3
$\begingroup$

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 enter image description here

$\endgroup$
  • $\begingroup$ hint given is using the approximation: e^x = 1 + x $\endgroup$ – Jeremy Jun 21 at 7:06
  • $\begingroup$ @Gordon I dont quite get the your simplification from step 1 to 2, why would the inner integral just disappear? $\endgroup$ – Jeremy Jun 23 at 17:13
  • 1
    $\begingroup$ @Jeremy Which textbook/chapter is this from? $\endgroup$ – Gabe Jun 23 at 21:32
  • $\begingroup$ 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. $\endgroup$ – Hans Jun 26 at 3:12
  • $\begingroup$ @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. $\endgroup$ – Hans Jun 26 at 3:14
4
$\begingroup$

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*}

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

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jeremy Jun 23 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.