In this module, we're going to look at a simple CDO example, where there's just one period in the model. We're going to assume that all the bonds in the reference portfolio have the same risk neutral probability of default. We'll also assume that all the pair-wise default events have the same correlation parameter rho. So these simplifying assumptions will enable us to easily calculate the loss distribution in the bond portfolio, and we will use it to compute the expected tranche loss in some tranches of a CDU. So in this simple example we're going to consider it is going to be a one period CDO. CDOs in practice are over multiple periods, but here we're just going to consider a one period CDO. The maturity is one year, or if you'd like one period. We'll assume that we've got 125 bonds in the reference portfolio. in case you're wondering why are choosing a number like 125, and not a, a, a st, more standard number like 100. Well, that's maybe a legacy of what goes on in the synthetic CDO market which we will discuss later. In those portfolios you often get 125 bonds in the so called reference portfolio, and that's why we're using 125 here. We're going to assume each bond pays a coupon of one unit. After one year, and that is if it has not defaulted. The recovery rate on each bond is 0, so in fact, if a bond defaults at any time over the next year, then the bond will pay 0. If it doesn't default then the bond will pay one unit. In one year's time, at the end of the CDO. We will assume that there are three tranches of interest. The equity, mezzanine and senior tranches. So the equity tranche is always the tranche which has a lower attachment point of zero. The senior tranche often has an upper attachment point of 100. But here we're actually going to assume that the senior tranche is exposed to default numbers seven up as far as nine. In fact, maybe a super senior tranche would be represented by these dots here, which would contain the defaults, or be exposed to the defaults. Num-, from numbers 10 up to 125. I will mention at this point, that this example we're considering is taken from this paper, the Devil is in Tails, Actuarial Mathematics and the Subprime Mortgage Crisis, by Donnelly and Embrechts. This is a very nice paper. It discusses CDOs, subprime CDOs, the crisis. It's a little bit technical in part, but much of it is readable for an undergraduate audience, and I'd certainly recommend those of you who are interested in learning more to go ahead and take a look at this paper. It can be easily found online. We're going to make the simple assumption that the probability q of defaulting within one year is identical across all bonds. We're also going to assume that the correlation between each pair of default events is identical. So these assumptions plus the fact that the maturity is one year and that this is just a one-period CDO is what makes this example particularly easy to analyze. As before, we assume X i is the normalized asset value of the ith credit. Or the ith bond, and now we're going to write that X i is equal to the square root of rho times M plus the square root of 1 minus rho times Z i. Where M, Z 1 up to Z n, are IID normal random variables. And recall that this assumption also implies that X i is also standard normal mean zero variance one. Recall that the ith credit defaults if X i is less than or equal to little x i bar. Therefore, the probability that the ith main defaults, is equal to the probability, we write it up here that X i is less than or equal to little x i bar. This must be equal to phi of little x i bar, because x i is the standard normal random variable, and phi's the standard normal CDF, so this is true. But we're all assuming that this is also equal to q. This is the probability of defaulting within one year, so this is equal to q, and from that, we see that if I take phi inverse across both sides of this expression here, I see x i bar equals phi inverse q. And that's what we down here. So the probability that x i defaults conditional on M is going to equal to the probability that X i is less than or equal to little x i bar. I can substitute equation three in down here. Replace x i bar with phi inverse q. Take the M term over to the right hand side, divided by the square root of 1 minus rho, and I get this expression here. This is what we're going to call q subscript M, this is going to be the risk neutral probability that X i defaults, conditional on knowing the value of the random variable M. What is also clear, is that this is equal to the standard normal CDF evaluated at this value here. Because Z i is a standard normal random variable. Therefore, conditional on knowing the value M, the total number of defaults is actually binomially distributed with parameters n, the number of names in the portfolio, the number of credits, the number of bonds and probability q M. Why is that? Well if you think about it, conditional on knowing M, we've already seen what's the probability of default is, it's q M. But also conditional on knowing M, all of these default events become independent. Because X i depends on M and Z i, and Z i, the Z i's are IID normal random variables. So, if I condition on the value of M, then the X i's become independent, which in particular means that the default events of all of the names in the portfolio are also independent. The probability of default conditional on M is q M. So therefore, the total number of defaults in the portfolio condition on M has a binomial distribution with parameters of M and q M. So therefore the probability of l defaults, and note I've dropped the dependence on t here which is fine. Because we're just considering one period with t equals one year. So therefore the probability of l defaults, out of the n names in the portfolio. Or n credits in the portfolio, conditional on the value of M is binomially distributed. And this is equal to N choose l times q M to the power of l, times 1 minus q M to the power of N minus l. Where phi we said in the previous slide is the standard normal CDF, and we know this expression for q M as well. Recall the q the risk neutral probability of a single name default within one year. So just be clear. You shouldn't get confused between q and q M. Q is the risk-neutral probability of a single name defaulting within one year. And qM is the same probability that is the risk neutral probability of a single name defaulting within one year, conditional on the value of M. So here are two questions. The first question is the following, when correlations on default probabilities are not identical, why is four no longer valid? Well, that's because, if the correlations an default probabilities are not identical, then these qM's are not the same for all of the names in the portfolio. You end up having a different qM, for each of the N names. So this quantity here is no longer a binomial. Random variable or a binomial probability. Next question, how do we calculate P superscript N l M in that case. Well, we'll just do it as we did in the last module, we saw that we had an iterative algorithm, remember we had the two nested for loops. That was the way, how we calculate P N of l in general. In this example, we don't need to run through those two for loops, because here we can just use the fact that the qMs are identical for all names I equals 1 to capital N. And therefore this is a binomial probability and we know this expression. [SOUND] So at this point, in this example, we actually know this quantity here. This quantity is a binomial probability. Assuming we can calculate q M, and we can because q M can be calculated as this expression here. We know these binomial probabilities, and so in particular we know the risk neutral probability of l defaults occurring conditional on the random variable M. How can we use this knowledge? Well, we can use it to compute the expected tranche losses. And that's what we're going to do in this slide. First of all, maybe I should go down to equation five first. So remember, what we're going to say is the following. The probability of l losses. Is equal to the probability of l losses given M times phi M and integrate that with respect to M. That's just using this trick we mentioned before from our undergraduate probability course. If I want to know the probability that X is equal to little x, well I can do that by summing over y the probability that X equals little x, Y equals little y. Which in turn is equal to the sum over little y of the probability that X equals little x given Y equals little y, by the probability that Y equals little y. So here we're identifying M with the random variable Y, and instead of a summation, we have an integral. But that's all we're doing here. So we know how to compute these quantities, we saw that on the previous slide, that these are binomial probabilities. So these are binomial probabilities and therefore we can do a numerical integration to calculate p of l. For any value of l equals 0 up to capital N. So we know that p of l's, we can do it very easily by integrating this expression numerically. So now let's turn to the issue of computing the expected tranche losses. In the equity tranche, we can actually lose either zero, one, two, or three. So the expecwe, expected equity tranche loss will be, k equals 1 times the probability of one default by year end. Plus k equals 2 by the probability of two defaults by year end. Plus k equals 3 by the probability of three or more defaults by year end. Remember the equity tranche is on the hook, if you like, for the first three losses in the portfolio. So, the most that the equity tranche can lose is actually three. But it will lose three if we have three or more defaults by year end. And so that's how we get the expected equity tranche loss. Let's turn, for example, to the expected senior tranche loss. Well, the senior tranche can only lose one, two, or three. You see back here that the senior tranche is on the hook for defaults seven to nine, so the most it can lose is nine minus, actually six, which is seven minus one, is three. So it's on the hook for these three defaults, seven, eight, and nine. So some that hope for three units, the most it can lose is three units. It will lose nothing if there are less than or equal to six losses in the portfolio. It will lose one, if there are exactly seven losses in the portfolio. It will lose two if there are exactly eight losses in the portfolio, and it will lose three. If there are nine or more losses in the portfolio. And that's represented over here in this expression for the expected senior tranche loss. So you should make sure that you're very comfortable and understand where these expressions come from. The equity, the equity tranche is on the hook for the first three losses. The Mezzanine tranche is on the hook for the next three losses. So, those losses will be lost as four, five and six. The Senior tranche is then on the hook for the next three losses, which would be losses seven, eight, and nine. And so these are the expected tranche losses for each of those three tranches.
