9  Credit Risk modelling 1: a single obligor

9.1 Why??

Finally in this course we would like to focus on one particular prominent category of risks we’ve already touched upon earlier, namely credit risk. Imagine that some party has an obligation to make one or more payments to you, for instance as part of a business deal, a financial construction like a loan or mortgage, an insurance contract or whatever other reason. This party is then an obligor (to you). Best case scenario for you is that you actually get these payments in full as agreed, but of course there is always the possibility, in some cases more likely than in others, that this does not happen and that you end up with only part of even none of these payments. In this situation we say that the obligor defaults on these obligations.

Obviously as risk manager it is crucially important that you are aware of this type of risk, and where needed (i.e. where default is not either essentially impossible1 or of little financial consequence to you2) put some thought/analysis in place to try and quantify what risk you are exactly running here. A key challenge is to assess how likely an obligor is to default i.e. what their default probability is — after all this will normally heavily depend on their overall financial state, and especially if this is not too great then they will very likely advertise this as little as possible…

1 For instance, lending money to a (stable, trustworthy) government, so-called bonds, is generally considered the gold standard of safety

2 For instance if the amount is small enough

The total amount that is at stake for you, i.e. the amount that you would lose if the obligor pays you nothing at all, we call your Exposure At Default (EAD). Further if default were to happen, then we call the fraction (i.e. a number in \([0,1]\)) of the EAD that you actually end up losing your Loss Given Default (LGD). The LGD may be a determistic number, but it may also well be a random variable rather — for instance in cases where, if default happens, the amount that you still receive back is subject to some uncertainty.

In our risk management context, the risk/loss (which we’ll again model by a random variable) due to possible default is called the credit risk. We hence have that \[\text{Payment you receive} = \text{EAD} - \text{credit risk}, \tag{9.1}\] and that the credit risk takes the value \(0\) if no default happens (clearly the best case scenario!), otherwise it equals \(\text{LGD} \cdot \text{EAD}\).

See Exercise 9.1–Exercise 9.3 for some examples in which all this terminology is used!

In line with our earlier discussion on the topics of sharing and aggregating risks, typically a risk manager deals with more than one obliger, and so your overall default risk profile (if you will) is formed by aggregating the individual default risks. Let’s do a nice and simple example with this.

Example 9.1 A zero-coupon bond is a contract in which a company promises to pay a certain amount of money (the face value) at some set point in the future (the due date). The buyer of such a contract is called the bond holder. For a company, issuing such bonds is a means of gathering capital which can be used for investing in improving its operations, anticipating that this will result in enough extra income that they’ll be able to make the face value payment when the time comes. For a bond buyer, this is an opportunity to secure the face value payment in due time, in return for a (smaller) investment in the company now.

Suppose that you are holding bonds for companies A and B, both due in say a month from now. Further suppose that:

  • The bond for company A has a face value of \(£100{,}000\) (which is (hence) your EAD for this contract). Company A has a default probability of \(0.02\) and an LGD of 40%.
  • The bond for company B has a face value of \(£200{,}000\) (your EAD for this contract). Company B has a default probability of \(0.01\) and an LGD of 70%.
  • You assess that the joint default probability i.e. the probability that both companies default is \(0.004\) (so apparently you consider these default events not to be independent — for instance because both companies are active in the same industry, so that a general downturn in that industry could affect both companies).

Denoting by \(X\) the credit risk you run by holding these two bonds, we see that \[X=\begin{cases} 0 & \text{if neither company defaults} \\ 0.4 \cdot 100{,}000 = 40{,}000 & \text{if company A defaults but B doesn't} \\ 0.7 \cdot 200{,}000= 140{,}000 & \text{if company B defaults but A doesn't} \\ 180{,}000 & \text{if both companies default.} \end{cases}\] To work out the distribution of \(X\), denoting by \(A\) resp. \(B\) the event that company A resp. B defaults, we’re given that \[\P(A)=0.02, \quad \P(B)=0.01 \quad \text{and} \quad \P(A \cap B)=0.004.\] So the probability that A defaults but not B is \[\P(A \cap B^c)=\P(A)-\P(A \cap B) =0.02-0.004=0.016\] and similarly the probability that B defaults but not A is \[\P(A^c \cap B)=\P(B)-\P(A \cap B) =0.01-0.004=0.006,\] which leaves, using the complement rule, the probability that neither default equal to \[\P(A^c \cap B^c)=1- \left( 0.016+0.006+0.004 \right)=0.974.\] So the credit risk \(X\) looks more explicitly as follows: \[X=\begin{cases} 0 & \text{with prob. } 0.974 \\ 40{,}000 & \text{with prob. } 0.016 \\ 140{,}000 & \text{with prob. } 0.006 \\ 180{,}000 & \text{with prob. } 0.004. \end{cases} \tag{9.2}\]

Remark 9.1. Note that \(X\) in Equation 9.2 has some specific properties that you tend to see commonly in credit risk: a very large probability of \(0\), yet also some quite large values with small probabilities.

Recalling for a moment the VaR from Chapter 6, note that for instance \(\VaR_{0.95}(X)=0\) which is clearly not very informative or even deceptive — it suggests that \(X\) is not very risky at all but tell that to your colleague risk manager over lunch who happened to have just suffered a credit risk loss of £180,000!

Generally the VaR by its very nature is not a great risk measure to use in the context of credit risk.

Remark 9.2. It’s nice to think for a moment about what the price \(P\) for an \(n\)-year zero-coupon bond with face value (say) \(F\) as in Example 9.1 should be (assuming a “perfect” market with rational agents etc.). There’s effectively only two factors feeding into it: the interest rate and the risk of default. If there is (essentially) no risk of default (e.g. when the issuer is a well trusted government), then only the interest rate remains and by no-arbitrage arguments (don’t worry of you’re not familiair with this term) it is not hard to see that \[P=(1+i)^{-n} F, \tag{9.3}\] where \(i\) is the annual interest rate. We could also invert this way of thinking: if we look up what the market price \(P\) of such an instrument is we can work out \(i\), which represents what the agents currently expect the (average) annual interest rate for the coming \(n\) years will turn out to be.

If a more default risky party issues such a bond, then the price agents are willing to pay will be lower than Equation 9.3 — rational agents will “price in” the risk of default, at least to the extent that they can judge it. We mentioned above already that generally it’s a challenge to correctly assess default probabilities due to a lack of (publicly available) information, but if we were to look up market prices for such bonds and compare them with Equation 9.3 then we can at least extract information about how “the market” judges the default risk!

Basically our credit risk journey is guided by answering two main questions:

  1. How do we come up with a sensible estimate for the probability of default for a single company in isolation? We mentioned above already that we have only limited insight into any company’s financial situation, namely only what is publicly available, so there is some thinking to do at this front. We’ll discuss some prominent models in the remainder of this chapter.
  2. What exactly is the perspective of a risk manager dealing with multiple obligors i.e. parties/companies that could each default? This also brings the important question in play of how these potential defaults depend on each other/influence each other. We’ll leave this for Chapter 10.

9.2 Scoring models

A classical and not-too-much-nonsense approach is to calculate a credit score for a company from its publicly available quarterly or annual financial reports, according to a certain rule, and then to link credits scores to default probabilities (for instance on the basis of historical default data for similar companies). To get some feel, here is a nice example.

Example 9.2 In 1968, Altman introduced the following simple scoring model for manufacturing companies (see e.g. Altman (2018) for a recent, wider discussion). He defined the following ratios, whose values can all be computed from a company’s financial reports: \[\begin{align*} x_1 &= \frac{\text{working capital}}{\text{total assets}}= \frac{\text{total assets}-\text{total liabilities}}{\text{total assets}}, \\ x_2 &= \frac{\text{retained earnings}}{\text{total assets}}, \\ x_3 &= \frac{\text{earnings before interests and taxes}}{\text{total assets}}, \\ x_4 &= \frac{\text{market value of equity}}{\text{total liabilities}}, \\ x_5 &= \frac{\text{sales}}{\text{total assets}}. \end{align*} \] These ratios are then given specific weights to come to a total so-called “\(Z\)-score”: \[Z=1.2 x_1 + 1.4 x_2 + 3.3 x_3 + 0.6 x_4 + 1.0 x_5,\] and the following assessment is proposed:

  • if \(Z>2.99\) then the company is unlikely to default within the next year,
  • if \(Z<1.81\) then the company is likely to default within the next year,
  • if \(1.81 \leq Z \leq 2.99\) then the company is in what Altman calls “the grey area”.

Although the above example is simple, it nicely illustrates the general structure of a scoring model:

  1. Fix \(n\) relevant factors i.e. quantities that we can measure from publicly available data and that we believe to have some bearing on/predictive power of the default probability,
  2. Choose weights to boil the \(n\) measurements down to single score,
  3. Design a rule to map scores to default probabilities.

Where steps 2 & 3 are mainly mathematical in nature, the question which factors to use for the first step is typically mostly informed by experience with business (analysis). More specifically, we could write down the following

Definition 9.1 A scoring model consists of a choice of \(n\) factors for some \(n=1,2,\ldots\), some constants/weights \(w_0,\ldots,w_n \in \R\), and a cdf \(F\). The model is then applied as follows.

For any company:

  1. the \(n\) factors are measured, yielding values \(x_1, \ldots x_n \in \R\),
  2. the score \(s\) of the company is computed as \[s=w_0+w_1 x_1 + \ldots + w_n x_n,\]
  3. the default probablity of the company (for the coming year, say) is estimated to be \[p=F(-s)=F(-w_0-w_1 x_1 - \ldots - w_n x_n).\]

Some notes:

  • The specific form that step iii has, using that cdf to transform scores into default probabilities, may seem to come out of nowhere. It is indeed somewhat arbitrary really to do it in this way, many other rules would have worked equally well in principle. It’s just that this form naturally arises from a GLM based way of thinking about (finding the optimal value of) these weights.
  • The reason that we set \(p=F(-s)\) rather than simply \(p=F(s)\) is that this choice preserves the intuition established by Altman’s model in Example 9.2: the larger the score \(s\), the smaller the default probability is (since any cdf is non-decreasing, cf. Section A.1.1).

Example 9.3 Two popular choices for the cdf \(F\) in the scoring model from Definition 9.1 are the cdf of the standard normal distribution (in which case the term probit scoring model is used) and the cdf \[F(x)=\frac{1}{1+e^{-x}} \quad \text{for all } x \in \R\] (the logit scoring model).

9.2.1 Implementing & calibrating scoring models

To implement a scoring model from Definition 9.1, you start with making choices for the \(n\) factors (as mentioned, mainly based on experience though some advanced Stats techniques could be useful tools as well) and the cdf \(F\) (any cdf that is strictly increasing and continuous on the whole of \(\R\) will do really — the choices mentioned in Example 9.3 are particularly prominent because they’re mathematically convenient in the calibration we’ll briefly discuss in this section). The remaining ingredient of the model is choosing the weights \(w_0, \ldots, w_n\).

The best techniques we have available for that task come from Stats: find past data, and then, assuming that the past data is a good representation of what the future will look like, “fit” your weights to that data. Somewhat of a problem is that typically you won’t have enough data by looking at the past of the particular company you’re interested in only, so rather you look at past data for a whole group of companies that are all “similar enough” — normally that means at least all operating in the same sector of the broader economy.

So putting on our Stats hat, the data we’re talking about here is more precisely past observations, say \(K\) in total, of this/similar enough companies where the \(n\) factors are measured at the start of a year and at the end of the year we observe whether or not the company has defaulted. For \(k=1,\ldots,K\), we denote by \[x^{(k)}_1, x^{(k)}_2, \ldots, x^{(k)}_n \tag{9.4}\] the measurements of the \(n\) factors and by \[y_k \in \{0,1\} \tag{9.5}\] the observed default state: \(y_k=0\) if the company didn’t default, \(y_k=1\) if it did.

Under the assumption that our model is correct, these \(K\) data points are \(K\) observed realisations/runs of our model i.e. for \(k=1,\ldots,K\), the company observed had a default probability of \[p_k=F \left( -w_0-w_1 x_1^{(k)} - \ldots - w_n x_n^{(k)} \right) \tag{9.6}\] and hence \(y_k\) is a realisation of a \(\text{Bernoulli}(p_k)\) distributed random variable \(Y_k\): \[\P(Y_k=0)=1-p_k, \quad \P(Y_k=1)=p_k.\] The likelihood function for the \(k\)-th observation (recall: the probability of observing our actual observation i.e. \(y_k\) if we run the model once with the given factor values, considered as a function of the unknown parameters we’re looking to estimate) is then \[\begin{align*} L_k(w_0,\ldots,w_n) &= \P(Y_k=y_k) \\ &= \begin{cases} p_k & \text{if } y_k=1 \\ 1-p_k & \text{if } y_k=0 \end{cases} \\ &= p_k^{y_k} \left( 1-p_k \right)^{1-y_k}. \end{align*} \] Further, assuming independence conditional on the factor values3, the (total) likelihood (i.e. the probability of actually observing \(y_1, \ldots, y_K\) if we run the model \(K\) times, each time with one of the sets of factor values) is \[\begin{align*} L(w_0,\ldots,w_n) &= \P(Y_1=y_1, \ Y_2=y_2, \ldots, Y_k=y_k) \\ &= \prod_{k=1}^K \P(Y_k=y_k) \\ &= \prod_{k=1}^K L_k(w_0,\ldots,w_n) \\ &= \prod_{k=1}^K p_k^{y_k} \left( 1-p_k \right)^{1-y_k}. \end{align*} \tag{9.7}\] Note that the above right hand side is indeed a function of \(w_0,\ldots,w_n\) since the \(p_k\)’s are (cf. Equation 9.6) and the \(y_k\)’s are known numbers (cf. Equation 9.5).

3 Which is a standard assumption in GLMs and essentially means the following: the factors we have chosen capture all the “structural” aspects contributing to default, what remains is pure chance

Finally, our beloved maximum likelihood principle entails that as estimates for the weights in our model we choose the maximiser of the likelihood function, or equivalently, the maximiser of the log-likelihood function \[\begin{align*} l(w_0,\ldots,w_n) &:= \log \big( L(w_0,\ldots,w_n) \big) \\ &= \sum_{k=1}^K y_k \log \left( p_k \right) + \left( 1-y_k \right) \log \left( 1-p_k \right) \end{align*} \tag{9.8}\] (key benefit: easier to analyse than the original likelihood function).

Note that what happens here is that we choose as estimates for the weights those values that make the actual observed data most likely. Don’t forget that this is nothing more than a principle, a philosophy4 — what it has going for it is that it is intuitively appealing and that it has a proven track record of doing a very good estimation job a lot of the time.

4 Fundamentally in Stats, any finite set of data does not contain enough information to derive the exact value of unknown parameter(s), nor indeed to prove or disprove that a suggested model is the correct model for the data at hand!

Proposition 9.1 Take a scoring model as defined in Definition 9.1 with the \(n\) factors and cdf \(F\) already chosen, and with \(K\) observations in the form Equation 9.4 & Equation 9.5 available. Following the principle of maximum likelihood we estimate the weights \(w_0,\ldots,w_n\) by taking the maximiser, say \(\widehat{w}_0,\ldots,\widehat{w}_n\), of the likelihood function Equation 9.7 or equivalently of the log-likelihood function Equation 9.8.

The model is then completely specified & calibrated, and dictates that the default probability for a company (similar enough to the ones on the data) with values \(x_1,\ldots,x_n\) for the \(n\) factors is given by \[p=F \left( -\widehat{w}_0-\widehat{w}_1 x_1 - \ldots - \widehat{w}_n x_n \right).\]

Remark 9.3. Those of you who have done the course Generalised Linear Models will recognise what we did in this section as essentially a very bare bones GLM type analysis. Indeed you could (and ideally should) do a proper full blown analysis with all the fancy tools from that course to do the data fitting and verifying the fit!

9.3 Firm value models

Firm value models feel (and generally are) more recent than the scoring models from the above sections. They try to model the “value” a company represents more directly and generally focus on publicly traded companies i.e. those whose shares can be bought and sold on publicly accessible stock exchanges.

9.3.1 The Merton model

The Merton model is the prototype of a firm value model. It is relatively simple and you could rightly question some of its assumptions, nevertheless it illustrates the way of thinking in such models very well and is still powerful enough to handle many situations well.

Imagine a company that comes into existence at time \(t=0\). We denote by \(V_t\) the firm value at time \(t \geq 0\): if at time \(t\) the company were liquidated i.e. all machinery, inventory, intellectual property etc. sold, then that would yield the amount \(V_t\).

At time \(t=0\). At time \(t=0\), the company raises capital from two sources, as follows.

  • Issuing shares. It offers a certain amount of shares on the stock market. An investor who buys \(x\)% of shares, becomes owner of \(x\)% of the company. We denote by \(S_0\) the total price paid by investors and hence cash raised by the company (so \(S_0\) is the price/value per share multiplied by the number of shares issued). As these shares can now be (further) traded on the stock market, their price/value will normally change over time: we denote by \(S_t\) the price/value of all shares together at any time \(t \geq 0\).
  • Issuing a zero-coupon bond with face value \(F\) and due date \(T\). Recall that we discussed such instruments in Example 9.1 already: the company sells the promise to pay the holder of the bond the amount \(F\) at time \(T\). The buyer of this bond pays an amount \(B_0\) for it at time \(t=0\) (normally less than \(F\)). Such a bond is also a tradable instrument whose price/value will change over time, due to the remaining period until time \(T\) getting shorter, changes in interest rates, changes in the market perception of whether the company will default on this bond or not etc.: we denote by \(B_t\) the price/value at any time \(t \in [0,T)\).

Note that both \(S_0\) and \(B_0\) are determined by “the market” (or at least, must be offered at a price “the market” is willing to pay), a reflection of how investors appraise the company’s financial future. At time \(t=0\), the company consists of nothing yet but plans for the future and the total amount of capital raised namely \(S_0+B_0\). So \[V_0=S_0+B_0. \tag{9.9}\]

At time \(t \in [0,T)\). The company embarks on its adventure and starts creating & selling its goods and/or services. Still, the relationship Equation 9.9 remains valid: \[V_t=S_t+B_t \quad \text{for all } t \in [0,T). \tag{9.10}\] The rationale is that at any time \(t \in [0,T)\), the value for the shareholders of owning the company is the (net) value of its assets i.e. \(V_t\) minus the value of the zero-coupon bond \(B_t\) (think: this is the value and hence also the amount the bondholder would rationally require to annul the bond at that point in time)5, i.e. \(S_t=V_t-B_t\).

5 Of course, this argument makes some assumptions, for instance on the information that agents on the stock market have available and how they act on that information — if you’re interested in digging out this point more: it is a consequence of the Modigliani–Miller theorem

At time \(t=T\). This is the time for the company to wake up and smell the coffee: the bond holder knocks on the door to come claim his amount \(F\). As long as \(V_T \geq F\) this claim can be honoured i.e. the company does not default, if needs be by liquidating (part of) the company. The remaining value is left for the shareholders: \(S_T=V_T-F\). Otherwise, if \(V_T<F\), then the company is necessarily liquidated and the amount \(V_T\) is paid to the bondholder. Since the bondholder gets less than \(F\) the company now does default, and the shares become completely worthless. So in particular: the company defaults if and only if \(V_T<F\).

We now have a full description of what is going in the model and have worked out when default happens, but in order to turn that into a default probability we need a sensible model for \(V_T\). For this we use the seminal Black & Scholes model6, which models the evolution of \(V_t\) through time as a geometric Brownian motion. If you haven’t encountered the Black & Scholes model before then don’t worry, it simply entails that we model \(V_T\) as a random variable of the following form7: \[V_T=V_0 \exp \left( \sigma \sqrt{T} Z + \left( \mu-\frac{\sigma^2}{2} \right) T \right) \tag{9.11}\] where:

6 Indeed, the “Merton” of the model we’re discussing here is Robert C. Merton, who made some major contributions to mathematical finance and in particular also in support of/around the Black & Scholes model

7 Note that we can rewrite this expression a little bit to the form \(V_T=e^Y\), where \(Y\) has a Normal distribution. This form shows that \(V_T\) has a log-normal distribution

  • \(V_0>0\) is a constant (recall its context, cf. Equation 9.9),
  • \(\mu \in \R\) and \(\sigma>0\) are constants (new to us, these should be chosen to “best fit” the company at hand),
  • \(Z \sim \mathcal{N}(0,1)\).

Recalling our observation from above, that in the Merton model default happens if and only if \(V_T<F\), we can now conclude that the default probability in the Merton model is given by \[\begin{align*} \P(V_T<F) &= \P \left( V_0 \exp \left( \sigma \sqrt{T} Z + \left( \mu-\frac{\sigma^2}{2} \right) T \right) <F\right) \\ &= \P \left( Z< \frac{\log(F/V_0)-(\mu-\sigma^2/2)T}{\sigma \sqrt{T}} \right) \\ &= \Phi \left( \frac{\log(F/V_0)-(\mu-\sigma^2/2)T}{\sigma \sqrt{T}} \right), \end{align*} \tag{9.12}\] where \(\Phi\) is the cdf of a standard Normal distribution.

Remark 9.4. So the Merton model gives us Equation 9.12 as probability of default, which is a straightforward enough expression. The only difficulty is determining the parameters \(\mu\) and \(\sigma\). In the Black & Scholes model, where a geometric Brownian motion is used to model the evolution of the share price \(S_t\) over time and the same two parameters need to be estimated, this is relatively straightforward as there are (normally) plenty of past observations from share prices available (as this is public information) to fit these parameters to. However (accurate enough) past observations from firm values are generally a lot more rare. There are some nice ideas to get around this problem though, see e.g. Section 8.2 in McNeil, Frey, and Embrechts (2015) if you’re interested.

9.4 Other models

We conclude this chapter by briefly mentioning some other models.

The KMV (Kealhofer–McQuown–Vasicek) model is a firm value model in similar spirit as the Merton model from Section 9.3.1. It is the in-house model of the famous Moody’s Analytics, one of the world leading providers of credit risk data, software etc. They publish their assessments using a letter based scale: from the most safe “Aaa” (“triple A”) down to the most risky “C”. A downwards change in their rating of an important entity like a government or a large international bank often sends shivers down the spine of many risk managers worldwide as it may very well, directly or indirectly, affect the “credit worthyness” of the risks they manage. For instance, a company investing funds on behalf of clients often guarantee their clients to only invest in opportunities with a sufficiently high rating by Moody’s.

Not all details of the KMV model are public knowledge, but we do know that the default probability is ultimately computed as \[G \left( - \frac{V_0-F}{\sigma V_0} \right),\] where \(G\) is some cdf and the other ingredients are the same as in the Merton model from Section 9.3.1. It is worth pointing out that Moody’s influence and authority is not so much based on a very sophisticated/complicated model, but rather on the decades of proprietary data that they have managed to collect which allows them to test and calibrate their model really well.

Finally a word about Credit Migration models, which form a different category from the ones we have so far discussed. These use a Markov chain setup (which you may well have seen in the course Markov Processes or the course Mortality Modelling in Insurance) in which it is assumed that each company fits into one of a finite number of credit states (think: from “as safe as it gets” via a number of steps to “as risky as it gets”) and probabilities are found/estimated for a company to move from one state to another state within a certain time period. See e.g. Section 8.2 in McNeil, Frey, and Embrechts (2015) if you’d like to know more.

9.5 Some exercises

NoteAbout the exercises

Each exercise has a (rough) indication of its difficulty, as follows:

* easier: can be solved by (almost) only using relevant definitions/results,
** medium: in addition to relevant definitions/results, needs a limited amount of work/creativity,
*** harder: in addition to relevant definitions/results, needs a larger amount of work/serious creativity,
💀 warning: might make your brain hurt! These are mainly intended to provide some extra challenge for those of you keen on that and are generally quite hard. You don't need to worry about these too much for exam purposes.

The exam consists of mostly ** and *** level questions, some *, and possibly at most a few marks worth of 💀.

A bit of preaching: it is an incredibly important part of the study process to try and work on the exercises as much as possible. To become a better mathematician/learn new maths (and also to get a good exam mark ;)), above all you need to do it. And yes, of course that includes falling over things, and making mistakes, and getting stuck, and getting frustrated — all part of the game and what you’re supposed to be doing! Your lecturers have done that as well and still do it. What matters is that you don’t let that discourage you and that you make good use of the help and resources available to help you develop your skills. As part of that, many exercises have a hint in a block like this:

Hint!

These are trying to help you on your way if you don’t know where to start or to provide some ideas if you get stuck. In spirit of the above, always have a look at these first and try again before you look at the full solution. (These hints are an extra service that won’t be available in the exam I’m afraid ;).)

Of course, we have our classes and there’s office hours, email etc. as well — I’m at any time very happy to help you with any questions you may have, and you should please never feel that any question is “too dumb” to ask!

Full/detailed solutions for the exercises will become available, just immediately below the exercises, immediately after our tutorial hour (you may have to refresh the page).

Exercise 9.1 [*] You’re going to get a pizza with your friend. As they’re a bit short of money you lend them £10 for their pizza. In principle they’ll pay you back tomorrow, but knowing your friend, there is a \(10\%\) chance that they will completely forget and then you won’t hear about it ever again.

  1. Who is the obligor in this situation, you or your friend?
  2. What does it mean for your friend to default on their loan?
  3. What is the default probability?
  4. What is your exposure at default?
  5. What is your loss given default?
  6. Let \(X\) denote your credit risk. What is the distribution of \(X\)? Also compute \(\E[X]\).

For parts i–v, these should all be one line answers essentially, just interpreting/applying the terminology we introduced in Section 9.1.

For part vi, recall that the credit risk \(X\) is the (random) amount you (potentially) lose in this deal. Observe that it is either £0 or £10.

For parts i–v, these are all referring to the terminology we introduced in Section 9.1:

  1. Since you friend owes you money, your friend is the obligor.
  2. In general it means that the obligor does not fully fulfil their financial obligations. In this case, it means that your friend does not repay the full £10. And in particular, here either your friend fully repays you or doesn’t repay you at all, so if your friend defaults on this loan then you get nothing back at all.
  3. \(10\%\) chance, i.e. probability \(0.1\).
  4. Your exposure at default is the amount that you have lent your friend i.e. £10.
  5. Your loss given default is the fraction of that £10 that you end up losing if your friend does default. In this case, if default happens you won’t get anything back, so your loss given default is \(1\) (or \(100\%\) if you prefer).

For part vi, the credit risk \(X\) is the (random) amount you (potentially) lose, so it is either £0 (if you get repaid, with probability \(0.9\)) or it is £10 (if you don’t get repaid, with probability \(0.1\)). Alternatively you could use Equation 9.1. That is to say: \[ X=\begin{cases} 10 & \text{with prob } 0.1 \\ 0 & \text{with prob } 0.9. \end{cases} \tag{9.13}\] Clearly (see e.g. Equation A.10 otherwise) \[\E[X]=10 \cdot 0.1 + 0 \cdot 0.9=1.\]

Note: if you’re confused about how to answer a question like “What is the distribution of \(X\)?”, then in general it means that you need to specify either its cdf (because cdf’s uniquely characterise distributions, cf. Section A.1.1). Alternatively if \(X\) is discrete you could specify its pmf (because that determines its cdf, cf. Section A.1.2) or if \(X\) is continuous you could specify its pdf (because that determines its cdf, cf. Section A.1.3).

Further alternative equivalent forms are:

  • stating that \(X\) has one of the well known “named” distributions, e.g. \(X \sim \text{Poisson}(10)\) or \(X \sim \text{Exp}(1)\) for instance. That is equivalent because it (indirectly) specifies the cdf.
  • if \(X\) is discrete, explicitly listing the values that \(X\) can take with corresponding probs, like we did in Equation 9.13. That is equivalent because it fully specifies its pmf.

Exercise 9.2 [*/**] Let’s look again at the situation in Exercise 9.1. Suppose now that one of three things can happen: either you get the full £10 back tomorrow, or you get £7 back within a week (with probability \(0.1\)), or you never hear anything about it again (with probability \(0.05\)). Now answer the same questions as in Exercise 9.1 again.

Hint: be slightly careful with the loss given default, because it is the loss given default i.e. the loss given that/conditional on default happens!

For parts i, ii, iv, does the answer change at all??

For part v, observe that given that default happens, you either get £7 back (so the fraction of the EAD that you lose is \(0.3\)) or you get nothing back (so the fraction of the EAD that you lose is \(1\)). So your loss given default, LGD, is now a random variable rather than a fixed number, and its possible values are \(0.3\) and \(1\). For the probabilities, if you’re not sure how to obtain those, then consider defining the events \[\begin{align*} A &= \{ \text{You get £10 repaid} \} \\ B &= \{ \text{You get £7 repaid} \} \\ C &= \{ \text{You get nothing repaid} \} \end{align*} \] and work out the probs in terms of these events.

For part vi, should be pretty straightfoward — observe that \(X\) can now take three values rather than two in Exercise 9.1!

For parts i, ii, iv, the answer is unchanged from Exercise 9.1.

For part iii, the probability of default i.e. the probability that you do not get fully repaid is now \(0.1+0.05=0.15\).

For part v, given that default happens, you either get £7 back (so your loss is £3, and the fraction of the EAD that you lose is \(0.3\)) or you get nothing back (so your loss is £10, and the fraction of the EAD that you lose is \(1\)). So your loss given default, LGD, is now a random variable rather than a fixed number, and its possible values are \(0.3\) and \(1\). But with what probabilities? More intuitively you could think about this as “ratios”: out of the total default probability of \(0.15\) (cf. part iii), we have \(0.1\) for a fractional loss of \(0.3\) and \(0.05\) for a loss of \(1\). That gives: \[\P(\text{LGD}=0.3)=\frac{0.1}{0.15}=\frac{2}{3} \quad \text{and} \quad \P(\text{LGD}=1)=\frac{0.05}{0.15}=\frac{1}{3}.\]

Alternatively, more mathematically, you could use conditional probs (recall from Equation A.42 e.g.): denoting the events \[\begin{align*} A &= \{ \text{You get £10 repaid} \} \\ B &= \{ \text{You get £7 repaid} \} \\ C &= \{ \text{You get nothing repaid} \} \end{align*} \] we have that \(\P(A)=0.85\), \(\P(B)=0.1\) and \(\P(C)=0.05\). Further the event that default happens is \(B \cup C\). So: \[\begin{align*} \P(\text{LGD}=0.3) &= \P(B \, | \, B \cup C) \\ &= \frac{\P(B \cap (B \cup C))}{\P(B \cup C)} \\ &= \frac{\P(B)}{\P(B \cup C)} \\ &= \frac{\P(B)}{\P(B)+\P(C)} \\ &= \frac{0.1}{0.1+0.05} \\ &= \frac{2}{3} \end{align*} \] and similarly (or by just using that necessarily \(\P(\text{LGD}=0.3)+\P(\text{LGD}=1)=1\)) we get that \(\P(\text{LGD}=1)=1/3\).

So in conclusion: \[\text{LGD}=\begin{cases} 0.3 & \text{with prob } 2/3 \\ 1 & \text{with prob } 1/3. \end{cases} \]

For part vi, the credit risk \(X\) i.e. the (random) amount that you (potentially) lose, is now given by \[ X=\begin{cases} 0 & \text{with prob } 0.85 \\ 3 & \text{with prob } 0.1 \\ 10 & \text{with prob } 0.05 \end{cases} \] with (again using Equation A.10) \[\E[X]=0 \cdot 0.85 + 3 \cdot 0.1 + 10 \cdot 0.05=0.8\] (not unexpected given the changes we made, a bit smaller than \(\E[X]\) was in Exercise 9.1).

Exercise 9.3 [**] Let’s go back to Exercise 9.1 one more time! Suppose now that with probability \(0.9\), you do get an amount back tomorrow and otherwise you never hear anything about it again. However by tomorrow, all your friend will have been able to do is to gather a few pounds here and there from between the sofa and the pocket of an old pair of jeans: the amount that they’ll give you back is random with a (say) \(\text{Unif}(0,10)\) distribution. Again answer the same questions as in Exercise 9.1!

Hint: both the loss given default and the credit risk \(X\) are more complicated random variables this time: they are neither discrete nor continuous. Don’t worry trying to characterise their distributions. However you can still compute \(\E[X]\) — for this, consider using the Law of Total Expectation from Equation A.47.

For part iii, can the probability of default equal \(1\)? Well yes, why not?

For part v, the best you can really do is describe the loss given default as a random variable consisting of two cases, one if you get something back tomorrow and one if you get noting back at all.

For part vi, observe that in this case, where the probability of default equals \(1\), we have that the credit risk and the loss gievn default are equal: \(X\) is also a random variable that we can only really characterise by considering two different cases. However, for \(\E[X]\), that’s exactly a problem Equation A.47 is there to solve! As long as you can compute what the expectation of \(X\) would be in the first case as well as in the second case, then you can put these bits together to get to \(\E[X]\) by using Equation A.47!

For parts i, ii, iv, the answer is (again) unchanged from Exercise 9.1.

For part iii, the probability of default i.e. the probability that you do not get fully repaid is now actually \(1\)! Indeed either you get nothing at all back, or you get back a random amount that follows a \(\text{Unif}(0,10)\) distribution, which has range/set of possible values \((0,10)\) (cf. Equation A.18) so whichever sample/realisation of that random amount we end up getting, it is always going to be less than £10!

For part v, given that default happens, you either get nothing back (so your loss is £10, with prob \(0.1\)) or you get a random amount back, say \(U \sim \text{Unif}(0,10)\) (so your loss is \(£10-U\), with probability \(0.9\)). The resulting loss given default is hence a random variable that is a mix of a “discrete part” (it takes the value \(1\) with prob \(0.1\), recall that continuous random variables cannot have such behaviour, cf. Equation A.14) with a “continuous part” (with prob \(0.9\) it equals \((10-U)/10\), which is a continuous rv). Besides this observation there is not much more we can do with this!

For part vi, the credit risk \(X\) i.e. the (random) amount that you (potentially) lose, is now equal to the loss given default (because default is guaranteed to happen): with prob \(0.1\) you get nothing back i.e. \(X=10\) and with prob \(0.9\) you get \(U \sim \text{Unif}(0,10)\) back and hence \(X=10-U\). So as mentioned in part v: very annoying distribution. However, as the question points out, it is actually not very hard to nevertheless compute \(\E[X]\)! Observe that in the case that \(X=10\) we can easily compute its expectation, namely (indeed!) \(10\); and also in the case that \(X=10-U\): \(10-\E[U]=5\) (cf. Appendix B if needs be). The difficulty is how we should put these two different values for the expectation together, if you willl. Given that the probs are \(0.1\) and \(0.9\) resp., a very appealing guess would be to think that we should have \[\E[X]=0.1 \cdot 10 + 0.9 \cdot 5=5.5.\]

Indeed this is correct, and the rigorous translation of this way of thinking is to use the Law of Total Probability from Equation A.47: define the event \[A=\{ \text{Your friend never pays you back} \}.\] Then given/conditional on \(A\), we have that \(X=10\) and hence \(\E[X \, | \, A]=10\), while given/conditional on \(A^c\) we have that \(X=10-U\) and hence \(\E[X \, | \, A^c]=\E[10-U]=10-\E[U]=5\). Now Equation A.47 (with \(h(x)=x\), \(n=2\), \(B_1=A\), \(B_2=A^c\)) indeed reads as: \[\E[X]=\E[X \, | \, A] \P(A) + \E[X \, | \, A^c] \P(A^c)=10 \cdot 0.1 + 5 \cdot 0.9=5.5.\]

Exercise 9.4 [*] Suppose that two companies, A and B, each issue a three year zero-coupon bond at time \(t=0\), with face value £100,000. Further the firm value at time \(t\) of company A resp. company B is modelled as \(V^{(A)}_t\) with parameters \(V^{(A)}_0=110{,}000\), \(\mu_A=0.1\) and \(\sigma^2_A=0.1\); resp. \(V^{(B)}_t\) with parameters \(V^{(B)}_0=120{,}000\), \(\mu_B=0.1\) and \(\sigma^2_B=0.1\);

  1. Using the Merton model, compute the default probability of either company.

    Hint: don’t forget about the R function pnorm() from Section 1.4.4!

  2. You should have found that the default probability for company A is larger than for company B. How could you have guessed this already just from the information we are given?

Listing 9.1: You can do any coding for this question here

Just plug things into Equation 9.12. Life is simple like that sometimes! :).

For part i, we simply need to evaluate Equation 9.12, with \(F=100{,}000\), \(T=3\) and \(\mu=\sigma^2=0.1\) (for both companies). We can for instance do that in R, as in Listing 9.2 below.

For part ii, the only difference between the parameters for both companies is that the firm value for company A is lower than for company B. That is, company A starts at a lower value, while the random dynamics after time \(0\) is the same for both. So we should that the probability that A ends up with a firm value at time \(T\) that is smaller than \(F\) is larger than the probability that A ends up with a firm value at time \(T\) that is smaller than \(F\), i.e. the default prob is larger for company A (cf. Equation 9.12).

Listing 9.2: Solution code – click “Run Code” to see the ouput