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I'm trying to take a closer look to option pricing in a risky environment.

Let's say a firm $A$ sells me an (European) option on an underlying $S$ (which of course can be any other financial product than the firm $A$) with payoff $h(S_T)$.

Let's take a r.v $\tau:\Omega \rightarrow \mathbb{R}_+$ to model the time at which the firm $A$ default (without really going into details of what "making default" means, we can say that it's the time the payoff of the option goes to $0$). A way to characterize the distribution of $\tau$ is to express it in function of the "hazard rate" $\lambda_t$ cf. Survival analysis. We can show that:

$$ \mathbb{P}\left(\tau > t\right) = e^{-\int_0^t\lambda_sds}. $$

In a second time, due to Feynman-Kac theorem, we can now (do we?) write the option price as:

$$P_t(T,K)=e^{-\int_t^Tr_sds}\mathbb{E}\left[h(S_T)\mathbb{1}_{\{\tau>T\}}|\mathcal{F}_t\right].$$

Before going any further, are $S_t$ and $\{\tau>t\}$ defined on the same filtered probability space? I cannot get an answer for this, because for me, intuitively (and I'm not saying my intuitions are good), we're most likely to have two filtrations $\mathcal{F}_t:=\sigma(S_s,s\leq t)$ and $\mathcal{G}_t:=\sigma(A_s,s\leq t)$ (with $A$ being the firm who sells the option) such that the option price is more likely to be something like:

$$P_t(T,K)=e^{-\int_t^Tr_sds}\mathbb{E}\left[h(S_T)\mathbb{1}_{\{\tau>T\}}|\mathcal{F}_t\vee\mathcal{G}_t\right],$$

if it does mean something...

I'm seeing it like something is horribly wrong in what I'm writing, but there is not so much literature for this, and I don't want to go any further before being sure of the basics in such an environment.

Thanks.

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    $\begingroup$ It's Feynman-Kac. (pronounced "Katz") $\endgroup$ – Raskolnikov Jan 28 '18 at 20:39
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    $\begingroup$ You do not specify any dependence structure between say the Brownian motion driving the spot price of the risky asset $S$ and the Poisson counting process driving the occurrence of a default event? So either you tie them under some measure and you can answer your question. Or you consider that they are independent and again this answers the question, right? Usually the latter is done. Have a look at this related question quant.stackexchange.com/questions/33031/… $\endgroup$ – Quantuple Jan 29 '18 at 8:07

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