Simplifying an expectation function of default time and rates

Question

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.

Answer 1

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.