Credit risk and contagion via self-exciting default intensity

Abstract

Recent empirical evidences indicate that default rates are influenced not only by the observable or latent risk factors, but also depend on the history of past defaults. Motivated by this empirical finding, we consider in this paper a reduced-form, intensity-based credit risk model, which allows for both frailty and default contagion, using a so-called “self-exciting” intensity, in the sense that the default intensity varies not only with the risk factors, but also depends on the previous default history of all the firms. With “self-exciting” default intensity, we are able to obtain closed-form expressions for the pricing of credit derivative securities in our model. The estimation of parameters using the EM algorithm is considered as well.

Appendices

Appendix 1

Proof of Theorem 1

We have

$$\begin{aligned} \varLambda _t&=1+\sum _{i=1}^N\int _0^t\varLambda ^i_{u-}\big (\lambda ^i_u-1 \big )\big (dZ^i_u-{\mathbf {1}}_{\{u\le \tau _i\}}du\big ). \end{aligned}$$

Then

$$\begin{aligned} \varLambda _tX_t&=X_0+\sum _{i=0}^N\int _0^t\varLambda _{u-}X_{u-} \big (\lambda ^i_u-1\big )\big (dZ^i_u-{\mathbf {1}}_{\{u\le \tau _i\}}du \big )\\&\quad +\int _0^t\varLambda _{u-}A_{u}X_{u}du+\int _0^t\varLambda _{u-}dM_u. \end{aligned}$$

As $\varLambda$ has independent increments under , taking a conditional expectation,

As $\lambda ^i$ is a scalar function of the state $X_{u-}$, we can write

$$\begin{aligned} X_{u-}\big (\lambda ^i_u-1\big ) =\big (\lambda ^i \big (u,X_{u-},N_{(\cdot )}\big )-1\big )X_{u-} ={{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big ), \end{aligned}$$

where ${{\mathrm{diag}}}(\lambda ^{(i)}-1)$ is the diagonal matrix with $(k,k)$-entry $\lambda ^i_{k,u}=\lambda ^i(u,e_k,N_{(\cdot )})$. This implies that ${{\mathrm{diag}}}(\lambda ^{(i)}-1)$ is ${\mathcal {F}}^Z_u$-measurable. Substituting this in the equation for $q$, we have

$$\begin{aligned} q_t&=q_0+\sum _{i=0}^N\int _0^t {{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big )q_{u-}\big (dZ^i_u -{\mathbf {1}}_{\{u\le \tau _i\}}du\big )+\int _0^tA_uq_udu\\&=q_0+\sum _{i=0}^N\int _0^t {{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big )q_{u-}dZ^i_u\\&\quad -\sum _{i=0}^N\int _0^{t\wedge \tau _i} {{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big )q_{u-}du +\int _0^tA_uq_udu. \end{aligned}$$

$\square$

Proof of Theorem 2

The proof is similar to that of Theorem 1. We start with $\overline{{\mathbb {P}}}$. Using Girsanov’s Theorem for jump processes, define a process $\varLambda _t$ by

$$\begin{aligned} \varLambda _t&=\prod _{i=1}^N\exp \left( -\int _0^t\int _EX_{u-} \big (h^i_u(y)-1\big ){\mathbf {1}}_{\{u\le \tau _i\}}f(y)dudy\right. \nonumber \\&\left. \quad +\int _0^t\int _EX_{u-}\log h^i_u(y)\mu ^i(du,dy)\right) \nonumber \\&=\exp \left( -\sum _{i=1}^N\int _0^t\int _EX_{u-}\big (h^i_u(y)-1 \big ){\mathbf {1}}_{\{u\le \tau _i\}}f(y)dudy\right. \nonumber \\&\left. \quad +\sum _{i=1}^N\int _0^t\int _EX_{u-}\log h^i_u(y)\mu ^i(du,dy)\right) . \end{aligned}$$

(2)

Using Itô’s rule, we also have

$$\begin{aligned} \varLambda _t=1+\sum _{i=1}^N\int _0^t\int _E\varLambda _{u-}X_{u-} \big (h^i_u(y)-1\big )\big (\mu ^i(dy,du) -{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydu\big ), \end{aligned}$$

or

$$\begin{aligned} d(\varLambda _t)=\varLambda _{t-}\sum _{i=1}^N\int _EX_{t-} \big (h^i_u(y)-1\big )\big (\mu ^i(dy,dt) -{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydt\big ). \end{aligned}$$

Since

$$\begin{aligned} d\left( \varLambda _tX_t \right)&=d\left( \varLambda _t\right) X_t +\varLambda _td(X_t)\\&=\varLambda _{t-}\left( \sum _{i=1}^N\int _EX_{t-} \big (h^i_u(y)-1\big )\big (\mu ^i(dy,dt\big ) -{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydt)\right) X_{t-}\\&\quad +\varLambda _{t-}\left( AX_t+dM_t\right) , \end{aligned}$$

we have

$$\begin{aligned} \varLambda _tX_t&=\varLambda _0X_0+\sum _{i=1}^N\int _0^t \varLambda _{u-}X_{u-}\int _EX_{u-}\big (h^i_u(y)-1\big )\\&\quad \times \big (\mu ^i(dy,du)-{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydu\big )\\&\quad +\int _0^t\varLambda _uAX_udu+\int _0^t\varLambda _{u-}dM_u. \end{aligned}$$

Write $q_t:=\overline{{\mathbb {E}}}[\varLambda _tX_t|{\mathcal {F}}_t^M]$. Then, a version of Bayes’ Theorem gives

$$\begin{aligned} {\mathbb {E}}\left[ X_t|{\mathcal {F}}_t^M\right] =\frac{\overline{{\mathbb {E}}} \left[ \varLambda _tX_t|{\mathcal {F}}_t^M\right] }{\overline{{\mathbb {E}}} \left[ \varLambda _t|{\mathcal {F}}_t^M\right] } =\frac{q_t}{\langle q_t, {\mathbf {1}}\rangle }. \end{aligned}$$

Conditioning on ${\mathcal {F}}_t^Z$ under $\overline{{\mathbb {P}}}$, using the Fubini Theorem in Hajek and Wong (1985), we have

$$\begin{aligned} q_t&=q_0+\int _0^tAq_udu+\sum _{i=1}^N\int _0^t\int _E\sum _{k=1}^K\langle q_{u-},e_k\rangle \big (h^i_k(y)-1\big )\\&\quad \times \big (\mu ^i(dy,du) -{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydu\big )e_k \end{aligned}$$

Write $H^i_k(u):=h^i_k(y_u)$. Then

$$\begin{aligned} q_t&=q_0+\int _0^tAq_udu+\sum _{i=1}^N\int _0^t\int _E\sum _{k=1}^K \langle q_{u-},e_k\rangle \big (h^i_k(y)-1\big )\\&\quad \times \big (\mu ^i(dy,du)-{\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydu\big )e_k\\&=q_0+\!\int _0^tAq_udu\!+\!\sum _{i=1}^N\int _0^t\int _E{{\mathrm{diag}}}\big (h^i_1(y)-1,\ldots ,h^i_K(y)-1\big )q_{u-}\mu ^i(dy,du)\\&\quad -\sum _{i=1}^N\int _0^t\int _E{{\mathrm{diag}}}\big (h^i_1(y)-1,\ldots ,h^i_K(y)-1\big ) q_{u-}\left( {\mathbf {1}}_{\{u\le \tau _i\}}f(y)dydu\right) \\&=q_0+\int _0^tAq_udu+\sum _{i=1}^N\int _0^t{{\mathrm{diag}}}\big (H^i_1(u)-1,\ldots ,H^i_K(u)-1\big )q_{u-}\\&\quad \times \int _E\mu ^i(dy,du)-\sum _{i=1}^N\int _0^t{{\mathrm{diag}}}\big (\lambda ^i_1-1,\ldots ,\lambda ^i_K-1 \big ){\mathbf {1}}_{\{u\le \tau _i\}}q_udu\\&=q_0+\int _0^tAq_udu+\sum _{i=1}^N\int _0^t{{\mathrm{diag}}}\big (H^i_1(u)-1,\ldots ,H^i_K(u)-1\big )q_{u-}dN^i_u\\&\quad -\sum _{i=1}^N\int _0^{t\wedge \tau _i}{{\mathrm{diag}}}\big (\lambda ^i_1-1,\ldots ,\lambda ^i_K-1\big )q_udu. \end{aligned}$$

$\square$

Proof of Lemma 2

By Itô’s product rule,

$$\begin{aligned} d(\gamma ^{-1}_k(t))=d(\gamma ^{-1}_k(t))&=-\gamma ^{-1}_k(t-)\sum _{i=1}^N\Big (\left( 1-\lambda ^i_{k,t}\right) {\mathbf {1}}_{\{t\le \tau _i\}}dt+\log (H^i_{k,t})\varDelta N^i_t\Big )\\&\quad +\gamma ^{-1}_k(t-)\sum _{i=1}^N\left( \left( \frac{1}{H^i_{k,t}}-1\right) \varDelta N^i_t+\log (H^i_{k,t})\varDelta N^i_t\right) \\&=-\gamma ^{-1}_k(t-)\sum _{i=1}^N\Big (\left( 1-\lambda ^i_{k,t}\right) {\mathbf {1}}_{\{t\le \tau _i\}}dt+\log (H^i_{k,t})\varDelta N^i_t\Big )\\&\quad +\gamma ^{-1}_k(t-)\sum _{i=1}^N\left( \left( \frac{1-H^i_{k,t}}{H^i_{k,t}}\right) \varDelta N^i_t+\log (H^i_{k,t})\varDelta N^i_t\right) \\&=\gamma ^{-1}_k(t-)\sum _{i=1}^N\left( 1-\lambda ^i_{k,t}\right) {\mathbf {1}}_{\{t\le \tau _i\}}dt\\&\quad +\gamma ^{-1}_k(t-)\sum _{i=1}^N\left( \frac{1-H^i_{k,t}}{H^i_{k,t}}\right) dN^i_t. \end{aligned}$$

Thus

$$\begin{aligned} d\big (\varGamma ^{-1}_t\big )=\sum _{i=1}^N{\mathbf {1}}_{\{t\le \tau _i\}}{{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big )\varGamma ^{-1}_tdt +\sum _{i=1}^N{{\mathrm{diag}}}\left( \frac{1-H^{(i)}}{H^{(i)}}\right) \varGamma ^{-1}_{t-}dN^i_t. \end{aligned}$$

$\square$

Proof of Theorem 3

By Itô’s product rule,

$$\begin{aligned} d(\bar{q}_t)&=d\left( \varGamma _t^{-1}q_t\right) \\&=\varGamma _t^{-1}d(q_t)+d\left( \varGamma _t^{-1}\right) q_t+d\left[ \varGamma ^{-1},q\right] _t\\&=\varGamma _t^{-1}\left( A_tq_tdt+\sum _{i=0}^N{{\mathrm{diag}}}(H^i_1(t)-1,\ldots ,H^i_K(t)-1)q_{t-}dN^i_t\right. \\&\left. \quad -\sum _{i=0}^N{\mathbf {1}}_{t\wedge \tau _i}{{\mathrm{diag}}}(\lambda ^{(i)}-1)q_{t-}dt\right) \\&\quad +\left( \sum _{i=1}^N{\mathbf {1}}_{\{t\le \tau _i\}}{{\mathrm{diag}}}\big (\lambda ^{(i)}-1\big )\varGamma ^{-1}_tdt +\sum _{i=1}^N{{\mathrm{diag}}}\left( \frac{1}{H^{(i)}}-1\right) \varGamma ^{-1}_{t-}dN^i_t\right) q_t\\&\quad +\sum _{i=1}^N{{\mathrm{diag}}}\left( \frac{1}{H^{(i)}}-1\right) \varGamma ^{-1}_{t-}{{\mathrm{diag}}}(H^{(i)}-1)q_{t-}dN^i_t\\&=\varGamma _t^{-1}A_tq_tdt+\sum _{i=1}^N\left( {{\mathrm{diag}}}(H^{(i)}-1)+{{\mathrm{diag}}}\left( \frac{1}{H^{(i)}}-1\right) \right. \\&\quad \left. +{{\mathrm{diag}}}\left( \frac{1}{H^{(i)}}-1\right) {{\mathrm{diag}}}(H^{(i)}-1)\right) \varGamma _t^{-1}q_{t-}dN^i_t\\&=\varGamma _t^{-1}A_tq_tdt. \end{aligned}$$

Thus

$$\begin{aligned} \bar{q}_t&=\bar{q}_0+\int _0^t\varGamma _u^{-1}A_uq_udu\\&=\bar{q}_0+\int _0^t\varGamma _u^{-1}A_u\varGamma _u\bar{q}_udu. \end{aligned}$$

$\square$

Appendix 2

In this appendix, we give a closed-form expression for the survival probability in our model with self-exciting intensities. We show that, if we exclude some uninteresting technical cases, a closed-form expression for the survival probability can be obtained. In our model, $\lambda ^i(t,X_{t-},N_{(\cdot )})$ depends on $X_{t-}$, and $N^j_{[0,t-]}$, for all $j\ne i$. We further assume the following, to exclude some uninteresting cases.

Assumption

$\lambda ^i(t,X_{t-},N_{(\cdot )})$ does not depend on $N^i_{[0,t-]}$.⁵

We make this assumption as we use $\lambda ^i(t,X_{t-},N_{(\cdot )})$ to model the contagion and frailty effects from other firms, so $\lambda ^i(t,X_{t-},N_{(\cdot )})$ does not depend on its own default history. Also, when $N^i_{t-}=0$, $N^i_{[0,t-]}$ does not influence $\lambda ^i(t,X_{t-},N_{(\cdot )})$, and once $N^i_{\tau _i}=1$ for $\tau _i<t$, firm $i$ has already defaulted. Thus, it is sensible to assume that $\lambda ^i(t,X_{t-},N_{(\cdot )})$ does not depend on $N^i_{[0,t-]}$.

With this assumption, we now prove the survival probability. By definition of stochastic intensity (see, e.g., Duffie and Singleton 2003 p. 361; Bielecki and Rutkowski 2002 p. 155), the compensator of $N^i_t$ is $\int _0^{t\wedge \tau }\lambda ^i(u,X_{u-},N_{(\cdot )})du$, i.e.,

$$\begin{aligned} N^i_t-\int _0^{t\wedge \tau _i}\lambda ^i \left( u,X_{u-},N_{(\cdot )}\right) du \end{aligned}$$

(3)

is an ${\mathcal {F}}$-martingale under measure ${\mathbb {P}}$. We need to find the survival probability ${\mathbb {P}}(\tau _i\ge t|{\mathcal {F}}_t)$.

The filtration $\{{\mathcal {F}}^{X}_{t},t\ge 0\}$ is defined by ${\mathcal {F}}^{X}_{t}:{=}\sigma \{X_{s}:s\le t\}$. Similarly, we write ${\mathcal {F}}^{Z}_{t}=\sigma (Z_{s};0\le s\le t)$. Note that in our setting, ${\mathcal {F}}_{t}=\sigma (X_{s},Z_{s};0\le s\le t)$.

We have

$$\begin{aligned} {\mathbb {E}}[N^i_t|{\mathcal {F}}_s]&={\mathbb {E}}\left[ I_{\{t\ge \tau _i\}}\Big |{\mathcal {F}}_s\right] \nonumber \\&={\mathbb {E}}\left[ \int _0^{t\wedge \tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big |{\mathcal {F}}_s\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text {(by (1))}\nonumber \\&={\mathbb {E}}\left[ I_{\{t\le \tau _i\}}\!\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du \!+\!I_{\{t\ge \tau _i\}}\!\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big |{\mathcal {F}}_s\right] \nonumber \\&={\mathbb {E}}\left[ I_{\{t\le \tau _i\}}\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big |{\mathcal {F}}_s\right] \nonumber \\&\quad +{\mathbb {E}}\left[ I_{\{t\ge \tau _i\}}\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big |{\mathcal {F}}_s\right] . \end{aligned}$$

(4)

Now we calculate the first conditional expectation in equation (4). Since $\lambda ^i(u,X_{u-},N_{(\cdot )})$ does not depend on $N^i$ (or $\tau _i$) conditional on $\tau _i\ge t$, we have

$$\begin{aligned} {\mathbb {E}}\left[ I_{\{t\le \tau _i\}}\!\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) \!du\Big |{\mathcal {F}}_s\right] \!&=\!\int _t^\infty \!\!I_{\{t\le \tau _i\}}\!\int _0^{t}\!\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,d{\mathbb {P}}(\tau _i|{\mathcal {F}}_s)\\&=\int _t^\infty \int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,d{\mathbb {P}}(\tau _i|{\mathcal {F}}_s)\\&=\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,\int _t^\infty d{\mathbb {P}}(\tau _i|{\mathcal {F}}_s)\\&=\!\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big (1\!-\!{\mathbb {P}}(\tau _i< t|{\mathcal {F}}_s)\Big ). \end{aligned}$$

Now we calculate the conditional expectation in the second term in Eq. (4). For $\tau _i\le t$, $\int _0^{\tau _i}\lambda ^i(u,X_{u-},N_{(\cdot )})du$ depends on $X_{u-}$, $N^j_{[0,u]}$, $j\ne i$, and $\tau _i$. Then

$$\begin{aligned}&{\mathbb {E}}\left[ I_{\{t\ge \tau _i\}}\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\Big |{\mathcal {F}}_s\right] \\&\quad =\int _0^t I_{\{t\ge \tau _i\}}\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,d{\mathbb {P}}(\tau _i|{\mathcal {F}}_s)\\&\quad =\int _0^t\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,d{\mathbb {P}}(s<\tau _i|{\mathcal {F}}_s).\\ \end{aligned}$$

Thus from Eq. (4) we have

$$\begin{aligned} {\mathbb {P}}(t|{\mathcal {F}}_s)&= \int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\big (1-{\mathbb {P}}(t|{\mathcal {F}}_s)\big )\nonumber \\&\quad + \int _0^t\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,d{\mathbb {P}}(\tau _i|{\mathcal {F}}_s) \end{aligned}$$

(5)

Denoting $p(t)={\mathbb {P}}(t|{\mathcal {F}}_s)={\mathbb {P}}(s<t|{\mathcal {F}}_s)$, we have from Eq. (5)

$$\begin{aligned} p(t)= \int _0^{t}\lambda ^i\big (u,X_{u-},N_{(\cdot )}\big )du\,\big (1-p(t)\big )+ \int _0^t\int _0^{\tau _i}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,dp(\tau _i). \end{aligned}$$

Differentiating the above equation, we have

$$\begin{aligned} dp(t)&=\lambda ^i_tdt\,\big (1-p(t)\big )-\int _0^{t}\lambda ^i\left( u,X_{u-},N_{(\cdot )}\right) du\,dp(t)\\&\quad +\int _0^{t}\lambda ^i\big (u,X_{u-},N_{(\cdot )}\big )du\,dp(t)\\&=\lambda ^i_tdt\,\big (1-p(t)\big ). \end{aligned}$$

The above differential equation is equivalent to

$$\begin{aligned} d\big (1-p(t)\big )=-\lambda ^i_tdt\,\big (1-p(t)\big ), \end{aligned}$$

which yields

$$\begin{aligned} 1-p(t)=\exp \left( -\int _0^t\lambda ^i_udu\right) . \end{aligned}$$

Thus

$$\begin{aligned} {\mathbb {P}}\left( \tau _i>t\big |{\mathcal {F}}_s\right) =1-p(t)= \exp \left( -\int _0^t\lambda ^i_udu\right) . \end{aligned}$$

This proves the desired probability. $\square$