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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Gordon
    Feb 17 '21 at 16:26
  • $\begingroup$ 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 $\tau$? $\endgroup$
    – Leoncino
    Feb 17 '21 at 16:54
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gordon
    Feb 17 '21 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.