# Measurability of contingent claim in State-variable approach

I'd like to know, if we have the filtrations $$\mathbb{F}$$ and $$\mathbb{G}$$ with $$\mathcal{F}_t\subset\mathcal{G}_t\subset \mathcal{F}_t\vee \sigma(\eta)$$, for $$\eta$$ being independent of $$\mathcal{F}_\infty$$, in a market where the default time and its associated intensity process are determined by a state variable process $$Y$$, does a general contingent claim $$X$$ with maturity $$T$$ need to be $$\mathcal{F}_T$$-measurable or $$\mathcal{G}_T$$-(but not $$\mathcal{F}_T$$)-measurable.

My problem is, that the default time $$\tau$$ is not $$\mathbb{F}$$ adapted, but $$\mathbb{G}$$ adapted and $$\mathbb{G}=\mathbb{H}\vee\mathbb{F}$$ is supposed to be generated from the filtration generated by $$\tau$$ (i.e. $$\mathbb{H}$$) and the reference filtration $$\mathbb{F}$$.

The logic behind this should be, whether I can say that the payoff $$X$$ at time $$T$$ is determined by the information I get from the market (i.e. by $$\mathcal{F}_T$$) or if I need it to depend on the default time. There are many examples where authors of papers assumed $$X$$ to be $$\mathcal{F}_T$$ measurable so I'm quite confused.

• As the payoff depends on the default time, you can not say it to be $\mathscr{F}_T$ measurable. However, if you define $X$ as the pre-default payoff, you can say it is $\mathscr{F}_T$ measurable, and the actual payoff is then defined by $\pmb{1}_{\tau > T}X$. See also this question. Feb 17, 2021 at 16:26
• In your answer to the question you refer to Lando, who (in one of his papers) assumes that $X$ is $\mathcal{F}_T$ measurable although the default time is not $\mathcal{F}_T$ measurable. How is this possible if, whether the payoff takes place or not, depends on $\tau$? Feb 17, 2021 at 16:54
• That is why I have put the indicator $\pmb{1}_{\tau > T}$ there. In general, we assume that the pre-default payoff $X$, rather than the actual payoff $\pmb{1}_{\tau>T}X$, is $\mathscr{F}_T$ measurable. Feb 17, 2021 at 20:48