Tempered stable structural model in pricing credit spread and credit default swap

Abstract

In this paper, we explore the features of a structural credit risk model wherein the firm value is driven by normal tempered stable (NTS) process belonging to the larger class of Lévy processes. For the purpose of comparability, the calibration to the term structure of a corporate bond credit spread is conducted under both NTS structural model and Merton structural model. We find that NTS structural model provides better fit for all credit ratings than Merton structural model. However, it is noticed that probabilities of default derived from the calibration of the term structure of a bond credit spread might be overestimated since the bond credit spread could contain non-default components such as illiquidity risk or asymmetric tax treatment. Hence, considering CDS spread as a reflection of the pure credit risk for the reference entity, we calibrate it in order to obtain more reasonable probability of default and obtain valid results in calibration of the market CDS spread with NTS structural model.

Appendices

Appendix A

Suppose that from Eq. (10)

$$\begin{aligned} V(t) = \frac{V(0) \exp \Big ( rt + Z(t)\Big )}{E \Big [ \exp \Big (Z(t) \Big ) \Big ]} = V(0) \exp (X(t)) . \end{aligned}$$

(19)

As we have discussed before, bond price at time t under the risk neutral probability measure is

$$\begin{aligned} B(t,T)&= E \Bigg [ e^{-r(T-t)} \bigg ( D - \max \Big (D - V(T),\ 0 \Big ) \bigg ) \Biggm \vert \ \mathcal {F}(t) \Bigg ] \\&= B_{DF}(t, T) - e^{-R_{DF} (t,T)(T-t)} \\&\qquad \,E \Bigg [ \max \bigg ( D - V(t) \exp \Big (X(T) - X(t) \Big ),\ 0 \bigg ) \ \Biggm \vert \ \mathcal {F}(t) \Bigg ] . \end{aligned}$$

Let f(x) be the p.d.f of the random variable $X(T)-X(t)$. $X(T)-X(t)$ has the same distribution as $X(T-t)$ by the stationary property of the NTS process. By complex inverse formula (Doetsch 1970), the p.d.f f(x) is given by:

$$\begin{aligned} f(x) = \frac{1}{2\pi } \int _{-\infty }^{\infty } e^{-i(u+i\rho )x} \phi _{X(T-t)} (u+i\rho ) du . \end{aligned}$$

Thus, we have

$$\begin{aligned}&E \bigg [ \max \bigg ( D - V(t) \exp \Big ( X(T) - X(t) \Big ),\ 0 \bigg ) \ \Bigm \vert \ \mathcal {F}(t) \bigg ] \\&\quad = \int _{-\infty }^{\log \frac{D}{V(t)}} \Big ( D - \exp (x + \log V(t)) \Big ) f(x) dx \\&\quad = \frac{1}{2\pi } \int _{-\infty }^{\log {\frac{D}{V(t)}}} \Big (D - \exp \big (x + \log V(t)\big )\Big ) \\&\qquad \times \,\int _{-\infty }^{\infty } e^{-i(u+i\rho )x} \phi _{X(T-t)}(u+i\rho ) du dx . \end{aligned}$$

If $Im(z) > -1$ for $z \in \mathbb {C}$, the following holds:

$$\begin{aligned} \int _{-\infty }^{k} e^{(1-iz)x} dx = \frac{e^{k(1-iz)}}{1-iz} , \end{aligned}$$

and if $Im(z) > 0$,

$$\begin{aligned} \int _{-\infty }^{k} e^{-izx} dx = -\frac{e^{-ikz}}{iz} . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned}&\int _{-\infty }^{\log \frac{D}{V(t)}} \Big (D - e^{(x + \log V(t))}\Big ) e^{-i(u+i\rho )x} dx \\&\quad = D \int _{-\infty }^{\log {\frac{D}{V(t)}}} e^{-i(u+i\rho )x} dx - V(t) \int _{-\infty }^{\log {\frac{D}{V(t)}}} e^{(1-i(u+i\rho ))x} dx\\&\quad = -D \frac{\exp \Big (-i(u+i\rho ) \log {\frac{D}{V(t)}}\Big )}{i(u+i\rho )} - V(t) \frac{\exp \Big ((1-i(u+i\rho )) \log {\frac{D}{V(t)}}\Big )}{1-i(u+i\rho )} \\&\quad = \Big (\frac{D}{V(t)}\Big )^{\rho -iu} \frac{D}{(\rho -iu)(1+\rho -iu)} , \end{aligned}$$

where $\rho > 0$.

Also, we have

$$\begin{aligned} \phi _{X(T-t)}(u+i\rho )&= E \Big [ \exp \Big (i(u+i\rho )X(T-t)\Big ) \Big ] \\&= E \left[ \exp \left( i(u+i\rho ) \log \left( \frac{\exp \Big ( r(T-t) + Z(T-t) \Big ) }{E [\exp (Z(T-t))]} \right) \right) \right] \\&= \frac{e^{(iu-\rho ) R_{DF} (t,T) (T-t)} \phi _{Z(T-t)} (u+i\rho )}{\Big (\phi _{Z(T-t)}(-i)\Big )^{iu-\rho }} . \end{aligned}$$

Hence, the bond price can be computed by the following formula:

$$\begin{aligned} B(t, T)&= B_{DF}(t, T) \left( 1 - \frac{1}{\pi } \mathfrak {R}\int _0^\infty \left( \frac{D}{V(t)} \right) ^{\rho - iu}\right. \\&\quad \left. \times \,\frac{e^{(iu-\rho )R_{DF}(t,T)(T-t)} \phi _{Z(T-t)}(u+i\rho )}{(\rho -iu)(1+\rho -iu)(\phi _{Z(T-t)}(-i))^{iu- \rho }} du \right) , \end{aligned}$$

where $\rho > 0$.

In the same way, we can obtain stock price under the risk neutral probability measure as follows:

$$\begin{aligned} S(t,\ T)&= E \left[ e^{-r(T-t)} \max \Big (V(T) - D,\ 0 \Big ) \ \Bigm \vert \ \mathcal {F}(t) \right] \\&= e^{-R_{DF}(t, T)(T-t)} E \bigg [ \max \Big ( V(t) \exp (X(T) - X(t)) - D,\ 0 \Big ) \ \Bigm \vert \ \mathcal {F}(t) \bigg ] \\&= B_{DF}(t, T)\ \frac{1}{\pi } \mathfrak {R}\int _0^\infty \left( \frac{D}{V(t)}\right) ^{\eta - iu}\\&\quad \times \, \frac{e^{(iu-\eta )R_{DF}(t,T)(T-t)} \phi _{Z(T-t)}(u+i\eta )}{(\eta -iu)(1+\eta -iu)(\phi _{Z(T-t)}(-i))^{iu- \eta }}\ du , \end{aligned}$$

where $\eta < -1$.

Then, credit spread is

$$\begin{aligned} s_{\textit{credit}}\ (t, T)&= - \frac{1}{T-t} \log \Bigg ( \frac{B(t, T)}{B_{DF}(t, T)}\Bigg ) \\&= - \frac{1}{T-t} \log \Bigg ( 1- \frac{1}{\pi } \mathfrak {R}\int _{0}^{\infty } \Big (\frac{D}{V(t)}\Big )^{\rho -iu} \\&\quad \times \,\frac{ e^{(iu-\rho ) R_{DF} (t,T) (T-t) } \phi _{Z(T-t)} (u+i\rho )}{(\rho -iu)(1+\rho -iu) (\phi _{Z(T-t)}(-i))^{iu-\rho }} du \Bigg ) , \end{aligned}$$

where $\rho > 0$.

Appendix B

From (19),

$$\begin{aligned} \textit{RR}_{\textit{NTS}} (T)&= E \Big [\ \frac{V(T)}{D} \Bigm \vert V(T)< D\ \Big ] \\&= E \Big [\ \frac{V(0)}{D} e^{X(T)} \Bigm \vert X(T) < \log \Big (\frac{D}{V(0)} \Big )\ \Big ] \\&= \frac{(V(0)/D)}{\textit{PD}_{\textit{NTS}}(T)}\ \int _{-\infty }^{\log \frac{D}{V(0)}} e^x f(x) dx , \end{aligned}$$

where f(x) is p.d.f. of NTS process X(T), and $\textit{PD}_{\textit{NTS}}(T)$ is the probability of default.

By complex inverse formula,

$$\begin{aligned} \int _{-\infty }^{\log \frac{D}{V(0)}}&e^x f(x) dx = \int _{-\infty }^{\log \frac{D}{V(0)}}\ e^x\ \frac{1}{2 \pi }\ \int _{-\infty }^{\infty } e^{-i(u+i\delta )x} \phi _{X(T)} (u+i\delta )\ du\ dx \\&= \frac{1}{2\pi }\int _{-\infty }^{\infty } \phi _{X(T)} (u+i\delta )\ \int _{-\infty }^{\log {\frac{D}{V(0)}}} e^{(1+\delta -iu)x} dx\ du \\&= \frac{1}{\pi } \mathfrak {R}\int _{0}^{\infty } \Bigg ( \frac{(D/V(0))^{1+\delta -iu}}{(1+\delta -iu)} \Bigg )\ \Bigg ( \frac{e^{(iu-\delta )R_{DF}(0,T)T} \phi _{Z(T)} (u+i\delta )}{(\phi _{Z(T)} (-i))^{iu-\delta }} \Bigg ) du , \end{aligned}$$

where $\delta > -1$.

Hence, for $\delta > -1$, expected recovery rate $\textit{RR}_{\textit{NTS}}(T)$ under the risk neutral probability measure is given by:

$$\begin{aligned} \textit{RR}_{\textit{NTS}} (T)= & {} \frac{1}{\pi } \Bigg ( \frac{1}{\textit{PD}_{\textit{NTS}}(T)} \Bigg )\ \mathfrak {R}\int _{0}^{\infty } \left( \frac{D}{V(0)} \right) ^{\delta -iu} \\&\frac{e^{(iu-\delta ) R_{DF} (0,T) T} \phi _{Z(T)} (u+i\delta )}{(1+\delta -iu)(\phi _{Z(T)} (-i))^{iu-\delta }}\ du . \end{aligned}$$