Credit Default Swaps Pt II – Credit Spreads

As well as governments, companies also need to borrow money and one way of achieving this is to issue bonds. Like government bonds, these pay a fixed rate of interest each year. Of course, this rate of interest will typically be quite a bit higher than government bonds in order to justify the increased credit risk associated with them. The difference between the two rates of interest is called the credit spread.

As we did before, today I’m just going to look at pricing a ZCB, but coupon-bearing bonds can be constructed as a product of these so it’s a relatively straight-forward extension.

Consider a ZCB issued by a risky company. It’s going to pay £1 at expiry, and nothing in-between. However, if the company defaults in the mean-time we get nothing. In fact, we’ll extend this a bit – instead of getting nothing we say that we will receive a fraction R of £1, since the company’s assets will be run down and some of their liabilities will get paid. This also allows us to distinguish between different classes of debt based on their seniority (we’re getting dangerously close to corporate finance here so let’s steer back to the maths).

In order to protect against the risk of loss, we might enter a Credit Default Swap (CDS) with a third party. In exchange for us making regular payments of £K/year to them (here we’re assuming that the payment accrues continuously for mathematical ease), they will pay us if and when the counterparty defaults. There are a few varieties of CDS – they might take the defaulted bond from us and reimburse what we originally paid for it, or they might just pay out an amount equal to our loss. We’ll assume the second variety here, so in the event of default they pay us £(1-R), since this is what we lose (up to some discount factors, but we’re keeping it simple here!). Also, if default occurs we stop making the regular payments, since we no longer need protection, so this product is essentially an insurance policy.

A cartoon of the payment schedule for a Credit Default Swap. We make continuous payments to the insurer of K per unit time, and in return we receive a one-off payment of (1-R) should the counterparty default, in which case the contract finishes early and no more payments happen
A cartoon of the payment schedule for a Credit Default Swap. We make continuous payments to the insurer of K per unit time, and in return we receive a one-off payment of (1-R) should the counterparty default, in which case the contract finishes early and no more payments happen

We value this product in the normal way. The value of the fixed leg is just the discounted cash flow of each payment multiplied by the probability that the counterparty hasn’t defaulted (because we stop paying in that case) – and remember that we said the payments are accruing continuously, so we use an integral instead of a sum:

    \[{\rm fixed} = -K \int_0^T \delta(0,t) S(t) dt\]

and the value of the insurance leg is the value of the payment times the chance that it defaults in any given interval

    \[{\rm floating} = -(1-R) \int_0^T \delta(0,t) p(t) dt\]

where in keeping with our notation from before, p(t) is the probability of default at a given time and S(t) is the chance that the counterparty hasn’t defaulted by a given time.

The contract has a fair value if these two legs are of equal value, which happens when

    \[0 = \int \delta(0,t) \Bigl(-(1-R) {dS \over dt} - K\cdot S(t) \Bigr) dt\]

At this point we refer to our exponential default model from Part I of this post.

In the exponential default model described in the previous post, we postulated a probability of default p(t) = \lambda e^{-\lambda t} which led to Survival Function S(t) = e^{-\lambda t}. Substituting these in to the equation above we have

    \[0 = \int \delta(0,t) e^{-\lambda t} \Bigl((1-R) \lambda - K \Bigr) dt\]

which is only zero when the integrand is also zero, so

    \[K = (1-R)\lambda\]

This is called the credit triangle, and it tells us the price of insuring a risky bond against default if we have it’s hazard rate. If the expected lifetime of the firm increases (ie. \lambda decreases) then the spreads will fall, as insurance is less likely to be required. If the expected recovery after default increases (R increases) then spreads also fall, as the payout will be smaller if it is required.

Although a little care is needed when moving from ZCBs to coupon-bearing bonds, it can be seen that the payments K (normally called the spreads) paid to insure the bond against default should be essentially the difference in interest payments between government (‘risk-free’) bonds and the risky bonds we are considering.

The default probabilities we have used here can be calibrated from market observed values for spreads K between government and corporate bonds. This resembles the process used to calculate volatility – the observable quantity on the market is actually the price, not the volatility/hazard rate, and we back this out from market prices using simple products and then use these parameters to price more exotic products to give consistent prices. This is the market’s view of the default probabilities in the risk-neutral measure, which means that they may not actually bear any resemblance to real world default probabilities (eg. historical default probabilities) and in fact are usually larger and often several times larger. See for example this paper for more on the subject.

Of course, throughout this analysis we have ignored the risk of the insurer themselves defaulting – indeed, credit derivatives rapidly become a nightmare when considered in general, and the Lehman Brothers default was largely down to some fairly heroic approximations that were made in pricing credit products that ultimately didn’t hold up in extreme environments. I’ll explore some of this behaviour soon in a post on correlated defaults and the gaussian copula model.

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 dtwhere \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
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}