Sensitivity analysis for marked Hawkes processes: application to CLO pricing

Abstract

This paper deals with a model for pricing Collateralized Loan Obligations, where the underlying credit risk is driven by a marked Hawkes process, involving both clustering effects on defaults and random recovery rates. We provide a sensitivity analysis of the CLO price with respect to the parameters of the Hawkes process using a change of probability and a variational approach. We also provide a simplified version of the model where the intensity of the Hawkes process is taken as the instantaneous default rate. In this setting, we give a moment-based formula for the expected survival probability.

Appendices

Appendices

1.1 Proof of Proposition 2

Proof

Let $\delta \in ]0, \overline{\delta }]$. The first step consists in a Taylor expansion of second order (in $\delta$) of $g(\delta ) := L[\lambda +\delta h](T_n)$. We have

$$\begin{aligned} \vert g(\delta ) - g(0) - \delta \dot{g}(0) \vert \le \frac{\delta ^2}{2}\sup _{a \in [0, \overline{\delta }]} \vert \ddot{g}(a) \vert \end{aligned}$$

With these notations, we omit the dependency in $T_1, \dots , T_n$, that will be made clear in the next equations. Straightforward calculation yields

$$\begin{aligned} \begin{aligned} \dot{g}(\delta )&= \exp \left( -\int _0^{T_n}(\lambda (s) + \delta h(s)-1)ds\right) \sum _{k=1}^n\prod _{j\not = k} h(T_k)\left( \lambda (T_j) + \delta h(T_j)\right) \\&\quad - g(\delta ) \int _0^{T_n}h(s)ds \\ \ddot{g}(\delta )&= -2\left( \int _0^{T_n}h(s)ds\right) \dot{g}(\delta ) -\left( \int _0^{T_n}h(s)ds\right) ^2g(\delta ) \\&\quad +\exp \left( -\int _0^{T_n}(\lambda (s) + \delta h(s)-1)ds\right) \sum _{k=1}^nh(T_k)\sum _{j\not =k}h(T_j)\prod _{{\mathop {i \not = j}\limits ^{i\not = k}}}\left( \lambda (T_i) + \delta h(T_i)\right) \end{aligned} \end{aligned}$$

Now, we can turn to the uniform integrability of the second order derivative for $\delta \in [0, \overline{\delta }]$. For this purpose, we notice that $\lambda (\cdot ) + \delta h(\cdot )$ is an element of ${\mathcal {D}}(m^{'},a^{'}, b^{'})$, with $m^{'} := \lambda _0$, $a^{'}:=\lambda _0+a$ and $b^{'}:=\overline{\alpha } + b \overline{\delta }$. We are now in position to provide boundaries for g, $\dot{g}$ and, eventually, $\ddot{g}$. We obtain successively

$$\begin{aligned} \begin{aligned}&\sup _{\delta \in [0, \overline{\delta }]} \vert g(\delta ) \vert \le e^{-(\lambda _0-1) T_n}(a^{'}+b^{'}n)^n \\&\sup _{\delta \in [0, \overline{\delta }]} \vert \dot{g}(\delta ) \vert \le e^{-(\lambda _0-1) T_n}(T_n+1)(a^{'}+b^{'}n)^{n+1} \\&\sup _{\delta \in [0, \overline{\delta }]} \vert \ddot{g}(\delta ) \vert \le e^{-(\lambda _0-1) T_n}\left[ (3T_n^2+2T_n)(a^{'}+b^{'}n)^{n+2} +n(n-1)(a^{'}+b^{'}n)^{n} \right] \end{aligned} \end{aligned}$$

Now, we need to check that elements of the form $T_n^2e^{-(\lambda _0-1) T_n}$ are $\mathbb {P}$-integrable. The law of $T_n$ under $\mathbb {P}$ is a Gamma law $\Gamma (n,1)$, thus

$$\begin{aligned} \begin{aligned} \mathbb {E}^\mathbb {P}\left[ T_n^2e^{-(\lambda _0-1) T_n}\right]&= \int _0^{+\infty }x^2e^{-(\lambda _0-1) x}\frac{e^{-x}x^{n-1}}{n!}dx= \int _0^{+\infty }x^2\frac{e^{-\lambda _0x}x^{n-1}}{n!}dx\\&= \lambda _0^{n-1}(n^2+n) \end{aligned} \end{aligned}$$

where we have used the formula of the second order moment of a Gamma law. With these elements in hand, we can state that there exists a constant C, depending only on n, $m^{'}$, $a^{'}$, $b^{'}$ and $\lambda _0$, such that

$$\begin{aligned} \left| \mathbb {E}^\mathbb {P}\left[ g(\delta ) - g(0) - \delta \dot{g}(0)\right] \right| \le \frac{\delta ^2}{2} C \end{aligned}$$

Since f is uniformly bounded, this implies that the directional derivative of $I[\lambda ](f)$ (on the right) in the direction h exists and is given by (6) with

$$\begin{aligned} \begin{aligned} G(T_n)&= \exp \left( -\int _0^{T_n}(\lambda (s)-1)ds\right) \sum _{k=1}^nh(T_k)\prod _{j\not = k} \lambda (T_j)\\&\quad - L[\lambda ](T_n) \int _0^{T_n}h(s)ds \end{aligned} \end{aligned}$$

(17)

We factorize $L[\lambda ](T_n)$ in (17), which yields the exact form of $G(T_n)$ if we notice that $\sum _{k=1}^n\frac{h(T_k)}{\lambda (T_k)} = \int _0^{T_n}\frac{h(s)}{\lambda (s)}\nu (ds)$. $\square$

1.2 Proof of Lemma 3

We shall prove the result by recurrence on n. The case $n=1$ is given by Corollary 3. Assume that the property holds for any $1\le k\le n-1$. Now, set, for any $t \ge 0$,

$$\begin{aligned} \Upsilon _n(t) :=n \, \lambda _0\beta \, {m}_{n-1}(t)+ \sum _{k=1}^{n-1}{n \atopwithdelims ()k-1}{m}_k(t)\alpha ^{n-k+1} \end{aligned}$$

By recurrence, we know that the limit of $\Upsilon _n(t)$, when t tends to $+\infty$ exists. Let us denote it by $\overline{\Upsilon }_n$. By Lemma 2, we can step $m_n(t) = g(t)e^{n(\alpha -\beta )t}$ obtaining $\dot{g}(t) = \Upsilon _n(t) e^{-n(\alpha -\beta )t}$. By the recurrence property, we have

$$\begin{aligned} \frac{\dot{g}(t)}{e^{-n(\alpha -\beta )t}} \underset{t \rightarrow +\infty }{\longrightarrow } \overline{\Upsilon }_n \end{aligned}$$

Then, by L’Hôpital’s rule, we have

$$\begin{aligned} \frac{n(\beta -\alpha ){g}(t)}{e^{-n(\alpha -\beta )t}-1} \underset{t \rightarrow +\infty }{\longrightarrow } \overline{\Upsilon }_n \end{aligned}$$

Hence, we have the equivalence (15) and $\overline{m}_n:= \frac{\overline{\Upsilon }_n}{n(\beta -\alpha )}$.

Now, let us prove that $m_n$ is increasing. For this purpose, note

$$\begin{aligned} \frac{\dot{g}(t)}{g(t)} = \frac{\Upsilon _n(t)e^{-n(\alpha -\beta )t}}{\lambda _0^n + \int _0^t\Upsilon _n(s)e^{-n(\alpha -\beta )s}ds} \end{aligned}$$

By recurrence assumption, $\Upsilon _n$ is increasing as a sum of increasing functions (all the coefficients are positive). Then

$$\begin{aligned} \frac{\dot{g}(t)}{g(t)} > \frac{\Upsilon _n(t)e^{-n(\alpha -\beta )t}}{\lambda _0^n + \Upsilon _n(t) \frac{e^{-n(\alpha -\beta )t}-1}{n(\beta -\alpha )} } \end{aligned}$$

Hence, we have $\frac{\dot{g}(t)}{g(t)} > n(\beta -\alpha )$ if, and only if, $\Upsilon _n(t) > n(\beta -\alpha ) \lambda ^n_0$. As $\Upsilon _n$ is increasing, it is sufficient to show that the condition holds for $t=0$. In this case, we easily see that $\Upsilon _n(0)> n\beta \lambda _0^n> n(\beta -\alpha ) \lambda _0^n$. Hence, $\frac{\dot{g}(t)}{g(t)} > n(\beta -\alpha )$, which implies that $m_n(t)$ is increasing. $\Box$

1.3 Proof of Proposition 4

According to Lemma 5, we only need to prove that the limit of $u_n:=\frac{m_n(t)}{n!}$ is 0, when n goes to $\infty$. First, let us rewrite $u_n$ with Eq. (13). It yields

$$\begin{aligned} u_n = u_{n-1} \left[ \frac{\lambda _0\beta }{n(\beta -\alpha )} + \frac{n-1}{2n} \frac{\alpha ^2}{(\beta - \alpha )}\right] + \frac{1}{\beta -\alpha }\sum _{k=3}^n\frac{\alpha ^k}{k!} \frac{n-k+1}{n}u_{n-k+1} \end{aligned}$$

Set $\delta >0$, such that $\frac{\alpha ^2}{2(\beta - \alpha )} (1 + \delta +\frac{\alpha }{3}e^{\alpha } ) < 1$. There exists $n_0$, such that, for any $n \ge n_0$,

$$\begin{aligned} \frac{\lambda _0\beta }{n(\beta -\alpha )} + \frac{n-1}{2n} \frac{\alpha ^2}{(\beta - \alpha )}< \frac{\alpha ^2}{2(\beta - \alpha )} (1 + \delta ) < 1 \end{aligned}$$

Let us define $K_p:=\max \left\{ u_k \; \vert \; 1\le k\le p \right\}$, for any $p \ge 1$. For $n \ge n_0$, we have

$$\begin{aligned} \begin{aligned} u_n&\le u_{n-1} \left[ \frac{\lambda _0\beta }{n(\beta -\alpha )} + \frac{n-1}{2n} \frac{\alpha ^2}{(\beta - \alpha )} \right] + \frac{\alpha ^3}{6(\beta -\alpha )} e^{\alpha }K_{n-1}\\&\le \frac{\alpha ^2}{2(\beta - \alpha )} \left( 1 + \delta +\frac{\alpha }{3} e^{\alpha }\right) K_{n-1} \end{aligned} \end{aligned}$$

(18)

where we have used the Taylor development of the exponential function at order 2. From Eq. (18), we deduce that $K_n = K_{n-1}$. Therefore, the sequence $(K_p)_{p\ge n-1}$ is constant, equal to $K>0$. Set

$$\begin{aligned} H_{\delta }:= \frac{\alpha ^2}{2(\beta - \alpha )} (1 + \delta )\,\, \text {and} \,\, \overline{H}_{\delta }:= \frac{\alpha ^2}{2(\beta - \alpha )} \left( 1 + \delta +\frac{\alpha }{3} e^{\alpha }\right) \end{aligned}$$

By Eq. (18), we have, for any $n\ge n_0$,

$$\begin{aligned} u_{n} \le \overline{H}_{\delta } K. \end{aligned}$$

Set $n\ge n_0+p$, with p such that $\frac{K}{\beta -\alpha } \sum _{k=p+1}^{\infty }\frac{\alpha ^k}{k!} \le \frac{K\overline{H}_{\delta }^2}{n_0}$. Using again the Taylor expansion of the exponential and the inequality above, we have the following inequality, for any $n \ge n_0+p$,

$$\begin{aligned} \begin{aligned} u_n&\le u_{n-1}H_{\delta } + \frac{\overline{H}_{\delta }K}{(\beta -\alpha )} \sum _{k=3}^p \frac{\alpha ^k}{k!} + \frac{K\overline{H}_{\delta }^2}{n_0} \le \overline{H}_{\delta }H_{\delta }K + \frac{\overline{H}_{\delta }K}{6(\beta -\alpha )} e^{\alpha } + \frac{K\overline{H}_{\delta }^2}{n_0}\\&\le \left( 1 + \frac{1}{n_0}\right) \overline{H}_{\delta }^2K \end{aligned} \end{aligned}$$

Set $n_1:=n_0+p$. By iteration of this method, we can construct an increasing sequence of integer $(n_k)_{k\ge 1}$ such that, for any $i \ge n_k$,

$$\begin{aligned} u_i \le \left( 1 + \frac{1}{n_{k-1}}\right) \overline{H}_{\delta }^{k+1}K \end{aligned}$$

The limit of the right hand side is 0, which yields the result. $\square$