2. The model

We consider companies with deterministic expenses and random gains, its available capital can be described by the process $\{U(t)\}_{t\geq 0}$ (in the absence of dividends) defined via

(2.1) \begin{equation} U(t)=u-ct+S(t)=u-ct+\sum_{i=1}^{N(t)}X_{i},\quad t\geq0, \end{equation}

where $U(0)=u\geq 0$ is the initial surplus and $c>0$ is the constant expense rate per unit time. The premium number process $\{N(t)\}_{t\geq 0}$ is a homogeneous Poisson process with intensity $\lambda >0$, and the gain amounts $\{X_{i}\}_{i=1}^{\infty }$ are mutually independent and identically distributed (i.i.d.), and also independent of $\{N(t)\}_{t\geq 0}$. The time of ruin is given by $\tau _{0}=\inf \{t\geq 0;\ U(t)=0\}$.

Inspired by the articles mentioned in Section 1, this paper aims to propose a mixed dividend strategy with a Parisian implementation delays in the dual risk model (2.1), which extends the work by Cheung and Wong [Reference Cheung and Wong8]. Then, the surplus process is denoted by $\{U_{b}^{d}\}_{t\geq 0}$. At the observation times $\{\nu _{i}\}_{i=0}^{\infty },~\nu _{0}=0$, if the level of surplus $x$ is observed to exceed the previously given barrier $b$, then the dividend mixture begins. Defining $T_{i}=\nu _{i}-\nu _{i-1}$ for $i=1,2,\ldots,$ and assumed that the inter-observation times $\{T_{i}\}_{i=1}^{\infty }$ are i.i.d. with same distribution as $T$ and are independent of $\{N(t)\}_{t\geq 0}$ and $\{X_{i}\}_{i=1}^{\infty }$. We use $\{V_{i}\}_{i=1}^{\infty }$ to denote the $i$th Parisian implementation delay when $\{U_{b}^{d}\}_{t\geq 0}$ is observed for the $i$th time above $b$. It is assumed that the delays $\{V_{i}\}_{i=1}^{\infty }$ form a sequence of i.i.d. positive random variables that are independent of $\{N(t)\}_{t\geq 0}$, $\{T_{i}\}_{i=1}^{\infty }$ and $\{X_{i}\}_{i=1}^{\infty }$. Now, define the threshold dividend model $U_{b}(t)$ based on Model 2.1, and the auxiliary process $W_{i}(t)$, $i=1,2,\ldots$

\begin{align*} U_{b}(t)=\left\{ \begin{array}{ll} U(t), & 0\leq t<\nu_{1},\\ U(t), & \nu_{j}\leq t\leq\nu_{1}^{+},\ j=1,2,\ldots,\\ \displaystyle U(\nu_{1}^{+})-(c+c_{1})(t-\nu_{1}^{+})+\sum_{i=N(\nu_{1}^{+})+1}^{N(t)}X_{i}, & \nu_{1}^{+}< t<\eta_{1}, \end{array}\right.\\ W_{i}(t)=\left\{ \begin{array}{ll} U_{b}(t), & t\geq0,\ i=1,\\ \displaystyle b-c(t-\eta_{i-1})-c_{1}(t-\nu_{i}^{+})+\sum_{i=N(\eta_{i-1})+1}^{N(t)}X_{i}, & t\geq \eta_{i-1},\ i=2,3,\ldots, \end{array}\right. \end{align*}

where $\nu _{i}^{+}=\inf \{\nu _{j}\geq \eta _{i-1};\ W_{i}(\nu _{j})>b\}$, $i = 1,2,\ldots$ is the first time $\{W_{i}(t)\}_{t\geq \eta _{i-1}}$ is above the dividend barrier; $\eta _{i}=(\nu _{i}^{+}+V_{i})\wedge \theta _{i}$ for $i=1,2,\ldots$, with the starting point $\eta _{0}=0\ (x\wedge y=\min (x,~y))$; whereas $\theta _{i}=\inf \{t>\nu _{i}^{+};\ W_{i}(t)=b\}$, $i=1,2,\ldots$ represents the first time $\{W_{i}(t)\}_{t\geq \nu _{i}^{+}}$ down-crosses level $b$ due to the expense rate and threshold dividend. The surplus process $\{U_{b}^{d}(t)\}_{t\geq 0}$ can now be characterized by

(2.2) \begin{equation} U_{b}^{d}(t)=W_{i}(t),\quad \eta_{i-1}\leq t\leq\eta_{i},\quad i=1,2,\ldots. \end{equation}

From the previous hypothesis, we noticed that if $\{W_{i}(t)\}_{t\geq \nu _{i}^{+}}$ stays above the dividend barrier continuously for a period of $V_{i}$ such that $W_{i}(\nu _{i}^{+}+V_{i})>b$ (or equivalently, $\nu _{i}^{+}+V_{i}\leq \theta _{i}$ so that $\eta _{i}=\nu _{i}^{+}+V_{i})$, then a dividend of $W_{i}(\eta _{i})-b$ will be paid at time $\eta _{i}$, dragging the process $\{U_{b}^{d}(t)\}_{t\geq 0}$ back to level $b$. On the other hand, if $\{W_{i}(t)\}_{t\geq \nu _{i}^{+}}$ drops below $b$ within a period of length $V_{i}$ (i.e. $\theta _{i}\leq \nu _{i}^{+}+V_{i})$ so that $\eta _{i}=\theta _{i}$, then no Parisian dividend will be paid at time $\eta _{i}$. The time of ruin in this modified model $\{U_{b}^{d}(t)\}_{t\geq 0}$ is defined as $\tau =\inf \{\nu _{i};\ U_{b}^{d}(\nu _{i})\leq 0\}$. For convenience, we let $c+c_{1}=c_{2}$.

To illustrate the features of $\{U_{b}^{d}(t)\}_{t\geq 0}$, we plot a sample path of it in Figure 1, where “type 1” and “type 2” represent dividends generated by continuous dividend payments at rate $c_{1}$ and Parisian implementation delays dividend, respectively. In this paper, we are interested in the Gerber-Shiu expected discounted penalty function that is defined as (classical risk model)

$$\phi(u):=E^{u}[e^{-\delta \tau}\omega(U(\tau-),|U(\tau)|)I_{\{\tau<\infty\}}],\quad u\geq0,$$

where $\delta \geq 0$ is the Laplace transform argument, and $E^{u}$ is the expectation of the initial surplus $u$, $I_{\{\tau <\infty \}}$ is an indicator function, $\omega :[0,\infty )\times [0,\infty )\rightarrow [0,\infty )$ is a measurable penalty function of the $U(\tau -)$ and $|U(\tau )|$. It has become an important and standard risk measure in ruin theory since various quantities of interests in ruin theory can be obtained for different values of the discount factor $\delta$ and different penalty functions $\omega$. For recent research progress on the Gerber-Shiu function, we can refer to work by Lin et al. [Reference Lin, Willmot and Drekic18], Yuen et al. [Reference Yuen, Wang and Li32], Zhao and Yin [Reference Zhao and Yin35], Chi and Lin [Reference Chi and Lin10], Deng et al. [Reference Deng, Liu, Huang, Li and Zhou15], Zhang and Su [Reference Zhang and Su33], Xie and Zhang [Reference Xie and Zhang27], among others. In this paper, we consider the Gerber-Shiu expected discounted penalty function is simply (a constant multiple of) the Laplace transform of the time of ruin given by

(2.3) \begin{equation} E^{u}[e^{-\delta \tau}I_{\{\tau<\infty\}}]=E^{u}[e^{-\delta \tau}]=\phi_{b}(u) =\left\{\begin{array}{ll} \phi_{Lb}(u), & u<0,\\ \phi_{Mb}(u), & 0\leq u\leq b,\\ \phi_{Ub}(u), & b< u,\\ \end{array}\right. \end{equation}

where the subscripts “L,” “M” and “U” stand for “lower,” “middle” and “upper” layers, respectively. Note that we have omitted the indicator $I_{\{\tau <\infty \}}$ of the event $\{\tau <\infty \}$ in the definition (2.3) because ruin occurs with probability one in the presence of a barrier. In addition to the Laplace transform of the time of ruin, we also defined about the expected discounted dividend payments before ruin, which is

(2.4) \begin{align} & E^{u}\left[\sum_{i=1}^{\infty}e^{-\delta \eta_{i}}[U^{d}_{b}(\eta_{i}^{-})-b+c_{1}\int_{\eta_{i}^{-}}^{\nu_{i}^{+}}e^{-\delta t}\,dt]I_{\{\eta_{i}<\tau\}}\right]\nonumber\\ & \quad =V_{b}(u)=\left\{ \begin{array}{ll} V_{Lb}(u), & u<0,\\ V_{Mb}(u), & 0\leq u\leq b,\\ V_{Ub}(u), & b< u. \end{array}\right. \end{align}

Remark 2.1. Theoretically, the classical risk model and the dual risk model have great similarity, the two are mutual reflection in nature. Through the duality principle, studying a problem in one model can often provide ideas or even directly solve the problem in another model. Therefore, the classical risk model and some corresponding existing results are introduced below which will be applied to calculate Eqs. (2.3) and (2.4) later. We first introduce the classical compound Poisson insurance risk process $\{U_{S}(t)\}_{t\geq 0}$ which is defined by

(2.5) \begin{equation} U_{S}(t)=u+ct-S(t)=u+ct-\sum_{i=1}^{N(t)}X_{i},\quad t\geq0, \end{equation}

where $U_{S}(0)=u\geq 0$ is the initial surplus, $c>0$ is now the incoming premium rate per unit time, $S(t)$ represents the aggregate claim amounts by time $t$ and $\{X_{i}\}_{i=1}^{\infty }$ is interpreted as the sequence of insurance claims. Considering the periodic observation $\{T_{i}\}_{i=1}^{\infty }$, the model (2.5) becomes

(2.6) \begin{equation} U_{S}(\nu_{i})=U_{S}(\nu_{i-1})+cT_{i}-[S(\nu_{i})-S(\nu_{i-1})],\quad i=1,2,\ldots. \end{equation}

Figure 1. Sample path of $\{U_{b}^{d}(t)\}_{t\geq 0}$.

The time of ruin is defined by $\tau _{d}=\nu _{k_{b}}$, where $k_{b}=\inf \{k\geq 1:U_{S}(\nu _{k})<0\}$ is the number of observation intervals before ruin. Similarly, we define the moment when the surplus process first crosses the barrier $b$ as $\tau _{b}=\nu _{k_{b}^{\ast }}$, where $k_{b}^{\ast }=\inf \{k\geq 1:U_{S}(\nu _{k})\geq b\}$ is the number of observation intervals before the first crossing barrier $b$. To derive the Gerber-Shiu expected discounted penalty function $\phi _{b}(u)$ and the expected discounted dividend payments before ruin $V_{b}(u)$, we made the following assumption.

Let

(2.7) \begin{equation} \lambda[\hat{f}_{X}(\varepsilon)-1]+c\varepsilon=\delta+\beta, \end{equation}

where $f_{X}(x)$ is the density function of single claim quantity, let $\hat {f}_{X}(s)=\int _{0}^{\infty }e^{-sx}f_{X}(x)\,dx$ denote the Laplace transform of the claim size density. Therefore, Eq. (2.7) can be simplified to

(2.8) \begin{equation} \varepsilon^{2}+\left(a_{1}-\frac{\lambda+\beta+\delta}{c}\right)\varepsilon-\frac{(\beta+\delta)a_{1}}{c}=0. \end{equation}

By Eq. (2.8) in Albrecher et al. [Reference Albrecher, Cheung and Thonhauser2], we known that it has a unique negative solution $\rho _{1}<0$ and a positive solution $\rho _{2}>0$. Note that for $\beta =0$, Eq. (2.8) reduces to the well-known Lundberg fundamental equation of the compound Poisson risk process. There is also a unique negative root $\rho _{1}^{0}<0$, and the only positive root $\rho _{2}^{0}>0$.

We define the discounted density of the ruin deficit of $\{U_{S}(t)\}_{t\geq 0}$ at random observation to be $h_{\delta }^{+}(y\,|\,u),\ u\geq 0, h_{\delta }^{-}(y\,|\,u),\ u<0$. According to the Eqs. (2.16) and (2.17) of Albrecher et al. [Reference Albrecher, Cheung and Thonhauser2]

(2.9) \begin{equation} h_{\delta}^{+}(y\,|\,u)=(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}u+\rho_{1}y},\quad u\geq0, \end{equation}

(2.10) \begin{align} h_{\delta}^{-}(y\,|\,u)& =\frac{\beta(a_{1}+\rho_{2})(\rho_{1}^{0}-\rho_{1})}{c(\rho_{2}-\rho_{1})(\rho_{2}-\rho_{1}^{0})}e^{\rho_{2}u +\rho_{1}y}+\frac{\beta(a_{1}+\rho_{2})}{c(\rho_{2}-\rho_{1})}e^{\rho_{2}(u+y)}I _{\{y\leq{-}u\}}\nonumber\\& \quad +\frac{\beta(a_{1}-\rho_{1})}{c(\rho_{2}-\rho_{1})}e^{\rho_{1}(u+y)}I_{\{y>{-}u\}},\quad u<0. \end{align}

We calculate the dividend function $V_b(x)$ by analyzing the discounted density of the increment of the process $\{U_{S}(t)\}_{t\geq 0}$ between continuous observation time points (see [Reference Albrecher, Cheung and Thonhauser2]). Due to the Markovian structure of $\{U_{S}(t)\}_{t\geq 0}$, this sequence of pairs is i.i.d. with generic distribution $(T, \sum _{i=1}^{N(T)}X_{i}-cT)$ and joint Laplace transform is

(2.11) \begin{equation} E\left[e^{-\delta T-s\left(\sum_{i=1}^{N(T)}X_{i}-cT\right)}\right]=E[e^{-[\lambda+\delta-cs-\lambda M_{X}({-}s)]T}]=\int_{-\infty}^{\infty}e^{{-}sy}g_{\delta}(y)\,dy, \end{equation}

where $g_{\delta }(y)(-\infty < y<\infty )$ represents the discounted density of the increment $\sum _{i=1}^{N(T)}X_{i}-cT$ between successive observation times, $M_{X}(s)$ is the moment generating function of $X$, discounted at rate $\delta$ with respect to time $T$. According to the Example (4.1) of Albrecher et al. [Reference Albrecher, Cheung and Thonhauser2], they gives

(2.12) \begin{equation} g_{\delta}^{-}(y)=\frac{\beta(a_{1}+\rho_{2})}{c(\rho_{2}-\rho_{1})}e^{\rho_{2}y},\quad g_{\delta}^{+}(y)=\frac{\beta(a_{1}+\rho_{1})}{c(\rho_{2}-\rho_{1})}e^{\rho_{1}y},\quad y>0. \end{equation}

3. The Laplace transform of the time of ruin

In this section, we study the Laplace transform of the time of ruin $\phi _{b}(u)$. It can be seen from Eq. (2.3) that the form of $\phi _{b}(u)$ varies with the initial surplus $u$. Therefore, we first consider the case where the initial surplus $0\leq u\leq b$. The case where the initial surplus $u<0$ and $u>b$ will be resolved later.

For $0\leq u\leq b$, we need to distinguish whether ruin occurs before the process was first observed above or below level $b$ (i.e. ruin occurs before or after $\nu _{1}^{+}$)

(3.1) \begin{equation} \phi_{Mb}(u)=E^{u}[e^{-\delta\tau}I_{\{\tau<\nu_{1}^{+}\}}] +E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]\phi_{Mb}(b). \end{equation}

Similarly, let $u=b$ in the above equation and bring back to the above equation to get

(3.2) \begin{equation} \phi_{Mb}(u)=E^{u}[e^{-\delta\tau}I_{\{\tau<\nu_{1}^{+}\}}] +\frac{E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]E^{b}[e^{-\delta\tau}I_{\{\tau<\nu_{1}^{+}\}}]} {1-E^{b}[e^{-\delta\eta_{^{1}}}I_{\{\nu_{1}^{+}<\tau\}}]}. \end{equation}

It can be seen from Eq. (3.2) that the key to calculate the Laplace transform of the ruin time is to calculate $E^{u}[e^{-\delta \tau }I_{\{\tau <\nu _{1}^{+}\}}]$ and $E^{u}[e^{-\delta \eta _{1}}I_{\{\nu _{1}^{+}<\tau \}}]$ for $0\leq u\leq b$. Next, we will calculate these two expressions separately.

3.1. The discussion of $E^{u}[e^{-\delta \tau }I_{\{\tau <\nu _{1}^{+}\}}]$, $0\leq u\leq b$

When $0\leq u\leq b$, the ruin occurred before the first observed surplus was above $b$, the surplus process $\{U_{b}^{d}(t)\}_{t\geq 0}$ simply behaves like the process $\{U_{b}(t)\}_{t\geq 0}$ prior to time $\nu _{1}^{+}$. As a result, the quantity $E^{u}[e^{-\delta \tau }I_{\{\tau <\nu _{1}^{+}\}}]$ is independent of the distributional assumption on the Parisian implementation delay for $0\leq u\leq b$, so that we have Figure 2.

Theorem 1. When $0\leq u\leq b$, we have

(3.3) \begin{equation} E^{u}[e^{-\delta\tau}I_{\{\tau<\nu_{1}^{+}\}}] =\frac{(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0})[(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}(b-u)}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}(b-u)}]} {\rho_{2}(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b}-(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}},\quad 0\leq u\leq~b. \end{equation}

Figure 2. Sample path before and after folding.

Proof. From Figure 3 and Remark 2.1, we want to calculate the $E^{u}[e^{-\delta \tau }I_{\{\tau <\nu _{1}^{+}\}}]$, which can be observed that the event $\{\tau <\nu _{1}^{+}\}$ in the process $\{U_{b}^{d}(t)|U_{b}^{d}(0)=u\}_{t\geq 0}$ is equivalent to the event that $\{U_{S}(t)|U_{S}(0)=b-u\}_{t\geq 0}$ reaches level $b$ before observation dropping below zero. Under such an event, $\tau$ in $\{U_{b}^{d}(t)|U_{d}(0)=u\}_{t\geq 0}$ is simply $\tau _{b}$ in $\{U_{S}(t)|U_{b}(0)=b-u\}_{t\geq 0}$. We have

(3.4) \begin{equation} E^{u}[e^{-\delta\tau}I_{\{\tau<\nu_{1}^{+}\}}]=E[e^{-\delta \tau_{b}}I_{\{\tau_{b}<\tau_{d}\}}|U_{S}(0)=b-u],\quad 0\leq u\leq b, \end{equation}

which can be seen as the Laplace transform of $\tau _{b}$ when $\{U_{S}(t)\}_{t\geq 0}$ is above $b$ for the first time before ruin. In the spirit of Albrecher et al. [Reference Albrecher, Cheung and Thonhauser1], suppose a penalty function $\omega ^{\ast }(\cdot )$ is applied to the first overshoot of $\{U_{S}(t)\}_{t\geq 0}$ over level $b$ avoiding ruin until then and define the quantity

(3.5) \begin{equation} \chi(u)=E[e^{-\delta \tau_{b}}\omega^{{\ast}}(U_{S}(\tau_{b})-b)I_{\{\tau_{b}<\tau_{d}\}}|U_{S}(0)=u],\quad 0\leq u\leq b. \end{equation}

According to the Section 4 of Albrecher et al. [Reference Albrecher, Cheung and Thonhauser1], we have

(3.6) \begin{equation} \chi(u)=\int_{0}^{\infty}\omega^{{\ast}}(y)h_{\delta}(y\,|\,u)\,dy,\quad 0\leq u\leq b, \end{equation}

where $h_{\delta }(y\,|\,u)$ is the discounted density of the overshoot above level $b$ avoiding ruin. Assume again that both the claim sizes and the observation intervals are exponentially distributed with mean ${1}/{a_{1}}$ and ${1}/{\beta }$, we have

(3.7) \begin{equation} h_{\delta}(y\,|\,u)=\frac{e^{-\rho_{2}y}(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0}) [(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}u}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}u}]} {(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b}-(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}},\quad y>0;\ 0\leq u\leq b. \end{equation}

Therefore, when we consider the penalty function $\omega ^{\ast }(\cdot )=1$, we obtain

(3.8) \begin{equation} \chi(u)=\int_{0}^{\infty}\frac{e^{-\rho_{2}y}(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0}) [(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}u}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}u}]} {(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b}-(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}}\,dy,\quad 0\leq u\leq b. \end{equation}

Consolidating the above observations, we arrive at Eq. (3.3). So replace $u$ in Eq. (3.8) with $b-u$.

Figure 3. Sample path before and after folding.

3.2. The discussion of $E^{u}[e^{-\delta \eta _{1}}I_{\{\nu _{1}^{+}<\tau \}}], 0\leq u\leq b$

Theorem 2. When $0\leq u\leq b$, we have

(3.9) \begin{align} E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}] & =\int_{0}^{\infty}\left(h_{\delta}^{+}(y\,|\,b-u)-\int_{0}^{\infty} E[e^{-\delta\tau_{b}}I_{\{\tau_{b}<\tau_{d}\}};U_{S}(\tau_{b})\in b+dx\,|\,U_{S}(0)=b-u]\right.\nonumber\\ & \qquad \left.h_{\delta}^{+}(y\,|\,b+x)\vphantom{\int_{0}^{\infty}}\right) E^{b+y}[e^{-\delta\eta_{1}}]\,dy, \end{align}

where

(3.10) \begin{equation} h_{\delta}^{+}(y\,|\,u)=(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}u+\rho_{1}y},\quad y>0. \end{equation}

Proof. In model (2.1), when the level of surplus crosses barrier $b$, it can be analogous to the deficit at the moment of ruin in model (2.5). Therefore, the deficit of ruin of surplus process $\{U_{b}^{d}(t)\}_{t\geq 0}$ is equivalent to the part that the surplus process of $\{U_{S}(t)\}_{t\geq 0}$ exceeds level $b$. It can be known from Eq. (2.9) that

$$h_{\delta}^{+}(y\,|\,u)=(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}u+\rho_{1}y},\quad y>0.$$

Note that for surplus process $\{U_{S}(t)\}_{t\geq 0}$, we consider the discounted density of the ruin deficit $h_{\delta }^{+}(y\,|\,u)$. One of the cases is that the time when the surplus level exceeds $b$ is observed for the first time before the ruin, which is equivalent to that in the surplus process $\{U_{b}^{d}(t)\}_{t\geq 0}$, the ruin occurs before the surplus level exceeds $b$ is observed for the first time which needs to be deducted. The sample trajectories of surplus process $\{U_{b}^{d}(t)\}_{t\geq 0}$ and surplus process $\{U_{S}(t)\}_{t\geq 0}$ are shown in Figure 3.

Such contribution can be removed by subtracting $\int _{0}^{\infty }E[e^{-\delta \tau _{b}}I_{\{\tau _{b}<\tau _{d}\}};U_{S}(\tau _{b})\in b+dx\,|\,U_{S}(0)=b-u]h_{\delta }^{+}(y\,|\,b+x)$ from $h_{\delta }^{+}(y\,|\,b-u)$. Thus, we can obtain Eq. (3.9). Since the functions $\int _{0}^{\infty }E[e^{-\delta \tau _{b}}I_{\{\tau _{b}<\tau _{d}\}};U_{S}(\tau _{b})\in b+dx\,|\,U_{S}(0)=b-u]h_{\delta }^{+}(y\,|\,b+x)$ can be obtained from the calculation of $\chi (u)$, refer to Section 4 of Albrecher et al. [Reference Albrecher, Cheung and Thonhauser1] for details, it suffices to derive $E^{u}[e^{-\delta \eta _{1}}]$ for $u>b$ in order to have a full characterization of $\phi _{Mb}(u)$. Finally, we consider $E^{u}[e^{-\delta \eta _{1}}]$, $u>b$.

According to the previous definition, when $U_{b}^{d}(0)=u>b$, since the surplus was observed above $b$ at the initial moment, one has that $\eta _{1}=\theta _{1}\wedge d$. We found that during the delayed period, due to the threshold dividend, the surplus level can only be continuously observed until the surplus again recover to the obstacle level $b$. For the results of the $E^{u}[e^{-\delta \eta _{1}}]$, $u>b$, can directly be used, where Cheung and Wong [Reference Cheung and Wong8] Eqs. (4.5), (4.9) and (4.10)

(3.11) \begin{equation} E^{u}[e^{-\delta\eta_{1}}]=E^{u-b}[e^{-\delta(\tau_{0}\wedge d)}]. \end{equation}

If ruin occurs before the first gain, then $\tau _{0}=u/c_{2}$ with probability $e^{-\lambda (u/c_{2})}$. In contrast, if there is at least one gain before ruin, then $\tau _{0}>u/c_{2}$. From the Eq. (4.38) of Seal [Reference Seal23], we know that the density function $f_{U}(t\,|\,u)$ at the time of ruin $\tau _{0}$ is

(3.12) \begin{equation} f_{U}(t\,|\,u)=\sum_{k=1}^{\infty}\frac{\lambda^{k}t^{k-1}e^{-\lambda t}(u)}{k!}p^{{\ast} k}(c_{2}t-u),\quad t>u/c_{2}, \end{equation}

where $p^{\ast k}(\cdot )$ is the $k$-fold convolution density of the gain density $p(\cdot )$ with itself. when $u\geq c_{2}d$, ruin happens after $u/c_{2}$, which is $\tau _{0}\geq d$, and we arrive at

(3.13) \begin{equation} E^{u}[e^{-\delta(\tau_{0}\wedge d)}]=e^{-\delta d},\quad u\geq c_{2}d. \end{equation}

For $0< u< c_{2}d$, conditioning on $\tau _{0}$ leads to

(3.14) \begin{align} E^{u}[e^{-\delta(\tau_{0}\wedge d)}]& =E^{u}[e^{-\delta\tau_{0}}I_{\{\tau_{0}< d\}}]+E^{u}[e^{-\delta d}I_{\{\tau_{0}\geq d\}}]\nonumber\\ & =e^{-{(\delta+\lambda)u}/{c_{2}}}+\int_{{u}/{c_{2}}}^{d}e^{-\delta t}f_{U}(t\,|\,u)\,dt\nonumber\\ & \quad +e^{-\delta d}\left(1-e^{-{\lambda u}/{c_{2}}}-\int_{{u}/{c_{2}}}^{d}f_{U}(t\,|\,u)\,dt\right),\quad 0< u< c_{2}d. \end{align}

Note that the density function $p(y)=a_{1}e^{-a_{1}y}$, $p^{\ast k}(\cdot )$ is an Erlang($k$) density and hence Eq. (3.12) becomes

(3.15) \begin{equation} f_{U}(t\,|\,u)=\sum_{k=1}^{\infty}\frac{\lambda^{k}t^{k-1}e^{-\lambda t}u}{k!}\left(\frac{a_{1}^{k}(c_{2}t-u)^{k-1}e^{{-}a_{1}(c_{2}t-u)}}{(k-1)!}\right),\quad t>u/c_{2}. \end{equation}

Then, the first integral in Eq. (3.14) is found to be

(3.16) \begin{align} & \int_{{u}/{c_{2}}}^{d}e^{-\delta t}f_{U}(t\,|\,u)\,dt\nonumber\\ & \quad =\sum_{k=1}^{\infty}\frac{(\lambda a_{1})^{k}u}{k!(k-1)!}\int_{{u}/{c_{2}}}^{d}t^{k-1}e^{-(\lambda+\delta)t}(c_{2}t-u)^{k-1} e^{{-}a_{1}(c_{2}t-u)}\,dt\nonumber\\ & \quad = \sum_{k=1}^{\infty}\frac{(\lambda a_{1}/c_{2})^{k}}{k!}\sum_{i=0}^{k-1}\frac{(k+i-1)!}{(i!(k-i-1)!)(\frac{\lambda+\delta}{c_{2}}+a_{1})^{k+i}}\nonumber\\ & \qquad \times\left(u^{k-i}e^{-\frac{\lambda+\delta}{c_{2}}u}-\sum_{j=0}^{k+i-1} \frac{[(\frac{\lambda+\delta}{c_{2}}+a_{1})(c_{2}d-u)]^{j}e^{-({(\lambda+\delta)}/{c_{2}}+a_{1})c_{2}d}}{j!}u^{k-i}e^{a_{1}u}\right). \end{align}

The remaining integral $\int _{{u}/{c_{2}}}^{d}f_{U}(t\,|\,u)\,dt$ in Eq. (3.14) is simply a special case of the above expression with $\delta =0$, and because that

(3.17) \begin{align} & h_{\delta}^{+}(y\,|\,b-u)-\int_{0}^{\infty}E[e^{-\delta\tau_{b}}I_{\{\tau_{b}<\tau_{d}\}};U_{S}(\tau_{b})\in b+dx\,|\,U_{S}(0)=b-u]h_{\delta}^{+}(y\,|\,b+x)\nonumber\\ & \quad =(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}b+\rho_{1}y}\left(e^{-\rho_{1}^{0}u} +\frac{(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0})[(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}(b-u)}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}(b-u)}]} {\rho_{1}^{0}\rho_{2}(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b}-(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}}\right). \end{align}

Application of Eqs. (3.11), (3.13) and (3.14) to Eq. (3.9) leads to

(3.18) \begin{align} & E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]\nonumber\\ & \quad=(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}b}\left(e^{-\rho_{1}^{0}u} +\frac{(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0})[(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}(b-u)}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}(b-u)}]} {\rho_{1}^{0}\rho_{2}(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b}-(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}}\right)\nonumber\\ & \qquad \times\left\{\frac{1}{\frac{\lambda+\delta}{c_{2}}-\rho_{1}} \left(1-e^{-(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d}\right)-e^{-\delta d}\left[\frac{1}{\rho_{1}}+\frac{1}{\frac{\lambda}{c_{2}}-\rho_{1}}(1-e^{-(\frac{\lambda}{c_{2}}-\rho_{1})c_{2}d})\right]\right.\nonumber\\ & \qquad \left.+\int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{{y}/{c_{2}}}^{d}e^{-\delta t}f_{U}(t|y)\,dt\,dy-e^{-\delta d}\int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{\frac{y}{c_{2}}}^{d}f_{U}(t|y)\,dt\,dy\right\},\quad 0\leq u\leq b. \end{align}

Using Eq. (3.15), the first double integral above is evaluated as

(3.19) \begin{align} & \int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{{y}/{c_{2}}}^{d}e^{-\delta t}f_{U}(t|y)\,dt\,dy\nonumber\\ & \quad =\sum_{k=1}^{\infty}\frac{(-\frac{\lambda\rho_{1}}{c_{2}})^{k}}{k!}\sum_{i=0}^{k-1} \frac{(k+i-1)!}{i!(k-i-1)!}\frac{1}{(\frac{\lambda+\delta}{c_{2}}-\rho_{1})^{k+i}} \left\{\frac{(k-i)!}{(\frac{\lambda+\delta}{c_{2}}-\rho_{1})^{k-i+1}}\right.\nonumber\\ & \qquad \times\left(1-\sum_{j=0}^{k-i}\frac{[(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d]^{j} e^{-(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d}}{j!}\right) -\sum_{j=0}^{k+i-1}\left(\frac{\lambda+\delta}{c_{2}}-\rho_{1}\right)^{j}\nonumber\\ & \qquad \left.\times e^{-(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d} \frac{(c_{2}d)^{j+k-i+1}(k-i)!}{(j+k-i+1)!}\right\}, \end{align}

and the second double integral in Eq. (3.18) can be directly set as $\delta =0$ in Eq. (3.19). With the Eqs. (3.18) and (3.3) derived, $\phi _{Mb}(u),\ 0\leq u\leq b$ are determined by Eq. (3.2).

The following two Remarks respectively give the results satisfied by $\phi _{Ub}(u)$ and $\phi _{Lb}(u)$.

Remark 3.1. When $u>b$, at the time of the first observation, there are two scenarios for surplus $\phi _{Ub}(u)$

1. 1. When the first observation is made, surplus $U_{b}^{d}(t)$ is already ruin $(U_{b}^{d}(\nu _{1})<0)$,

   (3.20) \begin{equation} \chi_{U_{1}}(u)=\int^{\infty}_{u}g_{\delta}^{-}(y)\,dy. \end{equation}

2. 2. When the first observation is occurred, the surplus $U_{b}^{d}(t)$ is above 0 $(U_{b}^{d}(\nu _{1})>0)$,

   (3.21) \begin{align} \chi_{U_{2}}(u)& =\int^{u}_{u-b}g_{\delta}^{-}(y)\phi_{Mb}(u-y)\,dy +\int^{u-b}_{0}g_{\delta}^{-}(y)E^{u-y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy\nonumber\\ & \quad +\int^{\infty}_{0}g_{\delta}^{+}(y)E^{u+y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy. \end{align}
   So $\phi _{Ub}(u)$ can be expressed as
   (3.22) \begin{align} \phi_{Ub}(u)& =\chi_{U_{1}}(u)+\chi_{U_{2}}(u)\nonumber\\ & =\int^{u}_{u-b}g_{\delta}^{-}(y)\phi_{Mb}(u-y)\,dy +\int^{\infty}_{u}g_{\delta}^{-}(y)\,dy+\int^{u-b}_{0}g_{\delta}^{-}(y)\nonumber\\ & \quad \times E^{u-y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy\nonumber\\ & \quad +\int^{\infty}_{0}g_{\delta}^{+}(y)E^{u+y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy,\quad u>b. \end{align}

Remark 3.2. When $u<0$, at the time of the first observation, there are also two scenarios for surplus $\phi _{Lb}(u)$.

1. 1. When the first observation is made, surplus $U_{b}^{d}(t)$ is already ruin $(U_{b}^{d}(\nu _{1})<0)$,

   (3.23) \begin{equation} \chi_{L_{1}}(u)=\int^{\infty}_{0}g_{\delta}^{-}(y)\,dy+\int^{{-}u}_{0}g_{\delta}^{+}(y)\,dy. \end{equation}

2. 2. When the first observation is occurred, the surplus $U_{b}^{d}(t)$ is above 0 $(U_{b}^{d}(\nu _{1})>0)$,

   (3.24) \begin{equation} \chi_{L_{2}}(u)=\int^{b-u}_{{-}u}g_{\delta}^{+}(y)\phi_{Mb}(u+y)\,dy +\int^{\infty}_{b-u}g_{\delta}^{+}(y)E^{u+y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy. \end{equation}
   Then $\phi _{Lb}(u)$ can be expressed as
   (3.25) \begin{align} \phi_{Lb}(u)& =\chi_{L_{1}}(u)+\chi_{L_{2}}(u)\nonumber\\ & =\int^{\infty}_{0}g_{\delta}^{-}(y)\,dy +\int^{{-}u}_{0}g_{\delta}^{+}(y)\,dy+\int^{b-u}_{{-}u}g_{\delta}^{+}(y)\phi_{Mb}(u+y)\,dy\nonumber\\ & \quad +\int^{\infty}_{b-u}g_{\delta}^{+}(y)E^{u+y-b}[e^{-\delta(\tau_{0}\wedge d)}]\phi_{Mb}(b)\,dy,\quad u<0. \end{align}

4. The expected discounted dividends until ruin

In this section, we study the expected discounted dividend payments before ruin $V_{Mb}(u)$. Similar to Section 3, we first consider the case where the initial surplus $0\leq u\leq b$. Since it is a Parisian implementation delays in a dividend model with periodic observations, we note that dividend payment is possible only if the process $\{U_{b}^{d}(t)\}_{t\geq 0}$ is observed above barrier $b$ before ruin, which leads to

(4.1) \begin{align} V_{Mb}(u)& =E^{u}\left[e^{-\delta\eta_{1}}\left[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt\right]I_{\{\nu_{1}^{+}<\tau\}}\right]\nonumber\\ & \quad +E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]V_{Mb}(b),\quad 0\leq u\leq b. \end{align}

Besides, Eq. (4.1) at $u=b$ implies

(4.2) \begin{align} V_{Mb}(u)& =E^{u}\left[e^{-\delta\eta_{1}}\left[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt\right]I_{\{\nu_{1}^{+}<\tau\}}\right]\nonumber\\ & \quad +\frac{E^{u}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]E^{b}[e^{-\delta\eta_{1}}[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt]I_{\{\nu_{1}^{+}<\tau\}}]}{1-E^{b}[e^{-\delta\eta_{1}}I_{\{\nu_{1}^{+}<\tau\}}]}. \end{align}

The calculation of Eq. (4.2) is same as the calculation of Laplace transform of the time of ruin $\phi _{Mb}(u)$, and the final result is directly given here. The case where the initial surplus $u<0$ and $u>b$ will be resolved later. Since $E^{u}[e^{-\delta \eta _{1}}I_{\{\nu _{1}^{+}<\tau \}}]$ is given in Eq. (3.18), we only need to calculate $E^{u}[e^{-\delta \eta _{1}}[U_{b}^{d}(\eta _{1}^{-})-b+c_{1}\int _{\eta _{1}^{-}}^{\nu _{1}^{+}}e^{-\delta t}\,dt]I_{\{\nu _{1}^{+}<\tau \}}]$, $0\leq u\leq b$.

Theorem 3. When $0\leq u\leq b$, we have

\begin{align*} & E^{u}\left[e^{-\delta\eta_{1}}\left[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt\right]I_{\{\nu_{1}^{+}<\tau\}}\right]\nonumber\\ & \quad=\int_{0}^{\infty}\left(h_{\delta}^{+}(y\,|\,b-u)-\int_{0}^{\infty}E[e^{-\delta\tau_{b}}I_{\{\tau_{b}<\tau_{d}\}};U_{S}(\tau_{b})\in b+dx\,|\,U_{S}(0)=b-u]h_{\delta}^{+}(y\,|\,b+x)\right)\nonumber\\ & \qquad\times\left(E^{y}\left[e^{-\delta d}\left[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt\right]\right]-E^{y}\left[e^{-\delta\tau_{0}}c_{1}\int_{0}^{\tau_{0}}e^{-\delta t}\,dtI_{\{\tau_{0}< d\}}\right]\right.\nonumber\\ & \qquad\left.+E^{y}\left[e^{-\delta d}\left[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt\right]I_{\{\tau_{0}\geq d\}}\right]\right)dy. \end{align*}

Proof. Analogous to Eq. (3.9), the former quantity can be expressed in terms of the latter one via

(4.3) \begin{align} & E^{u}\left[e^{-\delta\eta_{1}}[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt]I_{\{\nu_{1}^{+}<\tau\}}\right]\nonumber\\ & \quad =\int_{0}^{\infty}\left(h_{\delta}^{+}(y\,|\,b-u)-\int_{0}^{\infty}E[e^{-\delta\tau_{b}}I_{\{\tau_{b}<\tau_{d}\}};U_{S}(\tau_{b})\in b+dx\,|\,U_{S}(0)=b-u]h_{\delta}^{+}(y\,|\,b+x)\right)\nonumber\\ & \qquad \times E^{b+y}\left[e^{-\delta\eta_{1}}[U_{b}^{d}(\eta_{1}^{-})-b +c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt]\right]dy. \end{align}

According to the definitions in Section 3.2, when $U_{b}^{d}(0)=U(0)=u>b$, it gives that $\eta _{1}=\theta _{1}\wedge d$ as the surplus was observed above $b$ at the initial moment. Hence, $E^{u}[e^{-\delta \eta _{1}}[U_{b}^{d}(\eta _{1}^{-})-b+c_{1}\int _{\eta _{1}^{-}}^{\nu _{1}^{+}}e^{-\delta t}\,dt]]$ can be written as

(4.4) \begin{align} & E^{u}\left[e^{-\delta\eta_{1}}[U_{b}^{d}(\eta_{1}^{-})-b+c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt]\right]\nonumber\\ & \quad =E^{u-b}\left[e^{-\delta(\tau_{0}\wedge d)}[U(\tau_{0}\wedge d)-c_{1}\int_{0}^{\tau_{0}\wedge d}e^{-\delta t}\,dt]\right],\quad u>b. \end{align}

In order to obtain $V_{Mb}(u)$, we need to compute $E^{u}[e^{-\delta (\tau _{0}\wedge d)}[U(\tau _{0}\wedge d)-c_{1}\int _{0}^{\tau _{0}\wedge d}e^{-\delta t}\,dt]],\ u>0$. It is similar to Eq. (3.13), ruin occurs after $u/c_{2}$ when $u\geq c_{2}d$, which is $\tau _{0}\geq d$, and we arrive at

(4.5) \begin{align} E^{u}\left[e^{-\delta(\tau_{0}\wedge d)}[U(\tau_{0}\wedge d)-c_{1}\int_{0}^{\tau_{0}\wedge d}e^{-\delta t}\,dt]\right]& =E^{u}\left[e^{-\delta d}[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt]\right]\nonumber\\ & =e^{-\delta d}\left[u+(\lambda E[X_{1}]-c)d-\frac{c_{1}}{\delta}(1-e^{-\delta d})\right]. \end{align}

For $0< u< c_{2}d$, conditioning on $\tau _{0}$ leads to

(4.6) \begin{align} E^{u}\left[e^{-\delta(\tau_{0}\wedge d)}[U(\tau_{0}\wedge d)-c_{1}\int_{0}^{\tau_{0}\wedge d}e^{-\delta t}\,dt]\right]& =E^{u}\left[e^{-\delta d}[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt]I_{\{\tau_{0}\geq d\}}\right]\nonumber\\ & \quad -E^{u}\left[e^{-\delta\tau_{0}}c_{1}\int_{0}^{\tau_{0}}e^{-\delta t}\,dtI_{\{\tau_{0}< d\}}\right]. \end{align}

For the first term to the right of Eq. (4.6), we have

(4.7) \begin{align} & E^{u}\left[e^{-\delta d}[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt]I_{\{\tau_{0}\geq d\}}\right]=E^{u}\left[e^{-\delta d}[U(d)-c_{1}\int_{0}^{d}e^{-\delta t}\,dt]\right]\nonumber\\ & \qquad -E^{u}\left[e^{-\delta d}[U(d)-c_{1}\int_{d-\tau_{0}}^{d}e^{-\delta t}\,dt]I_{\{\tau_{0}< d\}}\right]\nonumber\\ & \quad = e^{-\delta d}\left(u+(\lambda E[X_{1}]-c)\left\{d-\int_{{u}/{c_{2}}}^{d}(d-t)f_{U}(t\,|\,u)\,dt -\left(d-\frac{u}{c_{2}}\right)e^{-\lambda(\frac{u}{c_{2}})}\right\}\right.\nonumber\\ & \qquad \left.-\frac{c_{1}}{\delta}(1-e^{-\delta d})+c_{1}\int_{d-{u}/{c_{2}}}^{d}e^{-\delta t}e^{-\lambda\frac{u}{c_{2}}}\,dt+c_{1}\int_{{u}/{c_{2}}}^{d}f_{U}(t\,|\,u)\int_{d-t}^{d}e^{-\delta z}\,dz\,dt\right). \end{align}

The other term in Eq. (4.6)

(4.8) \begin{align} & E^{u}\left[e^{-\delta\tau_{0}}c_{1}\int_{0}^{\tau_{0}}e^{-\delta t}\,dtI_{\{\tau_{0}< d\}}\right]\nonumber\\ & \quad =e^{-\delta\frac{u}{c_{2}}}\frac{c_{1}}{\delta}(1-e^{-\delta\frac{u}{c_{2}}}) +\int_{{u}/{c_{2}}}^{d}e^{-\delta t}c_{1}\int_{0}^{t}e^{-\delta z}\,dzf_{U}(t\,|\,u)\,dt. \end{align}

Then

(4.9) \begin{align} & E^{u}\left[e^{-\delta\eta_{1}}\left[U_{b}^{d}(\eta_{1}^{-})-b +c_{1}\int_{\eta_{1}^{-}}^{\nu_{1}^{+}}e^{-\delta t}\,dt\right]I_{\{\nu_{1}^{+}<\tau\}}\right]\nonumber\\ & \quad=e^{-\delta d}(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}b}\left(e^{-\rho_{1}^{0}u} +\frac{(\rho_{2}-\rho_{2}^{0})(\rho_{2}-\rho_{1}^{0}) [(\rho_{2}^{0}-\rho_{1})e^{\rho_{2}^{0}(b-u)}-(\rho_{1}^{0}-\rho_{1})e^{\rho_{1}^{0}(b-u)}]} {\rho_{1}^{0}\rho_{2}(\rho_{2}^{0}-\rho_{1})(\rho_{2}-\rho_{1}^{0})e^{\rho_{2}^{0}b} -(\rho_{1}^{0}-\rho_{1})(\rho_{2}-\rho_{2}^{0})e^{\rho_{1}^{0}b}}\right)\nonumber\\ & \qquad\times\left\{\frac{c_{1}e^{-(\frac{\delta}{c_{2}}-\rho_{1})c_{2}d}-c_{1}} {\delta(\frac{\delta}{c_{2}}-\rho_{1})}+\frac{c_{1}-c_{1} e^{-(\frac{2\delta}{c_{2}}-\rho_{1})c_{2}d}}{\delta(\frac{2\delta}{c_{2}}-\rho_{1})} +\frac{e^{\rho_{1}c_{2}d}}{\rho_{1}}e^{-\delta d}\frac{c_{1}}{\delta}(1-e^{-\delta d})\right.\nonumber\\ & \qquad-\int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{\frac{y}{c_{2}}}^{d}e^{-\delta t}c_{1}\left(\frac{1}{\delta}-\frac{e^{-\delta}}{\delta}\right)f_{U}(t\,|\,u)\,dt\,dy -\int_{0}^{c_{2}d}e^{\rho_{1}y}\left\{\frac{c_{1}}{\delta}(1-e^{-\delta d})\right.\nonumber\\ & \qquad \left.+c_{1}\int_{0}^{d-\frac{y}{c_{2}}}e^{-\delta t}e^{-\lambda\frac{y}{c_{2}}}\,dt+c_{1}\int_{\frac{y}{c_{2}}}^{d}f_{U}(t\,|\,u)\int_{0}^{d-t}e^{-\delta z}\,dz\,dt\right\}dy+\frac{1}{\rho_{1}^{2}}-\left(\frac{\lambda}{\rho_{1}}+c_{2}\right)\nonumber\\ & \qquad\times\left[-\frac{d}{\rho_{1}}-d\int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{\frac{y}{c_{2}}}^{d}f_{U}(t|y)\,dt\,dy +\int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{\frac{y}{c_{2}}}^{d}tf_{U}(t|y)\,dt\,dy -\frac{1}{\lambda-c_{2}\rho_{1}}\right.\nonumber\\ & \qquad \left.\left.\times\left(c_{2}d-\frac{1}{\frac{\lambda}{c_{2}}-\rho_{1}} \left(1-e^{(\rho_{1}-\frac{\lambda}{c_{2}})c_{2}d}\right)\right)\right]\right\},\quad 0\leq u\leq b. \end{align}

Using Eq. (3.19), the double integral above is evaluated as

(4.10) \begin{align} & \int_{0}^{c_{2}d}e^{\rho_{1}y}\int_{\frac{y}{c_{2}}}^{d}te^{-\delta t}f_{U}(t|y)\,dt\,dy =\sum_{k=1}^{\infty}\frac{(-\lambda\rho_{1})^{k}}{(k-1)!c_{2}^{k+1}}\sum_{i=0}^{k} \frac{(k+i-1)!}{i!(k-i)!}\frac{1}{(\frac{\lambda+\delta}{c_{2}}-\rho_{1})^{k+i}}\nonumber\\ & \quad\times\left\{\frac{(k+1-i)!}{(\frac{\lambda+\delta}{c_{2}}-\rho_{1})^{k-i+2}} \left(1-\sum_{j=0}^{k+1-i}\frac{[(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d]^{j} e^{-(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d}}{j!}\right) -\sum_{j=0}^{k+i-1}\left(\frac{\lambda+\delta}{c_{2}}-\rho_{1}\right)^{j}\right.\nonumber\\ & \quad \left.\times e^{-(\frac{\lambda+\delta}{c_{2}}-\rho_{1})c_{2}d}\frac{(c_{2}d)^{j+k-i+2}(k-i)!}{(j+k-i+2)!}\right\}, \end{align}

and the second double integral in Equation (4.9) can be directly set as $\delta =0$ in Eq. (3.19), the third double integral in Eq. (4.9) can be directly set as $\delta =0$ in Eq. (4.10). With Eqs. (3.18) and (4.9) derived, $V_{Mb}(u), 0\leq u\leq b$ are determined by Eq. (4.2).

The following two Remarks respectively give the results satisfied by $V_{Ub}(u)$ and $V_{Lb}(u)$.

5. Numerical illustrations

This section aims at providing some numerical examples to study the effect of Parisian implementation delays under a mixed dividend strategy on the Gerber-Shiu expected discounted penalty function and the expected discounted dividend payments before ruin. The optimal dividend barrier that maximizes $V_{b}(u)$ with respect to b will also be discussed. In all examples, it is assumed that the constant expense rate is $c=0.5$, the fixed continuous dividend interest rate is $c_{1}=0.25$, the periodic observation with $\beta =1$ and the Laplace transform argument is $\delta =0.1$. All numbers are generated using the software package Mathematica (Specific reference Cheung and Wong [Reference Cheung and Wong8]).

The functions $\phi _{b}(u)$ and $V_{b}(u)$ are computed for the initial surplus levels $u=2, 6, 8, 10$ and fixed barrier level $b=10$, and that's the case of $0< u\leq b$. When the initial surplus $u$, the amount of income is subject to the exponential parameter $a_{1}$ and the deterministic delays $d=10$ are also fixed, the higher the revenue intensity parameter $\lambda$ of Poisson arrival rate, the higher the total $V_{b}(u)$ of the expected discounted dividend is. Because more positive earnings a company has, the more dividends it pays. Similarly, the expected discounted penalty function $\phi _{b}(u)$ will decrease as $\lambda$ increases.

The findings on Table 2 is same as that in Table 1. The income intensity parameter $\lambda$ in Table 1 is replaced by the exponential parameter $a_{1}$ of the single income amount, and then the relationship among $a_{1}$ and the expected discounted penalty function $\phi _{b}(u)$ and the expected discounted dividend function $V_{b}(u)$ is discussed. As it can be seen from the results in Tables 1 and 2, both $\lambda$ and $a_{1}$ influence the total amount of earnings $S(t)$ at time $t$. Therefore, the impact of the change in parameter $a_{1}$ on $\phi _{b}(u)$ and $V_{b}(u)$ of the single return is the same as that in Table 1.

Table 1. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $a_{1}=1$, $d=2$, $b=10$, $\delta =0.1$.

Table 2. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $\lambda =1$, $d=2$, $b=10$, $\delta =0.1$.

With the extension of the certainty time $d$ of the delay, the Parisian implementation delays dividend becomes more stringent, which leads to a later ruin time and hence a smaller value of $\phi _{b}(u)$. As the delay gets longer, there are three effects on the expected discounted dividend function $V_{b}(u)$:

1. 1. A longer delay time makes the duration of threshold dividend longer. However, when $d$ reaches a certain length, the surplus process has passed through the barrier $b$ and dropped below $b$, which will no longer affect the total threshold dividend.

2. 2. The longer the delay, the lower the probability of ruin. The longer the company survives, the more potential dividend opportunities will be generated by the company's long surplus growth.

3. 3. The longer delay means that early Parisian implementation delays dividend are more unlikely.

The numerical results in Table 3 suggest that the above first two effects dominates.

Table 3. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $a_{1}=1$, $\lambda =1$, $b=10$, $\delta =0.1$.

The left side of Tables 4–6 considers the case where the initial surplus are greater than the dividend barrier $b=10$. It can be seen from the table that similar findings of Tables 1–3 are still retained. In each of Tables, for fixed $\lambda, a_{1}$ and $d$, it is observed that $\phi _{b}(u)$ decreases and converges as $u$ increases. Intuitively, when u is considerably larger than the dividend barrier $b=10$, although the threshold dividend is being paid, the initial surplus is so large that it is likely the surplus process will remain above the dividend barrier $b$ during the entire delay period. In such a case, the surplus will drop to dividend barrier $b$ upon payment of a dividend at the end of the delay period, and the time remaining until ruin is just the time of ruin with initial surplus level $b$. Therefore, any further increase in $u$ virtually does not affect the ruin time but merely increases the amount of the first dividend.

Table 4. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $a_{1}=1$, $d=2$, $b=10$, $\delta =0.1$.

Table 5. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $\lambda =1$, $d=2$, $b=10$, $\delta =0.1$.

Table 6. $c=0.5$, $c_{1}=0.25$, $\beta =1$, $a_{1}=1$, $\lambda =1$, $b=10$, $\delta =0.1$.

Similarly, the right side of Tables 4– 6 considers the case where the initial surplus are less than the dividend barrier $b$, which is also shown in Tables 1–3. By comparing Tables 1 and 4, it can be seen that the smaller the initial surplus $u$ is, the smaller the impact of revenue intensity parameter $\lambda$ on the expected penalty function $\phi _{b}(u)$ is. Besides, from Tables 2 and 5, it can be seen that the smaller the initial earnings $u$ is, the smaller the impact of exponential parameter $a_{1}$ of the single income amount on the expected penalty function $\phi _{b}(u)$ is. This is because the smaller the initial surplus is, the more likely ruin will occur and it will converges to a constant of 1.

In addition to the effects of various parameters on the expected discounted penalty function $\phi _{b}(u)$ and the expected discounted dividend function $V_{b}(u)$, we are also interested in the optimal barrier maximizing the expected discounted dividend function $V_{b}(u)$. The results are given in Figures 4–6 and are explained as follows. The three cases of initial earnings $u=2$,$u=12$ and $u=-2$ are considered here. The value range of dividend barrier $b$ is 2 to 9, which is corresponding to three cases of initial surplus between 0 and $b$, greater than $b$ and less than 0, respectively. It can be seen from Figures 4–6 that when the initial earnings $u$ and the delay time $d$ are determined, it might not be the case that the smaller the dividend barrier $b$ is, the better. A smaller dividend barrier will make it easier to distribute the dividend, but it will also be more likely to cause a ruin of a company. Similarly, it is not the case that the bigger the dividend barrier $b$ is, the better. A large dividend barrier will reduce the probability of ruin, but the dividend will occur later, and the total dividend amount will be reduced.The relationship between all dividend barriers $b$ and the expected discounted total dividend $V_{b}(u)$ is shown in Figures 4–6.

Figure 4. $V(x)$ for u = 2 with parameters $c=0.5$, $c_1=0.25$, $\beta =1$, $a_{1}=1$, $\lambda =1$, $d=2$, $\delta =0.1$.

Figure 5. $V(u)$ for $u=12$ with parameters $c=0.5$, $c_1=0.25$, $\beta =1$, $a_{1}=1$, $\lambda =1$, $d=2$, $\delta =0.1$.

Figure 6. $V(u)\ {\rm for}\; u=-2$ with parameters $c=0.5$, $c_1=0.25$, $\beta =1$, $a_{1}=1$, $\lambda =1$, $d=2$, $\delta =0.1$.