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Let $X_t$ be an american style financial derivative with random exercise time $T$ where $t$ and $T$ belongs to some finite set $A$.
Buying this derivative requires the buyer to pay $p_t$ up to time $T$.
Let $\Omega$ be the sample space of $X_t$, $p=(p_t)_{t \in A}$ the price process and $B={\left(C^A \right)}^\Omega$ the value space of $p$ for some set $C \subset \mathbb{R}$.
Assume expectation are taken under the risk-neutral measure with $B_t$ as the risk-free discounting factor from times $0$ to time $t$.
Is the no-arbitrage pricing process of the derivative given by $$ \arg_{p \in B} \left( \sup_{T \in A} E(X_T B_T - \int_0^T p_t B_tdt)=0 \right) \text{ (1)} $$

when $B$ requires that $p_t(\omega) \ne 0, t>0$ for some $\omega \in \Omega$?

My knowledge of finance tells me the no-arbitrage price would be $$ \sup_{T \in A} E(X_T B_T) \text{ (2)} $$ when B is degenerated to $p_0=k$ for some $k \in \mathbb{R}$ and $p_t=0,t \ne 0$.
Intuitively, I would expect (1) to be the natural extension to (2).
But is it theoretically true?
I searched, but I couldn't find any source confirming my hypothesis.

Thanks for any help.

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closed as off topic by Andreas Blass, Deane Yang, unknown (google), Igor Rivin, Andrés E. Caicedo Jul 14 '12 at 7:09

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This isn't a math question. I suggest you ask on the forums on – Deane Yang Jul 12 '12 at 17:51
@Deane Yang I think you are right. Thank you for suggesting another site. Should I delete my question? – Nicolas Essis-Breton Jul 12 '12 at 18:24
I can't imagine any reasonable sense in which this is not a math question. – Steven Landsburg Jul 13 '12 at 0:37
@Steven Landsburg Your comment motivates me to keep the question alive. Thank you. – Nicolas Essis-Breton Jul 13 '12 at 1:02
Actually, I think this question is totally inappropriate. In addition to @Deane's suggestion, is a right place for such questions. – Igor Rivin Jul 13 '12 at 16:28

good MATH question ! so closely related to what people in the MATH departments at Columbia , Chicago , Rutgers , Carnegie Mellon etc etc are doing. Can't help you with the answer to this MATH question though, but I wonder , are you paying in shares , receiving $X_T$ shares ? or what is that factor of $B_T$ doing in your MATH equations ?

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@mike In my setting, $X_T$ is a lump sum paid to the derivative buyer. My $B_T$ factor should be understand as the risk-free discounting factor. Thank you for pointing this inconsistency. Also, thank you for sharing that you find my question appropriate. – Nicolas Essis-Breton Jul 13 '12 at 0:59

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