Credit spreads, endogenous bankruptcy and liquidity risk

Abstract

In this paper, we consider a bond valuation model with both credit risk and liquidity risk to show that credit spreads are not negligible for short maturities. We adopt the structural approach to model credit risk, where the default triggering barrier is determined endogenously by maximizing equity value. As for liquidity risk, we assume that bondholders may encounter liquidity shocks during the lifetime of corporate bonds, and have to sell the bond immediately at the price, which is assumed to be a fraction of the price in a perfectly liquid market. Under this framework, we derive explicit expressions for corporate bond, firm value and bankruptcy trigger. Finally, numerical illustrations are presented.

Appendix

Proof of Proposition 1 The bond value is given by

$$\begin{aligned} d^I(V;V_B,t)&= \int _0^{t}{e}^{-rs}c(t)[1-F(s;V,V_B)]\mathbf{Q }[N_s=0]\mathrm{d }s\nonumber \\&+\, e^{-rt}p(t)[1-F(t;V,V_B)]\mathbf{Q }[N_t=0]\nonumber \\&+\int _0^{t}{e}^{-rs}\rho (t)V_Bf(s;V,V_B)\mathbf{Q }[N_s=0]\mathrm{d }s \nonumber \\&+\, \alpha \mathbf{E }_{\mathbf{Q }}\int _0^{t}{e}^{-rs}d^L(V_s;V_B,t-s)[1-F(s;V,V_B)]\lambda e^{-\lambda s}\mathrm{d }s.\qquad \end{aligned}$$

(11)

Denote

$$\begin{aligned} I(r,\lambda , t):=\int _0^{t}{e}^{-(r+\lambda )s}f(s;V,V_B)\mathrm{d }s. \end{aligned}$$

For the first term of (11), integration by parts implies that

$$\begin{aligned} \begin{aligned}&\int _0^t e^{-rs}c(t)[1-F(s;V,V_B)]\mathbf{Q }[N_s=0]\mathrm{d }s\\&\quad =\int _0^te^{-(r+\lambda )s}c(t)[1-F(s;V,V_B)]\mathrm{d }s\\&\quad =-\frac{c(t)}{r+\lambda }\left[e^{-(r+\lambda )t}(1-F(t;V,V_B))-1+\int _0^te^{-(r+\lambda )s}f(s;V,V_B)\mathrm{d }s\right]\\&\quad =\frac{c(t)}{r+\lambda }-\frac{c(t)}{r+\lambda }[1-F(t;V,V_B)]e^{-(r+\lambda )t}-\frac{c(t)}{r+\lambda }I(r,\lambda , t). \end{aligned} \end{aligned}$$

(12)

As for the second term of (11), it holds that

$$\begin{aligned} e^{-rt}p(t)[1-F(t;V,V_B)]\mathbf{Q }[N_t=0]=p(t)[1-F(t;V,V_B)]e^{-(r+\lambda )t}. \end{aligned}$$

(13)

It is easy to show that the third term of (11) equals,

$$\begin{aligned} \begin{aligned} \int _0^te^{-rs}\rho (t)V_Bf(s;V,V_B)\mathbf{Q }[N_s=0]\mathrm{d }s&=\rho (t)V_B\int _0^te^{-(r+\lambda )s}f(s;V,V_B)\mathrm{d }s\\&=\rho (t)V_BI(r,\lambda , t). \end{aligned} \end{aligned}$$

(14)

Note that in a perfectly liquid markets, the following holds,

$$\begin{aligned} d^L(V;V_B,t)&= \int _0^se^{-ru}c(t)[1-F(u;V,V_B)]\mathrm{d }u\nonumber \\&+\int _0^s e^{-ru}\rho (t)V_Bf(u;V,V_B)\mathrm{d }u \nonumber \\&+\, e^{-rs}d^L(V_s;V_B,t-s)[1-F(s;V,V_B)]. \end{aligned}$$

(15)

Then using integration by parts, the last term of (11) becomes

$$\begin{aligned} \begin{aligned}&\alpha \mathbf{E }_{\mathbf{Q }}\int _0^{t}{e}^{-rs}d^L(V_s;V_B,t-s)[1-F(s;V,V_B)]\lambda e^{-\lambda s}\mathrm{d }s \\&\quad =\alpha \lambda \left\{ \int _0^te^{-\lambda s}\left[d^L(V;V_B,t)-\frac{c(t)}{r}+\frac{c(t)}{r}(1-F(s))e^{-rs}\right.\right.\\&\qquad -\left.\left.\left(\rho (t)V_B-\frac{c(t)}{r}\right)G(s)\right]\mathrm{d }s\right\} \\&\quad =\alpha \lambda \left\{ \int _0^te^{-\lambda s}\left(d^L(V;V_B,t) -\frac{c(t)}{r}\right)\mathrm{d }s+\frac{c(t)}{r}\int _0^{t}{e}^{-(\lambda +r) s}(1-F(s))\mathrm{d }s\right.\\&\qquad -\left.\left(\rho (t)V_B-\frac{c(t)}{r}\right)\int _0^{t}{e}^{-\lambda s}G(s)\mathrm{d }s\right\} \\&\quad =\alpha \lambda \left\{ \frac{ d^L(V;V_B,t) -\frac{c(t)}{r}}{\lambda }(1-e^{-\lambda t})+\frac{c(t)}{r}\int _0^te^{-(\lambda +r) s}(1-F(s))\mathrm{d }s\right.\\&\qquad -\left.\left(\rho (t)V_B-\frac{c(t)}{r}\right)\int _0^t e^{-\lambda s}G(s)\mathrm{d }s\right\} . \end{aligned} \end{aligned}$$

(16)

Moreover, we have that

$$\begin{aligned} \int _0^t e^{-\lambda s}G(s)\mathrm{d }s&= -\frac{1}{\lambda }\left[G(t)e^{-\lambda t} -\int _0^te^{-(\lambda +r) s}f(s)\mathrm{d }s\right]\nonumber \\ \\&= -\frac{1}{\lambda }G(T)e^{-\lambda T}+\frac{1}{\lambda }I(r,\lambda , T).\nonumber \end{aligned}$$

(17)

Now we just need to consider the integral $I(r,\lambda , t)$ so that we can get the explicit expressions of $d^I(V, V_B, t)$. From Harrison (1986), we can find the expressions for $f(t)$,

$$\begin{aligned} I(r,\lambda , t)&= \int _0^te^{-(r+\lambda )s}f(s;V,V_B)\mathrm{d }s\nonumber \\&= \int _0^t e^{-(r+\lambda ) s}\frac{b}{\sigma \sqrt{2\pi s^3}}e^{-\frac{1}{2}\left(\frac{b+(r-\delta -\frac{1}{2}\sigma ^2)s}{\sigma \sqrt{s}}\right)^2}\mathrm{d }s\nonumber \\&= e^{\frac{b(R-r)}{\sigma ^2}}\int _0^t e^{-R s}\frac{b}{\sigma \sqrt{2\pi s^3}}e^{-\frac{1}{2}\left(\frac{b+(R-\delta -\frac{1}{2}\sigma ^2)s}{\sigma \sqrt{s}}\right)^2}\mathrm{d }s\nonumber \\ \\&= e^{\frac{b(R-r)}{\sigma ^2}}\left\{ \left(\frac{V}{V_B}\right)^{-A+Z}\!\!\!N[Q_1(t)]+\left(\frac{V}{V_B}\right)^{-A-Z}\!\!\!N[Q_2(t)]\right\} ,\nonumber \end{aligned}$$

(18)

where

$$\begin{aligned}&R=\delta -\frac{\sigma ^2}{2}+\sqrt{2(r+\lambda )\sigma ^2+\left(r-\delta -\frac{\sigma ^2}{2}\right)^2-2\delta \sigma ^2},\\&Q_1(t)=\frac{-b-Z\sigma ^2t}{\sigma \sqrt{t}};\ \ \ \ \ \ Q_2(t)=\frac{-b+Z\sigma ^2t}{\sigma \sqrt{t}};\\&A=\frac{R-\delta -\frac{\sigma ^2}{2}}{\sigma ^2};\ \ \ \ \ b=\ln \left(\frac{V}{V_B}\right);\ \ \ \ \ Z=\frac{[(A\sigma ^2)^2+2R\sigma ^2]^{\frac{1}{2}}}{\sigma ^2}, \end{aligned}$$

and $N(\cdot )$ is the cumulative standard normal distribution. Here, we have used the fact that,

$$\begin{aligned} \int _0^t e^{-r s}\frac{b}{\sigma \sqrt{2\pi s^3}}e^{-\frac{1}{2}\left(\frac{b+(r-\delta -\frac{1}{2}\sigma ^2)s}{\sigma \sqrt{s}}\right)^2}\mathrm{d }s\!=\!\left(\frac{V}{V_B}\right)^{-a+z}\!\!\!N[q_1(t)]+\left(\frac{V}{V_B}\right)^{-a-z}\!\!\!N[q_2(t)], \end{aligned}$$

where

$$\begin{aligned}&h_1(t)=\frac{-b-a\sigma ^2t}{\sigma \sqrt{t}};\ \ \ \ \ \ h_2(t)=\frac{-b+a\sigma ^2t}{\sigma \sqrt{t}};\\&q_1(t)=\frac{-b-z\sigma ^2t}{\sigma \sqrt{t}};\ \ \ \ \ \ q_2(t)=\frac{-b+z\sigma ^2t}{\sigma \sqrt{t}};\\&a=\frac{r-\delta -\frac{\sigma ^2}{2}}{\sigma ^2};\ \ \ \ \ b=\ln \left(\frac{V}{V_B}\right);\ \ \ \ \ z=\frac{[(a\sigma ^2)^2+2r\sigma ^2]^{\frac{1}{2}}}{\sigma ^2}, \end{aligned}$$

and $N(\cdot )$ is the cumulative standard normal distribution. Therefore, (12)–(18) jointly yield that

$$\begin{aligned} d^I(V;V_B,t)=\frac{c(t)(\alpha \lambda +r)}{r(r+\lambda )}+K_1(t)I(r,\lambda , t)+K_2(t)(1-F(t))+K_3(t)G(t), \end{aligned}$$

(19)

where $K_1$, $K_2$ and $K_3$ are defined in Proposition 1.