Value (price) of defaultable zero coupon bond with credit risk involved

I'm trying to derivate the Value (price) of defaultable zero coupon bond, but there some steps (math) in between I can't figure out.

From the default process modelling, we have:

$$P(t ≤ \tau < t+dt | \tau > t ) ≈ h_tdt$$

and:

$$P( \tau > t ) = \exp\left(-\int_0^t h_s ds\right)$$

Hence combining both, the unconditional probability:

$$P(t ≤ \tau < t+dt ) = h_t\exp\left(-\int_0^t h_s ds\right)dt$$

Next proceed with the derivation of the value of a defaultable bond

\begin{align} &B(0,T) \\ =& \color{fuchsia}{\text{EV[non-default scenario]}} + \color{blue}{\text{EV[default scenario]}} \\ =& E\left[\color{fuchsia}{\exp\left(-\int_0^T r_t dt\right)·\mathbf{1}_{\{T<\tau\}}} + \color{blue}{\int_0^T RR · \exp\left(-\int_0^t r_s ds\right) · P(t ≤ \tau < t+dt )} \right]\\ =& E\left[\color{fuchsia}{\exp\left(-\int_0^T r_t dt\right)·\exp\left(-\int_0^T h_t dt\right)} + \color{blue}{\int_0^T RR · \exp\left(-\int_0^t r_s ds\right) · h_t\exp\left(-\int_0^t h_s ds\right)dt} \right] \\ =& E\left[\color{fuchsia}{\exp\left(-\int_0^T (r_t+h_t) dt\right)} + \color{blue}{\int_0^T RR · h_t· \exp\left(-\int_0^\color{red}{t} (r_s+h_s) ds\right) \color{red}{dt}} \right] \end{align}

I have derived up to here and come to the problem that I do not know how to integrate the blue part, with the upper bound of the inner integral as the integration variable of the outer integral (which I have colored red for clarity).

The textbook do provide the final result as the following, but i'm not sure how those are derived from my steps above • hint given is using the approximation: e^x = 1 + x Jun 21 '20 at 7:06
• @Gordon I dont quite get the your simplification from step 1 to 2, why would the inner integral just disappear? Jun 23 '20 at 17:13
• @Jeremy Which textbook/chapter is this from? Jun 23 '20 at 21:32
• You are using a different definition of the recovery rate from that given in the solution. You are assuming a constant recovery with respect to the value of the bond at any time while the solution assume a recovery proportional to the bond value at the time of the default.
– Hans
Jun 26 '20 at 3:12
• @Gordon: Your ad hoc derivation is not really right, is it? You are making the same recovery assumption as that of OP rather than that of the solution. You are assuming a constant recovery with respect to the value of the bond at any time while the solution assumes a recovery proportional to the bond value at the time of the default. You are not going to get the desired solution.
– Hans
Jun 26 '20 at 3:14

There are a few recovery mechanisms, for example, recovery of par (i.e., the notional), recovery of treasury (i.e., the recovery value is a constant fraction of the equivalent default-free bond), and recovery of market value (i.e., a fraction of its pre-default market value). Here, your formula, which is also called the Lando formula, assumes the recovery of market value mechanism.

Let $$V_t$$ be the pre-default value at time $$t$$ of the zero-coupon bond with maturity $$T$$ and unit face value (note that $$V_T=1$$). Moreover, let $$R$$ be the recovery rate, of the pre-default value $$V_{\tau}$$. Furthermore, let $$\tau$$ be the default time, $$H_t=\pmb{1}_{\{\tau \leq t\}}$$. Let $$\mathscr{F}_t$$ be the market information set at time $$t$$ (which roughly speaking includes all information other than the fact of default or survival). Moreover, let $$\mathscr{H}_t = \sigma(H_u,\, u \leq t)$$ and $$\mathscr{G}_t = \mathscr{F}_t \vee \mathscr{H}_t$$ be the enlarged information set. Here, we can assume that the default time $$\tau$$ is defined as the first jump time of an in-homogeneous Poisson process, where the intensity process $$\{h_t,\, t \ge 0\}$$ is deterministic, or a Cox process, where the intensity is stochastic (see Bielecki and Rutkowski for more details).

Generally, we assume that the $$\mathscr{H}$$-condition is satisfied, that is, $$\mathscr{H}_t$$ and $$\mathscr{F}_{\infty}$$ are independent conditioned on $$\mathscr{F}_t$$; in other words, for any $$\mathscr{H}_t$$-measurable random variable $$X$$ and $$\mathscr{F}_{\infty}$$ measurable random variable $$Y$$, \begin{align*} E(XY\,|\,\mathscr{F}_t) = E(X\,|\,\mathscr{F}_t)E(Y\,|\,\mathscr{F}_t). \end{align*}

The other key formula to use is the filtration switching formula (see the book Interest Rate Models - Theory and Practice): For any $$\mathscr{G}_{\infty}$$ measurable random variable $$Y$$, \begin{align*} E\left(\pmb{1}_{\{\tau > t\}}Y\,|\,\mathscr{G}_t\right) = \pmb{1}_{\{\tau > t\}}\frac{E\left(\pmb{1}_{\{\tau > t\}}Y\,|\,\mathscr{F}_t\right)}{E\left(\pmb{1}_{\{\tau > t\}}\,|\,\mathscr{F}_t\right)}.\tag 1 \end{align*}

Then, for $$0 \leq t < T$$, \begin{align*} \pmb{1}_{\{\tau > t\}}V_{t} &= E\bigg(\pmb{1}_{\{\tau > T\}}e^{-\int_{t}^{T} r_s ds} + \pmb{1}_{\{t< \tau \le T\}} R\, V_{\tau}e^{-\int_{t}^{\tau} r_s ds} \, \big|\, \mathscr{G}_{t}\bigg) \\ &=\pmb{1}_{\{\tau > t\}} E\bigg(e^{-\int_{t}^{T} (r_s+h_s) ds} + \int_{t}^{T}R\, V_{u}h_u e^{-\int_{t}^{u} (r_s+h_s) ds} du \, \big|\, \mathscr{F}_{t}\bigg) \tag 2 \\ &=\pmb{1}_{\{\tau > t\}} e^{\int_0^{t} (r_s+h_s) ds} E\bigg(e^{-\int_0^{T} (r_s+h_s) ds} + \int_{t}^{T}R\, V_{u} h_u e^{-\int_0^{u} (r_s+h_s) ds} du \, \big|\, \mathscr{F}_{t}\bigg). \nonumber \end{align*} Here, the $$\mathscr{H}$$-condition and the filtration switching formula are employed in the derivation of $$(2)$$.

Let \begin{align*} M_t = E\bigg(e^{-\int_0^{T} (r_s+h_s) ds} + \int_0^{T}R\, V_{u}h_u e^{-\int_0^{u} (r_s+h_s) ds} du \, \big|\, \mathscr{F}_t\bigg). \end{align*} Then, $$M_t$$ is a martingale. Moreover, \begin{align*} V_t = e^{\int_0^t (r_s+h_s) ds}\bigg(M_t - \int_0^{t} R\,V_{u}h_u e^{-\int_0^{u} (r_s+h_s) ds} du \bigg). \end{align*} By Ito's lemma, \begin{align*} d\Big(e^{-\int_0^t (r_s+(1-R)h_s) ds} V_t \Big) = e^{\int_0^t R\, h_s ds} dM_t. \end{align*} Since $$M_t$$ is a martingale, $$e^{-\int_0^t (r_s+(1-R)h_s) ds} V_t$$ is also a martingale over $$[0, T]$$. Then, for any $$0\le t \le u\le T$$, \begin{align*} e^{-\int_0^t (r_s+(1-R)h_s) ds} V_t = E\Big(e^{-\int_0^u (r_s+(1-R)h_s) ds} V_u \, \big|\, \mathscr{F}_t \Big). \end{align*} In particular, \begin{align*} V_0 = E\left(e^{-\int_0^{T} (r_s+(1-R)h_s) ds}\right). \end{align*}

• +1. This is a very nice answer. I wonder why the OP did not award his bounty to it.
– Hans
Jul 3 '20 at 2:08
• Great answers as always @Gordon. Sep 15 '20 at 15:49
• Thanks @DeepInTheQF. Sep 15 '20 at 16:34

It's hard for me to understand exactly what you are asking, but I will try to answer. If my answer misses the mark please clarify exactly it is what you don't understand and I will try again.

We have \begin{aligned} P(\tau \leq t + dt \vert \tau > t) &= \frac{P(t < \tau \leq t+dt)}{P(\tau > t)} \\ &= 1 - \exp \bigg(\int_t^{t+dt} h_u du \bigg) \\ &\approx h_tdt \end{aligned}

Where the approxomation comes from Taylor expansion of $$e$$ (the hint provided).

Furthermore (from the definition of the hazard rate), $$P(\tau > t) = \exp\bigg( -\int_0^t h_u du \bigg)$$

Is that enough? Maybe you can work it out from here.

• Hi Rayl, I've updated my question entirely and added in more steps and color for clarity. The 2 equation you've provided are just the early steps which i've incorporated them in my derivation. Can u help to take a look again see if you can help? Jun 23 '20 at 14:58