# Simplifying an expectation function of default time and rates

I have the following expectation to calculate :

$$\mathbf{E}\left[ e^{\int_{t_0}^{\tau} r_s ds} \mathbf{1}_{\{\tau < T\}}\right]$$

More precisely, I want to show that :

$$\mathbf{E}\left[ e^{\int_{t_0}^{\tau} r_s ds} \mathbf{1}_{\{\tau < T\}}\right] = \int_{t_0}^T P(s) dQ(s)$$

where $P(s) \equiv \mathbf{E}\left[ e^{\int_{t_0}^{s} r_u du} \right]$ and $Q(s) \equiv \mathbf{E}\left[ e^{\int_{t_0}^{s} \lambda_u du} \right]$, $\tau$ is the first jump of a Poisson process with default intensity $(\lambda_t)_t$.

The only hypothesis I have is that rates and default are independent $-$ rates don't follow a special process.

• Given your independence assumption I assume rates are stochastic in your model. Do they follow a specific process? – Daneel Olivaw Nov 14 '17 at 15:51
• I edited the question – EricFlorentNoube Nov 14 '17 at 15:59

For an in-homogeneous Poisson process, the intensity process $\lambda_t$ is assumed to be deterministic. More generally, we can define $\tau$ to be the first jump time of a Cox process, or a conditional Poisson process (see Chapter 6 of the book Credit Risk). We assume that $t_0=0$ is the valuation date. Then the intensity process $\lambda_t$ can be stochastic, and \begin{align*} Q(t) \equiv P(\tau > t) = E\left(e^{-\int_{t_0}^t \lambda_s ds} \right). \end{align*} Moreover, given the independence assumption, \begin{align*} E\left(e^{-\int_{t_0}^{\tau} r_u du} \pmb{1}_{t_0<\tau<T} \right) &= E\left(E\left(e^{-\int_{t_0}^{\tau} r_u du} \pmb{1}_{t_0<\tau<T} \,|\, \tau \right)\right)\\ &=E\left(P(\tau) \pmb{1}_{\tau<T}\right)\\ &=-\int_{t_0}^TP(s) dQ(s), \end{align*} where \begin{align*} P(s) = E\left(e^{-\int_{t_0}^s r_u du} \right) \end{align*} is the price of a zero-coupon bond with maturity $s$ and unit face value.
Comments: If we assume that $t_0>0$, then the whole question should be changed. For example, we need to define the filtration for the market information and the enlarged filtration generated by the process defining the default time as well as the market information. Moreover, certain martingale invariance property, or $\mathcal{H}$-Hypothesis, should also be assumed.
• What is the nature of $P(\tau)$ in your second line? As defined by the OP, $P(\cdot)$ is an expectation, not a conditional expectation. – Daneel Olivaw Nov 15 '17 at 15:34
• In your second line, don't we have: $P(\tau)=E\left(D(t_0,\tau)|\tau\right)$ where $D(s,t)$ is the discount factor from $t$ to $s<t$? – Daneel Olivaw Nov 15 '17 at 15:55
• Here, we assumed the independence between $r$ and $\tau$ for simplicity. Otherwise, it would be pretty involved as in my comments. – Gordon Nov 15 '17 at 15:59
• How would you proceed (and would it be still true) would we assume "only" that $r$ and $\lambda$ are independent processes. – EricFlorentNoube Nov 18 '17 at 11:48
• The independence of $r$ and $\lambda$ is not enough. I will later add some more comments for the special case with a Cox process. – Gordon Nov 20 '17 at 18:24