I am currently reading "Modelling single-name and multi-name credit derivatives" by Dom O'Kane but I struggle at one point that should be relatively easy.

Let us consider a Zero Recovery Risky Zero Coupon Bond, that is to say a bond that pays 1 in case there is no default $\tau > T$ (that is the time of default arives after maturity time)

The pricing formula for such a product is thus given by: $Z(0, T) = E\left[\exp\left(-\int_0^T r(t)dt\right) \cdot \mathbb{1}( \tau > T)\right]$.

We also know that: $P(\tau > T) = \exp\left(-\int_0^T \lambda(t)dt\right)$.

In order to simplify the writing of $Z(0, T)$, here are the steps that are presented: $Z(0, T) = E\left[\exp\left(-\int_0^T r(t)dt\right) \cdot \mathbb{1}( \tau > T)\right]$.

Using the law of iterated expectation, we have:

$Z(0, T) = E\left[E\left[\exp\left(-\int_0^T r(t) dt\right) \cdot \mathbb{1}(\tau > T) | {\lambda(t)}_{t \in [0,T]}\right]\right]$.

And as $E\left[I(\tau > T) | \mathcal{\lambda(t)}_{t \in [0,T]}\right] = \text{P}(\tau > T) = \exp\left(-\int_0^T \lambda(t)dt\right)$.

So one has: $Z(0, T) = \mathbb{E}\left[\exp\left(-\int_0^T (r(t) + \lambda(t)) dt\right)\right]$.

Question 1: Why is the filtration selected the set of $\lambda(t)$?

Question 2: How can we split and write this: $E\left[\exp\left(-\int_0^T r(t) \, dt\right) \cdot \mathbf{1}_{\tau > T} \,|\, \tau\right] = \exp\left(-\int_0^T r(t) \, dt\right) \cdot E\left[\mathbf{1}_{\tau > T} \,|\, \tau\right]$ if we did not make any assumption on the independence of $r$ and $\lambda$?


1 Answer 1


In the context of credit risk and stochastic calculus, the filtration of a stochastic process, denoted by {F_t}, represents the accumulated information up to time t. This concept allows for the consideration of new information as it is revealed over time. Now, let's get into the specifics of your questions.

Question 1: The filtration chosen is the set of λ(t) as λ(t) represents the intensity of the default process τ and therefore contains all information about the default risk up to time t. In the case of modelling credit derivatives, knowing the default intensity is crucial in pricing. The filtration {λ(t)} contains all the available information about this default intensity. This means it contains all the necessary information we have up until time t to decide whether or not a default will occur after time T.

Question 2: Regarding the expectation, there seems to be a misunderstanding here. What you're describing isn't the separation of the integral of r(t) and λ(t), but rather, the application of the law of total expectation (or iterated expectations).

The law of total expectation states that the expected value of a random variable can be calculated by taking the expected value of the conditional expected value of the variable on a smaller sigma algebra. This allows us to separate the probability of default event (τ>T) from the discounting factor. It's not about the independence between r and λ, but the application of the law of total expectation.

In the equation E[exp(−∫T0r(t)dt)⋅1(τ>T)|τ]=exp(−∫T0r(t)dt)⋅E[1(τ>T)|τ], the independence between r and τ is assumed. Therefore, we can treat r as deterministic when considering the expectation given τ. This allows us to move exp(−∫T0r(t)dt) out of the expectation operator, as it doesn't involve τ.

Do note that this sort of simplification can often be found in credit risk modelling as we usually consider the short rate r and the default intensity λ as independent, which simplifies the calculations while still capturing the core features of the credit derivative's behavior.

  • $\begingroup$ A huge thank you for your answer. I would like to ask you a little more question The law of iterated expectation states that E(X | F1) = E(E(X | F2) | F1) if F1 included in F2. If I go back to my first expression of Z(0,T), would the correct writing of it including its filtration be: Z(0,T)=E[exp(−∫T0r(t)dt)⋅1(τ>T) | F0] that would thus be equal using the law of iterated expectations to: Z(0,T)=E[E(exp(−∫T0r(t)dt)⋅1(τ>T) | Fτ) | F0] and as τ and r are independant, we get: Z(0,T)=E[exp(−∫T0r(t)dt)⋅E(1(τ>T) | Fτ)) | F0] and I then replace E(1(τ>T) | Fτ)) by P(τ > T) . $\endgroup$
    – cp123456
    Commented May 24, 2023 at 17:45
  • 1
    $\begingroup$ @cp123456, Yes, you're absolutely right with your logic. That's an excellent application of the law of iterated expectation, for more details I would suggest to open a new post. $\endgroup$
    – TourEiffel
    Commented May 25, 2023 at 7:49
  • $\begingroup$ Thank you again. I would have another follow-up question to make it clear. Please let me know if you would like me to open a new post for more clarity wrt this question. When we consider the inner expectation E(exp(−∫T0r(t)dt)⋅1(τ>T) | Fτ), we "take out" exp(−∫T0r(t)dt) of the expectation, thus considering it as deterministic. However, as 0<τ<T, considering all the information at time τ (symbolised by the Filtration Fτ), the integral −∫T0r(t)dt does not seem determinisitc to me. Even if assuming independance between r and τ, wouldn't the information at time τ be impactful to r? $\endgroup$
    – cp123456
    Commented May 26, 2023 at 10:22
  • $\begingroup$ Furthermore, after taking exp(−∫T0r(t)dt) out of the inner integral, we would have exp(−∫T0r(t)dt)⋅E(1(τ>T) | Fτ). But looking at E(1(τ>T) | Fτ), wouldn't 1(τ>T) be considered as deterministic having the filtration Fτ, that is "knowing all the information until time τ". And so we would have E(1(τ>T) | Fτ) = 1(τ>T) instead of E(1(τ>T) | Fτ) = E(1(τ>T)) that is obtained assuming independance between 1(τ>T) and Fτ? Sorry for all these questions, I really want to make it clear in my mind. Thank you so much for your help. $\endgroup$
    – cp123456
    Commented May 26, 2023 at 10:27
  • $\begingroup$ Answers can be found at the following paper: researchgate.net/publication/… Thank you Kurt G for your help. $\endgroup$
    – cp123456
    Commented May 30, 2023 at 15:54

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