Credit migration: Generating generators

Credit markets face uncertain times, and the current upset caused by the coronavirus pandemic is likely to result in deterioration of the credit fundamentals for many issuers. Although there has been a huge rally since the end of the Global Financial Crisis, which now seems a distant memory, there have been upsets in the Eurozone (2011), and in commodity and energy markets (2014–16), while in emerging markets there have been downgrades in the sovereigns of Brazil, Russia and Turkey, and even during the benign year of 2019, the great majority of LatAm corporate names on rating-watch were potential downgrades rather than upgrades²²2See e.g. Credit Suisse Latin America Chartbook.. It is worthwhile, we think, to revisit the subject of credit migration.

A standard modelling framework, introduced by Jarrow et al. [Jarrow97] over twenty years ago, uses a Markov chain, parametrised by a generator matrix, to represent the evolution of the credit rating of an issuer. A good overview of subsequent developments, including its extension to multivariate settings, is given by [Bielecki11]; in terms of how it relates to the credit markets and default history, an excellent series of annual reports is produced by Standard & Poors, of which the most recent is [SandP19]. In the taxonomy of credit models, the ratings-based Markov chain model is best understood as a sort of reduction of the credit barrier models, in the sense that the firm value, or distance to default, is replaced by the credit rating. This reduces the dimensionality from a continuum of possible default distances to a small number of rating states, a natural idea as credit rating is a convenient way of thinking about a performing issuer.

The Markov chain model can be used in a number of ways, making it of general interest:

• •

  Econometrically, to tie potential credit losses to econometric indicators and thence to bank capital requirements [Fei13];

• •

  Risk management of vanilla and exotic credit portfolios;

• •

  Valuation of structured credit products. It is worth bearing in mind that although the synthetic CDO market is a fraction of its former self, the CLO market is still thriving, with new CLOs being issued frequently. Further, the ‘reg-cap’ sector (trades providing regulatory capital relief for banks) is growing, and requires the pricing and risk management of contingent credit products;

• •

  Understanding the impact of rating-dependent collateral calls in XVA;

• •

  Stripping the credit curve from bond prices, as a way of constructing sensibly-shaped yield curves for different ratings and tenors.

The model can, therefore, be used in both subjective (historical) and market-implied (risk-neutral) modelling, and we shall consider both here. Nevertheless there are several difficulties that have not been discussed in the literature, detracting from the utility of the model and limiting its application hitherto:

• (I)

  The number of free parameters is uncomfortably high (with $n$ states plus default it is $n^{2}$);

• (II)

  Computation of transition probabilities requires the generator matrix to be exponentiated, but this is a trappy subject [Moler03];

• (III)

  There are algebraic hurdles associated with trying to make a generator matrix time-varying.

Additionally the following fundamental matters are insufficiently understood:

• (IV)

  What is the best way to calibrate to an empirical (historical) transition matrix, and what level of accuracy may reasonably be asked for?

• (V)

  Model calibration is one of the most important disciplines in quantitative finance, and yet no serious attempt has been made to answer the following: Can we uniquely identify the risk-neutral generator from the term structure of credit spreads for each rating? If not, what extra information is needed?

We address all of these here, solving most of the above problems using a new parametrisation in which a tridiagonal generator matrix is coupled with a Lévy stochastic time change (we call this TDST).

We consider an $n$-states-plus-default continuous-time Markov chain and write $G\in\mathbb{R}^{n+1,n+1}$ for the generator matrix with the $(n+1)$th state denoting default (we will write this state as $\maltese$). The off-diagonal elements of $G$ must be nonnegative and the rows sum to zero; the last row is all zero because default is an absorbing state. The probability of default in time $\tau$, conditional on starting at state $i$, is the $i$th element of the last column of the matrix $\exp(G\tau)$: in other words $[\exp(G\tau)]_{i\maltese}$.

Let us begin by writing the generator matrix $G\in\mathbb{R}^{n+1,n+1}$ in the form

where $H\in\mathbb{R}^{n,n}$, and $v\in\mathbb{R}^{n}$ is such that the rows of $G$ sum to zero; the offdiagonal elements of $H$, and those of $v$, must of course be nonnegative. The $\tau$-period transition matrix is given by

where the function $\mathcal{E}_{1}$ is defined by $\mathcal{E}_{1}(x)=(e^{x}-1)/x$. As is apparent, we have used a matrix exponential, and also the function $\mathcal{E}_{1}$, and it is worth making some general remarks about analytic functions of matrices.

Let $f$ be a function analytic on a domain $\mathcal{D}\subset\mathbb{C}$. Let $B(z_{0},\rho)$ denote the open disc of centre $z_{0}$ and radius $\rho$, and assume that this lies inside $\mathcal{D}$. Then $f$ possesses a Taylor series $\sum_{k=0}^{\infty}f_{k}(z-z_{0})^{k}$ convergent for $z\in B(z_{0},\rho)$. It is immediate, proven for example by reduction to Jordan normal form [Cohn84], that the expansion

enjoys the same convergence provided that all the eigenvalues of $A$ lie in $B(z_{0},\rho)$. We can therefore write $f(A)$ without ambiguity, and will use this notation henceforth. This construction establishes the existence of the matrix exponential ($f_{k}=1/k!$ and $\rho=\infty$), though it does not at all make it a good computational method, because roundoff errors can be substantial³³3For example in computing $e^{-z}$ at $z=20$: the roundoff error is far in excess of the correct answer.. Taylor series expansion is one of the many bad ways of calculating the matrix exponential [Higham08, Moler03]—a point that is routinely ignored. A better idea is

and in each case computation is easiest when $n$ is a power of two because it can be accomplished by repeated squaring. The second expression is slightly harder to calculate as it requires a matrix inverse, but it has an advantage: for $n\in\mathbb{N}$, $(I-A/n)^{-n}$ is a valid transition matrix whereas $(I+A/n)^{n}$ may not be.

Having discussed the matrix exponential, we also have something to say about its inverse. It is in principle possible to write down a $\tau$-period (commonly one year) transition matrix $M$ and ask if it comes from a generator $G$, which amounts to finding the matrix logarithm of $M$. If such a generator exists then $M$ can be raised to any real positive power, and is sometimes said to be embeddable⁴⁴4‘Embedding’: there exists a smooth map $\mathfrak{m}$ from $\mathbb{R}_{\geq 0}$ into the space of valid transition matrices, obeying $\mathfrak{m}(s+t)=\mathfrak{m}(s)\mathfrak{m}(t)$. It is, of course, $\mathfrak{m}(t)=M^{t}$.. Not all valid transition matrices are embeddable. By identifying a generator at the outset we avoid this problem.

Another important thread is the use of diagonalisation, implicit in the above discussion because we have already mentioned eigenvalues. If $A=PEP^{-1}$ for some invertible $P$ and diagonal $E$ then $f(A)=Pf(E)P^{-1}$. However, while a real matrix is generically⁵⁵5This means that if it isn’t, it is arbitrarily close to one that is. diagonalisable, it may be that $P$ is almost singular. There is also the minor irritation that $P,E$ may not be real. This method is another popular way of exponentiating matrices, but again one that is only reliable in certain contexts, for example symmetric matrices: more of this presently.

A simplification of the model is to assume that $H$ is tridiagonal: that is to say the only nonzero elements are those immediately above, below, or on the leading diagonal. We define a transitive generator to be one in which all the super- and sub-diagonal elements of $H$ are strictly positive. An intransitive generator does not permit certain transitions to neighbouring states to occur: informally this causes the rating to ‘get stuck’ and can be ruled out on fundamental grounds, though it does retain some interest as a limiting case. Also, we define a restricted generator to be one in which all the rows of $H$, except for the $n$th, sum to zero: equivalently, $v$ is zero except for its bottom element. Both these simplified models have a clear connection to credit modelling as discretisations of the structural (Merton) model, in which the firm value is replaced by the credit rating as a distance-to-default measure. A restricted tridiagonal generator corresponds to a Brownian motion hitting a barrier, while an unrestricted tridiagonal generator has an additional dynamic of the firm suddenly jumping to default. The transitivity condition simply ensures that the volatility always exceeds zero.

It is clear that the space of restricted tridiagonal generators has $2n-1$ degrees of freedom, and the unrestricted one $3n-2$. These are substantially lower than the $n^{2}$ needed to define a general model. Also, tridiagonality is useful not just for dimensionality reduction. A transitive generator can always be diagonalised, and has real eigenvalues. This is because $H=D\widetilde{H}D^{-1}$ with $D$ a positive diagonal matrix and $\widetilde{H}$ a symmetric tridiagonal matrix; then $\widetilde{H}$ can be diagonalised via an orthogonal change of basis as $\widetilde{H}=QEQ^{\prime}$ with ^′ denoting transpose. (The method is particularly fast for tridiagonal matrices: see e.g. [NRC, §11.3].) Thus

The right-hand column of the matrix, written as “$\ast$”, need not be computed directly because it can be obtained from the rows summing to unity. Note that the eigenvalues of $H$, i.e. the (diagonal) elements of $E$, are real and negative. This makes computation of the matrix exponential straightforward⁶⁶6Technically, the limit of an intransitive generator will cause $D$ to become singular, and so the construction then fails, but this seems to be a theoretical issue only.. As an aside we note the connection between tridiagonal matrices, orthogonal polynomials and Riemann-Hilbert problems [Deift00, Ch.2].

The problem with tridiagonal generators is that they do not permit multiple downgrades over an infinitesimal time period, and so the probability of such events over short (non-infinitesimal) periods, while positive, is not high enough. In this respect, the presence of a jump-to-default transition achieves little, because that is not the main route to default: far more likely is the occurrence of multiple downgrades, often by several notches at a time. So how do we make these sudden multiple downgrades occur?

To retain the structure above we employ a stochastic time change, replacing $\tau$ in the above equation by $\tau_{t}$ with $t$ denoting ‘real’ time and $\tau_{t}$ ‘business’ time. We are going to make $\tau_{t}$ a pure jump process, with the idea that business time occasionally takes a big leap, causing multiple transitions, and we call this model TDST (tridiagonal with stochastic time change). Formally the process $(\tau_{t})$ is to be a monotone-increasing Lévy process⁷⁷7See e.g. [Schoutens03] for a general introduction. described by the generator $\varphi$, i.e.:

The $T$-period transition matrix is obtained by integrating over all paths $\{\tau_{t}\}_{t=0}^{T}$. As the dependence on $\tau$ is through an exponential, this is straightforward, and the effect is to replace $H$ by $\varphi(H)$ in (1), adjusting the last column as necessary. (In context $\varphi$ will be regular in the left half-plane, so we can legitimately write the matrix function $\varphi(H)$.) Accordingly the generator and $T$-period transition matrix are

We can impose that $\tau_{t}/t$ have unit mean (as otherwise the model is over-specified because $H$ and $\tau$ can be scaled in opposite directions), so $\varphi^{\prime}(0)=1$. An obvious choice is the CMY process

which has as special cases the Inverse Gaussian process and the Gamma process, respectively ($\gamma=\frac{1}{2},0$):

This retains the connection with structural (Merton) modelling, in that we have a discretisation of a Lévy process hitting a barrier—a standard idea in credit risk modelling [Lipton02c, Martin09a, Martin10b, Dalessandro11]. An incidental remark about the Gamma process is that it allows the matrix exponential $\varphi(H)$ to be calculated in an alternative simple way, as noted in the Appendix.

It is perhaps worth emphasising that despite our using the term ‘stochastic time change’ this model specification still has a static generator. The stochastic time change serves only as a mechanism for generating multiple downgrades in an infinitesimal time period using a tridiagonal generator matrix. The number of parameters for this model is $2n-1+n_{\varphi}$, where $n_{\varphi}$ ($=2$ here) is the number needed to specify $\varphi$.

Israel et al. [Israel00] spend some time showing that empirical transition matrices are not necessarily embeddable (q.v.). In fact it is only necessary to test whether the empirical matrix is statistically likely to be so. Taking S&P’s data in [SandP19, Tables 21,23], we ask if we can find a TDST model that fits well enough.

There are a number of issues to consider. The first is the right measure of closeness of fit. From the perspective of statistical theory the most appropriate is the Kullback–Leibler divergence⁸⁸8A Taylor series expansion around $p=q$ gives the well-known chi-squared measure of “$(O-E)^{2}/E$, summed” ($O$ observed, $E$ expected)—this can also be used to define the fitting error, and it gives similar results.:

where $p_{ij}$ are the historical probabilities of transition from state $i$ to $j$ (with $n+1$ denoting default) and $q_{ij}$ the model ones. This is essentially the log likelihood ratio statistic, i.e. the logarithm of the ratio of the likelihood of the model compared with that of the maximum likelihood estimator [Cox74, Ch.4]. Note that $\mathsf{d}_{M}$ is not symmetrical in $p,q$: nor should it be. To see why, consider first the effect of letting $q_{ij}\to 0$ with $p_{ij}>0$: this means that the event has been observed but that the model probability is zero, an error that should be heavily penalised. On the other hand, $p_{ij}\to 0$ with $q_{ij}>0$ simply means that the model is attaching positive probability to an event that has not been observed, which is not a major objection.

Empirical transition data reveal a problem: the probability of transiting from rated to unrated (‘NR’) is quite high. A standard idea is to distribute this probability pro rata amongst the elements of each row, excepting the transition to default: doing so retains, as it must, the property that each row sum to unity. However, this is not the only way of making the adjustment, and although the ‘NR’ problem is unfortunate, it allows for, or cannot rule out, considerable leeway in the fitting—which is why attempts to exactly fit the matrix are misguided. Another issue is that the standard deviation of low empirical probabilities is proportionally quite high. The finiteness of the samples used to construct empirical transition matrices introduces considerable relative uncertainties.

We then minimise $\mathsf{d}_{M}$ with respect to the elements of $H$ (and the bottom element of $v$) and the parameters $\beta,\gamma$ in (4). Optimisation of the 7-state model takes a fraction of a second and the results are shown in Figure 2. It is seen that the fit is good, and furthermore the fitting errors are small by comparison with the uncertainties introduced by the ‘NR’ problem. The errors are also typically smaller than the standard errors given by S&P, which are in [SandP19, Table 21] but omitted here⁹⁹9One has to be careful in interpreting these, for a reason that we have already touched on. Certain transitions have never been observed, and for these S&P show the probability and standard error as zero. Now if the true frequency is zero, then the mean and standard deviation of the observed frequency will be zero. But what we really want is an estimate of the true frequency and the standard deviation of that estimate. Clearly if Bayesian inference is followed then the latter is not zero, because the true frequency may be $>0$..

If we pass to an 18-state model (AAA, AA+, AA, …, CCC+, CCC) we still have a large number of parameters even after the reduction to a TDST model (324 to around 36). It is possible to fit these, but some further simplification proves advantageous. Suppose that the free parameters in the matrix $H$ are the sub/superdiagonal elements for the following transitions only: $\mbox{AA}\to\mbox{AA}\pm$, $\mbox{A}\to\mbox{A}\pm$, $\mbox{BBB}\to\mbox{BBB}\pm$, $\mbox{BB}\to\mbox{BB}\pm$, $\mbox{B}\to\mbox{B}\pm$, $\mbox{CCC}\to\mbox{CCC}+$, and also the bottom element of $v$, giving $\mbox{CCC}\to\maltese$. (Hence there are 12 in all.) The values for intermediate states ($\mbox{AA}-\to\mbox{AA}/\mbox{A}+$ and so on) are found by logarithmic interpolation¹⁰¹⁰10For convenience $\mbox{AA+}\to\mbox{AAA}$ is taken to be the same as $\mbox{AA}\to\mbox{AA}+$, and $\mbox{AAA}\to\mbox{AA+}$ and $\mbox{AA}+\to\mbox{AA}$ the same as $\mbox{AA}\to\mbox{AA}-$.. A minor complication is that the S&P data group CCC+, CCC and lower ratings into one. We have ignored any rating lower than CCC because in practice transition through these states is very rapid, and default almost always ensues [SandP19, Chart 10]. Again the results are good: see Figure 2.