4
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.