2. Preliminaries on spectrally negative Lévy processes and Parisian ruin problems

We first briefly introduce spectrally negative Lévy processes, the associated scale functions, and some fluctuation identities. Write $X\equiv\{X(t);\,t\geq0\}$ , defined on a probability space with probability laws $\{\mathbb{P}_{x};\,x\in(\!-\infty,\infty)\}$ and natural filtration $\{\mathcal{F}_{t};\,t\geq0\}$ , for a spectrally negative Lévy process that is not a purely increasing linear drift or the negative of a subordinator. Denote its running supremum process by

\[\bar{X}(t)\coloneqq \sup\limits_{0\leq s\leq t}X(s), \,\,\, t\geq0.\]

The Laplace exponent of X is given by

\begin{eqnarray}\psi(\theta)\coloneqq \ln \mathbb{E}_{x}\!\left(\textrm{e}^{\theta (X(1)-x)}\right)=\gamma\theta+\frac{1}{2}\sigma^{2}\theta^{2}-\int_{(0,\infty)}\left(1-\textrm{e}^{-\theta x}-\theta x\mathbf{1}_{(0,1)}(x)\right)\nu(\textrm{d}x),\nonumber\end{eqnarray}

where the Lévy measure $\nu$ satisfies $\int_{(0,\infty)}\left(1\wedge x^{2}\right)\nu(\textrm{d}x)<\infty$ . It is known that $\psi(\theta)$ is finite for $\theta\in[0,\infty)$ , and it is strictly convex and infinitely differentiable. As in [Reference Bertoin3], the q-scale functions $\{W^{(q)};\ q\geq0\}$ of X are defined as follows. For each $q\geq0$ , $W^{(q)}\,{:}\,[0,\infty)\rightarrow[0,\infty)$ is the unique strictly increasing and continuous function with Laplace transform

\begin{eqnarray}\int_{0}^{\infty}\textrm{e}^{-\theta x}W^{(q)}(x)\textrm{d}x=\frac{1}{\psi(\theta)-q},\quad \mbox{for }\theta>\Phi_{q},\nonumber\end{eqnarray}

where $\Phi_{q}$ is the largest solution of the equation $\psi(\theta)=q$ . Further define $W^{(q)}(x)=0 $ for $x<0$ , and write W for the 0-scale function $W^{(0)}$ . Note that $W^{(q)}(0+\!)=0 $ if and only if the process X has sample paths of unbounded variation.

For $p, p+q\geq 0$ , $y>0$ , and $x\in(\!-\infty,\infty)$ , define two more scale functions as

\[W^{(p,q)}_{y}(x)\coloneqq W^{(p)}(x)+q\int_{y}^{x}W^{(p+q)}(x-w)W^{(p)}(w)\textrm{d}w\]

and

\[Z^{(p)}(x)\coloneqq 1+p\int_{0}^{x}W^{(p)}(w)\textrm{d}w.\]

For any $x\in\mathbb{R}$ and $\vartheta\geq0$ , define an exponential change of measure for the spectrally negative Lévy process by

\begin{eqnarray}\left.\frac{\textrm{d}\mathbb{P}_{x}^{\vartheta}}{\textrm{d}\mathbb{P}_{x}}\right|_{\mathcal{F}_{t}}=\textrm{e}^{\vartheta(X(t)-x)-\psi(\vartheta)t}.\nonumber\end{eqnarray}

Furthermore, note that under the probability measures $\mathbb{P}_{x}^{\vartheta}$ , the process X remains a spectrally negative Lévy process. From now on we denote by $W_{\vartheta}^{(q)}$ and $W_{\vartheta}$ the q-scale function and 0-scale function, respectively, under the measure $\mathbb{P}_{x}^{\vartheta}$ .

For the process X, define its first up-crossing time and down-crossing time of level $a\in(\!-\infty,\infty)$ by

\begin{eqnarray}\tau^{+}_{a}\coloneqq \inf\{t\geq0\,{:}\,X(t)>a\} \quad \text{and} \quad \tau_{a}^{-}\coloneqq \inf\{t\geq0: X(t)<a\},\nonumber\end{eqnarray}

respectively. It can be found in [Reference Kyprianou16] that

(1) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\tau_{0}^{-}\}}\right)= \frac{W^{(q)}(x)}{W^{(q)}(a)}, \quad x\in(\!-\infty,a].\end{eqnarray}

In addition, it follows from [Reference Zhou28] that

(2) \begin{eqnarray}\lim\limits_{x\rightarrow\infty}\frac{W^{(q)\prime}(x)}{W^{(q)}(x)}=\Phi_{q}\,\,\, \text{and} \,\,\, \lim\limits_{y\rightarrow\infty}\frac{W^{(q)}(x+y)}{W^{(q)}(y)}=\textrm{e}^{\Phi_{q}x}.\end{eqnarray}

A function $\xi\,{:}\, (\!-\infty,\infty)\rightarrow (\!-\infty,\infty)$ is called a draw-down function if $\xi(x)< x$ for all the values of x that are of concern. Define the $\xi$ -draw-down time $\tau_{\xi}$ of X as

\[\tau_{\xi}\coloneqq \inf\{t\geq0\,{:}\,X(t)<\xi(\bar{X}(t))\},\]

with the convention that $\inf\emptyset\coloneqq \infty$ . We call $\xi(\bar{X}(\tau_\xi))$ the associated draw-down level. By [Reference Li, Vu and Zhou19], for $\bar{\xi}(z)\coloneqq z-\xi(z)$ we have

(3) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q\tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\tau_{\xi}\}}\right)=\exp\left\{-\int_{x}^{a}\frac{W^{(q)\prime}(\bar{\xi}(z))}{W^{(q)}(\bar{\xi}(z))}\textrm{d}z\right\},\quad x\in(\!-\infty,a].\end{eqnarray}

For $r>0$ the Parisian ruin time is defined by

\[\kappa_{r}\coloneqq \inf\{t>r\,{:}\, t-g_{t}>r\} \,\,\text{with} \,\, \inf\emptyset\coloneqq \infty, \]

where

\[ g_{t}\coloneqq \sup\{0\leq s \leq t\,{:}\, X(s) \geq0\} \,\,\text{with} \,\, \sup\emptyset\coloneqq 0.\]

Given the draw-down function $\xi$ , we define the $\xi$ -draw-down Parisian ruin time of X as

\[\kappa_{r}^{\xi}\coloneqq \inf\{t>r\,{:}\, t-g_{t}^{\xi}>r\} \,\,\text{with} \,\, \inf\emptyset\coloneqq \infty,\]

where

\[g_{t}^{\xi}\coloneqq \sup\{0\leq s \leq t\,{:}\, X(s) \geq\xi(\bar{X}(s))\}\,\,\text{with} \,\, \sup\emptyset\coloneqq 0.\]

From [Reference Loeffen, Czarna and Palmowski21] and [Reference Czarna and Palmowski10], we have

\begin{eqnarray}\mathbb{P}_{x}\!\left(\kappa_{r}<\infty\right)=1-\mathbb{E}\!\left(X(1)\right)\frac{\int_{0}^{\infty}W(x+z)z\mathbb{P}\!\left(X(r)\in\textrm{d}z\right)}{\int_{0}^{\infty}z\mathbb{P}\!\left(X(r)\in\textrm{d}z\right)},\quad x\in(\!-\infty,\infty),\nonumber\end{eqnarray}

and

(4) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}\}}\right)=\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(a)},\quad x\in(\!-\infty, a],\end{eqnarray}

where

\begin{eqnarray}\ell_{r}^{(q)}(x)&\coloneqq &\int_{0}^{\infty}\textrm{e}^{-\Phi_{q}z+qr}W^{(q)}(x+z)\frac{z}{r}\mathbb{P}^{\Phi_{q}}\left(X(r)\in\textrm{d}z\right)\nonumber\\[3pt]&=&\int_{0}^{\infty} W^{(q)}(x+z)\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\!.\nonumber\end{eqnarray}

Note that $\ell_{r}^{(q)}$ can be treated as a scale function associated to the Parisian ruin.

Write $\ell_{r}\coloneqq \ell_{r}^{(0)}$ for simplicity. Then

\begin{equation}\mathbb{P}_{x}\!\left(\kappa_{r}<\infty\right)=1-\frac{\mathbb{E}\left(X(1)\right)}{\int_{0}^{\infty}\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)}\ell_{r}(x).\nonumber\end{equation}

For $b\in(0,\infty)$ , let

(5) \begin{eqnarray}D(t)\coloneqq \left(\bar{X}(t)- b\right)\vee0,\quad t\geq0,\end{eqnarray}

denote the accumulated amount of dividends paid until time t of the barrier strategy with barrier at level b.

In this paper, we are interested in the following fluctuation quantities related to the draw-down Parisian ruin time:

1. (i) The draw-down Parisian-ruin-time-related two-sided exit problem

   \[\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{\eta}\}}\right)\!, \quad x\in (\!-\infty,\infty),\,\,a\in[x,\infty),\]
   where $\eta$ is another draw-down function such that $\eta(z)<\xi(z)< z$ for all $z\leq a$ .

2. (ii) The joint Laplace transform involving the draw-down Parisian ruin time, the position of X at the draw-down Parisian ruin time, and its running supremum until the draw-down Parisian ruin time:

   \[\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi}-r\right)}\textrm{e}^{\lambda X(\kappa_{r}^{\xi})-\psi\left(\lambda\right)r}\varphi(\bar{X}(\kappa_{r}^{\xi}))\mathbf{1}_{\{\kappa_{r}^{\xi}<\tau_{a}^{+}\}}\right)\!,\quad x\in(\!-\infty,a],\,a\in(\!-\infty,\infty),\]
   where $\varphi\,{:}\,(\!-\infty,\infty)\rightarrow(\!-\infty,\infty)$ is an arbitrary bounded measurable function.

3. (iii) The potential measure of X involving the draw-down Parisian ruin time,

   \[\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t),\bar{X}(t)); \,t<\kappa_{r}^{\xi}\wedge \tau_{a}^{+}\right)\textrm{d}t,\quad x\in(\!-\infty,a],\,a\in(\!-\infty,\infty),\]
   where f is an arbitrary bounded bivariate function which is differentiable with respect to the first argument.

4. (iv) The kth moment of the following integral involving D, which can be interpreted as the accumulated amount of time-discounted dividends up to the draw-down Parisian ruin time:

   \[V_{k}^{\xi}(x;\ b)\coloneqq \mathbb{E}_{x}\!\left(\left[D_{b}\right]^{k}\right)\!,\quad x\in(\!-\infty,\infty),\,b\in(0,\infty),\]
   with
   \[D_{b}\coloneqq D_{\xi,b}\coloneqq \int_{0-}^{\kappa^{\xi}_{r}}\textrm{e}^{-q t}\,\textrm{d}D(t).\]
   Further let $V_{k}(x)\coloneqq V_{k}^{\xi}(x;\ x),\,\,x\in(\!-\infty,\infty)$ . For $\xi(x)=(x-b)\vee 0 $ with $b>0$ , $D_{b}$ can be interpreted as the accumulated discounted dividends paid according to the barrier dividend strategy with barrier at level b until the draw-down Parisian ruin time.

We assume the differentiability of $\ell_{r}^{(q)}$ whenever needed. In fact, by (2) and the definition of $\ell_{r}^{(q)}$ , $\ell_{r}^{(q)}$ inherits the same differentiability as $W^{(q)}$ . It is known that, when X has sample paths of unbounded variation, or when X has sample paths of bounded variation and the Lévy measure has no atoms, the scale function $W^{(q)}$ (and hence $\ell_{r}^{(q)}$ ) is continuously differentiable over $(0, \infty)$ (resp. $(\!-\infty,\infty)$ ). Moreover, if X has a nontrivial Gaussian component, then $W^{(q)}$ (and hence $\ell_{r}^{(q)}$ ) is twice continuously differentiable over $(0, \infty)$ (resp. $(\!-\infty,\infty)$ ). The interested reader is referred to [Reference Chan, Kyprianou and Savov4] and [Reference Kuznetsov, Kyprianou and Rivero15] for more detailed discussions on the smoothness of scale functions.

3. Excursion process and Parisian-ruin-related quantities under excursion measure

In this section, we briefly recall basic concepts in excursion theory for the reflected process $\{\bar{X}(t)-X(t);\ t\geq0\}$ , and we refer to [Reference Bertoin3] for more details. We also obtain new Parisian-ruin-related results on the excursion measure.

For $x\in(\!-\infty,\infty)$ , the process $\{L(t)\coloneqq \bar{X}(t)-x, t\geq0\}$ is a local time at 0 for the Markov process $\{\bar{X}(t)-X(t);\,t\geq0\}$ under $\mathbb{P}_{x}$ . The corresponding inverse local time is defined as

\[L^{-1}(t)\coloneqq \inf\{s\geq0\,{:}\, L(s)>t\}=\sup\{s\geq0\,{:}\, L(s)\leq t\}.\]

Further, let $L^{-1}(t-\!)\coloneqq \lim\limits_{s\uparrow t}L^{-1}(s)$ . Define a Poisson point process $\{(t, \varepsilon_{t});\ t\geq0\}$ by

(6) \begin{eqnarray}\varepsilon_{t}(s)\coloneqq X(L^{-1}(t))-X(L^{-1}(t-\!)+s), \qquad s\in(0,L^{-1}(t)-L^{-1}(t-\!)]\end{eqnarray}

whenever the lifetime of $\varepsilon_{t}$ is strictly positive, i.e. $L^{-1}(t)- L^{-1}(t-\!)>0$ . If $L^{-1}(t)-\break L^{-1} (t-\!)=0 $ , define $\varepsilon_{t}\coloneqq \Upsilon$ with $\Upsilon$ being an additional isolated point. It is known that $\varepsilon$ is a Poisson point process taking values in the space of excursion paths with characteristic measure n if $\{\bar{X}(t)-X(t);\ t\geq0\}$ is recurrent; otherwise, $\{\varepsilon_{t};\ t\leq L(\infty)\}$ is a Poisson point process stopped at the first excursion of infinite lifetime. Here, n is a $\sigma$ -finite measure on the space $\mathcal{E}$ of excursions, i.e. the space $\mathcal{E}$ of càdlàg functions f satisfying

\[f\,{:}\,(0,\zeta)\rightarrow (0,\infty)\,\,\text{ for some} \,\, \zeta\in(0,\infty] \,\, \text{ and} \,\, f(\zeta)\in (0, \infty) \,\, \text{if}\,\, \zeta<\infty,\]

where $\zeta\equiv\zeta(\,f)$ denotes the excursion length or lifetime; see Definition 6.13 of [Reference Kyprianou16] for the definition of $\mathcal{E}$ . Denote by $\varepsilon(\cdot)$ , or $\varepsilon$ for short, a generic excursion belonging to the space $\mathcal{E}$ of canonical excursions. The excursion height of a canonical excursion $\varepsilon$ is denoted by $\bar{\varepsilon}=\sup\limits_{t\in[0,\zeta]}\varepsilon(t)$ .

With a little abuse of notation, for $a\in(0,\infty)$ and $t\in[0,\zeta]$ , let

\[g_{t}^{a}(\varepsilon)\coloneqq \left\{\begin{array}{l@{\quad}l} \inf\{s\in[0,\zeta]\,{:}\, s\leq t, \varepsilon(w)\geq a \mbox{ for all }w\in[s,t]\}\quad &\mbox{if } \,\varepsilon(t)\geq a,\\ \\t & \mbox{otherwise};\end{array}\right.\]

and

\[d_{t}^{a}(\varepsilon)\coloneqq \left\{\begin{array}{l@{\quad}l}\sup\{s\in[0,\zeta]\,{:}\, s\geq t, \varepsilon(w)\geq a \mbox{ for all }w\in[t,s)\}\quad &\mbox{if } \,\varepsilon(t)\geq a,\\ \\[-7pt]t &\mbox{otherwise}.\end{array}\right.\]

Write $\zeta_{t}^{a}(\varepsilon)\coloneqq d_{t}^{a}(\varepsilon)-g_{t}^{a}(\varepsilon)$ for the length of the maximum time interval (containing t) when the canonical excursion $\varepsilon$ stays above the level a. Further define

\[\alpha_{a}^{+}(\varepsilon)\coloneqq \inf\{g_{t}^{a}(\varepsilon)\,{:}\, t\in[0,\zeta], \zeta_{t}^{a}(\varepsilon)>r\},\]

with the convention that $\inf\emptyset\coloneqq \zeta$ . Intuitively, $\alpha_{a}^{+}(\varepsilon)$ is the starting time of the first time interval of length more than r when the excursion path stays continuously above level a.

The following result gives the excursion measure of the event that there exists a time interval with length at least r during which either the excursion process continuously stays above level $z>0$ , or there is an excursion with height strictly greater than $z+y$ for some $y>0$ .

Proposition 3.1. For any $z,y\in(0,\infty)$ , we have

(7) \begin{eqnarray}n\!\left(\alpha_{z}^{+}(\varepsilon)<\zeta \mbox{ \emph{or} } \overline{\varepsilon}> z+y\right)=\frac{W^{\prime}(z)\phi(y,r)+\chi^{\prime}(z,y,r)}{W(z)\phi(y,r)+\chi(z,y,r)},\end{eqnarray}

where the derivative of $\chi$ is with respect to the first argument, and the Laplace transforms of $\phi(y,r)$ and $\chi(x,y,r)$ (in r) are given, respectively, by

\begin{eqnarray}\,\,\int_{0}^{\infty}\textrm{e}^{-\theta r}\chi(x,y,r)\,\textrm{d}r=\frac{1}{\theta}\left(\frac{W_{y}^{(\theta,-\theta)}(x+y)}{W^{(\theta)}(y)}-\frac{W(x)Z^{(\theta)}(y)}{W^{(\theta)}(y)}\right)\nonumber\end{eqnarray}

and

\begin{eqnarray}\int_{0}^{\infty}\textrm{e}^{-\theta r}\phi(y,r)\,\textrm{d}r=\frac{Z^{(\theta)}(y)}{\theta W^{(\theta)}(y)}.\nonumber\end{eqnarray}

Proof. It follows from Theorem 1 of [Reference Czarna and Renaud11] that

(8) \begin{eqnarray}1-\mathbb{P}_{x}\left(\kappa_{r}\wedge \tau_{-y}^{-}<\infty\right)=\mathbb{E}\left(X(1)\right)\left(W(x)+\frac{\chi(x,y,r)}{\phi(y,r)}\right)\end{eqnarray}

for any fixed positive y. By the strong Markov property, for $a>x$ we have

$$\eqalign{ & {_x}\left( {{\kappa _r} \wedge \tau _{ - y}^ - < \infty } \right){\mkern 1mu} \cr & = {\mkern 1mu} {_x}\left( {\tau _a^ + < {\kappa _r} \wedge \tau _{ - y}^ - < \infty } \right) + {_x}\left( {{\kappa _r} \wedge \tau _{ - y}^ - < \tau _a^ + } \right){\mkern 1mu} \cr & = {\mkern 1mu} {_x}\left( {\tau _a^ + < {\kappa _r} \wedge \tau _{ - y}^ - } \right){_a}\left( {{\kappa _r} \wedge \tau _{ - y}^ - < \infty } \right) + 1 - {_x}\left( {\tau _a^ + < {\kappa _r} \wedge \tau _{ - y}^ - } \right), \cr} $$

which together with (8) implies

(9) \begin{eqnarray}\mathbb{P}_{x}\!\left(\tau_{a}^{+}<\kappa_{r}\wedge \tau_{-y}^{-}\right)&\,{=}\,&\frac{1-\mathbb{P}_{x}\!\left(\kappa_{r}\wedge \tau_{-y}^{-}<\infty\right)}{1-\mathbb{P}_{a}\!\left(\kappa_{r}\wedge \tau_{-y}^{-}<\infty\right)}\nonumber\\[3pt]&\,{=}\,&\frac{W(x)\phi(y,r)+\chi(x,y,r)}{W(a)\phi(y,r)+\chi(a,y,r)}.\end{eqnarray}

Because $\{(t, \varepsilon_{t});\ t\geq0\}$ defined via (6) is a Poisson point process with intensity measure $\textrm{d}t\times\textrm{d}n$ , we have

(10) \begin{align}\mathbb{P}_{x}\!\left(\tau_{a}^{+}<\kappa_{r}\wedge \tau_{-y}^{-}\right)&=\mathbb{E}_{x}\!\left(\prod\limits_{t\leq a-x}\mathbf{1}_{\{\alpha_{x+t}^{+}(\varepsilon_{t})= \zeta(\varepsilon_{t}),\,\overline{\varepsilon}_{t}\leq x+t+y\}}\right)\nonumber\\[3pt]&=\exp\left(\!-\int_{0}^{a-x}n\!\left(\alpha_{x+t}^{+}(\varepsilon)<\zeta \mbox{ or } \overline{\varepsilon}> x+t+y\right)\,\textrm{d}t\right)\nonumber\\[3pt]&=\exp\left(\!-\int_{x}^{a} n\!\left(\alpha_{w}^{+}(\varepsilon)<\zeta \mbox{ or } \overline{\varepsilon}> w+y\right)\,\textrm{d}w\right)\!,\end{align}

where $\overline{\varepsilon}_{t}$ denotes the excursion height of $\varepsilon_{t}$ . Combining (9) and (10) yields (7).□

We next prove a version of Proposition 3.1 for $y=\infty$ .

Corollary 3.1. For any $x\in(0,\infty)$ , we have

\[ n\!\left(\alpha_{x}^{+}(\varepsilon)<\zeta\right)=\frac{\ell_{r}^{\prime}(x)}{\ell_{r}(x)}. \]

Proof. By definition we have

(11) \begin{eqnarray}\,\,\int_{0}^{\infty}\textrm{e}^{-\theta r}\left(W(x)\phi(y,r)+\chi(x,y,r)\right)\textrm{d}r=\frac{W_{y}^{(\theta,-\theta)}(x+y)}{\theta W^{(\theta)}(y)}.\end{eqnarray}

The definition of $W^{(\theta,-\theta)}_{y}(x+y)$ together with (2) yields

(12) \begin{align}\lim\limits_{y\uparrow\infty}\frac{W^{(\theta,-\theta)}_{y}(x+y)}{\theta W^{(\theta)}(y)}&=\textrm{e}^{\Phi_{\theta}x}\left(\frac{1}{\theta}-\int_{0}^{x}W(w)\,\textrm{e}^{-\Phi_{\theta}w}\,\textrm{d}w\right)\nonumber\\[4pt]&=\int_{0}^{\infty}W(x+w)\,\textrm{e}^{-\Phi_{\theta} w}\,\textrm{d}w,\end{align}

which coincides with the Laplace transform (in r) of $\ell_{r}(x)$ as follows:

(13) \begin{align}\int_{0}^{\infty}\textrm{e}^{-\theta r}\ell_{r}(x)\,\textrm{d}r&=\int_{0}^{\infty}\textrm{e}^{-\theta r}\int_{0}^{\infty}W(x+z)\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\,\textrm{d}r\nonumber\\[3pt]&=\int_{0}^{\infty}\textrm{e}^{-\theta r}\int_{0}^{\infty}W(x+z)\mathbb{P}(\tau_{z}^{+}\in\textrm{d}r)\textrm{d}z\nonumber\\[3pt]&=\int_{0}^{\infty}W(x+z)\mathbb{E}(\textrm{e}^{-\theta \tau_{z}^{+}})\textrm{d}z\nonumber\\[3pt]&=\int_{0}^{\infty}W(x+w)\,\textrm{e}^{-\Phi_{\theta} w}\,\textrm{d}w,\end{align}

where we have used Kendall’s identity,

\[\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\,\textrm{d}r=\mathbb{P}(\tau_{z}^{+}\in\textrm{d}r)\textrm{d}z,\qquad z,r\geq0.\]

By (11), (12), (13), and the continuity of Laplace transforms, we have

\[\lim\limits_{y\uparrow\infty}\left(W(x)\phi(y,r)+\chi(x,y,r)\right)=\ell_{r}(x).\]

In fact, the same arguments as above lead to

\[\lim\limits_{y\uparrow\infty}\left(W^{\prime}(x)\phi(y,r)+\chi^{\prime}(x,y,r)\right)=\ell_{r}^{\prime}(x).\]

Therefore,

\begin{eqnarray}n\!\left(\alpha_{x}^{+}(\varepsilon)<\zeta\right)&\,{=}\,&\lim\limits_{y\uparrow\infty}n\!\left(\alpha_{x}^{+}(\varepsilon)<\zeta \mbox{ \emph{or} } \overline{\varepsilon}> x+y\right)\nonumber\\[3pt]&\,{=}\,&\lim\limits_{y\uparrow\infty}\frac{W^{\prime}(x)\phi(y,r)+\chi^{\prime}(x,y,r)}{W(x)\phi(y,r)+\chi(x,y,r)}=\frac{\ell_{r}^{\prime}(x)}{\ell_{r}(x)},\nonumber\end{eqnarray}

which is the desired result. □

Remark 3.1. Applying Corollary 3.1 and a property of Poisson random measure, we have for $x<a$ that

\begin{equation*}\begin{split}\mathbb{P}_{x}\left(\tau_{a}^{+}<\kappa_{r}\right)&=\mathbb{P}_{x}\left( \alpha_{x+t}^{+}(\varepsilon_{t})= \zeta(\varepsilon_{t}), \,\, t\leq a-x \right)\\[3pt]&=\textrm{e}^{-\int_0^{a-x}n(\alpha_{x+t}^{+}(\varepsilon)<\zeta) \textrm{d}t} =\textrm{e}^{-\int_0^{a-x}\frac{\ell_{r}^{\prime}(x+t)}{\ell_{r}(x+t)} \textrm{d}t}=\frac{\ell_{r}(x)}{\ell_{r}(a)}.\end{split}\end{equation*}

Then (4) can be recovered via a change of measure.

Denote by $\epsilon_{g}$ the excursion (away from 0) for the reflected process $\{\bar{X}(s)-X(s);\, s\geq0\}$ starting at time g in the time scale for X, i.e. $g=L^{-1}(t-\!)<L^{-1}(t)$ for some $t\geq 0$ and $\epsilon_{g}\coloneqq \varepsilon_{t}$ . In addition, denote by $\zeta_{g}\coloneqq \zeta(\varepsilon_t )$ and $\bar{\epsilon}_{g}\coloneqq \bar{\varepsilon}_t$ its lifetime and its excursion height, respectively, and write $ \alpha_{a}^{+}(\epsilon_g)\coloneqq \alpha_{a}^{+}(\varepsilon_{t})$ ; see Section IV.4 of [Reference Bertoin3]. The following result gives the joint Laplace transform involving $\alpha_{a}^{+}$ under the excursion measure.

Proposition 3.2. For any $q,\lambda\in[0,\infty)$ and $a,r\in(0,\infty)$ , we have

(14) \begin{eqnarray}&&n\!\left(\textrm{e}^{-q \alpha_{a}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(a-\varepsilon\!\left(\alpha_{a}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{a}^{+}(\varepsilon) < \zeta\}}\right)\nonumber\\[3pt]&\,{=}\,&\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\left(\textrm{e}^{\lambda a}-\left(\psi(\lambda)-q\right)\left(\textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(a)\textrm{d}s\right)\right)\nonumber\\[3pt]&&-\,\lambda \textrm{e}^{\lambda a}+\left(\psi(\lambda)-q\right)\left(\lambda \textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+W^{(q)}(a)\right.\nonumber\\[3pt]&&\left.+\,\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\!.\end{eqnarray}

Proof. Given $q,\lambda\geq0$ , $r, a>0$ , and $a\geq x$ , by Theorem 3.1 in [Reference Loeffen, Palmowski and Surya22] we have

(15) \begin{eqnarray}&&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}-r\right)}\textrm{e}^{\lambda X(\kappa_{r})-\psi\left(\lambda\right)r}\mathbf{1}_{\{\kappa_{r}<\tau_{a}^{+}\}}\right)\nonumber\\[3pt]&\,{=}\,&\textrm{e}^{\lambda x}-\left(\psi(\lambda)-q\right)\left(\textrm{e}^{\lambda x}\int_{0}^{x}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(x)\textrm{d}s\right)\nonumber\\[3pt]&&-\,\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(a)}\left(\textrm{e}^{\lambda a}-\left(\psi(\lambda)-q\right)\left(\textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z\right.\right.\nonumber\\[3pt]&&\left.\left.+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(a)\textrm{d}s\right)\right)\!.\end{eqnarray}

By (4) and the compensation formula (see for example Corollary 4.11 of [Reference Bertoin3] or Theorem 4.4 of [Reference Kyprianou16]), one gets

\begin{eqnarray}&&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}-r\right)}\textrm{e}^{\lambda X(\kappa_{r})-\psi\left(\lambda\right)r}\mathbf{1}_{\{\kappa_{r}<\tau_{a}^{+}\}}\right)\nonumber\\[3pt]&\,{=}\,&\mathbb{E}_{x}\!\left(\sum_{g}\textrm{e}^{-q g}\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\textrm{e}^{-q \alpha_{x+L(g)}^{+}(\epsilon_{g})}\right.\nonumber\\[3pt]&&\left.\times\,\textrm{e}^{\lambda \left(x+L(g)-\epsilon_{g}\left(\alpha_{x+L(g)}^{+}(\epsilon_{g})+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{x+L(g)}^{+}(\epsilon_{g})< \zeta_{g}}\}\right)\nonumber\\[3pt]&\,{=}\,&\mathbb{E}_{x}\!\left(\int_{0}^{\infty}\textrm{e}^{-q w}\prod\limits_{h<w}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(w)\leq a\}}\int_{\mathcal{E}}\textrm{e}^{-q \alpha_{x+L(w)}^{+}(\varepsilon)}\right.\nonumber\\[3pt]&&\left.\times\,\textrm{e}^{\lambda \left(x+L(w)-\varepsilon\!\left(\alpha_{x+L(w)}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{x+L(w)}^{+}(\varepsilon)< \zeta\}}\,n(\,\textrm{d}\varepsilon)\,\textrm{d}L(w)\right)\nonumber\\[3pt]&\,{=}\,&\mathbb{E}_{x}\!\left(\int_{0}^{a-x}\textrm{e}^{-q L^{-1}(w-\!)}\prod\limits_{h<L^{-1}(w-\!)}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h}\}}\right.\nonumber\\[3pt]&&\left.\times\,\int_{\mathcal{E}}\textrm{e}^{-q \alpha_{x+w}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(x+w-\varepsilon\!\left(\alpha_{x+w}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{x+w}^{+}(\varepsilon)< \zeta\}}\,n(\,\textrm{d}\varepsilon)\,\textrm{d}w\right)\nonumber\\[3pt]&\,{=}\,&\int_{x}^{a}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{w}^{+}}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}\}}\right)\,n\!\left(\textrm{e}^{-q \alpha_{w}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(w-\varepsilon\!\left(\alpha_{w}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon)< \zeta\}}\right)\,\textrm{d}w\nonumber\\[3pt]&\,{=}\,&\int_{x}^{a}\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(w)}\,n\!\left(\textrm{e}^{-q \alpha_{w}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(w-\varepsilon\!\left(\alpha_{w}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon) < \zeta\}}\right)\,\textrm{d}w,\nonumber\end{eqnarray}

where $\epsilon_{h}(h\leq g)$ denotes the excursion (away from 0) with left endpoint h for the reflected process $\{\bar{X}(t)-X(t);\ t\geq0\}$ , and $\zeta_{h}$ and $\bar{\epsilon}_{h}$ denote its lifetime and excursion height, respectively. Note that by (15),

\begin{eqnarray}&&\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(a)}\,n\!\left(\textrm{e}^{-q \alpha_{a}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(a-\varepsilon\!\left(\alpha_{a}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{a}^{+}(\varepsilon) < \zeta\}}\right)\nonumber\\[3pt]&\,{=}\,&\frac{\ell_{r}^{(q)}(x)\,\ell_{r}^{(q)\prime}(a)}{\left(\ell_{r}^{(q)}(a)\right)^{2}}\bigg(\textrm{e}^{\lambda a}-\left(\psi(\lambda)-q\right)\bigg(\textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(a)\textrm{d}s\bigg)\bigg)\nonumber\\[3pt]&&-\,\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(a)}\left(\lambda \textrm{e}^{\lambda a}-\left(\psi(\lambda)-q\right)\left(\lambda \textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+W^{(q)}(a)\right.\right.\nonumber\\[3pt]&&\left.\left.+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\right)\!.\nonumber\end{eqnarray}

We thus obtain (14). □

The following result gives an expression for the potential measure of the excursion process until time $\alpha_{a}^{+}+r$ under the excursion measure.

Proposition 3.3. For any $q,\lambda\in[0,\infty)$ , $a,r\in(0,\infty)$ and any bounded differentiable function f, we have

(16) \begin{align}&\quad W^{(q)}(0+\!)\,\textrm{e}^{q r}f(a)+\,n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(a-\varepsilon\!\left(t\right))\mathbf{1}_{\{\alpha_{a}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\nonumber\\[3pt]&=\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{a}\left(\,f(X(s))\right)\!\textrm{d}s-\int_{0}^{a}W^{(q)}(a-z)\mathbb{E}_{z}\left(\,f(X(r))\right)\!\textrm{d}z\right.\nonumber\\[3pt]&\quad\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(a)\textrm{d}s\right)\nonumber\\[3pt]&\quad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f'(a+X(s))\right)\!\textrm{d}s+\int_{0}^{a}W^{(q)\prime}(a-z)\mathbb{E}_{z}\left(\,f(X(r))\right)\!\textrm{d}z\nonumber\\[3pt]&\quad+W^{(q)}(0+\!)\,\mathbb{E}_{a}\left(\,f(X(r))\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)\prime}(a)\textrm{d}s.\end{align}

Proof. Let $e_{q}$ be an exponentially distributed random variable with mean $1/q$ independent of X. For $q,\lambda\geq0$ and $r, b>0$ with $x\leq a$ , we have

(17) \begin{eqnarray}&&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t));\,t<\kappa_{r}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&\,{=}\,&\mathbb{E}_{x}\!\left(\int_{0}^{\kappa_{r}\wedge \tau_{a}^{+}}\textrm{e}^{-q \left(t-r\right)}f(X(t))\,\textrm{d}\left(\int_{0}^{t}\mathbf{1}_{\{X(s)=\bar{X}(s)\}}\textrm{d}s\right)\right)\nonumber\\[3pt]&&+\,\frac{1}{q}\mathbb{E}_{x}\!\left(\textrm{e}^{q r}f(X(e_{q}))\,\mathbf{1}_{\{e_{q}<\kappa_{r}\wedge \tau_{a}^{+}\}}\mathbf{1}_{\{X(e_{q})<\bar{X}(e_{q})\}}\right)\nonumber\\[3pt]&\coloneqq\, &h_{1}(x)+h_{2}(x).\end{eqnarray}

Note that

\[\int_{0}^{t}\mathbf{1}_{\{X(s)=\bar{X}(s)\}}\textrm{d}s=W^{(q)}(0+\!) \,\bar{X}(t)\]

and that $X(t)=\bar{X}(t)$ implies $t=L^{-1}(L(t))$ almost surely. Thus, the function $h_{1}(x)$ can be further expressed as follows:

(18) \begin{align}&W^{(q)}(0+\!)\,\mathbb{E}_{x}\!\left(\int_{0}^{\infty}\textrm{e}^{-q \left(L^{-1}(L(t))-r\right)}f(x+L(t))\mathbf{1}_{\{x+L(t)\leq a,\,L^{-1}(L(t))<\kappa_{r}\}}\textrm{d}L(t)\right)\nonumber\\[3pt]&=W^{(q)}(0+\!)\,\mathbb{E}_{x}\!\left(\int_{0}^{a-x}\textrm{e}^{-q \left(L^{-1}(w)-r\right)}f(x+w)\mathbf{1}_{\{L^{-1}(w)<\kappa_{r}\}}\textrm{d}w\right)\nonumber\\[3pt]&=W^{(q)(0+\!)}\,\textrm{e}^{q r}\int_{0}^{a-x}\mathbb{E}_{x}\!\left(\textrm{e}^{-q L^{-1}(w)}\mathbf{1}_{\{L^{-1}(w)<\kappa_{r}\}}\right)f(x+w)\textrm{d}w\nonumber\\[3pt]&=W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{0}^{a-x}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{x+w}^{+}}\mathbf{1}_{\{\tau_{x+w}^{+}<\kappa_{r}\}}\right)f(x+w)\textrm{d}w\nonumber\displaybreak\\[3pt]&=W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{x}^{a}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{w}^{+}}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}\}}\right)f(w)\textrm{d}w\nonumber\\[3pt]&=W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{x}^{a}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(w)}\,f(w)\textrm{d}w,\end{align}

where (4) is used in the final equality.

To further develop $h_{2}(x)$ , note that $X(e_{q})<\bar{X}(e_{q})$ if and only if there is an excursion with left endpoint g such that $e_{q}\in(g,g+\zeta_{g}).$ Hence, by the compensation formula and the memoryless property of the exponential random variable, $h_{2}(x)$ can be rewritten as

(19) \begin{eqnarray}&&\frac{1}{q}\mathbb{E}_{x}\Bigg(\!\sum_{g}\textrm{e}^{q r}\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\nonumber\\[3pt]&&\quad\qquad\times\ f(x+L(g)-\epsilon_{g}\left(e_{q}-g\right))\mathbf{1}_{\{\alpha_{x+L(g)}^{+}(\epsilon_{g})>e_{q}-g-r,\,0<e_{q}-g<\zeta_{g}\}}\Bigg)\nonumber\\[3pt]&\,{=}\,&\frac{1}{q}\mathbb{E}_{x}\Bigg(\!\sum_{g}\textrm{e}^{-q (g-r)}\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\nonumber\\[3pt]&&\quad\qquad\times\ f(x+L(g)-\epsilon_{g}(e_{q}))\mathbf{1}_{\{\alpha_{x+L(g)}^{+}(\epsilon_{g})>e_{q}-r,\,e_{q}<\zeta_{g}\}}\Bigg)\nonumber\\[3pt]&{=}\,&\frac{1}{q}\mathbb{E}_{x}\!\left(\!\int_{0}^{\infty}\textrm{e}^{-q (w-r)}\prod\limits_{h<w}\mathbf{1}_{\{\alpha_{x+L(h)}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(w)\leq a\}}\right.\nonumber\\[3pt]&&\quad\qquad\left.\times \int_{\mathcal{E}}f(x+L(w)-\varepsilon\!\left(e_{q}\right))\mathbf{1}_{\{\alpha_{x+L(w)}^{+}(\varepsilon)>e_{q}-r,\,e_{q}<\zeta\}}n(\,\textrm{d}\varepsilon)\,\textrm{d}L(w)\right)\nonumber\\[3pt]&\,{=}\,&\frac{1}{q}\mathbb{E}_{x}\!\left(\int_{x}^{a}\textrm{e}^{-q (\tau_{w}^{+}-r)}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}\}}\int_{\mathcal{E}}f(w-\varepsilon\!\left(e_{q}\right))\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon)> e_{q}-r,\,e_{q}<\zeta\}}n(\,\textrm{d}\varepsilon)\,\textrm{d}w\right)\nonumber\\[3pt]&\,{=}\,&\int_{x}^{a}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{w}^{+}}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}\}}\right)\,n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(w-\varepsilon\!\left(t\right))\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\,\textrm{d}w\nonumber\\[3pt]&\,{=}\,&\int_{x}^{a}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(w)}n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(w-\varepsilon\!\left(t\right))\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\textrm{d}w.\end{eqnarray}

It follows from (17), (18), and (19) that

(20) \begin{eqnarray}&&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t)); \,t<\kappa_{r}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&\,{=}\,&W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{x}^{a}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(w)}\,f(w)\textrm{d}w\nonumber\\[3pt]&&+\int_{x}^{a}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(w)}\,n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(w-\varepsilon\!\left(t\right))\mathbf{1}_{\{\alpha_{w}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\textrm{d}w.\end{eqnarray}

Meanwhile, by Theorem 4.4 of [Reference Loeffen, Palmowski and Surya22] we know that

(21) \begin{eqnarray}&&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t)); \,t<\kappa_{r}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&=&\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{x}\!\left(\,f(X(s))\right)\!\textrm{d}s-\int_{0}^{x}W^{(q)}(x-z)\mathbb{E}_{z}\left(\,f(X(r))\right)\!\textrm{d}z\nonumber\\[3pt]&&-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(x)\textrm{d}s\nonumber\\[3pt]&&-\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(a)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{a}\left(\,f(X(s))\right)\!\textrm{d}s-\int_{0}^{a}W^{(q)}(a-z)\mathbb{E}_{z}\left(\,f(X(r))\right)\!\textrm{d}z\right.\nonumber\\[3pt]&&\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(a)\textrm{d}s\right)\!.\end{eqnarray}

Combining (20) and (21), we obtain (16). □

Remark 3.2. For any $b\in(\!-\infty,\infty)$ , if we replace f(x) with f(x, b) in Proposition 3.3, by similar arguments we have

(22) \begin{eqnarray}&&W^{(q)}(0+\!)\,\textrm{e}^{q r}f(a,b)+\,n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(a-\varepsilon(t),b)\mathbf{1}_{\{\alpha_{a}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\nonumber\\[3pt]&\,{=}\,&\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{a}\left(\,f(X(s),b)\right)\!\textrm{d}s-\int_{0}^{a}W^{(q)}(a-z)\mathbb{E}_{z}\left(\,f(X(r),b)\right)\!\textrm{d}z\right.\nonumber\\[3pt]&&\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s),b)\right)\ell_{s}^{(q)}(a)\textrm{d}s\right)\nonumber\\[3pt]&&-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\frac{\partial}{\partial x}f(a+X(s),b)\right)\!\textrm{d}s+\int_{0}^{a}W^{(q)\prime}(a-z)\mathbb{E}_{z}\left(\,f(X(r),b)\right)\!\textrm{d}z\nonumber\\[3pt]&&\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{a}\left(\,f(X(r),b)\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s),b)\right)\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\!.\end{eqnarray}

Remark 3.3. Letting $f(x)\coloneqq \textrm{e}^{\lambda x-\psi\left(\lambda\right)r}$ in Proposition 3.3, we have

\begin{eqnarray}&&n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}\textrm{e}^{\lambda \left(a-\varepsilon\!\left(t\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{a}^{+}(\varepsilon)> t-r\}}\,\textrm{d}t\right)+W^{(q)}(0+\!)\,\textrm{e}^{q r}\textrm{e}^{\lambda a-\psi\left(\lambda\right)r}\nonumber\\[3pt]&\,{=}\,&\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\bigg(\frac{\textrm{e}^{\lambda a}\left(1-\textrm{e}^{-(\psi(\lambda)-q)r}\right)}{\psi(\lambda)-q}-\textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z-\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(a)\textrm{d}s\bigg)\nonumber\\[3pt]&&-\,\frac{\lambda \textrm{e}^{\lambda a}\left(1-\textrm{e}^{-(\psi(\lambda)-q)r}\right)}{\psi(\lambda)-q}+\lambda \textrm{e}^{\lambda a}\int_{0}^{a}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+W^{(q)}(a)\nonumber\\&&+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(a)\textrm{d}s.\nonumber\end{eqnarray}

4. Main results

In this section we present several results concerning the draw-down Parisian ruin. The first result below solves a draw-down Parisian-ruin-based two-sided exit problem. It generalizes Theorem 1 of [Reference Czarna and Renaud11].

Theorem 4.1. Given any a, let $\eta$ be another draw-down function such that $\eta(z)<\xi(z)< z$ for $z\leq a$ . For any $x\in (\!-\infty, a)$ , we have

\begin{eqnarray*}&&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{\eta}\}}\right)\nonumber\\\\[-7pt]&\,{=}\,&\exp\left(\!-\int_{x}^{a}\frac{W^{(q)\prime}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)-\eta(w),r)+\chi^{(q)\prime}(\overline{\xi}(w),\xi(w)-\eta(w),r)}{W^{(q)}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)-\eta(w),r)+\chi^{(q)}(\overline{\xi}(w),\xi(w)-\eta(w),r)}\,\textrm{d}w\right)\!,\nonumber\end{eqnarray*}

where the Laplace transforms of $\phi_{_{\Phi_{q}}}(y,r)$ and $\chi^{(q)}(x,y,r)\coloneqq \textrm{e}^{\Phi_{q}x}\chi_{_{\Phi_{q}}}(x,y,r)$ with respect to r are given, respectively, by

(23) \begin{align}\int_{0}^{\infty}&\!\textrm{e}^{-\theta r}\chi^{(q)}(x,y,r)\,\textrm{d}r\nonumber\\[3pt]&\!=\frac{1}{\theta}\left(\frac{W_{y}^{(\theta+q,-\theta)}(x+y)}{W^{(\theta+q)}(y)}-\frac{W^{(q)}(x)\textrm{e}^{\Phi_{q}y}\left(1+\theta\int_{0}^{y}\textrm{e}^{-\Phi_{q}w}W^{(\theta+q)}(w)\textrm{d}w\right)}{W^{(\theta+q)}(y)}\right)\end{align}

and

(24) \begin{eqnarray}\int_{0}^{\infty}\textrm{e}^{-\theta r}\phi_{_{\Phi_{q}}}(y,r)\,\textrm{d}r=\frac{\textrm{e}^{\Phi_{q}y}\left(1+\theta\int_{0}^{y}\textrm{e}^{-\Phi_{q}w}W^{(\theta+q)}(w)\textrm{d}w\right)}{\theta W^{(\theta+q)}(y)};\end{eqnarray}

here, $y\in(0,\infty)$ , the derivative of $\chi^{(q)}$ is taken on the first argument, and $\phi_{_{\Phi_{q}}}$ and $\chi_{_{\Phi_{q}}}$ play the roles of $\phi$ and $\chi$ for the process $(X,\mathbb{P}_{x}^{\Phi_{q}})$ .

Proof. By (7) and an argument similar to that of (10) we have

(25) \begin{eqnarray}&&\mathbb{P}_{x}\!\left(\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{\eta}\right)\nonumber\\\nonumber\\&\,{=}\,&\mathbb{E}_{x}\Bigg(\prod\limits_{t\leq a-x}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+t)}^{+}(\varepsilon_{t})= \zeta(\varepsilon_{t}),\,\overline{\varepsilon}_{t}\leq \overline{\eta}(x+t)\}}\Bigg)\nonumber\\\nonumber\\&\,{=}\,&\exp\left(\!-\int_{x}^{a}n\!\left(\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)<\zeta \mbox{ or } \overline{\varepsilon}> \overline{\xi}(w)+\xi(w)-\eta(w)\right)\,\textrm{d}w\right)\nonumber\\\nonumber\\&\,{=}\,&\exp\left(\!-\int_{x}^{a}\frac{W^{\prime}(\overline{\xi}(w))\phi(\xi(w)-\eta(w),r)+\chi^{\prime}(\overline{\xi}(w),\xi(w)-\eta(w),r)}{W(\overline{\xi}(w))\phi(\xi(w)-\eta(w),r)+\chi(\overline{\xi}(w),\xi(w)-\eta(w),r)}\,\textrm{d}w\right)\!,\end{eqnarray}

where $\overline{\eta}(w)\coloneqq w-\eta(w)$ . By (25) together with a change of measure, one has

\begin{align}&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{\eta}\}}\right)\nonumber\\\nonumber\\[-7pt]&=\textrm{e}^{-\Phi_{q}\left(a-x\right)}\mathbb{P}_{x}^{\Phi_{q}}\left(\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{\eta}\right)\nonumber\\\nonumber\\[-7pt]&=\textrm{e}^{-\Phi_{q}\left(a-x\right)}\exp\left(\!-\int_{x}^{a}n_{_{\Phi_{q}}}\left(\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)<\zeta \mbox{ or }\overline{\varepsilon}> \overline{\eta}(w)\right)\,\textrm{d}w\right)\nonumber\\\nonumber\displaybreak\\[-7pt]&=\textrm{e}^{-\Phi_{q}\left(a\,{-}\,x\right)\,{-}\,\int_{x}^{a}\frac{W_{\Phi_{q}}^{\prime}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)\,{-}\,\eta(w),r)\,{+}\,\chi_{_{\Phi_{q}}}^{\prime}(\overline{\xi}(w),\xi(w)\,{-}\,\eta(w),r)}{W_{\Phi_{q}}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)\,{-}\,\eta(w),r)\,{+}\,\chi_{_{\Phi_{q}}}(\overline{\xi}(w),\xi(w)\,{-}\,\eta(w),r)}\,\textrm{d}w}\nonumber\\\nonumber\\[-7pt]&=\textrm{e}^{-\int_{x}^{a}\frac{W^{(q)\prime}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)\,{-}\,\eta(w),r)\,{+}\,\chi^{(q)\prime}(\overline{\xi}(w),\xi(w)\,{-}\,\eta(w),r)}{W^{(q)}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(\xi(w)\,{-}\,\eta(w),r)\,{+}\,\chi^{(q)}(\overline{\xi}(w),\xi(w)\,{-}\,\eta(w),r)}\,\textrm{d}w},\nonumber\end{align}

where

\[W_{\Phi_{q}}(x)=\textrm{e}^{-\Phi_{q}x}W^{(q)}(x), \,\,\, \,\,\, \chi_{_{\Phi_{q}}}(x,y,r)=\textrm{e}^{-\Phi_{q}x}\chi^{(q)}(x,y,r),\]

and $n_{_{\Phi_{q}}}$ represents the excursion measure under the new probability measure $\mathbb{P}_{x}^{\Phi_{q}}$ . □

We next provide a version of Theorem 4.1 for $\eta\equiv-\infty$ .

Corollary 4.1. For $x\in (\!-\infty, a)$ , we have

(26) \begin{eqnarray}\mbox{\,\,\,\,\,}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\}}\right)=\exp\left(\!-\int_{x}^{a} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\,\textrm{d}w\right)\!.\end{eqnarray}

Proof. By definition we have

\[W^{(\theta+q,-\theta)}_{y}(x+y)=W^{(\theta+q)}(x+y)-\theta\int_{0}^{x}W^{(q)}(w)W^{(\theta+q)}(x+y-w)\,\textrm{d}w,\]

which together with (2) yields

(27) \begin{eqnarray}&&\lim\limits_{y\uparrow\infty}\frac{W^{(\theta+q,-\theta)}_{y}(x+y)}{\theta W^{(\theta+q)}(y)}\nonumber\\[4pt]&\,{=}\,&\textrm{e}^{\Phi_{\theta+q}x}\left(\frac{1}{\theta}-\int_{0}^{x}W^{(q)}(w)\,\textrm{e}^{-\Phi_{\theta+q}w}\,\textrm{d}w\right)\nonumber\\[4pt]&\,{=}\,&\textrm{e}^{\Phi_{\theta+q}x}\left(\frac{1}{\theta}-\int_{0}^{\infty}W^{(q)}(w)\,\textrm{e}^{-\Phi_{\theta+q}w}\,\textrm{d}w+\int_{x}^{\infty}W^{(q)}(w)\,\textrm{e}^{-\Phi_{\theta+q}w}\,\textrm{d}w\right)\nonumber\\[4pt]&=&\int_{0}^{\infty}W^{(q)}(x+w)\,\textrm{e}^{-\Phi_{\theta+q} w}\,\textrm{d}w.\end{eqnarray}

This coincides with the Laplace transform (in r) of $\,\textrm{e}^{-qr}\ell_{r}^{(q)}(x)$ as follows:

(28) \begin{eqnarray}\int_{0}^{\infty}\textrm{e}^{-\theta r}\textrm{e}^{-qr}\ell_{r}^{(q)}(x)\,\textrm{d}r&\,{=}\,&\int_{0}^{\infty}\textrm{e}^{-(\theta+q)r}\int_{0}^{\infty}W^{(q)}(x+z)\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\,\textrm{d}r\nonumber\\&\,{=}\,&\int_{0}^{\infty}\textrm{e}^{-(\theta+q)r}\int_{0}^{\infty}W^{(q)}(x+z)\mathbb{P}(\tau_{z}^{+}\in\textrm{d}r)\textrm{d}z\nonumber\\&\,{=}\,&\int_{0}^{\infty}W^{(q)}(x+z)\mathbb{E}(\textrm{e}^{-(\theta+q) \tau_{z}^{+}})\textrm{d}z\nonumber\\&\,{=}\,&\int_{0}^{\infty}W^{(q)}(x+w)\,\textrm{e}^{-\Phi_{\theta+q} w}\,\textrm{d}w,\end{eqnarray}

where we have used Kendall’s identity,

\[\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\,\textrm{d}r=\mathbb{P}(\tau_{z}^{+}\in\textrm{d}r)\textrm{d}z,\qquad z,r\geq0.\]

From (23) and (24) one knows that

(29) \begin{eqnarray}\int_{0}^{\infty}\textrm{e}^{-\theta r}\left(W^{(q)}(x)\phi_{_{\Phi_{q}}}(y,r)+\chi^{(q)}(x,y,r)\right)\textrm{d}r=\frac{W_{y}^{(\theta+q,-\theta)}(x+y)}{\theta W^{(\theta+q)}(y)}.\end{eqnarray}

Combining (27), (28), and (29), one can conclude that

\[\lim\limits_{y\uparrow\infty}\left(W^{(q)}(x)\phi_{_{\Phi_{q}}}(y,r)+\chi^{(q)}(x,y,r)\right)=\textrm{e}^{-qr}\ell_{r}^{(q)}(x).\]

By the same arguments, we have

\[\lim\limits_{y\uparrow\infty}\left(W^{(q)\prime}(x)\phi_{_{\Phi_{q}}}(y,r)+\chi^{(q)\prime}(x,y,r)\right)=\textrm{e}^{-qr}\ell_{r}^{(q)\prime}(x).\]

Hence, we have

\begin{eqnarray}&&\lim\limits_{c\uparrow\infty}\frac{W^{(q)\prime}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(c+\xi(w),r)+\chi^{(q)\prime}(\overline{\xi}(w),c+\xi(w),r)}{W^{(q)}(\overline{\xi}(w))\phi_{_{\Phi_{q}}}(c+\xi(w),r)+\chi^{(q)}(\overline{\xi}(w),c+\xi(w),r)}=\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}.\nonumber\end{eqnarray}

It then follows easily from Theorem 4.1 that

\begin{eqnarray}\mbox{\,\,\,\,\,}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\}}\right)&\,{=}\,&\lim\limits_{c\uparrow\infty}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\wedge \tau_{-c}^{-}\}}\right)\nonumber\\[4pt] &\,{=}\,&\exp\left(\!-\int_{x}^{a} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\,\textrm{d}w\right)\!,\nonumber\end{eqnarray}

which is the desired result. □

Remark 4.1. Letting $\xi\equiv0$ in (26), one recovers (18) of [Reference Czarna and Palmowski10]:

\begin{eqnarray*}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}\}}\right)&\,{=}\,&\exp\left(\!-\int_{x}^{a} \frac{\ell^{(q)\prime}_{r}(w)}{\ell_{r}^{(q)}(w)}\,\textrm{d}w\right)=\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(a)}.\nonumber\end{eqnarray*}

Letting $\xi(x)=kx-d$ with $k\in(\!-\infty,1)$ and $d\in (0,\infty)$ in (26), we have

\begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q\tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi}\}}\right)&=&\left(\frac{\ell_{r}^{(q)}((1-k)x+d)}{ \ell_{r}^{(q)}((1-k)a+d)}\right)^{\frac{1}{1-k}}.\nonumber\end{eqnarray}

By (26), one can also obtain the draw-down Parisian ruin probability

\begin{eqnarray}\mathbb{P}_{x}\left(\kappa_{r}^{\xi}<\infty\right)=1-\exp\left(\!-\int_{x}^{\infty} \frac{\ell_{r}^{\prime}(\overline{\xi}(z))}{\ell_{r}(\overline{\xi}(z))}\,\textrm{d}z\right)\!.\nonumber\end{eqnarray}

The following result presents the joint Laplace transform involving the draw-down Parisian ruin time, the position of X at the draw-down Parisian ruin time, and its running supremum until the draw-down Parisian ruin time. It generalizes Theorem 3.1 in [Reference Loeffen, Palmowski and Surya22].

Theorem 4.2. For any $q,\lambda\in[0,\infty)$ , $x\in(\!-\infty,\infty)$ , $a \geq x$ and any bounded measurable function $\varphi:\,(\!-\infty,\infty)\mapsto (\!-\infty,\infty)$ , we have

(30) \begin{eqnarray}&&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi}-r\right)}\textrm{e}^{\lambda X(\kappa_{r}^{\xi})-\psi\left(\lambda\right)r}\varphi(\bar{X}(\kappa_{r}^{\xi}))\mathbf{1}_{\{\kappa_{r}^{\xi}<\tau_{a}^{+}\}}\right)\nonumber\\[3pt]&=&\int_{x}^{a}\textrm{e}^{\lambda \xi(w)}\varphi(w)\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\left(\textrm{e}^{\lambda \overline{\xi}(w)}-\left(\psi(\lambda)-q\right)\right.\right.\nonumber\\[3pt]&&\times\left.\left.\left(\textrm{e}^{\lambda \overline{\xi}(w)}\int_{0}^{\overline{\xi}(w)}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(\overline{\xi}(w))\textrm{d}s\right)\right)\right.\nonumber\\[3pt]&&\left.-\,\lambda \textrm{e}^{\lambda \overline{\xi}(w)}+\left(\psi(\lambda)-q\right)\left(\lambda \textrm{e}^{\lambda \overline{\xi}(w)}\int_{0}^{\overline{\xi}(w)}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z\right.\right.\nonumber\\[3pt]&&\left.\left.+\,W^{(q)}(\overline{\xi}(w))+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(\overline{\xi}(w))\textrm{d}s\right)\right]\,\textrm{d}w.\end{eqnarray}

Proof. By (26) and the compensation formula, we have

\begin{align}&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi}-r\right)}\textrm{e}^{\lambda X(\kappa_{r}^{\xi})-\psi\left(\lambda\right)r}\varphi(\bar{X}(\kappa_{r}^{\xi}))\mathbf{1}_{\{\kappa_{r}^{\xi}<\tau_{a}^{+}\}}\right)\nonumber\\[3pt]&=\mathbb{E}_{x}\Bigg(\sum_{g}\textrm{e}^{-q g}\varphi(x+L(g))\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\textrm{e}^{-q \alpha_{\overline{\xi}(x+L(g))}^{+}(\epsilon_{g})}\nonumber\\[3pt]&\qquad\qquad\times\textrm{e}^{\lambda \left(x+L(g)-\epsilon_{g}\left(\alpha_{\overline{\xi}(x+L(g))}^{+}(\epsilon_{g})+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(g))}^{+}(\epsilon_{g})< \zeta_{g}\}}\Bigg)\nonumber\\[3pt]&=\mathbb{E}_{x}\!\left(\int_{0}^{\infty}\textrm{e}^{-q w}\varphi(x+L(w))\prod\limits_{h<w}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(w)\leq a\}}\int_{\mathcal{E}}\textrm{e}^{-q \alpha_{\overline{\xi}(x+L(w))}^{+}(\varepsilon)}\right.\nonumber\\[3pt]&\qquad\qquad\left.\times\textrm{e}^{\lambda \left(x+L(w)-\varepsilon\!\left(\alpha_{\overline{\xi}(x+L(w))}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(w))}^{+}(\varepsilon)< \zeta\}}\,n(\,\textrm{d}\varepsilon)\,\textrm{d}L(w)\right)\nonumber\\[3pt]&=\mathbb{E}_{x}\Bigg(\int_{0}^{a-x}\textrm{e}^{-q L^{-1}(w-\!)}\varphi(x+w)\prod\limits_{h<L^{-1}(w-\!)}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h}\}}\nonumber\\[3pt]&\qquad\qquad\times\int_{\mathcal{E}}\textrm{e}^{-q \alpha_{\overline{\xi}(x+w)}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(x+w-\varepsilon\!\left(\alpha_{\overline{\xi}(x+w)}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+w)}^{+}(\varepsilon)< \zeta\}}\,n(\,\textrm{d}\varepsilon)\,\textrm{d}w\Bigg)\nonumber\\[3pt]&=\int_{x}^{a}\varphi(w)\mathbb{E}_{x}\bigg(\textrm{e}^{-q \tau_{w}^{+}}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}^{\xi}\}}\bigg)\nonumber\\[3pt]&\qquad\qquad\times n\bigg(\textrm{e}^{-q \alpha_{\overline{\xi}(w)}^{+}(\varepsilon)}\textrm{e}^{\lambda \big(w-\varepsilon\big(\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)+r\big)\big)-\psi(\lambda)r}\mathbf{1}_{\{\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)< \zeta\}}\bigg)\,\textrm{d}w\nonumber\displaybreak\\[3pt]&=\int_{x}^{a}\textrm{e}^{\lambda \xi(w)}\varphi(w)\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\nonumber\\[3pt]&\qquad\qquad\times n\!\left(\textrm{e}^{-q \alpha_{\overline{\xi}(w)}^{+}(\varepsilon)}\textrm{e}^{\lambda \left(\overline{\xi}(w)-\varepsilon\!\left(\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)+r\right)\right)-\psi\left(\lambda\right)r}\mathbf{1}_{\{\alpha_{\overline{\xi}(w)}^{+}(\varepsilon) < \zeta\}}\right)\,\textrm{d}w,\nonumber\end{align}

which together with (14) yields (30). □

Remark 4.2. By Theorem 4.2, for any $q\in[0,\infty)$ , $x\in(\!-\infty,\infty)$ , and $a \in[x,\infty)$ we have

\begin{align}&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi}-r\right)}\mathbf{1}_{\{\kappa_{r}^{\xi}<\tau_{a}^{+}\}}\right)\nonumber\\[3pt]&\,\,\quad=\int_{x}^{a}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\left(Z^{(q)}(\overline{\xi}(w))+q\int_{0}^{r}\ell_{s}^{(q)}(\overline{\xi}(w))\textrm{d}s\right)\right.\nonumber\\[3pt]&\qquad\left.-\,q\left(W^{(q)}(\overline{\xi}(w))+\int_{0}^{r}\ell_{s}^{(q)\prime}(\overline{\xi}(w))\textrm{d}s\right)\right]\,\textrm{d}w.\nonumber\end{align}

Letting $a\uparrow\infty$ and then choosing $\xi(w)=w\vee y-a$ with $0\leq y-x<a<\infty$ gives

\begin{align}&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi}-r\right)}\mathbf{1}_{\{\kappa_{r}^{\xi}<\infty\}}\right)\nonumber\\[3pt]&\,\,\quad=\int_{x}^{y}\frac{\partial}{\partial w}\left[\frac{-\ell_{r}^{(q)}(x-y+a)}{\ell_{r}^{(q)}(w-y+a)}\left(Z^{(q)}(w-y+a)+q\int_{0}^{r}\ell_{s}^{(q)}(w-y+a)\textrm{d}s\right)\right]\textrm{d}w\nonumber\\[3pt]&\qquad+\frac{\ell_{r}^{(q)}(x-y+a)}{\ell_{r}^{(q)}(a)}\int_{y}^{\infty}\exp\left(\!-\int_{y}^{w} \frac{\ell^{(q)\prime}_{r}(a)}{\ell_{r}^{(q)}(a)}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\left(Z^{(q)}(a)+q\int_{0}^{r}\ell_{s}^{(q)}(a)\textrm{d}s\right)\right.\nonumber\\[3pt]&\qquad\left.-\,q\left(W^{(q)}(a)+\int_{0}^{r}\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\right]\,\textrm{d}w\nonumber\\[3pt]&\,\,\quad=\left[\frac{-\ell_{r}^{(q)}(x-y+a)}{\ell_{r}^{(q)}(w-y+a)}\left(Z^{(q)}(w-y+a)+q\int_{0}^{r}\ell_{s}^{(q)}(w-y+a)\textrm{d}s\right)\right]\Bigg|_{w=x}^{w=y}\nonumber\\[3pt]&\qquad+\frac{\ell_{r}^{(q)}(x-y+a)}{\ell_{r}^{(q)\prime}(a)}\left[\frac{\ell_{r}^{(q)\prime}(a)}{\ell_{r}^{(q)}(a)}\left(Z^{(q)}(a)+q\int_{0}^{r}\ell_{s}^{(q)}(a)\textrm{d}s\right)-q\left(W^{(q)}(a)+\int_{0}^{r}\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\right]\nonumber\\[3pt]&\quad\,\,=Z^{(q)}(x-y+a)+q\int_{0}^{r}\ell_{s}^{(q)}(x-y+a)\textrm{d}s-q\frac{\ell_{r}^{(q)}(x-y+a)}{\ell_{r}^{(q)\prime}(a)}\left(W^{(q)}(a)+\int_{0}^{r}\ell_{s}^{(q)\prime}(a)\textrm{d}s\right)\!,\nonumber\end{align}

which recovers Theorem 2.1 of [Reference Surya26]. It can then be checked via a change of measure that Proposition 2.3 of [Reference Surya26] can also be recovered from our Theorem 4.2.

The following result gives the potential measure of X involving the draw-down Parisian ruin time. It generalizes Theorem 4.4 in [Reference Loeffen, Palmowski and Surya22].

Theorem 4.3. For any $q,\lambda\geq0$ , $r>0$ , $a\geq x$ and any bounded bivariate function f(x,y) that is differentiable with respect to x, we have

\begin{align}&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t),\bar{X}(t)); \,t<\kappa_{r}^{\xi}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt] &\,\,\quad=\int_{x}^{a}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s),w)\right)\!\textrm{d}s\right.\right.\nonumber\\[3pt]&\qquad\left.-\int_{0}^{\overline{\xi}(w)}W^{(q)}(\overline{\xi}(w)-z)\mathbb{E}_{}\left(\,f(z+\xi(w)+X(r),w)\right)\!\textrm{d}z\right.\nonumber\\[3pt]&\qquad\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(\xi(w)+X(r-s),w)\right)\ell_{s}^{(q)}(\overline{\xi}(w))\textrm{d}s\right)\nonumber\\[3pt]&\qquad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\frac{\partial}{\partial x}f(w+X(s),w)\right)\!\textrm{d}s\nonumber\\[3pt]&\qquad+\int_{0}^{\overline{\xi}(w)}W^{(q)\prime}(\overline{\xi}(w)-z)\mathbb{E}_{}\left(\,f(z+\xi(w)+X(r),w)\right)\!\textrm{d}z\nonumber\\[3pt]&\qquad\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(w+X(r),w)\right)+\int_{0}^{r}\mathbb{E}_{}(\,f(\xi(w)+X(r-s),w))\ell_{s}^{(q)\prime}(\overline{\xi}(w))\textrm{d}s\right]\textrm{d}w.\nonumber\end{align}

Proof. For $q,\lambda\geq0$ and $r>0$ with $ a\geq x$ , we have

(31) \begin{align}&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t),\bar{X}(t));\,t<\kappa_{r}^{\xi}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\&\quad=\mathbb{E}_{x}\!\left(\int_{0}^{\kappa_{r}^{\xi}\wedge \tau_{a}^{+}}\textrm{e}^{-q \left(t-r\right)}f(X(t),\bar{X}(t))\,\textrm{d}\left(\int_{0}^{t}\mathbf{1}_{\{X(s)=\bar{X}(s)\}}\textrm{d}s\right)\right)\nonumber\\&\qquad+\,\frac{1}{q}\,\mathbb{E}_{x}\!\left(\textrm{e}^{qr}f(X(e_{q}),\bar{X}(e_{q}))\,\mathbf{1}_{\{X(e_{q})<\bar{X}(e_{q}), \,e_{q}<\kappa_{r}^{\xi}\wedge \tau_{a}^{+}\}}\right)\nonumber\\&\quad\coloneqq I_{1}(x)+I_{2}(x).\end{align}

Note that

\[\int_{0}^{t}\mathbf{1}_{\{X(s)=\bar{X}(s)\}}\textrm{d}s=W^{(q)}(0+\!) \,\bar{X}(t)\]

and that $X(t)=\bar{X}(t)$ implies $t=L^{-1}(L(t))$ almost surely. The function $I_{1}(x)$ can be rewritten as

(32) \begin{align}&W^{(q)}(0+\!)\,\mathbb{E}_{x}\!\left(\int_{0}^{\infty}\textrm{e}^{-q \left(L^{-1}(L(t))-r\right)}f(x+L(t),x+L(t))\right.\nonumber\\[3pt]&\quad\times\mathbf{1}_{\{x+L(t)\leq a,\,L^{-1}(L(t))<\kappa_{r}^{\xi}\}}\textrm{d}L(t)\bigg)\nonumber\\\nonumber\\[-9pt]&=W^{(q)}(0+\!)\,\mathbb{E}_{x}\!\left(\int_{0}^{a-x}\textrm{e}^{-q \left(L^{-1}(w)-r\right)}f(x+w,x+w)\mathbf{1}_{\{L^{-1}(w)<\kappa_{r}^{\xi}\}}\textrm{d}w\right)\nonumber\\\nonumber\displaybreak\\[-9pt]&=W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{0}^{a-x}\mathbb{E}_{x}\!\left(\textrm{e}^{-q L^{-1}(w)}\mathbf{1}_{\{L^{-1}(w)<\kappa_{r}^{\xi}\}}\right)f(x+w,x+w)\textrm{d}w\nonumber\\\nonumber\\[-9pt]&=W^{(q)}(0+\!)\,\textrm{e}^{q r}\int_{x}^{a}\textrm{e}^{-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z}\,f(\overline{\xi}(w)+\xi(w),\overline{\xi}(w)+\xi(w))\textrm{d}w,\end{align}

where (4) is used in the last equality. Using the compensation formula, the function $I_{2}(x)$ can be rewritten as

(33) \begin{eqnarray}&&\frac{1}{q}\mathbb{E}_{x}\Bigg(\sum_{g}\textrm{e}^{q r}\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\nonumber\\[5pt]&&\,\,\qquad\times\ f(x+L(g)-\epsilon_{g}\left(e_{q}-g\right)\!,x+L(g))\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(g))}^{+}(\epsilon_{g})>e_{q}-g-r,\,0<e_{q}-g<\zeta_{g}\}}\Bigg)\nonumber\\[5pt]&=&\frac{1}{q}\mathbb{E}_{x}\Bigg(\sum_{g}\textrm{e}^{-q (g-r)}\prod\limits_{h<g}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(g)\leq a\}}\nonumber\\[5pt]&&\,\,\qquad\times\ f(x+L(g)-\epsilon_{g}(e_{q}),x+L(g))\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(g))}^{+}(\epsilon_{g})>e_{q}-r,\,e_{q}<\zeta_{g}\}}\Bigg)\nonumber\\&=&\frac{1}{q}\mathbb{E}_{x}\!\left(\int_{0}^{\infty}\textrm{e}^{-q (w-r)}\prod\limits_{h<w}\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(h))}^{+}(\epsilon_{h})= \,\zeta_{h},\,x+L(w)\leq a\}}\right.\nonumber\\&&\,\,\qquad\left.\times \int_{\mathcal{E}}f(x+L(w)-\varepsilon\!\left(e_{q}\right)\!,x+L(w))\mathbf{1}_{\{\alpha_{\overline{\xi}(x+L(w))}^{+}(\varepsilon)>e_{q}-r,\,e_{q}<\zeta\}}n(\,\textrm{d}\varepsilon)\,\textrm{d}L(w)\right)\nonumber\\&=&\frac{1}{q}\mathbb{E}_{x}\!\left(\int_{x}^{a}\textrm{e}^{-q (\tau_{w}^{+}-r)}\mathbf{1}_{\{\tau_{w}^{+}<\kappa_{r}^{\xi}\}}\int_{\mathcal{E}}f(w-\varepsilon\!\left(e_{q}\right)\!,w)\mathbf{1}_{\{\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)> e_{q}-r,\,e_{q}<\zeta\}}n(\,\textrm{d}\varepsilon)\,\textrm{d}w\right)\nonumber\\&=&\int_{x}^{a}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\nonumber\\&&\ \ \ \quad\times\ \, n\!\left(\int_{0}^{\zeta}\textrm{e}^{-q (t-r)}f(\overline{\xi}(w)-\varepsilon\!\left(t\right)+\xi(w),\overline{\xi}(w)+\xi(w))\mathbf{1}_{\{\alpha_{\overline{\xi}(w)}^{+}(\varepsilon)> t-r\}}\textrm{d}t\right)\textrm{d}w.\end{eqnarray}

Combining (31), (32), (33), and (22) leads to the desired result. □

We have the following version of Theorem 4.3 when f is independent of y.

Corollary 4.2. For any $q,\lambda\geq0$ , $r>0$ , $a\geq x$ and any bounded differentiable function f, we have

\begin{align}&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t)); \,t<\kappa_{r}^{\xi}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&\quad=\int_{x}^{a}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s))\right)\!\textrm{d}s\right.\right.\nonumber\\[3pt]&\qquad\left.-\int_{0}^{\overline{\xi}(w)}W^{(q)}(\overline{\xi}(w)-z)\mathbb{E}_{}\left(\,f(z+\xi(w)+X(r))\right)\!\textrm{d}z\right.\nonumber\displaybreak\\[3pt]&\qquad\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(\xi(w)+X(r-s))\right)\ell_{s}^{(q)}(\overline{\xi}(w))\textrm{d}s\right)\nonumber\\[3pt]&\qquad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f'(w+X(s))\right)\!\textrm{d}s+\int_{0}^{\overline{\xi}(w)}W^{(q)\prime}(\overline{\xi}(w)-z)\mathbb{E}_{}\left(\,f(z+\xi(w)+X(r))\right)\!\textrm{d}z\nonumber\\[3pt]&\qquad\left.\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(w+X(r))\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(\xi(w)+X(r-s))\right)\ell_{s}^{(q)\prime}(\overline{\xi}(w))\textrm{d}s\right)\right]\,\textrm{d}w.\nonumber\end{align}

The following result gives the Laplace transform of the potential measure of X killed upon up-crossing $a\,(\!\geq x)$ or draw-down Parisian ruin.

Corollary 4.3. For any $q,\lambda\geq0$ , $r>0$ , and $a \geq x$ , we have

\begin{align}&\mathbb{E}_{x}\!\left(\int_{0}^{\kappa_{r}^{\xi}\wedge \tau_{a}^{+}}\textrm{e}^{-q \left(t-r\right)}\textrm{e}^{\lambda X(t)-\psi\left(\lambda\right)r}\textrm{d}t\right)\nonumber\\[3pt]&\quad=\int_{x}^{a}\textrm{e}^{\lambda \xi(w)}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(z))}{\ell_{r}^{(q)}(\overline{\xi}(z))}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\bigg(\frac{\textrm{e}^{\lambda \overline{\xi}(w)}\left(1-\textrm{e}^{-(\psi(\lambda)-q)r}\right)}{\psi(\lambda)-q}\right.\nonumber\\[3pt]&\qquad\left.-\,\textrm{e}^{\lambda \overline{\xi}(w)}\int_{0}^{\overline{\xi}(w)}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z-\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(\overline{\xi}(w))\textrm{d}s\bigg)\right.\nonumber\\[3pt]&\qquad\left.-\,\frac{\lambda \textrm{e}^{\lambda \overline{\xi}(w)}\left(1-\textrm{e}^{-(\psi(\lambda)-q)r}\right)}{\psi(\lambda)-q}+\lambda \textrm{e}^{\lambda \overline{\xi}(w)}\int_{0}^{\overline{\xi}(w)}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z\right.\nonumber\\[3pt]&\qquad\left.+\,W^{(q)}(\overline{\xi}(w))+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(\overline{\xi}(w))\textrm{d}s\right]\,\textrm{d}w.\nonumber\end{align}

Proof. Letting $f(x)\coloneqq \textrm{e}^{\lambda x-\psi\left(\lambda\right)r}$ in Theorem 4.3, or using the compensation formula together with Remark 3.3, one can get the desired result. □

Recall the definition of $V_k^{\xi}\left( x;\ b \right)$ at the end of Section 2. The following result generalizes (20) in [Reference Czarna and Palmowski10] and Propositions 1 and 2 in [Reference Renaud and Zhou25].

Theorem 4.4. For any $q\geq0$ and $k\geq1$ , we have

\begin{eqnarray}&&{V_k^{\xi}}\left( x;\ b \right)= \int_{b}^{\infty}kV_{k-1}(z)\exp \left( { - \int_x^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \!\left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\!\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right)\!\textrm{d}z,\quad x \in \left( {-\infty,b} \right],\nonumber\end{eqnarray}

where

\begin{align}V_k\!\left( x \right) = \int_{x}^{\infty}kV_{k-1}(z)\exp \left( { - \int_x^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right)\!\textrm{d}z,\qquad x\in(\!-\infty,\infty),\nonumber\end{align}

with $V_0\left( x \right)\equiv1$ .

Proof. For $\epsilon>0$ and any integer $n\geq1$ , we have

(34) \begin{equation}{\mathbb{E}_b}\left( {{{\left( {\int_0^{\tau _{b + \epsilon }^ + } {{\textrm{e}^{ - q s}}D\left( s \right)\!\textrm{d}s} } \right)}^n}{{\mathbf{1}}_{\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \}}}} \right) = o\!\left( { \epsilon } \right)\end{equation}

and

(35) \begin{equation}{\mathbb{E}_b}\left( {{{\left( {\int_0^{{\kappa^{\xi}_{r} }} {{\textrm{e}^{ - q s}}\textrm{d}D\left( s \right)} } \right)}^n}{{\mathbf{1}}_{\{ {{\kappa^{\xi}_{r} } < \tau _{b +\epsilon }^ + } \}}}} \right) = o\!\left(\epsilon \right)\!.\end{equation}

Actually, $X\left( {\tau _{b + \epsilon }^ + } \right) = b + \epsilon $ implies $ D(s)\leq \epsilon$ for all $s\in[0,{\tau _{b + \epsilon }^ + }]$ . Hence, the left-hand side of (34) is less than

\begin{align} &\ { \epsilon^n }{\mathbb{E}_b}\left[ {{{\left( {\int_0^{\tau _{b + \epsilon }^ + } {{\textrm{e}^{ - q s}}\textrm{d}s} } \right)}^n}{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \right\}}}} \right] \nonumber \\[3pt]\leq &\frac{\epsilon^{n}}{q^{n}}\left( {{\mathbb{E}_b}\left[ {{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon}^ + < {\kappa^{\xi}_{r} }} \right\}}}} \right] - {\mathbb{E}_b}\left[ {{\textrm{e}^{ - q \tau _{b + \epsilon}^ + }}{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon}^ + < {\kappa^{\xi}_{r} }} \right\}}}} \right]} \right) \nonumber \\[3pt]=& \frac{\epsilon^{n}}{q^{n} }\left( {\exp \left( { - \int_b^{b + \epsilon } {\frac{{\ell_{r}^\prime \left( {\bar \xi \left( w \right)} \right)}}{{\ell_{r}\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right) - \exp \left( { - \int_b^{b + \epsilon } {\frac{{{\ell_{r}^{\left( q \right)\prime}} \left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( q \right)}}\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right)} \right) = o\!\left( { \epsilon } \right)\!.\nonumber\end{align}

which gives (34). By integration by parts, the left-hand side of (35) can be rewritten as

\begin{align}&\ {\mathbb{E}_b}\left[ {{{\left( {{\textrm{e}^{ - q {\kappa^{\xi}_{r} }}}D( {{\kappa^{\xi}_{r} }} ) + q \int_0^{{\kappa^{\xi}_{r} }} {{\textrm{e}^{ - q s}}D\left( s \right)\!\textrm{d}s} } \right)}^{n}}{\mathbf{1}_{\{ {{\kappa^{\xi}_{r} } < \tau _{b + \epsilon }^ + } \}}}} \right]\ \nonumber \\[3pt] \le &\ {\mathbb{E}_b}\left[ {{{\left( {{\epsilon \textrm{e}^{ - q {\kappa^{\xi}_{r} }}} + \epsilon \int_0^{{\kappa^{\xi}_{r} }} { {q\textrm{e}^{ - q s}} \textrm{d}s} } \right)}^{n}}{\mathbf{1}_{\{ {{\kappa^{\xi}_{r} } < \tau _{b + \epsilon }^ + } \}}}} \right]\nonumber \\ =&\ {{\epsilon ^{n}}\left( {1 - \exp \left( { - \int_b^{b + \epsilon} {\frac{{\ell_{r}^\prime \left( {\bar \xi \left( w \right)} \right)}}{{\ell_{r}\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right)} \right)} = o\!\left( { \epsilon } \right)\!,\nonumber\end{align}

which gives (35). By (35), one has

\begin{align*}{V_k}\!\left( b \right) = {\mathbb{E}_b}\left( {\left[D_{b}\right]^{k}{\mathbf{1}_{\{ {{\tau _{b +\epsilon }^{+}<\kappa^{\xi}_{r} } } \}}}} \right) + o\!\left( { \epsilon } \right)\!.\end{align*}

Using the strong Markov property and the binomial theorem, one can rewrite the term ${\mathbb{E}_b}\left( {\left[D_{b}\right]^{k}{\mathbf{1}_{\{ \tau _{b +\epsilon }^{+}<\kappa^{\xi}_{r}\}}}} \right)$ as

(36) \begin{align}&{\mathbb{E}_b}\left[ {{{\sum\limits_{i = 0}^k {C_k^i\left( {\int_0^{\tau _{b + \epsilon }^ + } {{\textrm{e}^{ - q s}}\textrm{d}D\left( s \right)} } \right)} }^i}{{\left( {\int_{\tau _{b + \epsilon }^ + }^{{\kappa^{\xi}_{r} }} {{\textrm{e}^{ - q s}}\textrm{d}D\left( s \right)} } \right)}^{k - i}}{{\mathbf{1}}_{\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \}}}} \right] \nonumber \\[3pt] &\quad = {\mathbb{E}_b}\left[ {{{\sum\limits_{i = 0}^k {C_k^i\left( {\int_0^{\tau _{b + \epsilon }^ + } {{\textrm{e}^{ - q s}}\textrm{d}D\left( s \right)} } \right)} }^i}{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \right\}}}{{\left( {{\textrm{e}^{ - q \tau _{b + \epsilon }^ + }}} \right)}^{k - i}}{V_{k - i}}( b+\epsilon )} \right] \nonumber \\[3pt] &\quad = \sum\limits_{i = 0}^k {C_k^i} V_{k - i}( b+\epsilon )\mathbb{E}_b\left[ \left( \int_0^{\tau _{b + \epsilon }^ + } \textrm{e}^{ - q s}\textrm{d}D\left( s \right) \right)^i \textrm{e}^{ - (k - i)q \tau _{b + \epsilon }^ +} \mathbf{1}_{\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \}} \right] \nonumber \displaybreak\\[3pt] &\quad= \sum\limits_{i = 0}^k {C_k^i} V_{k - i}( b+\epsilon ){\mathbb{E}_b}\Bigg[ {\textrm{e}^{ - (k - i)q \tau _{b + \epsilon }^ +}}\sum\limits_{j = 0}^i C_i^j \epsilon^{j} {\textrm{e}^{ -j q \tau _{b + \epsilon }^ + }} \nonumber \\ &\qquad \left.\times {{\left( {q \int_0^{\tau _{b + \epsilon }^ + } {{\textrm{e}^{ - q s}}D\left( s \right)\!\textrm{d}s} } \right)}^{i - j}} {{\mathbf{1}}_{\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \}}} \right] \nonumber \\ &\quad = \sum\limits_{i = 0}^k {C_k^i} V_{k - i}( b+\epsilon )\sum\limits_{j = 0}^i {C_i^j} {\epsilon ^j}{\mathbb{E}_b}\Bigg[ {\textrm{e}^{ - q \left( {k - i + j} \right)\tau _{b + \epsilon}^ + }}{q ^{i - j}} \nonumber \\ &\qquad \left.\times {{\left( {\int_0^{\tau _{b + \epsilon}^ + } {{\textrm{e}^{ - q s}}D\left( s \right)\!\textrm{d}s} } \right)}^{i - j}}{{\mathbf{1}}_{\{ {\tau _{b + \epsilon}^ + < {\kappa^{\xi}_{r} }} \}}} \right],\end{align}

where the identity

\begin{align}&\mathbb{E}_b\left(\left.\left(\int_{\tau _{b + \epsilon }^ + }^{{\kappa^{\xi}_{r} }} {{\textrm{e}^{ - q s}}\textrm{d}D\left( s \right)}\right)^{i}\right|\mathcal{F}_{\tau _{b + \epsilon }^ +}\right)\nonumber\\[3pt] &\quad =\mathbb{E}_b\left(\left.\left(\int_{\tau _{b + \epsilon }^ + }^{{\kappa^{\xi}_{r} }} {{\textrm{e}^{ - q s}}\textrm{d}\left( (\bar{X}(s)-(b+\epsilon))\vee 0 \right)}\right)^{i}\right|\mathcal{F}_{\tau _{b + \epsilon }^ +}\right)\nonumber \\[3pt] &\quad=\,\textrm{e}^{ - iq \tau _{b + \epsilon }^ + }\, V_{i}( b+\epsilon ),\quad i\geq0,\nonumber\end{align}

is used for the first equality in (36). By (34) and (35), one can keep only those summands with $j=i=1$ or $j=i=0$ in (36), and then

\begin{align}V_k\!\left( b \right)=& {V_k}\left( b+\epsilon \right){\mathbb{E}_b}\left[ {{\textrm{e}^{ - kq \tau _{b + \epsilon }^ + }}{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \right\}}}} \right]\nonumber\\[3pt]&+ k{V_{k - 1}}\left( b+\epsilon \right){\mathbb{E}_b}\left[ { \epsilon {\textrm{e}^{ - kq \tau _{b + \epsilon }^ + }}{{\mathbf{1}}_{\left\{ {\tau _{b + \epsilon }^ + < {\kappa^{\xi}_{r} }} \right\}}}} \right]+ o\!\left( { \epsilon } \right) \nonumber \\[3pt] =& \left( {{V_k}\left( b+\epsilon \right) + k\epsilon{V_{k - 1}}\left( b+\epsilon \right)} \right) \exp \left( { - \int_b^{b + \epsilon } {\frac{{{\ell_{r}^{\left( {kq } \right)\prime}} \left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \!\left( w \right)} \right)}}\textrm{d}w} } \right) + o\!\left( { \epsilon } \right)\!,\nonumber\end{align}

which can be rearranged as

(37) \begin{align}0=&\left(\frac{V_k\left( b+\epsilon \right)-V_k\left( b \right)}{\epsilon}+ k V_{k - 1}\left( b+\epsilon \right)\right) \exp \left( { - \int_b^{b + \epsilon } {\frac{{{\ell_{r}^{\left( {kq } \right)\prime}} \left( {\bar \xi \!\left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \!\left( w \right)} \right)}}\textrm{d}w} } \right) \nonumber\\[3pt] &+V_k\left( b\right)\frac{-1+\exp \left( { - \int_b^{b + \epsilon } {\frac{{{\ell_{r}^{\left( {kq } \right)\prime}} \left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \left( w \right)} \right)}}\textrm{d}w} } \right) }{\epsilon} + o( 1 ).\end{align}

Letting $\epsilon\rightarrow0$ in (37) we get

(38) \begin{align}0=V_k^{\prime}\left( b \right)+ k V_{k - 1}\left( b\right)-V_k\left( b\right)\frac{{{\ell_{r}^{\left( {kq } \right)\prime}} \left( {\bar \xi \!\left( b \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \!\left( b \right)} \right)}}.\end{align}

By the standard method of variation of constants, one can obtain the solution of (38) with boundary conditions ${V_k}\left(\infty\right) = 0$ and $V_{0}(x)=1$ as

\begin{align}V_k\left( b \right) = \int_{b}^{\infty}kV_{k-1}(z)\exp \left( { - \int_b^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( {\bar \xi \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \left( w \right)} \right)}}} \textrm{d}w} \right)\!\textrm{d}z.\nonumber\end{align}

For $x \in \left(\!-\infty, b\right]$ and $k\geq1$ , we have

\begin{align}&V_k^{\xi}\left(x;\ b \right) =\mathbb{E}_{x}\left[\left(\int_{\tau _{b}^{+}}^{\kappa^{\xi}_{r}}\textrm{e}^{-q s}\,\textrm{d}\left(\bar{X}(s)-b\right)\right)^{k}\mathbf{1}_{\{\tau _{b}^{+}<\kappa^{\xi}_{r} \}}\right] \nonumber\\[3pt] =&\mathbb{E}_{x}\!\left( \textrm{e}^{-k q \tau _{b}^{+}}\mathbf{1}_{\{\tau _{b}^{+}<\kappa^{\xi}_{r} \}}\right)V_{k}(b) \nonumber\\[3pt] =&\exp \left( { - \int_x^b {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( {\bar \xi\! \left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\!\left( {\bar \xi \!\left( w \right)} \right)}}} \textrm{d}w} \right) \int_{b}^{\infty}kV_{k-1}(z)\exp \left( { - \int_b^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( {\bar \xi \!\left( w \right)} \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( {\bar \xi \!\left( w \right)} \right)}}} \textrm{d}w} \right)\!\textrm{d}z. \nonumber \end{align}

The proof is complete. □

Remark 4.4. For $k=1$ we present an alternative, more transparent argument. Given $a>b $ , by (26) we have

(39) \begin{align}V_{1}(b)&=\mathbb{E}_{b} \left(\int_{0}^{\infty}\mathbf{1}_{\{L(t)<L(\tau_{a}^{+}\wedge \kappa^{\xi}_{r})\}}\textrm{e}^{-q t}\,\textrm{d}L(t)\right)\nonumber\\[3pt]&=\int_{0}^{\infty}\mathbb{E}_{b}\left(\textrm{e}^{-q L^{-1}(y)}\mathbf{1}_{\{y<L(\tau_{a}^{+})=a-b, \,y<L(\kappa^{\xi}_{r})\}}\right)\,\textrm{d}y\nonumber\\[3pt]&=\int_{0}^{a-b}\mathbb{E}_{b}\left(\textrm{e}^{-q L^{-1}(y)}\mathbf{1}_{\{L^{-1}(y)<\kappa^{\xi}_{r}\}}\right)\,\textrm{d}y\nonumber\\[3pt]&=\int_{b}^{a}\mathbb{E}_{b}\left(\textrm{e}^{-q \tau_{z}^{+}}\mathbf{1}_{\{\tau_{z}^{+}<\kappa^{\xi}_{r}\}}\right)\,\textrm{d}z\nonumber\\[3pt]&=\int_{b}^{a}\exp\left(\!-\int_{b}^{z} \frac{\ell^{(q)\prime}_{r}(\overline{\xi}(w))}{\ell_{r}^{(q)}(\overline{\xi}(w))}\,\textrm{d}w\right)\,\textrm{d}z.\end{align}

5. Application of draw-down Parisian ruin results to a spectrally negative Lévy process reflected at its past supremum

Recall the dividend process D defined in (5). Let the corresponding risk process with dividends deducted according to the barrier strategy with barrier level b be defined as

\[Y(t)\coloneqq X(t)-D(t), \quad t\geq0.\]

For fixed $b\in(0,\infty)$ , if we choose the general draw-down function $\xi$ such that

\[\xi(z)\coloneqq \xi_{b}(z)=\left(z- b\right)\vee0,\quad z\in(\!-\infty,\infty),\]

then we have

\[\kappa_{r}^{\xi}\coloneqq \inf\{t>r: t-g_{t}^{\xi}>r\}, \,\,\mbox{ where }\,\, g_{t}^{\xi}\coloneqq \sup\{0\leq s \leq t\,{:}\, Y(s)\geq0\};\]

i.e., $\kappa_{r}^{\xi}=\kappa_{r}^{\xi_{b}}$ degenerates to the classical Parisian ruin time for the risk process Y. In addition, by the definition of $\tau_{\xi}$ we have

\[\tau_{\xi_{b}}\coloneqq \inf\{t\geq0\,{:}\, Y(t)\leq0\},\]

which is the ruin time for the risk process Y.

The following result gives the potential measure of Y upon the up-crossing time of level a or the Parisian ruin time of Y.

Corollary 5.1. For $b\in(0,\infty)$ , $\xi=\xi_{b}$ , $q,\lambda\geq0$ , $r>0$ , $a\geq x$ , and bounded differentiable function f, we have

\begin{align}&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f\left(Y(t)\right)\!; \,t<\kappa_{r}^{\xi_{b}}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\&\quad\,{=}\,\bigg[-\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(w)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s))\right)\!\textrm{d}s\right.-\int_{0}^{w}W^{(q)}(w-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\nonumber\\&\qquad\left.\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(w)\textrm{d}s\right)\right]\Bigg|_{w=x}^{w=a\wedge b}\nonumber\\&\qquad\,{+}\,\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)\prime}(b)}\bigg(1-\textrm{e}^{-\frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}(a-b)}\bigg)\left[\frac{\ell_{r}^{(q)\prime}(b)}{\ell_{r}^{(q)}(b)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(b+X(s))\right)\!\textrm{d}s\right.\right.\nonumber\\&\qquad\left.-\int_{0}^{b}W^{(q)}(b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(b)\textrm{d}s\right)\nonumber\\&\qquad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f^{\prime}(b+X(s))\right)\!\textrm{d}s+\int_{0}^{b}W^{(q)\prime}(b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\nonumber\\&\qquad\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(b+X(r))\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)\prime}(b)\textrm{d}s\right]\mathbf{1}_{(b,\infty)}(a).\nonumber\end{align}

Proof. Replacing $\xi$ and f(x,y) respectively with $\xi_{b}$ and $f(x-(y-b)\vee0)$ in Theorem 4.3 yields

\begin{align}&\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f\left(Y(t)\right)\!; \,t<\kappa_{r}^{\xi_{b}}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&\quad=\int_{x}^{a}\exp\left(\!-\int_{x}^{w} \frac{\ell^{(q)\prime}_{r}(z\wedge b)}{\ell_{r}^{(q)}(z\wedge b)}\,\textrm{d}z\right)\left[\frac{\ell_{r}^{(q)\prime}(w\wedge b)}{\ell_{r}^{(q)}(w\wedge b)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w\wedge b+X(s))\right)\!\textrm{d}s\right.\right.\nonumber\\[3pt]&\qquad\left.-\int_{0}^{w\wedge b}W^{(q)}(w\wedge b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\right.\nonumber\\[3pt]&\qquad\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(w\wedge b)\textrm{d}s\right)\nonumber\displaybreak\\[3pt]&\quad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f^{\prime}(w\wedge b+X(s))\right)\!\textrm{d}s+\int_{0}^{w\wedge b}W^{(q)\prime}(w\wedge b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\nonumber\\[3pt]&\quad\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(w\wedge b+X(r))\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)\prime}(w\wedge b)\textrm{d}s\right]\textrm{d}w\nonumber\\[3pt]&=\int_{x}^{a}\frac{\partial}{\partial w}\left[\frac{-\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(w)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s))\right)\!\textrm{d}s\right.-\int_{0}^{w}W^{(q)}(w-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\right.\nonumber\\[3pt]&\quad\left.\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(w)\textrm{d}s\right)\right]\textrm{d}w\,\mathbf{1}_{[x,b]}(a)\nonumber\\[3pt]&\quad+\int_{x}^{b}\frac{\partial}{\partial w}\left[\frac{-\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(w)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s))\right)\!\textrm{d}s\right.-\int_{0}^{w}W^{(q)}(w-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\right.\nonumber\\[3pt]&\quad\left.\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(w)\textrm{d}s\right)\right]\textrm{d}w\,\mathbf{1}_{(b,\infty)}(a)\nonumber\\[3pt]&\quad+\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(b)}\int_{b}^{a}\textrm{e}^{-\frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}(w-b)}\,\textrm{d}w\left[\frac{\ell_{r}^{(q)\prime}(b)}{\ell_{r}^{(q)}(b)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(b+X(s))\right)\!\textrm{d}s\right.\right.\nonumber\\[3pt]&\quad\left.-\int_{0}^{b}W^{(q)}(b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)}(b)\textrm{d}s\right)\nonumber\\[3pt]&\quad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f^{\prime}(b+X(s))\right)\!\textrm{d}s+\int_{0}^{b}W^{(q)\prime}(b-z)\mathbb{E}_{}\left(\,f(z+X(r))\right)\!\textrm{d}z\nonumber\\[3pt]&\quad\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(b+X(r))\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s))\right)\ell_{s}^{(q)\prime}(b)\textrm{d}s\right]\mathbf{1}_{(b,\infty)}(a),\nonumber\end{align}

which gives the desired result. □

The following result gives the joint Laplace transform involving the Parisian ruin time of Y.

Corollary 5.2. For $q,\lambda\in[0,\infty)$ , $a \in(\!-\infty,\infty)$ , and $x\in(\!-\infty,a)$ , we have

\begin{eqnarray}&&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \left(\kappa_{r}^{\xi_{b}}-r\right)}\textrm{e}^{\lambda Y(\kappa_{r}^{\xi_{b}})}\mathbf{1}_{\{\kappa_{r}^{\xi_{b}}<\tau_{a}^{+}\}}\right)=\textrm{e}^{\psi\left(\lambda\right)r}\left[\!-\frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)}(w)}\left(\textrm{e}^{\lambda w}-\left(\psi(\lambda)-q\right)\right.\right.\nonumber\\[3pt]&&\left.\left.\left.\times\left(\textrm{e}^{\lambda w}\int_{0}^{w}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(w)\textrm{d}s\right)\right)\right.\right]\Bigg|_{w=x}^{w=a\wedge b}\nonumber\\[3pt]&&+\,\textrm{e}^{\psi\left(\lambda\right)r} \frac{\ell_{r}^{(q)}(x)}{\ell_{r}^{(q)\prime}(b)}\bigg(1-\textrm{e}^{-\frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}(a-b)}\bigg)\,\mathbf{1}_{(b,\infty)}(a)\left[\frac{\ell_{r}^{(q)\prime}(b)}{\ell_{r}^{(q)}(b)}\left(\textrm{e}^{\lambda b}-\left(\psi(\lambda)-q\right)\right.\right.\nonumber\\[3pt]&&\left.\left.\times\left(\textrm{e}^{\lambda b}\int_{0}^{b}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)}(b)\textrm{d}s\right)\right)\right.\nonumber\\[3pt]&&-\lambda \textrm{e}^{\lambda b}+(\psi(\lambda)-q)\bigg(\lambda \textrm{e}^{\lambda b}\int_{0}^{b}W^{(q)}(z)\textrm{e}^{-\lambda z}\textrm{d}z+W^{(q)}(b)+\int_{0}^{r}\textrm{e}^{-\psi(\lambda) s}\,\ell_{s}^{(q)\prime}(b)\textrm{d}s\bigg)\bigg].\nonumber\end{eqnarray}

Proof. Letting $\xi(w)=\xi_{b}(w)$ and $\varphi(w)=\textrm{e}^{-\lambda \xi_{b}(w)+\psi(\lambda)r}$ in Theorem 4.2 and using a similar argument as the proof of Corollary 5.1, one arrives at the desired result. □

The following result, on the kth moment of the discounted total dividends paid according to the barrier strategy with barrier b until the Parisian ruin time for Y, is a direct consequence of Theorem 4.4. In particular, the corresponding result with $k=1$ recovers the identity (20) in [Reference Czarna and Palmowski10].

Corollary 5.3. For $b\in(0,\infty)$ , $\xi=\xi_{b}$ , $q\geq0$ , and $k\geq1$ , we have

\begin{eqnarray}&&{V_k^{\xi_{b}}}\left( x;\ b \right)= k!\,\frac{{\ell_{r}^{\left( {kq } \right)}}\left(x\right)}{{\ell_{r}^{\left( {kq } \right)}}\left(b\right)}\prod\limits_{i=1}^{k}\frac{{{\ell_{r}^{\left( {iq } \right)}}\left( b \right)}}{{{\ell_{r}^{\left( iq \right)\prime}} \left( b \right)}},\qquad x \in \left( {-\infty,b} \right],\nonumber\end{eqnarray}

and

\begin{eqnarray}V_k\left( x \right)&=&{\ell_{r}^{\left( {kq } \right)}}\left(x\right)\left(\int_{x}^{b}\frac{kV_{k-1}(z)}{{\ell_{r}^{\left( {kq } \right)}}(z)}\textrm{d}z+\frac{k!}{{\ell_{r}^{\left( kq \right)\prime}} \left(b\right)}\prod\limits_{i=1}^{k-1}\frac{{{\ell_{r}^{\left( {iq } \right)}}\left( b \right)}}{{{\ell_{r}^{\left( iq \right)\prime}} \left( b \right)}}\right)\!,\qquad x\in(\!-\infty,b],\nonumber\end{eqnarray}

with $V_0\left( x \right)\equiv1$ . In particular, for $k=1$ we have

\begin{eqnarray}V_{1}^{\xi_{b}}(x;\ b)&=&\frac{\ell_{r}^{(q)}(x)}{\ell^{(q)\prime}_{r}(b)},\qquad x\in(\!-\infty,b].\nonumber\end{eqnarray}

Proof. For $x\in(\!-\infty,b]$ , we have

\begin{align}V_k\left( x \right) &= \int_{x}^{b}kV_{k-1}(z)\exp \left( { - \int_x^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( w \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( w \right)}}} \textrm{d}w} \right)\!\textrm{d}z\nonumber\\[3pt]&\quad +\,\exp \left( { - \int_x^b {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( w \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( w \right)}}} \textrm{d}w} \right)\int_{b}^{\infty}kV_{k-1}(z)\exp \left( { - \int_b^z {\frac{{{\ell_{r}^{\left( kq \right)\prime}} \left( b \right)}}{{{\ell_{r}^{\left( {kq } \right)}}\left( b \right)}}} \textrm{d}w} \right)\!\textrm{d}z\nonumber\\[3pt]&={\ell_{r}^{\left( {kq } \right)}}\left(x\right)\left(\int_{x}^{b}\frac{kV_{k-1}(z)}{{\ell_{r}^{\left( {kq } \right)}}(z)}\textrm{d}z+\frac{k!}{{\ell_{r}^{\left( kq \right)\prime}} \left(b\right)}\prod\limits_{i=1}^{k-1}\frac{{{\ell_{r}^{\left( {iq } \right)}}\left( b \right)}}{{{\ell_{r}^{\left( iq \right)\prime}} \left( b \right)}}\right)\!.\nonumber\end{align}

The proof is thus complete. □

Let

\[D_{\tau_{\xi_{b}}}\coloneqq \int_{0}^{\tau_{\xi_{b}}}\textrm{e}^{-qt}\textrm{d}D(t),\]

represent the present value of the accumulated dividends paid until the time of ruin for Y. In addition, for each $k\geq1$ , we introduce the kth moment of $D_{\tau_{\xi_{b}}}$ as

\[U_{k}(x;\ b)\coloneqq \mathbb{E}_{x}\!\left(\left[D_{\tau_{\xi_{b}}}\right]^{k}\right)\!.\]

The following result recovers Propositions 1 and 2 in [Reference Renaud and Zhou25].

Corollary 5.4. For $b\in(0,\infty)$ , $q\geq0$ , and $k\geq1$ , we have

\begin{eqnarray}&&U_{k}(x;\ b)= k!\,\frac{{W^{\left( {kq } \right)}}\left(x\right)}{{W^{\left( {kq } \right)}}\left(b\right)}\prod\limits_{i=1}^{k}\frac{{{W^{\left( {iq } \right)}}\left( b \right)}}{{{W^{\left( iq \right)\prime}} \left( b \right)}},\qquad x \in \left( {-\infty,b} \right].\nonumber\end{eqnarray}

In particular, when $k=1$ we have

\begin{eqnarray}U_{1}(x;\ b)&=&\frac{W^{(q)}(x)}{W^{(q)\prime}(b)},\qquad x\in(\!-\infty,b].\nonumber\end{eqnarray}

Proof. Note that $\tau_{\xi_{b}}=\lim\limits_{r\rightarrow0}\kappa_{r}^{\xi_{b}}$ with probability 1. Note from (26) that

(40) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{b}^{+}}\mathbf{1}_{\{\tau_{b}^{+}<\kappa_{r}^{\xi_{b}}\}}\right)=\exp\left(\!-\int_{x}^{b} \frac{\ell^{(q)\prime}_{r}(w)}{\ell_{r}^{(q)}(w)}\,\textrm{d}w\right)= \frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(b)},\qquad x\in(\!-\infty,b],\end{eqnarray}

and

(41) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\kappa_{r}^{\xi_{b}}\}}\right)&\,{=}\,&\exp\left(\!-\int_{x}^{b} \frac{\ell^{(q)\prime}_{r}(w)}{\ell_{r}^{(q)}(w)}\,\textrm{d}w\right)\exp\left(\!-\int_{b}^{a} \frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}\,\textrm{d}w\right)\nonumber\\&\,{=}\,&\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(b)}\,\textrm{e}^{-\frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}(a-b)},\qquad x\in(\!-\infty,b], a\in(b,\infty).\end{eqnarray}

By (40) together with (1), we have

(42) \begin{eqnarray}&&\lim\limits_{r\rightarrow0}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(b)}=\lim\limits_{r\rightarrow0}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{b}^{+}}\mathbf{1}_{\{\tau_{b}^{+}<\kappa_{r}^{\xi_{b}}\}}\right)\nonumber\\&=&\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{b}^{+}}\mathbf{1}_{\{\tau_{b}^{+}<\tau_{\xi_{b}}\}}\right)=\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{b}^{+}}\mathbf{1}_{\{\tau_{b}^{+}<\tau_{0}^{-}\}}\right)= \frac{W^{(q)}(x)}{W^{(q)}(b)}, \quad x\in(\!-\infty,b].\end{eqnarray}

By (3) we have

(43) \begin{eqnarray}\mathbb{E}_{x}\!\left(\textrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+}<\tau_{\xi_{b}}\}}\right)&\,{=}\,&\exp\left(\!-\int_{x}^{a} \frac{W^{(q)\prime}(\bar{\xi}_{b}(w))}{W^{(q)}(\bar{\xi}_{b}(w))}\,\textrm{d}w\right)\nonumber\\&\,{=}\,&\frac{W^{(q)}(x)}{W^{(q)}(b)}\,\textrm{e}^{-\frac{W^{(q)\prime}(b)}{W^{(q)}(b)}(a-b)},\quad x\in(\!-\infty,b], a\in(b,\infty).\end{eqnarray}

Combining (41), (42), and (43), we arrive at

(44) \begin{eqnarray}\lim\limits_{r\rightarrow0}\frac{\ell^{(q)\prime}_{r}(b)}{\ell_{r}^{(q)}(b)}=\frac{W^{(q)\prime}(b)}{W^{(q)}(b)}.\end{eqnarray}

The desired results follow from a combination of (42), (44), and Corollary 5.3. □

If we choose $\xi$ such that $\xi\equiv0$ , then the draw-down Parisian ruin time $\kappa_{r}^{\xi}$ of X degenerates to the Parisian ruin time $\kappa_{r}$ of X. The following result gives a generalized version of the potential measure for the process X killed upon up-crossing level $a\,(\!\geq x)$ or the Parisian ruin time of X.

Corollary 5.5. For $\xi\equiv0$ , $q,\lambda\geq0$ , $r>0$ , $a\geq x$ , and bounded bivariate function f(x,y) which is differentiable with respect to x, we have

\begin{align}&\quad\int_{0}^{\infty}\textrm{e}^{-q \left(t-r\right)}\mathbb{E}_{x}\!\left(\,f(X(t),\bar{X}(t)); \,t<\kappa_{r}\wedge \tau_{a}^{+}\right)\textrm{d}t\nonumber\\[3pt]&=\int_{x}^{a}\frac{\ell^{(q)}_{r}(x)}{\ell_{r}^{(q)}(w)}\left[\frac{\ell_{r}^{(q)\prime}(w)}{\ell_{r}^{(q)}(w)}\left(\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\,f(w+X(s),w)\right)\!\textrm{d}s\right.\right.\nonumber\\[3pt]&\quad\left.-\int_{0}^{w}W^{(q)}(w-z)\mathbb{E}_{}\left(\,f(z+X(r),w)\right)\!\textrm{d}z\right.\nonumber\displaybreak\\[3pt]&\quad\left.-\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s),w)\right)\ell_{s}^{(q)}(w)\textrm{d}s\right)\nonumber\\[3pt]&\quad-\int_{0}^{r}\textrm{e}^{q(r-s)}\mathbb{E}_{}\left(\frac{\partial}{\partial x}f(w+X(s),w)\right)\!\textrm{d}s+\int_{0}^{w}W^{(q)\prime}(w-z)\mathbb{E}_{}\left(\,f(z+X(r),w)\right)\!\textrm{d}z\nonumber\\[3pt]&\quad\left.\left.+\,W^{(q)}(0+\!)\,\mathbb{E}_{}\left(\,f(w+X(r),w)\right)+\int_{0}^{r}\mathbb{E}_{}\left(\,f(X(r-s),w)\right)\ell_{s}^{(q)\prime}(w)\textrm{d}s\right)\right]\textrm{d}w.\nonumber\end{align}

Proof. This is a direct consequence of Theorem 4.3. □

The following result gives the solution for the kth moment of the accumulated discounted dividend payout until the Parisian ruin time $\kappa_{r}$ for X.

Corollary 5.6. For $\xi\equiv0$ , $q\geq0$ , and $k\geq1$ , we have

\begin{eqnarray}&&{V_k^{\xi}}\left( x;\ b \right)= {{\ell_{r}^{\left( kq \right)}} \left( {x} \right)}\int_{b}^{\infty}\frac{kV_{k-1}(z)}{{{\ell_{r}^{\left( {kq } \right)}}\left( {z} \right)}}\textrm{d}z,\qquad x \in \left( {-\infty,b} \right],\nonumber\end{eqnarray}

where

\begin{align}V_k\left( x \right) ={{\ell_{r}^{\left( kq \right)}} \left( {x} \right)} \int_{x}^{\infty}\frac{kV_{k-1}(z)}{{{\ell_{r}^{\left( {kq } \right)}}\left( {z} \right)}}\textrm{d}z,\qquad x\in(\!-\infty,\infty),\nonumber\end{align}

with $V_0\left( x \right)\equiv1$ .

Proof. The desired result is a direct application of Theorem 4.4, letting $\xi\equiv0$ . □

6. Examples

To illustrate how to apply the main results in Section 4 to obtain expressions or numerical values for quantities related to the draw-down Parisian ruin, we compute in this section the functions $\ell_{r}^{(q)}$ , $\chi^{(q)}$ , and $\phi_{_{\Phi_{q}}}$ for two examples of spectrally negative Lévy processes: the drifted Brownian motion and the Cramér–Lundberg risk model with exponential claims.

6.1. Small claims: Brownian motion

If $X(t)=\mu t+\sigma B(t)$ is a Brownian motion with drift $\mu\in(\!-\infty,\infty)$ and volatility $\sigma\in(0,\infty)$ , then

\[W^{(q)}(x)=\frac{1}{\sigma^{2}\delta_{q}}\big[\textrm{e}^{(\!-w+\delta_{q})x}-\textrm{e}^{-(w+\delta_{q})x}\big],\]

with $\delta_{q}\coloneqq \sigma^{-2}\sqrt{\mu^{2}+2q\sigma^{2}}$ and $w\coloneqq \mu/\sigma^{2}$ . Hence,

\[\mu+(\!-w+\delta_{q})\sigma^{2}=-\mu+(w+\delta_{q})\sigma^{2}=\sqrt{\mu^{2}+2q\sigma^{2}}\]

and

\begin{align}\ell_{r}^{(q)}(x)&=\int_{0}^{\infty} W^{(q)}(x+z)\frac{z}{r}\mathbb{P}\left(X(r)\in\textrm{d}z\right)\nonumber\\[3pt]&=\frac{1}{\sigma^{2}\delta_{q} r}\frac{1}{\sqrt{2\pi\sigma^{2}r}}\int_{0}^{\infty}\big[\textrm{e}^{(\!-w+\delta_{q})(x+z)}-\textrm{e}^{-(w+\delta_{q})(x+z)}\big]z\textrm{e}^{-\frac{(z-\mu r)^{2}}{2\sigma^{2}r}}\textrm{d}z\nonumber\\[3pt]&=\frac{1}{\sigma^{2}\delta_{q} r}\frac{1}{\sqrt{2\pi\sigma^{2}r}}\bigg(\textrm{e}^{(\!-w+\delta_{q})x}\int_{0}^{\infty}z\textrm{e}^{-\frac{(z-\mu r-(\!-w+\delta_{q})\sigma^{2}r)^{2}+\mu^{2}r^{2}-(\mu +(\!-w+\delta_{q})\sigma^{2})^{2}r^{2}}{2\sigma^{2}r}}\textrm{d}z\nonumber\displaybreak\\[3pt]&\quad-\textrm{e}^{-(w+\delta_{q})x}\int_{0}^{\infty}z\textrm{e}^{-\frac{(z-\mu r+(w+\delta_{q})\sigma^{2}r)^{2}+\mu^{2}r^{2}-(\mu -(w+\delta_{q})\sigma^{2})^{2}r^{2}}{2\sigma^{2}r}}\textrm{d}z\bigg)\nonumber\\[3pt]&=\frac{1}{\sigma^{2}\delta_{q} r}\frac{1}{\sqrt{2\pi\sigma^{2}r}}\bigg(\textrm{e}^{(\!-w+\delta_{q})x}\int_{0}^{\infty}z\textrm{e}^{-\frac{(z-\sqrt{\mu^{2}+2q\sigma^{2}}r)^{2}-2q\sigma^{2}r^{2}}{2\sigma^{2}r}}\textrm{d}z\nonumber\\[3pt]&\quad-\textrm{e}^{-(w+\delta_{q})x}\int_{0}^{\infty}z\textrm{e}^{-\frac{(z+\sqrt{\mu^{2}+2q\sigma^{2}}r)^{2}-2q\sigma^{2}r^{2}}{2\sigma^{2}r}}\textrm{d}z\bigg)\nonumber\\[3pt]&=\frac{\textrm{e}^{qr}}{\sigma^{2}\delta_{q} r}\bigg[\textrm{e}^{(\!-w+\delta_{q})x}\bigg(\frac{\sigma\sqrt{r}}{\sqrt{2\pi}}\textrm{e}^{-\frac{\mu^{2}+2q\sigma^{2}}{2\sigma^{2}}r}+\sqrt{\mu^{2}+2q\sigma^{2}}r\mathcal{N}\bigg(\frac{\sqrt{\mu^{2}+2q\sigma^{2}}\sqrt{r}}{\sigma}\bigg)\bigg)\nonumber\\[3pt]&\quad-\textrm{e}^{-(w+\delta_{q})x}\bigg(\frac{\sigma\sqrt{r}}{\sqrt{2\pi}}\textrm{e}^{-\frac{\mu^{2}+2q\sigma^{2}}{2\sigma^{2}}r}-\sqrt{\mu^{2}+2q\sigma^{2}}r\mathcal{N}\bigg(\frac{-\sqrt{\mu^{2}+2q\sigma^{2}}\sqrt{r}}{\sigma}\bigg)\bigg)\bigg],\nonumber\end{align}

where $\mathcal{N}$ is the cumulative distribution function of a standard normal random variable.

Notice that to compute the draw-down Parisian ruin probability (see Remark 4.1), we only need the expression for $\ell_{r}\coloneqq \ell_{r}^{(0)}$ . To find the Laplace transform associated to the two-sided exit solution (26), the kth moment of dividends (see Theorem 4.4), or the potential measure of X (see Theorem 4.3) involving the draw-down Parisian ruin time, we also need the expression for $\ell_{r}^{(q)}$ . If we want to compute the probability density of the draw-down Parisian ruin time, we must first use the expression for $\ell_{r}^{(q)}$ to compute the right-hand side of (30), then numerically invert it using the algorithms in [Reference Cohen7] or the method of Fourier series expansion proposed by [Reference Moorthy23], which has been proven to be an efficient method for the numerical inversion of Laplace transforms. Furthermore, if we want to compute the Laplace transform associated to the generalized two-sided exit solution (see Theorem 4.1), we need to numerically compute $\chi^{(q)}$ and $\phi_{_{\Phi_{q}}}$ via the algorithms of Laplace inversion in [Reference Cohen7] or [Reference Moorthy23]. To this end, we need to compute the right-hand sides of (23) and (24). In fact, one has $\Phi_{q}=\delta_{q}-w$ and

\begin{eqnarray}&&\int_{y}^{x+y}W^{(q)}(x+y-z)W^{(\theta+q)}(z)\textrm{d}z\nonumber\\&\,{=}\,&\frac{1}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\int_{y}^{x+y}\big[\textrm{e}^{(\!-w+\delta_{q})(x+y-z)}-\textrm{e}^{-(w+\delta_{q})(x+y-z)}\big]\big[\textrm{e}^{(\!-w+\delta_{\theta+q})z}-\textrm{e}^{-(w+\delta_{\theta+q})z}\big]\textrm{d}z\nonumber\\&\,{=}\,&\frac{\textrm{e}^{-w(x+y)}}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\bigg(\textrm{e}^{\delta_{q}(x+y)}\bigg(\frac{\textrm{e}^{(\delta_{\theta+q}-\delta_{q})(x+y)}-\textrm{e}^{(\delta_{\theta+q}-\delta_{q})y}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-(\delta_{\theta+q}+\delta_{q})(x+y)}-\textrm{e}^{-(\delta_{\theta+q}+\delta_{q})y}}{\delta_{\theta+q}+\delta_{q}}\bigg)\nonumber\\&&+\,\textrm{e}^{-\delta_{q}(x+y)}\bigg(\!-\frac{\textrm{e}^{(\delta_{\theta+q}+\delta_{q})(x+y)}-\textrm{e}^{(\delta_{\theta+q}+\delta_{q})y}}{\delta_{\theta+q}+\delta_{q}}-\frac{\textrm{e}^{-(\delta_{\theta+q}-\delta_{q})(x+y)}-\textrm{e}^{-(\delta_{\theta+q}-\delta_{q})y}}{\delta_{\theta+q}-\delta_{q}}\bigg)\bigg)\nonumber\\&\,{=}\,&\frac{\textrm{e}^{-w(x+y)}}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\bigg(\frac{\textrm{e}^{\delta_{\theta+q}(x+y)}-\textrm{e}^{\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-\delta_{\theta+q}(x+y)}-\textrm{e}^{-\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}\nonumber\\&&-\frac{\textrm{e}^{\delta_{\theta+q}(x+y)}-\textrm{e}^{\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}-\frac{\textrm{e}^{-\delta_{\theta+q}(x+y)}-\textrm{e}^{-\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}\bigg)\nonumber\\&\,{=}\,&\frac{\textrm{e}^{-w(x+y)}}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\bigg(\frac{\sqrt{\mu^{2}+2q\sigma^{2}}}{\theta}\bigg(\textrm{e}^{\delta_{\theta+q}(x+y)}-\textrm{e}^{-\delta_{\theta+q}(x+y)}\bigg)-\frac{\textrm{e}^{\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}-\frac{\textrm{e}^{-\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}\nonumber\\&&+\,\frac{\textrm{e}^{\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}+\frac{\textrm{e}^{-\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}\bigg),\nonumber\end{eqnarray}

which implies

\begin{eqnarray}&&\quad W^{(\theta+q,-\theta)}_{y}(x+y)\nonumber\\[3pt]&&=W^{(\theta+q)}(x+y)-\theta\int_{y}^{x+y}W^{(q)}(x+y-w)W^{(\theta+q)}(w)\textrm{d}w\nonumber\\[3pt]&&=\frac{1}{\sigma^{2}\delta_{\theta+q}}\bigg(\textrm{e}^{(\!-w+\delta_{\theta+q})(x+y)}-\textrm{e}^{-(w+\delta_{\theta+q})(x+y)}\bigg)-\frac{\theta\,\textrm{e}^{-w(x+y)}}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\bigg(\frac{\textrm{e}^{\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}+\frac{\textrm{e}^{-\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}\nonumber\\[3pt]&&\,\,\,\,\,\,\,\,+\,\frac{\sqrt{\mu^{2}+2q\sigma^{2}}}{\theta}\bigg(\textrm{e}^{\delta_{\theta+q}(x+y)}-\textrm{e}^{-\delta_{\theta+q}(x+y)}\bigg)-\frac{\textrm{e}^{\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}-\frac{\textrm{e}^{-\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}\bigg).\nonumber\end{eqnarray}

One also finds that

\begin{eqnarray}1+\theta\int_{0}^{y}\textrm{e}^{-\Phi_{q}z}W^{(\theta+q)}(z)\textrm{d}z&\,{=}\,&1+\frac{\theta}{\sigma^{2}\delta_{\theta+q}}\int_{0}^{y}\textrm{e}^{(w-\delta_{q})z}\big[\textrm{e}^{(\!-w+\delta_{\theta+q})z}-\textrm{e}^{-(w+\delta_{\theta+q})z}\big]\textrm{d}z\nonumber\\[3pt]&\,{=}\,&1+\frac{\theta}{\sigma^{2}\delta_{\theta+q}}\bigg(\frac{\textrm{e}^{(\delta_{\theta+q}-\delta_{q})y}-1}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-(\delta_{\theta+q}+\delta_{q})y}-1}{\delta_{\theta+q}+\delta_{q}}\bigg)\nonumber\\[3pt]&\,{=}\,&\frac{\theta}{\sigma^{2}\delta_{\theta+q}}\bigg(\frac{\textrm{e}^{(\delta_{\theta+q}-\delta_{q})y}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-(\delta_{\theta+q}+\delta_{q})y}}{\delta_{\theta+q}+\delta_{q}}\bigg),\nonumber\end{eqnarray}

which implies

\begin{eqnarray}&&W^{(q)}(x)\textrm{e}^{\Phi_{q}y}\left(1+\theta\int_{0}^{y}\textrm{e}^{-\Phi_{q}w}W^{(\theta+q)}(w)\textrm{d}w\right)\nonumber\\[3pt]&\,{=}\,&\frac{\theta}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\big[\textrm{e}^{(\!-w+\delta_{q})x}-\textrm{e}^{-(w+\delta_{q})x}\big]\bigg(\frac{\textrm{e}^{(\delta_{\theta+q}-w)y}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-(\delta_{\theta+q}+w)y}}{\delta_{\theta+q}+\delta_{q}}\bigg)\nonumber\\[3pt]&\,{=}\,&\frac{\theta}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\textrm{e}^{-w(x+y)}\big[\textrm{e}^{\delta_{q}x}-\textrm{e}^{-\delta_{q}x}\big]\bigg(\frac{\textrm{e}^{\delta_{\theta+q}y}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-\delta_{\theta+q}y}}{\delta_{\theta+q}+\delta_{q}}\bigg)\nonumber\\[3pt]&\,{=}\,&\frac{\theta\,\textrm{e}^{-w(x+y)}}{\sigma^{4}\delta_{q}\delta_{\theta+q}}\bigg(\frac{\textrm{e}^{\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}-\frac{\textrm{e}^{\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}-\delta_{q}}+\frac{\textrm{e}^{-\delta_{\theta+q}y+\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}-\frac{\textrm{e}^{-\delta_{\theta+q}y-\delta_{q}x}}{\delta_{\theta+q}+\delta_{q}}\bigg).\nonumber\end{eqnarray}

Therefore, the right-hand side of (23) becomes

\begin{eqnarray}&&\frac{W^{(\theta+q)}(x+y)-\dfrac{\textrm{e}^{-w(x+y)}}{\sigma^{2}\delta_{\theta+q}}\big(\textrm{e}^{\delta_{\theta+q}(x+y)}-\textrm{e}^{-\delta_{\theta+q}(x+y)}-\textrm{e}^{\delta_{\theta+q}y-\delta_{q}x}+\textrm{e}^{-\delta_{\theta+q}y-\delta_{q}x}\big)}{\theta W^{(\theta+q)}(y)}\nonumber\\[3pt]&\,{=}\,&\frac{\dfrac{\textrm{e}^{-(w+\delta_{q})x}}{\sigma^{2}\delta_{\theta+q}}\big(\textrm{e}^{\delta_{\theta+q}y-w y}-\textrm{e}^{-\delta_{\theta+q}y-w y}\big)}{\theta \dfrac{1}{\sigma^{2}\delta_{\theta+q}}\big(\textrm{e}^{(\!-w+\delta_{\theta+q})y}-\textrm{e}^{-(w+\delta_{\theta+q})y}\big)}=\frac{1}{\theta}\textrm{e}^{-(w+\delta_{q})x},\nonumber\end{eqnarray}

which together with (23) yields

\[\chi^{(q)}(x,y,r)=\textrm{e}^{-(w+\delta_{q})x},\qquad x\in(\!-\infty,\infty), \quad y\in(0,\infty), \quad r\in(0,\infty).\]

In addition, the right-hand side of (24) can be re-expressed as

\[\dfrac{\dfrac{1}{\delta_{\theta+q}-\delta_{q}}\textrm{e}^{(\delta_{\theta+q}-w)y}+\dfrac{1}{\delta_{\theta+q}+\delta_{q}}\textrm{e}^{-(\delta_{\theta+q}+w)y}}{\textrm{e}^{(\!-w+\delta_{\theta+q})y}-\textrm{e}^{-(w+\delta_{\theta+q})y}},\]

which can be numerically inverted (to get $\phi_{_{\Phi_{q}}}$ ) using the algorithms in [Reference Cohen7] or the method of Fourier series expansion in [Reference Moorthy23].

6.2. Big claims: the Cramér–Lundberg risk model

Let $X(t)=ct-\sum_{n=1}^{N(t)}C_{n}$ be a linear drift with $c>0$ minus a compound Poisson process with jump intensity $\eta$ and independent, identically exponentially distributed jump sizes $(C_{n})_{n\geq 1}$ with mean $1/\alpha$ . Then

\[W^{(q)}(x)=c^{-1}\big(A^{+}\textrm{e}^{q^{+}(q)x}-A^{-}\textrm{e}^{q^{-}(q)x}\big),\]

with

\[A_{\pm}\coloneqq \frac{\alpha+q^{\pm}(q)}{q^{+}(q)-q^{-}(q)}\quad\text{ and}\quad q^{\pm}(q)\coloneqq \frac{q+\eta-\alpha c\pm\sqrt{(q+\eta-\alpha c)^{2}+4cq\alpha}}{2c}.\]

One can verify that

\begin{eqnarray}\ell_{r}^{(q)}\left( x \right)&\,{=}\,&\textrm{e}^{-\eta r}\left( A_{+}\textrm{e}^{q^{+}(q)(x+cr)}-A_{-}\textrm{e}^{q^{-}(q)(x+cr)} \right)+\frac{A_{+}\textrm{e}^{q^{+}(q)(x+cr)-\eta r}}{cr}\sum_{m=0}^{\infty}\frac{\left( \frac{\alpha\eta r}{q^{+}(q)+\alpha} \right)^{m+1}}{m!(m+1)!}\nonumber\\&&\times\left[ cr\Gamma\left( m+1,cr(q^{+}(q)+\alpha) \right)-\frac{\Gamma(m+2,cr(q^{+}(q)+\alpha))}{q^{+}(q)+\alpha}\right]-\frac{A_{-}\textrm{e}^{q^{-}(q)(x+cr)-\eta r}}{cr}\nonumber\\&&\times\sum_{m=0}^{\infty}\frac{\left( \frac{\alpha\eta r}{q^{-}(q)+\alpha} \right)^{m+1}}{m!(m+1)!}\left[ cr\Gamma\left( m+1,cr(q^{-}(q)+\alpha) \right)-\frac{\Gamma(m+2,cr(q^{-}(q)+\alpha))}{q^{-}(q)+\alpha}\right],\nonumber\end{eqnarray}

where $\Gamma(\beta,x)=\int_{0}^{x}\textrm{e}^{-w}w^{\beta-1}\textrm{d}w$ is the gamma function. The Laplace transforms of $\chi^{(q)}$ and $\phi_{_{\Phi_{q}}}$ in r, i.e. the right-hand sides of (23) and (24), can also be derived accordingly. The corresponding computations are omitted because they are fairly lengthy.