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 Answer 1


In explicitly wording my own question yesterday and naming my doubts, I think I may have stumbled upon the explanation:

  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 $t^\star=s^\star \text{a.s.}$.

Therefore solving the optimal stopping problem (to which optimal stopping theory lends itself nicely as a tool) solves the deterministic formulation too.

I will leave this answer for now to gather feedback and comments.


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