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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?

cp123456
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    amdopt was kind enough to convert your unreadable previous post to mathJax. This time I'd say this is your task. – Kurt G. May 26 '23 at 11:41
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    The seminal article where you can find those proofs in D. Lando, On Cox processes and credit risky securities. link. – Kurt G. May 27 '23 at 18:09
  • 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.

     – cp123456 May 30 '23 at 14:59
  • 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: $$E\left(\mathbb{1}{{\tau \geq T}} , \bigg| , \mathcal{G}_T \cup \mathcal{H}_t\right) = \mathbb{1}{{\tau > t}} \cdot E\left(\mathbb{1}_{{\tau \geq T}} , \bigg| , \mathcal{G}_T \cup \mathcal{H}_t\right)$$

     – cp123456 May 30 '23 at 15:07
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    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. – Kurt G. May 30 '23 at 15:33
  • Thank you very much, I was not sure about that!

    I just rewrite here my second question as it had a small error on it.

    I am just wondering how to go from: $$ \mathbb{1}{{\tau > t}} \cdot E\left(\mathbb{1}{{\tau \geq T}} , \bigg| , \mathcal{G}_T \cup \mathcal{H}_t\right) $$

    To: $$ \mathbb{1}_{{\tau \geq T}} \cdot \frac{P({\tau \geq T} \cap {\tau > t} ,|, \mathcal{G}_T)}{P({\tau > t} ,|, \mathcal{G}_T)} $$

    I tried to turn it into all directions but can't see any "obvious" reason for this equality

     – cp123456 May 30 '23 at 15:49
  • Would anyone have an answer on this? Otherwise I will create another subject to try and have an answer on it. – cp123456 Jun 09 '23 at 12:27
  • Not sure why anyone should have the obligation to have an answer on this. Possible hint: ${\tau>t}$ is an atom of ${\cal H}t,.$ It looks like Lando uses an analogy by which the conditional expectation of $1_B$ w.r.t. a finitely generated $\sigma$-algebra $\sigma(A_1,\dots,A_n)$ can be written as the finite sum $$\sum{i=1}^n1_{A_i}\mathbb P(B|A_i)$$ where $\mathbb P(B|A_i)=\mathbb P(B\cap A_i)/\mathbb P(A_i),.$ – Kurt G. Jun 09 '23 at 12:46
  • "Not sure why anyone should have the obligation to have an answer on this.": Never said or suggested anything like that but thank you very much for your help. – cp123456 Jun 09 '23 at 14:06

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