Let us consider a product paying an amount of 1 if default (τ<T that is the time of default arrives before maturity time), and 0 otherwise.

The payoff of such a product would be given by:

$$D(0, T) = E\left(\exp\left(-\int_0^\tau r(t)dt\right) \cdot \mathbb{1}_{\tau \leq T}\right)$$

Knowing that:

$$\text{Probability}(T \leq \tau \leq T + dT) = \lambda(T) \cdot \exp\left(-\int_0^{T} \lambda(t) dt\right) \cdot dT$$

Could someone explain how it is possible to arrive to: $$D(0,T) = E\left(\int_0^T \lambda(t) \cdot \exp\left(-\int_0^t (r(s) + \lambda(s)) ds\right) dt\right)$$

I tried to use the law of iterated expectation but struggled to find a way out.


1 Answer 1


The quantity you're trying to derive is the price at time 0 of a defaultable zero coupon bond which pays off 1 if default occurs before the maturity time T (i.e., τ<T), and 0 otherwise.

This is a bit tricky but the key here is to understand that we can represent this price as an integral over the possible default times. The intuition is that you're integrating the payoffs across all possible times of default.

Since default could happen at any time t in [0, T], we write D(0,T) as an integral from 0 to T.

At any given time t, the payoff is the present value of 1 discounted back to time 0 if default occurs, weighted by the probability that default occurs at that time. The discounting term is $$\exp\left(-\int_{0}^{t} r(s)ds\right)$$ , and from your given information, the default probability density function is $$\lambda(t) \cdot \exp\left(-\int_{0}^{t} \lambda(s)ds\right) $$.

Therefore, we can write D(0,T) as:

$$ D(0,T) = \int_{0}^{T} E[ \exp(-\int_{0}^{t}r(s)ds) \cdot \lambda(t) \cdot \exp(-\int_{0}^{t}\lambda(s)ds) \mid \mathcal{F}_t ] dt $$

Note that we're using the filtration F_t as it would represent the information available at time t, which would contain both r(s) and λ(s) for all s≤t.

Since r and λ are assumed to be independent, we can treat r as deterministic when taking the conditional expectation: $$ D(0,T) = \int_{0}^{T} \exp(-\int_{0}^{t}r(s)ds) \cdot E[ \lambda(t) \cdot \exp(-\int_{0}^{t}\lambda(s)ds) \mid \mathcal{F}_t ] dt $$

Which simplifies to: $$ D(0,T) = \int_{0}^{T} \lambda(t) \cdot \exp(-\int_{0}^{t} (r(s) + \lambda(s)) ds) dt $$ In this last expression, the $$\lambda(t) \cdot \exp\left(-\int_{0}^{t} \lambda(s)ds\right)$$ term represents the probability of default at time t and $$\exp\left(-\int_{0}^{t} r(s)ds\right) $$ is the discount factor. We integrate this product over [0, T] to get the expected discounted payoff of the bond.

It's important to remember that the law of iterated expectations and the assumptions of independence between certain variables are key to simplifying these kind of expressions in the credit risk context.

  • $\begingroup$ Thank you for your answer but I am bit struggling to follow it with the necessary rigour. Let us start from: D(0,T)=E(exp(−∫τ0r(t)dt)⋅1τ≤T) that can be written as D(0,T)=E(exp(−∫τ0r(t)dt)⋅1τ≤T | F0). As we integrate from 0 to T, one has: D(0,T)=E(∫T0exp(−∫t0r(s)ds)⋅1t≤T)dt | F0). After this step, should I use the law of iterated expectation? If yes, I am not sure how, as I could use the filtration Ft as F0 is included in Ft. But it seems to me it won't help as I won't be able to separate exp(−∫t0r(s)ds) from 1t≤T as both depend on t Hoping this question is not to annoying! $\endgroup$
    – cp123456
    May 24, 2023 at 18:39
  • $\begingroup$ @cp123456 Please open a new post, because the answer can't be posted in there (too long). best. $\endgroup$
    – TourEiffel
    May 25, 2023 at 7:47
  • $\begingroup$ You will find it just here! quant.stackexchange.com/questions/75669/… Thank you very much Eiffel $\endgroup$
    – cp123456
    May 26, 2023 at 10:34
  • $\begingroup$ Please convert your formulas to mathJax. $\endgroup$
    – Kurt G.
    May 26, 2023 at 11:42

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