Structural Pricing of CoCos and Deposit Insurance with Regime Switching and Jumps

Abstract

In this article, we develop a semi-analytical solution for a structural model that combines jump and regime switching risk. We use an Esscher transform that is applicable to regime switching double exponential jump diffusion to move from the historical world to the risk-neutral world. Further, we define and implement a matrix Wiener–Hopf factorization associated with the latter process, allowing us to price the various components of balance sheet. We illustrate the model with a study of a bank that issues contingent convertible bonds (CoCos). Thus, we obtain valuation formulas for the bank’s equity, debt, deposits, CoCos, and deposit insurance. We also show in an illustration the respective effects of the jump risk and of regime switching on the values of all of a bank’s balance sheet components.

Appendices

Appendix A

1.1 Proof of Proposition 2.1

Proof

$$\begin{aligned} \begin{array}{cl} V_0&{}=E_{{\mathbb {Q}}}\left( e^{-\int \limits _0^t r_u \mathrm {d}u} V_t\right) \\ &{}=V_0 E_{{\mathbb {P}}}\left( \text {exp}\left( {\int \limits ^t_0 (\mu _s-r_s) ds+\int \limits ^t_0 (1+\xi _s) \sigma _s \mathrm {d}W_s-\displaystyle \frac{1}{2} \int \limits ^t_0 \xi _s^2\sigma _s^2 \mathrm {d}s+\int \limits ^t_0 (1+\xi _s)\mathrm {d}N_s-\int \limits _0^t\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(\varvec{{\widehat{\xi }}}) \rangle \mathrm {d}s}\right) \right) \\ &{}=V_0 E_{{\mathbb {P}}}\left( \text {exp}\left( \int \limits _0^t \left( \mu _s-r_s+\displaystyle \frac{1}{2}{\sigma _s^2+\xi _s \sigma _s^2}+\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(1+\varvec{{\widehat{\xi }}}) \rangle -\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(\varvec{{\widehat{\xi }}}) \rangle \right) \mathrm {d}s\right) \right) \\ &{}=V_0 \varvec{J}_0 \text {exp}\left( {\left( Q+\text {diag}\left( {\widehat{\mu }}_i-{\widehat{r}}_i+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2+\psi _i(1+{\widehat{\xi }}_{i})-\psi _i({\widehat{\xi }}_{i})\right) \right) t}\right) {\mathbf {1}}_n. \end{array} \end{aligned}$$

Let $\varvec{d}$ be a null vector. Inserting it into Lemma 2.1, we readily have:

$$\begin{aligned} \varvec{J}_0e^{Qt}{\mathbf {1}}_n=1. \end{aligned}$$

(A.1)

The sufficient condition of the martingale condition is then obvious. Denote

$$\begin{aligned} f(t)=\varvec{J}_0 \text {exp}\left( {\left( Q+\text {diag}\left( {\widehat{\mu }}_i-{\widehat{r}}_i+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2+\psi _i(1+{\widehat{\xi }}_{i})-\psi _i({\widehat{\xi }}_{i})\right) \right) t}\right) {\mathbf {1}}_n. \end{aligned}$$

Assume that the martingale condition is satisfied:

$$\begin{aligned} f(t)=1 \quad \forall t. \end{aligned}$$

Then,

$$\begin{aligned} f'(0)=\varvec{J}_0{\left( Q+\text {diag}\left( {\widehat{\mu }}_i-{\widehat{r}}_i+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2+\psi _i(1+{\widehat{\xi }}_{i})-\psi _i({\widehat{\xi }}_{i})\right) \right) }{\mathbf {1}}_n=0. \end{aligned}$$

Similarly, eq. (A.1) means

$$\begin{aligned} \varvec{J}_0 Q{\mathbf {1}}_n=0. \end{aligned}$$

Therefore,

$$\begin{aligned} \varvec{J}_0 {\left( \text {diag}\left( {\widehat{\mu }}_i-{\widehat{r}}_i+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2+\psi _i(1+{\widehat{\xi }}_{i})-\psi _i({\widehat{\xi }}_{i})\right) \right) }{\mathbf {1}}_n=0. \end{aligned}$$

Since this equation is satisfied for any initial vector of states $\varvec{J}_0$,

$$\begin{aligned} {\widehat{\mu }}_i-{\widehat{r}}_i+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2+\psi _i(1+{\widehat{\xi }}_{i})-\psi _i({\widehat{\xi }}_{i})=0 \quad \forall i=1,2,\ldots ,n. \end{aligned}$$

$\square$

1.2 Proof of Proposition 2.2

Proof

Because the Markov chain $\varvec{J}$ is not changed under ${\mathbb {Q}}$ and

$$\begin{aligned} \begin{array}{cl} E_{{\mathbb {Q}}}(e^{uX_t})=&{}E_{{\mathbb {P}}}\left( \text {exp}\left( {u \int \limits ^t_0 \mu _s ds+\int \limits ^t_0 (u+\xi _s) \sigma _s \mathrm {d}W_s-\displaystyle \frac{1}{2}\int \limits ^t_0 \xi _s^2 \sigma _s^2 \mathrm {d}s+\int \limits ^t_0 (u+\xi _s)\mathrm {d}N_s-\int \limits _0^t\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(\varvec{{\widehat{\xi }}}) \rangle \mathrm {d}s}\right) \right) \\ &{}=E_{{\mathbb {P}}}\left( \text {exp}\left( \int \limits _0^t \left( \mu _s u+\displaystyle \frac{1}{2}\sigma _s^2 u^2+\xi _s \sigma _s^2 u +\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(\varvec{{\widehat{\xi }}}+u) \rangle -\langle \varvec{J}_s,\varvec{{\widehat{\psi }}}(\varvec{{\widehat{\xi }}}) \rangle \right) \mathrm {d}s\right) \right) \\ &{}={\varvec{J}_0}\text {exp}\left( {\left( Q+\text {diag}\left( {\widehat{\mu }}_i u+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2 u^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2 u +\psi _i({\widehat{\xi }}_{i}+u)-\psi _i({\widehat{\xi }}_{i})\right) \right) t}\right) {\mathbf {1}}_n, \end{array} \end{aligned}$$

the Laplace exponent $\varphi _i^*(u)$ of $X^i$ under ${\mathbb {Q}}$ is

$$\begin{aligned} \varphi _i^*(u)={\widehat{\mu }}_i u+\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2 u^2 +{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2 u +\psi _i({\widehat{\xi }}_{i}+u)-\psi _i({\widehat{\xi }}_{i}). \end{aligned}$$

The term $\displaystyle \frac{1}{2}{{\widehat{\sigma }}_i}^2 u^2+{\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2 u$ corresponds to ${\widehat{\sigma }}_i W_1\sim N({\widehat{\xi }}_{i}{{\widehat{\sigma }}_i}^2,{{\widehat{\sigma }}_i}^2)$. Then, we have that $W^*$ defined by $W^*_t=W_t-\int \limits _0^t \xi _{s}\sigma _s \mathrm {d}s$ is a standard Brownian motion under ${\mathbb {Q}}$ with respect to ${\mathcal {F}}$.

Denote

$$\begin{aligned} \omega _i=\displaystyle \frac{{\widehat{p}}_i{\widehat{\eta }}_{i}}{{\widehat{\eta }}_{i}-{\widehat{\xi }}_{i}}+\displaystyle \frac{{\widehat{q}}_i{\widehat{\theta }}_{i}}{{\widehat{\theta }}_{i}+{\widehat{\xi }}_{i}}. \end{aligned}$$

After simple computations, we obtain:

$$\begin{aligned} \varphi _i^*(u)=\left( {\widehat{\mu }}_i+{\widehat{\xi }}_i {\widehat{\sigma }}_i^2\right) u+\displaystyle \frac{1}{2}{\widehat{\sigma }}_i^2 u^2+{\widehat{\lambda }}_i \omega _i \left( \displaystyle \frac{1}{\omega _i}\left( \displaystyle \frac{{\widehat{p}}_i{\widehat{\eta }}_{i}}{{\widehat{\eta }}_{i}-{\widehat{\xi }}_i}\right) \displaystyle \frac{{\widehat{\eta }}_{i}-{\widehat{\xi }}_i}{{\widehat{\eta }}_{i}-{\widehat{\xi }}_i-u} +\displaystyle \frac{1}{\omega _i}\left( \displaystyle \frac{{\widehat{q}}_i {\widehat{\theta }}_i}{{\widehat{\theta }}_{i}+{\widehat{\xi }}_i}\right) \displaystyle \frac{{\widehat{\theta }}_{i}+{\widehat{\xi }}_i}{{\widehat{\theta }}_{i}+{\widehat{\xi }}_i+u}-1 \right) . \end{aligned}$$

This expression shows that $X^i$ remains a double exponential jump diffusion process under ${\mathbb {Q}}$ where

$$\begin{aligned} \left\{ \begin{aligned} {\widehat{\mu }}^*_i&={\widehat{\mu }}_i+{\widehat{\xi }}_i {\widehat{\sigma }}_i^2 \\ {\widehat{\sigma }}^*_i&={\widehat{\sigma }}_i \\ {\widehat{\lambda }}^*_i&={\widehat{\lambda }}_i \omega _i \\ p_i^*&=\displaystyle \frac{1}{\omega _i}\left( \displaystyle \frac{{\widehat{p}}_i{\widehat{\eta }}_{i}}{{\widehat{\eta }}_{i}-{\widehat{\xi }}_{i}}\right) \\ q_i^*&=\displaystyle \frac{1}{\omega _i}\left( \displaystyle \frac{{\widehat{q}}_i {\widehat{\theta }}_i}{{\widehat{\theta }}_{i}+{\widehat{\xi }}_i}\right) \\ {\widehat{\eta }}_{i}^*&={\widehat{\eta }}_{i}-{\widehat{\xi }}_{i} \\ {\widehat{\theta }}_{i}^*&={\widehat{\theta }}_{i}+{\widehat{\xi }}_{i}\\ \end{aligned} \right. . \end{aligned}$$

From $E_{{\mathbb {Q}}}(e^{X_t})=e^{\int \limits _0^t r_s \mathrm {d}s}$, we see that ${\widehat{\mu }}^*_i$ satisfies

$$\begin{aligned} {\widehat{\mu }}^*_i={\widehat{r}}_i-\displaystyle \frac{1}{2} {{\widehat{\sigma }}_i^{*^2}}-{\widehat{\lambda }}^*_i\left( \displaystyle \frac{{\widehat{p}}_i^*{\widehat{\eta }}_{i}^*}{{\widehat{\eta }}_{i}^*-1}+\displaystyle \frac{{\widehat{q}}_i^*{\widehat{\theta }}_{i}^*}{{\widehat{\theta }}_{i}^*+1}-1\right) . \end{aligned}$$

Therefore, the structure of the regime switching double exponential jump diffusion process is unchanged under the new measure ${\mathbb {Q}}$. $\square$

1.3 Proof of Proposition 3.1

Proof

Let the $2n \times 1$ vector $\varvec{{\bar{h}}}=({\bar{h}}_1,\ldots ,{\bar{h}}_{2n})'$ and $\varvec{h}^-_{\tau }=\langle \varvec{Y}_{\tau },\varvec{{\widehat{h}}^-} \rangle$ where the $1 \times 3n$ vector $\varvec{{\widehat{h}}^-} =(0,\ldots ,0,\varvec{{\bar{h}}}').$ From the Markov chain theory, we have:

$$\begin{aligned} \begin{array}{cl} E\left( e^{-\int \limits _0^{\tau } a_s \mathrm {d}s}\varvec{h}^-_{\tau }\right)&=\varvec{Y}_0W^{(\varvec{a},-)}e^{Q^{(\varvec{a},-)}(x-b)}\varvec{{\bar{h}}}. \end{array} \end{aligned}$$

When $\varvec{Y}_{\tau } \in E^0$, $X_{\tau }=b$. When $\varvec{Y}_{\tau } \in E^-$ and $\varvec{Y}_{\tau }=\varvec{s}_{2n+i}$, the overshoot $|X_{\tau }-b|$ is independent of $\tau$ and has an exponential distribution with parameter ${\widehat{\theta }}_i$. Then, we have:

$$\begin{aligned} \begin{array}{cl} E\left( e^{-\int \limits _0^{\tau } a_s \mathrm {d}s+wX_{\tau }}\varvec{h}_{\tau }\right) &{}=\sum \limits _{i=1}^{n}\left( E\left( e^{-\int \limits _0^{\tau }a_s \mathrm {d}s +w b}\mathbb {1}_{\{\varvec{Y}_{\tau }=\varvec{s}_{n+i}\}}{\widehat{h}}_i\right) \right. \\ &{}\quad \left. +E\left( e^{-\int \limits _0^{\tau }a_s \mathrm {d}s +w b}\mathbb {1}_{\{\varvec{Y}_{\tau }=\varvec{s}_{2n+i}\}}{\widehat{h}}_i\right) E\left( e^{w(X_{\tau }-b)} |\varvec{Y}_{\tau }=\varvec{s}_{2n+i}\right) \right) . \end{array} \end{aligned}$$

Since

$$\begin{aligned} \begin{array}{cl} E\left( e^{w(X_{\tau }-b)} |\varvec{Y}_{\tau }=\varvec{s}_{2n+i}\right) &{}=\int \limits _{-\infty }^{0} e^{wx} {\widehat{\theta }}_{i} e^{{\widehat{\theta }}_{i}x} \mathrm {d}x\\ &{}=\displaystyle \frac{{\widehat{\theta }}_{i}}{w+{\widehat{\theta }}_{i}}, \end{array} \end{aligned}$$

we have:

$$\begin{aligned} \begin{array}{cl} E\left( e^{-\int \limits _0^{\tau } a_s \mathrm {d}s+wX_{\tau }}\varvec{h}_{\tau }\right) =\varvec{Y}_0W^{(\varvec{a},-)}e^{Q^{(\varvec{a},-)}(x-b)+wb I_{2n}}\varvec{{\widetilde{h}}}. \end{array} \end{aligned}$$

$\square$

1.4 Proof of Lemma 3.2

Proof

The $Q^{(\varvec{a},+)}$ and $Q^{(\varvec{a},-)}$ are the generator matrices. Then, we have

$$\begin{aligned} e^{Q^{(\varvec{a},+)}}{\mathbf {1}}_{2n}\le {\mathbf {1}}_{2n}, \qquad e^{Q^{(\varvec{a},-)}}{\mathbf {1}}_{2n}\le {\mathbf {1}}_{2n}. \end{aligned}$$

The eigenvalues of $e^{Q^{(\varvec{a},+)}}$ and $e^{Q^{(\varvec{a},-)}}$ are $e^{{\widetilde{\beta }}_1},\ldots ,e^{{\widetilde{\beta }}_{2n}}$ and $e^{{\bar{\beta }}_1},\ldots ,e^{{\bar{\beta }}_{2n}}$, respectively. From the Perron-Frobenius theorem, we obtain

$$\begin{aligned} \max \{|e^{{\widetilde{\beta }}_i}|,\ i=1,\ldots ,2n \}\le 1, \qquad \max \{|e^{{\bar{\beta }}_j}|,\ j=1,\ldots ,2n \}\le 1, \end{aligned}$$

where |z| is the modulus of $z \in {\mathbb {C}}$. Then,

$$\begin{aligned} \mathfrak {R}({\widetilde{\beta }}_i)\le 0,\ i=1,\ldots ,2n, \qquad \mathfrak {R}({\bar{\beta }}_j)\le 0,\ j=1,\ldots ,2n. \end{aligned}$$

The

$$\begin{aligned} f({\widetilde{\beta }}_i)=0,\ i=1,\ldots ,2n, \qquad f(-{\bar{\beta }}_j)=0,\ j=1,\ldots ,2n \end{aligned}$$

gives the results. $\square$

Appendix B

The valuation of balance sheet components is conducted under the risk-neutral measure ${\mathbb {Q}}$. It makes use of Lemmas 2.1 and 3.1, Corollary 3.1 and Proposition 3.1.

1.1 Valuation of Straight Bonds

We have:

$$\begin{aligned} \begin{array}{cl} B=D_1\int \limits _0^{\infty } me^{-mt}B(t)\mathrm {d}t \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{cl} B&{}=D_1E\left( \int \limits _0^{\infty } me^{-mt}\int \limits _0^{\min (t,\tau _2)}e^{-\int \limits _0^s r_u \mathrm {d}u}c_1 \mathrm {d}s \mathrm {d}t\right) + D_1 E\left( \int \limits _0^{\infty } me^{-mt} \mathbb {1}_{\{t\ge \tau _2\}}e^{-\int \limits _0^{\tau _2} r_u \mathrm {d}u}\pi _1 \mathrm {d}t\right) \\ &{}\quad +D_1 E\left( \int \limits _0^{\infty } me^{-mt} \mathbb {1}_{\{t< \tau _2\}}e^{-\int \limits _0^t r_u \mathrm {d}u} \mathrm {d}t \right) . \end{array} \end{aligned}$$

Then, by the Fubini theorem, we exchange the order of integration for t and s and obtain:

$$\begin{aligned} \begin{array}{cl} B&{}=D_1 E\left( \int \limits _0^{\tau _2} m c_1e^{-\int \limits _0^s r_u \mathrm {d}u} \int \limits _s^{\infty }e^{-mt}\mathrm {d}t \mathrm {d}s \right) + \pi _1 D_1 E\left( e^{-\int \limits _0^{\tau _2} (r_u+m)\mathrm {d}u} \right) \\ &{}\quad +mD_1 E\left( \int \limits _0^{\tau _2} e^{-\int \limits _0^t (r_u+m)\mathrm {d}u} \mathrm {d}t \right) . \end{array} \end{aligned}$$

Further, we obtain:

$$\begin{aligned} \begin{array}{cl} B&=\underbrace{(c_1+m) D_1E\left( \int \limits _0^{\tau _2} e^{-\int \limits _0^s (r_u+m) \mathrm {d}u}\mathrm {d}s \right) }_{B_1}+ \underbrace{\pi _1 D_1 E\left( e^{-\int \limits _0^{\tau _2} (r_u+m)\mathrm {d}u} \right) }_{B_2}. \end{array} \end{aligned}$$

From Corollary 3.1, we obtain:

$$\begin{aligned} \begin{array}{cl} B_1&=(c_1+m) D_1 \left( \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+m,-)}e^{Q^{(\varvec{{\widehat{r}}}+m,-)}\left( x-x_C\right) }H(0)-\varvec{J}_0\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n. \end{array} \end{aligned}$$

Then, we compute:

$$\begin{aligned} B_2=\pi _1 D_1 E\left( e^{-\int \limits _0^{\tau _2} \left( r_u+m\right) \mathrm {d}u} \right) = \pi _1 D_1 \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+m,-)}e^{Q^{(\varvec{{\widehat{r}}}+m,-)}\left( x-x_C\right) }{\mathbf {1}}_{2n}. \end{aligned}$$

Finally, we have:

$$\begin{aligned} \begin{array}{cl} B&{}=(c_1+m) D_1 \left( \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+m,-)}e^{Q^{(\varvec{{\widehat{r}}}+m,-)}\left( x-x_C\right) }H(0)-\varvec{J}_0\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad +\pi _1 D_1 \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+m,-)}e^{Q^{(\varvec{{\widehat{r}}}+m,-)}\left( x-x_C\right) }{\mathbf {1}}_{2n}. \end{array} \end{aligned}$$

1.2 Valuation of Deposit Insurance

We compute:

$$\begin{aligned} \begin{array}{cl} E\left( \left. \int \limits _0^{\infty }ke^{-kt} \int \limits _{\tau _2}^{\tau _2+t}e^{-\int \limits _{\tau _2}^s r_u \mathrm {d}u}c_2D_2 \mathrm {d}s \mathrm {d}t\right| {\mathcal {F}}_{\tau _2}\right) &{}=E\left( \left. k c_2 D_2\int \limits _{\tau _2}^{\infty }\int \limits _{s-\tau _2}^{\infty } e^{-kt}e^{-\int \limits _{\tau _2}^s r_u \mathrm {d}u} \mathrm {d}t\mathrm {d}s \right| {\mathcal {F}}_{\tau _2}\right) \\ &{}=c_2 D_2 E\left( \left. \int \limits _{\tau _2}^{\infty }e^{-\int \limits _{\tau _2}^s \Big (k+r_u\Big )\mathrm {d}u}\mathrm {d}s \right| {\mathcal {F}}_{\tau _2}\right) \\ &{}=-c_2 D_2 \varvec{J}_{\tau _2}\Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n \\ \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{cl} E\left( \left. \int \limits _0^{\infty }ke^{-kt} e^{-\int \limits _{\tau _2}^{\tau _2+t} r_u \mathrm {d}u}D_2 \mathrm {d}t\right| {\mathcal {F}}_{\tau _2}\right) &{}=\int \limits _0^{\infty } kD_2 \varvec{J}_{\tau _2} e^{\Big (Q-\text {diag}\big (k+\varvec{{\widehat{r}}}\big )\Big )t}{\mathbf {1}}_n \mathrm {d}t\\ &{}=-kD_2 \varvec{J}_{\tau _2} \Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n. \end{array} \end{aligned}$$

Then,

$$\begin{aligned} \begin{array}{cl} DI&=\sum \limits _{i=1}^n E\left( e^{-\int \limits _0^{\tau _2} r_u \mathrm {d}u}\mathbb {1}_{\varvec{J}_{\tau _2}=\varvec{e}_i} \bigg (-(c_2+k) D_2 {\varvec{e}_i}\Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n-\pi _2 D_2\bigg )^+\right) . \end{array} \end{aligned}$$

Finally, we define the $n \times 1$ vector $\varvec{K}$ for simple representation and obtain:

$$\begin{aligned} \begin{array}{cl} DI&=\varvec{Y}_0W^{(r,-)}e^{Q^{(r,-)}\left( x-x_C\right) }H(0)\varvec{K}. \end{array} \end{aligned}$$

1.3 Valuation of CoCos

We have:

$$\begin{aligned} \begin{array}{cl} C=D_3\int \limits _0^{\infty }le^{-lt}C(t)\mathrm {d}t \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{cl} C&{}=D_3E\left( \int \limits _0^{\infty }le^{-lt} \int \limits _0^{\min (t,\tau _1)}e^{-\int \limits _0^s r_u \mathrm {d}u}c_3 \mathrm {d}s \mathrm {d}t \right) +(1-\rho )E\left( \int \limits _0^{\infty }le^{-lt}\mathbb {1}_{\{{\tau _1}\le t\}}e^{-\int \limits _0^{\tau _1}r_u \mathrm {d}u }S_{\tau _1} \mathrm {d}t \right) \\ &{}\quad +D_3E\left( \int \limits _0^{\infty }le^{-lt}\mathbb {1}_{\{{\tau _1}>t\}}e^{-\int \limits _0^t r_u \mathrm {d}u} \mathrm {d}t \right) . \end{array} \end{aligned}$$

Further, we obtain:

$$\begin{aligned} \begin{array}{cl} C&{}=c_3 D_3E\left( \int \limits _0^{\tau _1} e^{-\int \limits _0^s (l+r_u) \mathrm {d}u}\mathrm {d}s \right) +\Bigg (1-\rho \Bigg )E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u }S_{\tau _1} \right) +l D_3 E\left( \int \limits _0^{\tau _1}e^{-\int \limits _0^t (r_u+l) \mathrm {d}u} \mathrm {d}t \right) \\ \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{cl} C&=\underbrace{D_3(c_3+l)E\left( \int \limits _0^{\tau _1} e^{-\int \limits _0^s (l+r_u) \mathrm {d}u}\mathrm {d}s \right) }_{C_1}+\underbrace{\Bigg (1-\rho \Bigg )E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u }S_{\tau _1} \right) }_{C_2}. \end{array} \end{aligned}$$

From Corollary 3.1, we have:

$$\begin{aligned} \begin{array}{cl} C_1&=D_3(c_3+l)\left( \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }H(0)-\varvec{J}_0\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+l)\Big )^{-1}{\mathbf {1}}_n. \end{array} \end{aligned}$$

We compute $S_{\tau _1}$ by subtracting the value of straight bonds and deposits from the bank’s value at conversion time $\tau _1$. The bank’s value at conversion time $\tau _1$ is

$$\begin{aligned} \begin{array}{cl} BV_{\tau _1}&{}=V_{\tau _1}+\underbrace{\gamma E\left( \left. \int \limits _{\tau _1}^{\tau _2}e^{-\int \limits _{\tau _1}^s r_u \mathrm {d}u} \bigg (c_1D_1+c_2D_2\bigg ) \mathrm {d}s\right| {\mathcal {F}}_{\tau _1}\right) }_{\text {tax benefits}}+DI_{\tau _1}\\ &{}\quad -\underbrace{E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u}\left( \left( 1-\kappa \right) V_{\tau _2}-\pi _1D_1-\pi _2D_2\right) \right| {\mathcal {F}}_{\tau _1}\right) }_{\text {bankruptcy costs}}-\underbrace{E\left( \left. \int \limits _{\tau _1}^{\tau _2} e^{-\int \limits _{\tau _1}^{s} r_u \mathrm {d}u}\varsigma D_2\mathrm {d}s\right| {\mathcal {F}}_{\tau _1}\right) }_{\text {deposit insurance premiums}} \end{array} \end{aligned}$$

so that

$$\begin{aligned} \begin{array}{cl} BV_{\tau _1}&{}=V_{\tau _1}+(\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2) \left( \left. E\left( e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \varvec{J}_{\tau _2}\right| {\mathcal {F}}_{\tau _1}\right) -\varvec{J}_{\tau _1}\right) \left( Q-\text {diag}(\varvec{{\widehat{r}}})\right) ^{-1}{\mathbf {1}}_n\\ &{}\quad +DI_{\tau _1}-(1-\kappa )V_0 E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} e^{X_{\tau _2}}\right| {\mathcal {F}}_{\tau _1}\right) +(\pi _1D_1+\pi _2D_2)E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \right| {\mathcal {F}}_{\tau _1}\right) . \end{array} \end{aligned}$$

Then, the value of the bank’s equity $S_{\tau _1}$ is

$$\begin{aligned} S_{\tau _1}=BV_{\tau _1}-B_{\tau _1}-D_{\tau _1}. \end{aligned}$$

Then, we have:

$$\begin{aligned} \begin{array}{cl} S_{\tau _1}&{}=V_{\tau _1}+\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big ) \left( E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \varvec{J}_{\tau _2}\right| {\mathcal {F}}_{\tau _1}\right) -\varvec{J}_{\tau _1}\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad +E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u}\bigg (-(c_2+k) D_2 \varvec{J}_{\tau _2}\Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n-\pi _2 D_2\bigg )^+ \right| {\mathcal {F}}_{\tau _1}\right) \\ &{}\quad -(1-\kappa )V_0 E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} e^{X_{\tau _2}}\right| {\mathcal {F}}_{\tau _1}\right) +(\pi _1D_1+\pi _2D_2)E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \right| {\mathcal {F}}_{\tau _1}\right) \\ &{}\quad -(c_1+m)D_1\left( E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2}(r_u+m)\mathrm {d}u}\varvec{J}_{\tau _2}\right| {\mathcal {F}}_{\tau _1}\right) -\varvec{J}_{\tau _1}\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad -\pi _1 D_1 E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} \Big (r_u+m\Big )\mathrm {d}u} \right| {\mathcal {F}}_{\tau _1} \right) \\ &{}\quad -(c_2+k)D_2\left( E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2}(r_u+k)\mathrm {d}u}\varvec{J}_{\tau _2}\right| {\mathcal {F}}_{\tau _1}\right) -\varvec{J}_{\tau _1}\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+k)\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad -\pi _2 D_2 E\left( \left. e^{-\int \limits _{\tau _1}^{\tau _2} \Big (r_u+k\Big )\mathrm {d}u} \right| {\mathcal {F}}_{\tau _1}\right) . \end{array} \end{aligned}$$

Further, we have:

$$\begin{aligned} \begin{array}{cl} C_2&{}=(1-\rho )\left( V_0 e^{x_B} \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }\varvec{{\widetilde{I}}}\right. \\ &{}\quad +\underbrace{\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big ) \left( E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \varvec{J}_{\tau _2} \right) -E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}\varvec{J}_{\tau _1} \right) \right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n}_{T_1}\\ &{}\quad +\underbrace{E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u}\bigg (-(c_2+k) D_2 \varvec{J}_{\tau _2} \Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n-\pi _2 D_2\bigg )^+\right) }_{U_1}\\ &{}\quad -\underbrace{(1-\kappa )V_0 E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} e^{X_{\tau _2}}\right) }_{E_1}\\ &{}\quad +\underbrace{(\pi _1D_1+\pi _2D_2)E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}e^{-\int \limits _{\tau _1}^{\tau _2} r_u \mathrm {d}u} \right) }_{R_1}\\ &{}\quad -\underbrace{(c_1+m)D_1\left( E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}e^{-\int \limits _{\tau _1}^{\tau _2}(r_u+m)\mathrm {d}u}\varvec{J}_{\tau _2}\right) -E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}\varvec{J}_{\tau _1}\right) \right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n}_{T_2}\\ &{}\quad -\underbrace{\pi _1 D_1 E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}e^{-\int \limits _{\tau _1}^{\tau _2} \Big (r_u+m\Big )\mathrm {d}u} \right) }_{R_2}\\ &{}\quad -\underbrace{(c_2+k)D_2\left( E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}e^{-\int \limits _{\tau _1}^{\tau _2}(r_u+k)\mathrm {d}u}\varvec{J}_{\tau _2}\right) -E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}\varvec{J}_{\tau _1}\right) \right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+k)\Big )^{-1}{\mathbf {1}}_n}_{T_3}\\ &{}\quad -\left. \underbrace{\pi _2 D_2 E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}e^{-\int \limits _{\tau _1}^{\tau _2} \Big (r_u+k\Big )\mathrm {d}u} \right) }_{R_3}\right) . \end{array} \end{aligned}$$

Note that the computation of $T_2, T_3$ and $R_2, R_3$ are the same as that of $T_1$ and $R_1$, respectively. Thus, we focus on the calculation of $T_1, U_1, E_1, R_1$. Then, we have:

• $$\begin{aligned} \begin{array}{cl} T_1&{}=\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big ) \left( E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \varvec{J}_{\tau _1} \right) \right. \\ &{}\quad +E\left( \mathbb {1}_{\{X_{\tau _1}> x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \varvec{Y}_{\tau _1} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }H(0)\right) \\ &{}\quad \left. -\varvec{Y}_0 W^{(\varvec{{\widehat{r}}} +l,-)}e^{Q^{( \varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n, \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} U_1&{}=E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \bigg (-(c_2+k) D_2 \varvec{J}_{\tau _1} \Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n-\pi _2 D_2\bigg )^+\right) \\ &{}\quad +E\left( \mathbb {1}_{\{X_{\tau _1}> x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \varvec{Y}_{\tau _1} W^{( \varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }H(0)\varvec{K}\right) , \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} E_1&{}=(1-\kappa )V_0\left( E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} e^{X_{\tau _1}}\right) \right. \\ &{}\quad \left. +E\left( \mathbb {1}_{\{X_{\tau _1}> x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} e^{x_C}\varvec{Y}_{\tau _1} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }\varvec{{\widetilde{I}}}\right) \right) , \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} R_1&{}=(\pi _1D_1+\pi _2D_2)\left( E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}\right) \right. \\ &{}\quad +\left. E\left( \mathbb {1}_{\{X_{\tau _1}> x_C\}} e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \varvec{Y}_{\tau _1} W^{( \varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }{\mathbf {1}}_{2n}\right) \right) . \end{array} \end{aligned}$$

Denote by $E_i=E^+_i\cup E^0_i \cup E^-_i=\{\varvec{s}_i\}\cup \{\varvec{s}_{n+i}\} \cup \{\varvec{s}_{2n+i}\}, i=1,\ldots ,n$, the subset of E where $E^0_i, E^+_i$ and $E^-_i$ correspond to the state $\varvec{e}_i$ where X moves as a pure diffusion, makes a positive jump and makes a negative jump, respectively. The first passage time across a lower barrier can only happen when X moves as a pure diffusion or makes a negative jump, that is, $\varvec{Y}_{\tau _1}\in E^0 \cup E^-$. When $\varvec{Y}_{\tau _1}\in E^0_i$, $X_{\tau _1}=x_B$. When $\varvec{Y}_{\tau _1}\in E^-_i$, the overshoot $|X_{\tau _1}-x_B|$ is independent of $\tau _1$ and has an exponential distribution with parameter ${\widehat{\theta }}_{i}$. Since the specific distribution of $|X_{\tau _1}-x_B|$ depends on $\varvec{Y}_{\tau _1}$ and $X_{\tau _1}<x_C$ can only happen when $\varvec{Y}_{\tau _1}\in E^-$, we have:

• $$\begin{aligned} \begin{array}{cl} T_1&{}=\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big )\left( \sum \limits _{i=1}^{n}E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u}\mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\}}\right) \underbrace{E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}}|\varvec{Y}_{\tau _1} =\varvec{s}_{2n+i}\right) }_{Z_1}\varvec{e}_i\right. \\ &{}\quad +\sum \limits _{i=1}^{2n} E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \mathbb {1}_{\{ \varvec{Y}_{\tau _1} =\varvec{s}_{n+i}\}}\right) \underbrace{E\left( \varvec{s}_{n+i} W^{( \varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }\mathbb {1}_{\{X_{\tau _1}> x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\right) }_{Z_2}H(0) \\ &{}\quad \left. - \varvec{Y}_0 W^{( \varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n, \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} U_1&{}=\sum \limits _{i=1}^{n}E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u } \mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\}}\right) E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\right) \\ &{}\qquad \bigg (-(c_2+k) D_2 \varvec{e}_i \Big (Q-\text {diag}(k+\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n-\pi _2 D_2\bigg )^+\\ &{}\quad +\sum \limits _{i=1}^{2n}E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u } \mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\}}\right) E\left( \varvec{s}_{n+i} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }\mathbb {1}_{\{X_{\tau _1}> x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\right) H(0)\varvec{K}, \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} E_1&{}=(1-\kappa )V_0 \left( \sum \limits _{i=1}^{n} E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\}}\right) \underbrace{E\left( e^{X_{\tau _1}}\mathbb {1}_{\{X_{\tau _1}\le x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\right) }_{Z_3} \right. \\ &{}\quad \left. +e^{x_C}\sum \limits _{i=1}^{2n} E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u} \mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\}} \right) E\left( \varvec{s}_{n+i} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }\mathbb {1}_{\{X_{\tau _1}> x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\right) \varvec{{\widetilde{I}}} \right) , \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} R_1&{}=(\pi _1D_1+\pi _2D_2)\left( \sum \limits _{i=1}^{n} E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u }\mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\}} \right) E\left( \mathbb {1}_{\{X_{\tau _1}\le x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{2n+i}\right) \right. \\ &{}\quad \left. +\sum \limits _{i=1}^{2n} E\left( e^{-\int \limits _0^{\tau _1}(l+r_u) \mathrm {d}u } \mathbb {1}_{\{\varvec{Y}_{\tau _1}=\varvec{s}_{n+i}\}} \right) E\left( \varvec{s}_{n+i} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}\left( X_{\tau _1}-x_C\right) }\mathbb {1}_{\{X_{\tau _1}> x_C\}}|\varvec{Y}_{\tau _1}=\varvec{s}_{n+i} \right) {\mathbf {1}}_{2n}\right) . \end{array} \end{aligned}$$

Next, we compute three unknown terms $Z_1, Z_2, Z_3$. Since $X_{\tau _1}=x_B$ when $\varvec{Y}_{\tau _1}\in E^0$, we only need calculate $Z_2$ in the case that $\varvec{Y}_{\tau _1}\in E^-$. Then, we have:

• $$\begin{aligned} \begin{array}{cl} Z_1&=e^{{\widehat{\theta }}_{i}(x_C-x_B)}, \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} Z_2&{}=\varvec{s}_{2n+i} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}(x_B-x_C)}\int \limits _{(x_C-x_B)}^0 e^{Q^{(\varvec{{\widehat{r}}},-)x}}{\widehat{\theta }}_{i} e^{{\widehat{\theta }}_{i}x} \mathrm {d}x\\ &{}={\widehat{\theta }}_{i} \varvec{s}_{2n+i} W^{(\varvec{{\widehat{r}}},-)}e^{Q^{(\varvec{{\widehat{r}}},-)}(x_B-x_C)}(Q^{(\varvec{{\widehat{r}}},-)}+{\widehat{\theta }}_{i}I_{2n})^{-1}\left( I_{2n}-e^{(Q^{(\varvec{{\widehat{r}}},-)}+{\widehat{\theta }}_{i}I_{2n})(x_C-x_B)}\right) , \end{array} \end{aligned}$$
• $$\begin{aligned} \begin{array}{cl} Z_3&{}=e^{x_B}\int \limits _{-\infty }^{(x_C-x_B)} e^x {\widehat{\theta }}_{i} e^{{\widehat{\theta }}_{i}x} \mathrm {d}x\\ &{}=\displaystyle \frac{{\widehat{\theta }}_{i}}{{\widehat{\theta }}_{i}+1}e^{({\widehat{\theta }}_{i}+1)x_C-{\widehat{\theta }}_{i} x_B}. \end{array} \end{aligned}$$

Then, we define the matrices ${\mathcal {G}}_1,{\mathcal {G}}_2,{\mathcal {G}}_3, {\mathcal {M}}$ and the vector $\varvec{N}$ for simple representation and have:

$$\begin{aligned} \begin{array}{cl} C_2&{}=(1-\rho )\varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }\Bigg (V_0 e^{x_B} \varvec{{\widetilde{I}}}\\ &{}\quad +\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big ) \left( {\mathcal {M}}+{\mathcal {G}}_1 H(0)-H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad +\left( {\mathcal {M}}+{\mathcal {G}}_1 H(0)\right) \varvec{K} -(1-\kappa )V_0 \left( \varvec{N}+e^{x_C} {\mathcal {G}}_1 \varvec{{\widetilde{I}}}\right) \\ &{}\quad +\left( \pi _1D_1\left( {\mathcal {G}}_1-{\mathcal {G}}_2\right) + \pi _2D_2 \left( {\mathcal {G}}_1-{\mathcal {G}}_3\right) \right) {\mathbf {1}}_{2n}\\ &{}\quad -(c_1+m) D_1 \left( {\mathcal {M}}+{\mathcal {G}}_2 H(0)-H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n \\ &{}\quad -(c_2+k) D_2 \left( {\mathcal {M}}+{\mathcal {G}}_3 H(0)-H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+k)\Big )^{-1}{\mathbf {1}}_n \Bigg ). \end{array} \end{aligned}$$

Finally, by recalling that $C=C_1+C_2$, we have:

$$\begin{aligned} \begin{array}{cl} C&{}=D_3(c_3+l)\left( \varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }H(0)-\varvec{J}_0\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+l)\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad +(1-\rho )\varvec{Y}_0 W^{(\varvec{{\widehat{r}}}+l,-)}e^{Q^{(\varvec{{\widehat{r}}}+l,-)}\left( x-x_B\right) }\Bigg (V_0 e^{x_B} \varvec{{\widetilde{I}}}\\ &{}\quad +\Big (\gamma (c_1 D_1+c_2 D_2)-\varsigma D_2\Big ) \left( {\mathcal {M}}+{\mathcal {G}}_1 H(0)-H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}})\Big )^{-1}{\mathbf {1}}_n\\ &{}\quad +\left( {\mathcal {M}}+{\mathcal {G}}_1 H(0)\right) \varvec{K} -(1-\kappa )V_0 \left( \varvec{N}+e^{x_C} {\mathcal {G}}_1 \varvec{{\widetilde{I}}}\right) \\ &{}\quad +\left( \pi _1D_1\left( {\mathcal {G}}_1-{\mathcal {G}}_2\right) + \pi _2D_2 \left( {\mathcal {G}}_1-{\mathcal {G}}_3\right) \right) {\mathbf {1}}_{2n}\\ &{}\quad -(c_1+m) D_1 \left( {\mathcal {M}}+{\mathcal {G}}_2 H(0)-H(0)\right) \Big (Q-\text {diag}(\varvec{{\widehat{r}}}+m)\Big )^{-1}{\mathbf {1}}_n \\ &{}\quad -(c_2+k) D_2 \left( {\mathcal {M}}+{\mathcal {G}}_3 H(0)-H(0)\right) \Big (Q-{\text {diag}}(\varvec{{\widehat{r}}}+k)\Big )^{-1}{\mathbf {1}}_n \Bigg ). \end{array} \end{aligned}$$