In this section, we study the moment-generating function \(M(u,y,b)\), which has been discussed in various surplus processes; for example, see Albrecher *et al.* [11], Cheung (2008) *etc.* Similarly, we can analyze the moments of \(D(u,b)\) through \(M(u,y,b)\); since \(M(u,y,b)\) has different paths for \(-\frac{c}{\beta}< u\leq b\), we define

$$ M(u,y,b)=\left \{ \textstyle\begin{array}{l@{\quad}l} M_{1}(u,y,b), & -\frac{c}{\beta}< u< 0, \\ M_{2}(u,y,b), & 0\leq u< \triangle, \\ M_{3}(u,y,b), & \triangle\leq u\leq b. \end{array}\displaystyle \right . $$

### Theorem 2.1

\(M_{1}(u,y,b)\), \(M_{2}(u,y,b)\), *and*
\(M_{3}(u,y,b)\)
*satisfy the following system of integro*-*differential equations*:

$$\begin{aligned}& (\beta u+c)\frac{\partial M_{1}}{\partial u}(u,y,b)-\alpha y\frac {\partial M_{1}}{ \partial y}(u,y,b)-\lambda M_{1}(u,y,b)+\lambda\overline{F}\biggl(u+\frac {c}{\beta}\biggr) \\& \quad {}+\lambda \int_{0}^{u+\frac{c}{\beta}}M_{1}(u-x,y,b)\, dF(x)=0, \quad -\frac{c}{\beta}< u< 0, \end{aligned}$$

(2.1)

$$\begin{aligned}& c\frac{\partial M_{2}}{\partial u}(u,y,b)-\alpha y\frac{\partial M_{2}}{ \partial y}(u,y,b)-\lambda M_{2}(u,y,b)+\lambda \int _{0}^{u}M_{2}(u-x,y,b)\, dF(x) \\& \quad {}+\lambda \int_{u}^{u+\frac{c}{\beta}}M_{1}(u-x,y,b)\, dF(x)+ \lambda \overline{F}\biggl(u+\frac{c}{\beta}\biggr)=0, \quad 0\leq u< \triangle, \end{aligned}$$

(2.2)

$$\begin{aligned}& {\bigl[r(u-\triangle)+c\bigr]}\frac{\partial M_{3}}{\partial u}(u,y,b)-\alpha y \frac{\partial M_{3}}{ \partial y}(u,y,b)-\lambda M_{3}(u,y,b) \\& \quad {}+\lambda \int _{0}^{u-\triangle}M_{3}(u-x,y,b)\, dF(x) + \lambda \int_{u-\triangle}^{u}M_{2}(u-x,y,b)\, dF(x) \\& \quad {}+\lambda \int_{u}^{u+\frac{c}{\beta}}M_{1}(u-x,y,b)\, dF(x) + \lambda\overline{F}\biggl(u+\frac{c}{\beta}\biggr)=0,\quad \triangle\leq u\leq b, \end{aligned}$$

(2.3)

*with boundary conditions*:

$$\begin{aligned}& M_{1}\biggl(-\frac{c}{\beta},y,b\biggr)=1, \end{aligned}$$

(2.4)

$$\begin{aligned}& {\biggl.\frac{\partial M_{3}(u,y,b)}{\partial u}\biggr|_{u=b}}=yM_{3}(b,y,b), \end{aligned}$$

(2.5)

$$\begin{aligned}& M_{1}(0-,y,b)=M_{2}(0,y,b),\qquad M_{2}( \triangle-,y,b)=M_{3}(\triangle,y,b), \end{aligned}$$

(2.6)

$$\begin{aligned}& \begin{aligned} &{\biggl.\biggl(c \frac{\partial M_{1}(u,y,b)}{\partial u}-\alpha y \frac{\partial M_{1}(u,y,b)}{\partial y}\biggr)\biggr|_{u=0-}} \\ &\quad ={\biggl.\biggl(c\frac{\partial M_{2}(u,y,b)}{\partial u}-\alpha y \frac{\partial M_{2}(u,y,b)}{\partial y}\biggr)\biggr|_{u=0}}, \\ &{\biggl.\biggl(c\frac{\partial M_{2}(u,y,b)}{\partial u}-\alpha y \frac{\partial M_{2}(u,y,b)}{\partial y}\biggr)\biggr|_{u=\triangle-}} \\ &\quad = {\biggl.\biggl(c\frac{\partial M_{3}(u,y,b)}{\partial u}-\alpha y \frac{\partial M_{2}(u,y,b)}{\partial y}\biggr)\biggr|_{u=\triangle}}. \end{aligned} \end{aligned}$$

(2.7)

### Proof

When \(-\frac{c}{\beta}< u<0\), we consider the infinitesimal time from 0 to *t*, and three distinct events can happen: no claim in \((0,t)\), a claim in \((0,t)\) without occurring ruin, a claim in \((0,t)\) with occurring ruin. Conditioning on the time and amount of the first claim, we obtain that

$$\begin{aligned} M_{1}(u,y,b) =&(1-\lambda t)M_{1}\biggl(ue^{\beta t}+c \frac{e^{\beta t}-1}{\beta},y e^{-\alpha t},b\biggr)+\lambda t \int_{ue^{\beta t}+c\frac{e^{\beta t}-1}{\beta}+\frac{c}{\beta }}^{\infty}dF(x) \\ &{}+\lambda t \int_{0}^{ue^{\beta t}+c\frac{e^{\beta t}-1}{\beta}+\frac{c}{\beta }}M_{1}\biggl(ue^{\beta t}+c \frac{e^{\beta t}-1}{\beta}-x,y e^{-\alpha t},b\biggr)\,dF(x)+o(t). \end{aligned}$$

(2.8)

Letting \(h_{\beta}(t,u)=ue^{\beta t}+c\frac{e^{\beta t}-1}{\beta}-u=(c+\beta u)\frac{e^{\beta t}-1}{\beta}\), we observe that \(h_{\beta}(t,u)\rightarrow0\) as \(t\rightarrow0\). By Taylor expansion we have

$$\begin{aligned}& M_{1}\biggl(ue^{\beta t}+c\frac{e^{\beta t}-1}{\beta},y e^{-\alpha t},b\biggr) \\& \quad =M_{1}(u,y,b)+(\beta u+c)t\frac{\partial M_{1}(u,y,b)}{\partial u}-\alpha yt \frac{\partial M_{1}(u,y,b)}{\partial y}+o(t). \end{aligned}$$

Substituting this expression into (2.8), dividing both sides of (2.8) by *t*, and letting \(t\rightarrow0\), we get (2.1). Similarly, we obtain (2.2) and (2.3).

When \(u=-\frac{c}{\beta}\), absolute ruin is immediate, namely, no dividend is paid, and we obtain (2.4).

When \(u=b\),

$$\begin{aligned} M_{3}(b,y,b) =&(1-\lambda t)e^{y[c+r(b-\triangle)]t}M_{3}\bigl(b,y e^{-\alpha t},b\bigr) \\ &{}+\lambda t \int_{0}^{b-\triangle}M_{3}\bigl(b-x,y e^{-\alpha t},b\bigr)\,dF(x) \\ &{}+\lambda t \int_{b-\triangle}^{b}M_{2}\bigl(b-x,y e^{-\alpha t},b\bigr)\,dF(x) \\ &{}+\lambda t \int_{b}^{b+\frac{c}{\beta}} M_{1}\bigl(b-x,y e^{-\alpha t},b\bigr)\,dF(x) \\ &{}+\lambda t \int_{b+\frac{c}{\beta}}^{\infty}dF(x)+o(t). \end{aligned}$$

(2.9)

By similar methods we obtain the following equation from (2.9):

$$\begin{aligned} \alpha y \frac{\partial M_{3}(u,y,b)}{\partial y} =& \bigl[y \bigl(r(b-\triangle)+c \bigr)-\lambda \bigr]M_{3}(b,y,b)+\lambda \overline{F}\biggl(b+\frac{c}{\beta} \biggr) \\ &{}+\lambda \int_{0}^{b-\triangle}M_{3}(b-x,y, b)\,dF(x) + \lambda \int_{b-\triangle}^{b}M_{2}(b-x,y,b)\,dF(x) \\ &{}+\lambda \int_{b}^{b+\frac {c}{\beta}} M_{1}(b-x,y,b) \,dF(x). \end{aligned}$$

(2.10)

Letting \(u\uparrow b\) in (2.3) and comparing it with (2.10), we obtain (2.5).

Next, we prove condition (2.6). Here we only prove \(M_{2}(\triangle-,y,b)=M_{3}(\triangle,y,b)\). For \(0\leq u<\triangle\), let \(\tau_{\triangle}\) be the time that the surplus reaches △ for the first time from \(0\leq u<\triangle\). As before, we know that \(t_{1}\) is the time that the surplus reaches △ for the first time from \(0\leq u<\triangle\) with no claims. Then, by the Markov property of \(U_{b}(t)\),

$$\begin{aligned} M_{2}(u,y,b) =&E\bigl[I\bigl(\tau_{\triangle}< T_{u}^{b} \bigr)e^{yD_{u,b}}\bigr]+ E\bigl[I\bigl(\tau_{\triangle}\geq T_{u}^{b}\bigr)e^{yD_{u,b}}\bigr] \\ =&E\bigl[I\bigl(\tau_{\triangle }< T_{u}^{b} \bigr)M_{3}\bigl(\triangle,ye^{-\alpha\tau_{\triangle} },b\bigr)\bigr]+P\bigl( \tau_{\triangle}\geq T_{u}^{b}\bigr) \\ =&M_{3}(\triangle,y,b)E\bigl[e^{-\alpha \tau_{\triangle}}I\bigl( \tau_{\triangle}< T_{u}^{b}\bigr)\bigr]+P\bigl( \tau_{\triangle }\geq T_{u}^{b}\bigr) \\ \leq& M_{3}(\triangle,y,b)+P\bigl(\tau_{\triangle}\geq T_{u}^{b}\bigr). \end{aligned}$$

(2.11)

On the other hand, we have

$$\begin{aligned} M_{2}(u,y,b) \geq& E\bigl[I\bigl(\tau_{\triangle}< T_{u}^{b}, \tau_{\triangle}=t_{1}\bigr)e^{yD_{u,b}}\bigr]+ E\bigl[I\bigl( \tau_{\triangle}< T_{u}^{b}\bigr)e^{yD_{u,b}}\bigr] \\ =&E\bigl[I\bigl(\tau_{\triangle}< T_{u}^{b}, \tau_{\triangle}=t_{1}\bigr)M_{3}\bigl( \triangle,ye^{-\alpha t_{1} },b\bigr)\bigr]+P\bigl(\tau_{\triangle}\geq T_{u}^{b}\bigr) \\ =&M_{3}(\triangle,y,b)e^{-\alpha t_{1}}P(T_{1}>t_{1})+P \bigl(\tau_{\triangle}\geq T_{u}^{b}\bigr) \\ \geq& e^{-(\lambda+\alpha)t_{1}}M_{3}(\triangle,y,b)+P\bigl( \tau_{\triangle }\geq T_{u}^{b}\bigr), \end{aligned}$$

(2.12)

where \(T_{1}\) is the first claim time. As \(u\uparrow\triangle\), \(\tau_{\triangle}\) and \(t_{1}\) both tend to zero, and \(\lim_{u\uparrow\triangle}P(\tau_{\triangle}\geq T_{u}^{b})=0\); letting \(u\uparrow\triangle\) in (2.11) and (2.12), we get \(M_{2}(\triangle-,y,b)=M_{3}(\triangle,y,b)\).

Further, letting \(u\uparrow0\) in (2.1), \(u\downarrow0\) in (2.2), and using (2.6), and then letting \(u\uparrow\triangle\) in (2.2), \(u\downarrow\triangle\) in (2.3), and using (2.6), we get (2.7). □

### Remark 2.1

When \(\triangle=0\), the conclusions are consistent with Wang *et al.* [4]. Write

$$ V_{n}(u,y,b)=\left \{ \textstyle\begin{array}{l@{\quad}l} V_{n1}(u,y,b), & -\frac{c}{\beta}< u< 0, \\ V_{n2}(u,y,b), & 0\leq u< \triangle, \\ V_{n3}(u,y,b), & \triangle\leq u\leq b. \end{array}\displaystyle \right . $$

### Theorem 2.2

*The moment of the discounted dividend payments until absolute ruin satisfies the following integro*-*differential equations*:

$$\begin{aligned}& (\beta u+c)V_{n1}'(u,b)-(\lambda+n\alpha)V_{n1}(u,b) \\& \quad {}+\lambda \int_{0}^{u+\frac{c}{\beta}}V_{n1}(u-x,b)\,dF(x)=0,\quad - \frac{c}{\beta}< u< 0, \end{aligned}$$

(2.13)

$$\begin{aligned}& cV_{n2}'(u,b)-(\lambda+n\alpha)V_{n2}(u,b)+ \lambda \int _{0}^{u}V_{n2}(u-x,b)\,dF(x) \\& \quad {} +\lambda \int_{u}^{u+\frac{c}{\beta}}V_{n1}(u-x,b)\,dF(x) =0 \end{aligned}$$

(2.14)

*for*
\(0\leq u<\triangle\), *and*, *for*
\(\triangle\leq u\leq b\),

$$\begin{aligned}& \bigl[r(u-\triangle)+c\bigr]V_{n3}'(u,b)-(\lambda+n\alpha )V_{n3}(u,b)+\lambda \int_{0}^{u-\triangle}V_{n3}(u-x,b)\,dF(x) \\& \quad {}+\lambda \int_{u-\triangle}^{u}V_{n2}(u-x,b)\,dF(x) +\lambda \int_{u}^{u+\frac{c}{\beta}}V_{n1}(u-x,b) \,dF(x)=0, \end{aligned}$$

(2.15)

*with the following conditions*:

$$\begin{aligned}& V_{n1}\biggl(-\frac{c}{\beta},b\biggr)=0, \end{aligned}$$

(2.16)

$$\begin{aligned}& V_{n3}'(u,b)|_{u=b}=nV_{(n-1)3}(b,b), \end{aligned}$$

(2.17)

$$\begin{aligned}& V_{n1}(0-,b)=V_{n2}(0,b),\qquad V_{n2}( \triangle-,b)=V_{n3}(\triangle,b), \end{aligned}$$

(2.18)

$$\begin{aligned}& V_{n1}'(0-,b)=V_{n2}'(0,b),\qquad V_{n2}'(\triangle-,b)=V_{n3}'( \triangle,b). \end{aligned}$$

(2.19)

### Proof

The proof is obvious and we omit it here. □

### Corollary 2.1

*For*
\(n=1\), *we retain the risk process*, *and indeed* (2.13), (2.14), *and* (2.15) *can be simplified to*

$$\begin{aligned}& (\beta u+c)V_{11}'(u,b)-(\lambda+\alpha)V_{11}(u,b) \\& \quad {}+\lambda \int_{0}^{u+\frac{c}{\beta}}V_{11}(u-x,b)\,dF(x)=0,\quad - \frac{c}{\beta}< u< 0, \end{aligned}$$

(2.20)

$$\begin{aligned}& cV_{12}'(u,b)-(\lambda+\alpha)V_{12}(u,b)+ \lambda \int _{0}^{u}V_{12}(u-x,b)\,dF(x) \\& \quad {}+\lambda \int_{u}^{u+\frac {c}{\beta}}V_{11}(u-x,b)\,dF(x) =0 \end{aligned}$$

(2.21)

*for*
\(0\leq u<\triangle\), *and for*
\(\triangle\leq u\leq b\),

$$\begin{aligned}& \bigl[r(u-\triangle)+c\bigr]V_{13}'(u,b)-( \lambda+\alpha )V_{13}(u,b)+\lambda \int_{0}^{u-\triangle}V_{13}(u-x,b)\,dF(x) \\& \quad {}+\lambda \int_{u-\triangle}^{u}V_{12}(u-x,b)\,dF(x) +\lambda \int_{u}^{u+\frac{c}{\beta}}V_{11}(u-x,b)\,dF(x)=0. \end{aligned}$$

(2.22)

*Correspondingly*, *the boundary condition can be simplified to*

$$\begin{aligned}& V_{11}\biggl(-\frac{c}{\beta},b\biggr)=0, \end{aligned}$$

(2.23)

$$\begin{aligned}& V_{13}'(u,b)|_{u=b}=1, \end{aligned}$$

(2.24)

$$\begin{aligned}& V_{11}(0-,b)=V_{12}(0,b),\qquad V_{12}( \triangle-,b)=V_{13}(\triangle,b), \end{aligned}$$

(2.25)

$$\begin{aligned}& V_{11}'(0-,b)=V_{12}'(0,b),\qquad V_{12}'(\triangle-,b)=V_{13}'( \triangle,b). \end{aligned}$$

(2.26)

### Remark 2.2

When \(\triangle=0\), (2.20) and (2.22) are reduced to (2.1) and (2.2) of Yuen *et al.* [3], and (2.23)-(2.26) are reduced to (A_{1})-(A_{4}) of Yuen *et al.* [3], respectively.