal{F}_0\right] = \mathbb{E}\left[Z_{\tau^\star}|\mathcal{F}_0\right] = \mathbb{E}\left[Z_{t^\star}|\mathcal{F}_0\right].$$ It remains to demonstrate that our stopping time derived exercise moment $$t^\star$$ is indeed equal to the one from the second formulation. To that end, let us assume that $$s^\star$$ is the optimal exercise time derived from solving the deterministic formulation: $${\text{ess}\sup}_{s\in[0,T]}\mathbb{E}\left[Z_s|\mathcal{F}_0\right] = \mathbb{E}\left[Z_{s^\star}|\mathcal{F}_0\right].$$ On the one hand we have: $$\mathbb{E}\left[Z_{s^\star}|\mathcal{F}_0\right] = {\text{ess}\sup}_{s\in[0,T]}\mathbb{E}\left[Z_s|\mathcal{F}_0\right] \leq {\text{ess}\sup}_{\tau\in\mathrm{T}_{[0,T]}}\mathbb{E}\left[Z_\tau|\mathcal{F}_0\right] = \mathbb{E}\left[Z_{t^\star}|\mathcal{F}_0\right].$$ On the other, trivially (by definition of $$\text{ess}\sup$$ and $$s^\star$$ being the value that realizes it) we have: $$\mathbb{E}\left[Z_{t^\star}|\mathcal{F}_0\right] \leq \mathbb{E}\left[Z_{s^\star}|\mathcal{F}_0\right].$$ These two together gives us: $$\mathbb{E}\left[Z_{t^\star}|\mathcal{F}_0\right] = \mathbb{E}\left[Z_{s^\star}|\mathcal{F}_0\right]$$ which implies $$t^\star=s^\star \text{a.s.}$$.