Two-name credit

Two companies whose default intensities share a hidden credit cycle, and the default correlation the cycle creates.

The model

Two companies have default intensities $x_1$ and $x_2$. Each follows its own square-root (CIR) process, driven by independent Brownian motions, but both take their long-run level from one hidden regime $y_t \in \{1, 2\}$, a credit cycle:

$$dx_{j} = \kappa\,\big(\theta_{j,\,y_t} - x_j\big)\,dt + \sigma\sqrt{x_j}\,dW_j, \qquad j = 1, 2 .$$

Regime 1 is a crisis and regime 2 is normal times. The cycle switches at rate $\lambda$ in each direction.

Each company defaults when its accumulated intensity first reaches an exponential clock:

$$\tau_j = \inf\Big\{t : \int_0^t x_j(s)\,ds \ge E_j\Big\}, \qquad j = 1, 2,$$

where $E_1$ and $E_2$ are independent unit exponentials, independent of the Brownian motions and of the cycle. Given the intensity paths the two defaults are then independent, so the shared cycle is the only source of default dependence.

The example assumes that the cycle starts from its stationary law, $(\tfrac12, \tfrac12)$, independent of $x_0$. The effect of another prior is shown below.

There are two time scales: each intensity reverts over $1/\kappa$, and the cycle turns every $1/\lambda$.

Parameters: $\kappa = 2$, $\sigma = 0.2$; the first company has $\theta_1 = (0.8,\ 0.005)$, the second $\theta_2 = (0.6,\ 0.005)$; $x_0 = (0.05,\ 0.05)$; horizon $T = 3$.

A CIR intensity stays strictly positive when

$$2\kappa\theta \ge \sigma^2,$$

the Feller condition.

Here $\sigma^2 = 0.04$, while $2\kappa\theta$ is 3.2 and 2.4 in the crisis regime and 0.02 for both companies in normal times. The condition fails in normal times, where an intensity can touch zero and the hazard of that company vanishes for a moment. Zero is reflecting, so the intensities stay nonnegative and are valid default intensities. The survival formulas below use only the Laplace transform of the integrated CIR process, which holds whether or not the condition is met.

The quantity

With the exponential clocks, the probability that every company in $c$ survives to $T$ is

$$S_c(T) = \mathbb{E}\Big[e^{-\int_0^T (c_1 x_1 + c_2 x_2)\,dr}\Big], \qquad c = (1,0),\ (0,1),\ (1,1).$$

The default probabilities follow from them:

$$p_1 = 1 - S_{(1,0)}, \qquad p_2 = 1 - S_{(0,1)}, \qquad p_{12} = 1 - S_{(1,0)} - S_{(0,1)} + S_{(1,1)} .$$

The correlation of the two default indicators is

$$\operatorname{corr} = \frac{p_{12} - p_1 p_2}{\sqrt{p_1(1 - p_1)\,p_2(1 - p_2)}} = \frac{S_{(1,1)} - S_{(1,0)}S_{(0,1)}}{\sqrt{p_1(1 - p_1)\,p_2(1 - p_2)}} .$$

A shared clock, $E_1 = E_2$, would keep each company's survival but replace the joint survival by the expectation of $e^{-\max(A_1, A_2)}$, with $A_j = \int_0^T x_j$. The dependence would then not vanish even in the averaged model.

Averaging

When the cycle switches infinitely fast, each intensity is a CIR process with the averaged level, and the two are independent. With independent clocks the averaged model has no default dependence at all.

Reduction to a linear system

Conditional on the regime path the two intensities are independent CIR processes. Try $u_i = e^{-B(t)(c\cdot x)}a_i(t)$. The Riccati equation for $B$,

$$B' = 1 - \kappa B - \tfrac12\sigma^2B^2, \qquad B(0) = 0,$$

involves $\kappa$ and $\sigma$, which are shared, so one $B$ serves both companies and both regimes. What remains is $a' = (Q + \operatorname{diag} g)a$ with

$$g_i(t) = -\kappa\,\big(c_1\theta_{1,i} + c_2\theta_{2,i}\big)\,B(t) .$$

This is the CIR example with the levels of the chosen companies added, and its expansion applies directly.

The survival probabilities written out

Given the regime path the two intensities are independent CIR processes, and each has the same coefficient $B$. With the stationary start the memory terms cancel, and the survival of any set of names is the CIR formula with the levels added:

$$S_c(T) = \exp\Big(-B\,(c\cdot x_0) - \kappa\,\bar\theta_c\,I_1 + \frac{\varepsilon}{2}\,\kappa^2\tilde\theta_c^2\,I_2 - \frac{\varepsilon^2}{8}\,\kappa^2\tilde\theta_c^2\,B^2\Big) + O(\varepsilon^3),$$ $$\begin{aligned} \bar\theta_c &= c_1\bar\theta_1 + c_2\bar\theta_2, \qquad \tilde\theta_c = c_1\tilde\theta_1 + c_2\tilde\theta_2, \qquad \bar\theta_j, \tilde\theta_j = \frac{\theta_{j,1} \pm \theta_{j,2}}{2}, \\ B &= \frac{2(e^{hT} - 1)}{(h + \kappa)(e^{hT} - 1) + 2h}, \qquad h = \sqrt{\kappa^2 + 2\sigma^2}, \\ I_1 &= \frac{2}{\sigma^2}\Big(\log\big((h + \kappa)(e^{hT} - 1) + 2h\big) - \log 2h - \frac{(\kappa + h)\,T}{2}\Big), \\ I_2 &= \frac{4}{h}\Big(\frac{hT}{(h - \kappa)^2} + \frac{a_2}{h + \kappa}\log\frac{(h + \kappa)e^{hT} + h - \kappa}{2h} \\ &\qquad\quad - \frac{a_3}{h + \kappa}\Big(\frac{1}{(h + \kappa)e^{hT} + h - \kappa} - \frac{1}{2h}\Big)\Big), \\ a_2 &= \frac{1}{h + \kappa} - \frac{h + \kappa}{(h - \kappa)^2}, \qquad a_3 = -2 - \frac{2(h + \kappa)}{h - \kappa} - a_2\,(h - \kappa). \end{aligned}$$

Since $\tilde\theta_{(1,1)}^2 - \tilde\theta_1^2 - \tilde\theta_2^2 = 2\tilde\theta_1\tilde\theta_2$, the joint survival exceeds the product of the marginals by

$$\log\frac{S_{(1,1)}(T)}{S_{(1,0)}(T)\,S_{(0,1)}(T)} = \varepsilon\,\kappa^2\,\tilde\theta_1\tilde\theta_2\,I_2 - \frac{\varepsilon^2}{4}\,\kappa^2\,\tilde\theta_1\tilde\theta_2\,B^2 + O(\varepsilon^3).$$

With the independent default clocks this is the whole of the dependence. The averaged model has none, and the first term is the Green–Kubo covariance of the two hazards.

At $\lambda = 2$, so that each regime lasts about half a year:

quantity	value
$B$, $I_1$, $I_2$	0.496357, 1.246232, 0.558945
$\tilde\theta_1$, $\tilde\theta_2$	0.3975, 0.2975
dependence $\log(S_{12}/S_1S_2)$, second order	0.124914
dependence, numerical	0.117901
$S_{(1,0)}$, $S_{(0,1)}$, $S_{(1,1)}$, second order	0.388841, 0.480925, 0.211884
$S_{(1,0)}$, $S_{(0,1)}$, $S_{(1,1)}$, numerical	0.387984, 0.480400, 0.209711

The starting regime

The stationary start is a choice of prior for the hidden cycle. With probability $q$ of starting in the crisis regime,

$$S_c(T) = e^{-B\,(c\cdot x_0)}\,\big(q\,a_1(T) + (1 - q)\,a_2(T)\big).$$

To first order this multiplies the stationary-start survival by

$$1 + (2q - 1)\,\frac{\varepsilon}{2}\,\tilde g_c(T), \qquad \tilde g_c = -\kappa\,\tilde\theta_c\,B .$$

If $x_0$ is an observation of a system already running, the right prior is the posterior of the regime given $x_0$, which is in general not $(\tfrac12, \tfrac12)$. The prior moves the correlation at $\lambda = 2$:

prior crisis probability $q$	default correlation, numerical
0.1	0.101692
0.5	0.095799
0.9	0.083268

Results

The correlation of the two default indicators over three years, with the stationary start, against the numerical solution. The averaged model gives exactly zero at every switching rate.

switching rate	default probabilities	numerical	Monte Carlo	order 1	order 2	order 4	order 6	first to five digits, or best
1	0.587, 0.502	0.18268	0.18213 ± 0.00032	0.26468	0.22560	0.18872	0.20072	0.17954 (order 7)
2	0.612, 0.520	0.09580	0.09582 ± 0.00023	0.10923	0.10215	0.09593	0.09601	0.09580 (order 7)
4	0.627, 0.530	0.04745	0.04748 ± 0.00011	0.04977	0.04827	0.04745	0.04745	0.04745 (order 4)
8	0.634, 0.535	0.02332	0.02336 ± 0.00006	0.02377	0.02343	0.02332	0.02332	0.02332 (order 4)

The numerical column solves the reduced linear system. The Monte Carlo column checks it with no time discretization: given a simulated regime path the level is piecewise constant, each CIR survival is exact, and only the regime path is sampled, 400,000 times. The two agree within the Monte Carlo error. Certificate: mc_credit_exact.py.

The higher orders come from the engine. With regimes lasting three months or less ($\lambda \ge 4$) the expansion reproduces the correlation to all five digits by order 4. With half-year regimes ($\lambda = 2$) it reaches all five digits at order 7. With regimes lasting a year ($\lambda = 1$) the series comes within about 0.003 before its terms start to grow.

That is the behaviour of an asymptotic series. When the switching time is not small against the time scale of the forcing, the terms first shrink and then grow, and the best answer comes from stopping at the smallest term.