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

Question

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).

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The textbook do provide the final result as the following, but i'm not sure how those are derived from my steps above

Answer 1

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*}

Answer 2

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.