This question is a follow-up of this question Fixed Payment at Default - Pricing for more clarity.

Starting from the following expression of payoff:

$$ D(0,T) = E(\exp(-\int_{0}^{\tau} r(t)dt) \cdot \mathbb{1}_{\{\tau \leq T\}}) $$

and knowing that $$ \text{Probability}(T \leq \tau \leq T+dT) = \lambda(T) \cdot \exp(-\int_{0}^{T} \lambda(t)dt) \cdot dT $$ we aim to arrive to:

$$ D(0,T) = E(\int_{0}^{T} \lambda(t) \cdot \exp(-\int_{0}^{t} (r(s)+\lambda(s))ds)dt) $$

The idea as stated is to integrate the payoff over all possible times of default, thus form 0 to T.

We thus have: $$ D(0,T) = E(\int_{0}^{T} \exp(-\int_{0}^{t} r(s)ds) \cdot \mathbb{1}_{\{t \leq T\}}dt \mid \mathcal{F}_0) $$ But I am struggling from this step to develop the calculus. Would the law of iterated expectation be needed?

  • 3
    $\begingroup$ amdopt was kind enough to convert your unreadable previous post to mathJax. This time I'd say this is your task. $\endgroup$
    – Kurt G.
    Commented May 26, 2023 at 11:41
  • 1
    $\begingroup$ The seminal article where you can find those proofs in D. Lando, On Cox processes and credit risky securities. link. $\endgroup$
    – Kurt G.
    Commented May 27, 2023 at 18:09
  • $\begingroup$ Ty for your answer! As we know $$ [ \mathcal{F}_t = \mathcal{G}_t \cup \mathcal{H}_t ] $$ Where processes relevant in determining values for spot rate and hazard rate of default are adapted to $$ \mathcal{G}_t$$ and $$ \mathcal{H}_t $$ hold the information of wether there has been a default at time t. $\endgroup$
    – cp123456
    Commented May 30, 2023 at 14:59
  • $\begingroup$ I don't get the proof 3.4: $$E\left(\mathbb{1}_{\{\tau \geq T\}} \, \bigg| \, \mathcal{G}_T \cup \mathcal{H}_t\right) = \mathbb{1}_{\{\tau > t\}} \cdot \exp\left(-\int_t^T \lambda_s \, ds\right)$$ What is the point between $$\{\tau > t\}$$ being an atom, and the following equality: $$E\left(\mathbb{1}_{\{\tau \geq T\}} \, \bigg| \, \mathcal{G}_T \cup \mathcal{H}_t\right) = \mathbb{1}_{\{\tau > t\}} \cdot E\left(\mathbb{1}_{\{\tau \geq T\}} \, \bigg| \, \mathcal{G}_T \cup \mathcal{H}_t\right)$$ $\endgroup$
    – cp123456
    Commented May 30, 2023 at 15:07
  • 1
    $\begingroup$ You can pull out $1_{\{\tau>t\}}$ from the conditional expectation because it is ${\cal H}_t$-measurable and $1_{\{\tau\ge T\}}=1_{\{\tau\ge T\}}1_{\{\tau>t\}}\,.$ A lot of insight into Lando's approach is gained from the fact that if we knew the entire path up to $T$ of the ${\cal G}_T$-measurable random variable $\exp(-\int_t^T\lambda_s\,ds)$ then we knew the survival probability until $T$ conditional on survival until $t\,.$ Lando's equations are a formalization of this. $\endgroup$
    – Kurt G.
    Commented May 30, 2023 at 15:33


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