Credit Default Swaps Pt I: A Default Model for Firms

In today’s post I’m going to discuss a simple model for default of a firm, and in Part II I’ll discuss the price of insuring against losses caused by the default. As usual, the model I discuss today will be a vast over-simplification of reality, but it will serve as a building block for development. Indeed, there are many people in the credit derivatives industry who take these things to a much higher level of complexity.

Modelling the default of a firm is an interesting challenge. Some models are based around ratings agencies, giving firms a certain grade and treating transitions between grades as a Markov chain (similar to a problem I discussed before). I’m going to start with a simpler exponential default model. This says that the event of a given firm defaulting on all of its liabilities is a random event obeying a Poisson process. That is, it is characterised by a single parameter $\inline \lambda$ which gives the likelihood of default in a given time period. We make a simplifying assumption that this is independent of all previous time periods, so default CAN’T be foreseen (this may be a weakness of the model, but perhaps not… discuss!).

Also, the firm can’t default more than once, so the process stops after default. A generalisation of the model will treat $\inline \lambda$ as a function of time $\inline \lambda(t)$ and even potentially a stochastic variable, but we won’t think about that for now.

Mathematically, the probability of default time tau occurring in the small window
$\inline [t,t+dt]$ is

$\lim_{dt \to 0} p( \tau < t + dt\ |\ \tau > t ) = \lambda dt$

If we start at $\inline t=0$ with the firm not yet having defaulted, this tells us that

$p ( \tau < dt\ |\ \tau > 0 ) = \lambda dt$

and in a second and third narrow window, using the independence of separate time windows in the Poisson process and taking a physicist’s view on the meaning of $\inline 2dt$ and similar terms,

$\begin{matrix} p ( dt < \tau < 2dt ) & = & p ( \tau < 2dt\ |\ \tau > dt )\cdot p(\tau > dt ) \\ &=& \lambda dt \cdot (1- \lambda dt) \end{matrix}$ $\begin{matrix} p (\ 2dt < \tau < 3dt\ ) &=& \lambda dt \cdot \bigl(1 - \lambda dt \cdot(1 - \lambda dt) - \lambda dt \bigr)\\ &=& \lambda dt \cdot (1 - 2\lambda dt - \lambda^2 dt^2)\\ &=& \lambda dt \cdot (1 - \lambda dt)^2\end{matrix}$

and in general

$p (\ t < \tau < t+dt\ ) = (1- \lambda dt)^n\cdot \lambda dt$ where $\inline n = {t \over dt}}$ and we must take the limit that $\inline dt \to 0$ , which we can do by making use of the identity

$\lim_{a\to \infty} (1 + {x \over a})^a = e^x$

we see that

$\begin{matrix} \lim_{dt \to 0}p (\ t < \tau < t+dt\ ) & = & \lambda e^{-\lambda t}\\ & = & p(\tau = t) \end{matrix}$

and its cumulative density function is

${\rm F}(t) = \int_0^t p(\tau = u) du = 1 - e^{-\lambda t}$ which gives the probability that default has happened before time $\inline t$ (the survival function $\inline {\rm S}(t) = 1 - {\rm F}(t)$ gives the reverse – the chance the firm is still alive at time $\inline t$ ) *another derivation of $\inline p(\tau=t)$ is given at the bottom

A few comments on this result. Firstly, note that the form of $\inline p(\tau=t)$ is the Poisson distribution for $\inline n=1$ , which makes a lot of sense since this was what we started with! It implies a mean survival time of $\inline \lambda^{-1}$ , which gives us some intuition about the physical meaning of lambda. The CDF (and consequently the survival function) are just exponential decays, I’ve plotted them below.

The survival/default probabilities for a firm in the exponential default model with various values of decay parameter

Having characterised the default probability of the firm, in Part II I will think about how it affects the price of products that they issue.

*Another derivation of $\inline p(\tau=t)$ is as follows:

$\lambda dt = p ( \tau < t+dt\ |\ \tau > t ) = {p(t < \tau < t+dt)\over p(\tau > t)}$

the first of these terms is approximately $\inline p(\tau=t)dt$ , while the second is simply the survival function $\inline S(t)$ , which by definition obeys

$p(\tau=t) = -{\partial S \over \partial t}$

combining this we have

$\lambda = -{1\over S}\cdot {\partial S \over \partial t}$

and integrating gives

$-\lambda t = \ln {S(t) \over S(0)}$ from which we get the result above, that

$S(t) = e^{-\lambda t}$

Credit Default Swaps Pt I: A Default Model for Firms

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