2.1. Large deviations in reinsurance

In [Reference Rhee18], the large deviations results (2) were derived in the Skorokhod ${J_1}$ topology. Correspondingly, we let ${\mathds{D}}= {\mathds{D}} ([0,1],{\mathds{R}})$ be a Skorokhod space, i.e. a space of real-valued càdlàg (right continuous with left limits) functions on [0, 1], equipped with the ${J_1}$ metric defined by

\begin{equation*} {d({\xi},{\zeta})} = \inf_{{h} \in {\Lambda}} \{ { \| {h} - {id} \| } \vee { \| {\xi} - {\zeta} \circ {h} \| } \}, \qquad ({\xi},{\zeta}) \in {\mathds{D}^{2}},\end{equation*}

where ${\Lambda}$ denotes the set of all strictly increasing continuous bijections from [0, 1] to itself, ${id}$ denotes the identity mapping, and ${ \| \cdot \| }$ denotes the uniform (sup) norm on [0, 1]. Thus, ${A}$ and ${ C_{j} }$ in (2) are a measurable set and a measure on ${\mathds{D}}$ , respectively. Furthermore, if ${\phi}\, :\, {\mathds{D}} \to {\mathds{R}}$ is a continuous functional on ${\mathds{D}}$ and ${ B } \in { { \mathcal{B} }({\mathds{R}}) }$ is a Borel set such that ${A} = {\phi}^{-1}({ B })$ , where ${\phi}^{-1}$ stands for the inverse of ${\phi}$ , we have

\begin{equation*} {\mathbb{P}} ( {\phi} ({\bar{Y}_{n}}) \in { B }) ={\mathbb{P}} ( {\bar{Y}_{n}} \in {\phi}^{-1}({ B }) ) ={\mathbb{P}} ( {\bar{Y}_{n}} \in {A} ).\end{equation*}

The above relation portrays how it is possible to use the result (2) to study continuous functionals of ${\bar{Y}_{n}}$ . To connect this to our ruin problem, we define $ {\bar{S}_{n}}\coloneqq \{ {\bar{S}_{n}(t)},\, t \in [0,1] \}$ as the centered and scaled process

\begin{equation*} {\bar{S}_{n}(t)} = \frac1n {S(nt)} - {\lambda} \mathbb{E} ( {{X}}) t= \frac1n \sum_{i=1}^{{N(nt)}} {{X}_{i}} - {\lambda} \mathbb{E} ( {{X}}) t, \qquad t \geqslant 0.\end{equation*}

Moreover, we assume that the capital ${u}$ increases linearly in n, i.e. there exists an ${a}>0$ such that ${u} = n {a}$ . We now formulate the large deviations problem to estimate the probabilities

\begin{align*} {\mathbb{P}} \bigg( \sup_{t\in[0,1]}& \{ {S(nt)} - {p_{\text D}} nt - {R(nt)} \} \geqslant n {a} \bigg) \notag \\ &= {\mathbb{P}} \bigg( \sup_{t\in[0,1]} \{ {S(nt)} - {\lambda} \mathbb{E} ( {{X}}) nt- ({p_{\text D}} - {\lambda} \mathbb{E}( {{X}})) nt - {R(nt)} \} \geqslant n {a} \bigg) \notag \\ &= {\mathbb{P}} \bigg( \sup_{t\in[0,1]} \{ n {\bar{S}_{n}(t)} - {c} nt - {R(nt)} \} \geqslant n{a} \bigg) \notag \\ &= {\mathbb{P}} \bigg( \sup_{t\in[0,1]} \{ {\bar{S}_{n}(t)} - {c} t - \frac1n {R(nt)} \} \geqslant {a} \bigg),\end{align*}

where ${c} = {p_{\text D}} - {\lambda} \mathbb{E}( {{X}})$ . As a next step, we must identify a continuous functional ${\phi}$ such that

(5) \begin{equation} \sup_{t\in[0,1]} \Big\{ {\bar{S}_{n}(t)} - {c} t - \frac1n {R(nt)} \Big\} = {\phi} ({\bar{S}_{n}}),\end{equation}

so that we can write

(6) \begin{equation}{\mathbb{P}} \bigg( \sup_{t\in[0,1]} \Big\{ {\bar{S}_{n}(t)} - {c} t - \frac1n {R(nt)} \Big\} \geqslant {a} \bigg)= {\mathbb{P}} ( {\phi} ({\bar{S}_{n}}) \geqslant {a} )= {\mathbb{P}} ( {\bar{S}_{n}} \in {\phi}^{-1}( [{a}, \infty)) ).\end{equation}

However, it is not immediately obvious from (5) what the functional ${\phi}$ looks like because ${R(nt)}$ is not expressed in terms of ${\bar{S}_{n}}$ . We focus first on the LCR treaty and observe that

\begin{equation*} \frac1n {R(nt)} =\frac1n {L_{{r}}(nt)} =\frac1n \sum_{i=1}^{r} {{X}^\star_{i,{N(nt)}}} =\max_{\substack{ (s_1,\dots,s_{r})\in[0,t]^{r}\\s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^{r} ( {\bar{S}_{n}(s_i)}-{\bar{S}_{n}(s_i^-)} ), \quad t \in [0,1],\end{equation*}

i.e. ${L_{{r}}(nt)}/n$ can be expressed as the sum of the ${r}$ biggest jumps of the process ${\bar{S}_{n}(t)}$ . For every ${\xi} \in {\mathds{D}}$ and $m \in {\mathds{N}}$ , we define, for $t\in(0,1]$ ,

(7) \begin{equation} {{\mathfrak J}^{m}_{{\xi}}(t)} = \sup_{\substack{ (s_1,\dots,s_m)\in[0,t]^m\\s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^m ( {{\xi}(s_i)}-{{\xi}(s_i^-)} ) = \max_{\substack{ (s_1,\dots,s_m)\in[0,t]^m\\s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^m ( {{\xi}(s_i)}-{{\xi}(s_i^-)} ) \end{equation}

as the supremum of the sum of the m largest jumps of the function ${\xi}$ . Naturally, ${{\mathfrak J}^{m}_{{\xi}}(0)} =0$ . Consequently, the functional ${\phi}$ we are looking for is a mapping ${\phi_{r}}\,:\,{\mathds{D}} \to {\mathds{R}}$ defined for every ${\xi} \in {\mathds{D}}$ as

(8) \begin{equation} {{\phi_{r}}(\xi)} = \sup_{t \in [0,1]} \big\{ {{\xi}(t)} - {c} t - {{\mathfrak J}^{{r}}_{{\xi}}(t)}\big\}.\end{equation}

Moreover, we denote the pre-image of $[{a},\infty)$ under ${\phi_{r}}$ as ${{A}^{r}_{{c},{a}}} = {\phi_{r}}^{-1}\big( [{a},\infty) \big)$ , where

(9) \begin{equation} {{A}^{r}_{{c},{a}}} = \bigg\{ {\xi} \in {\mathds{D}}\,: \sup_{t \in [0,1]} \Big\{ {{\xi}(t)} - {c} t - {{\mathfrak J}^{{r}}_{{\xi}}(t)}\Big\} \geqslant {a} \bigg\}.\end{equation}

By comparing (3) and (4), we observe that the relation between the reinsured amounts of the two treaties is

\begin{equation*} {E_{{r}}(t)} = {L_{{r}}(t)} - {r} {{X}^\star_{{r}+1,{N(t)}}} = ({r} + 1){L_{{r}}(t)} - {r} ( {L_{{r}}(t)} + {{X}^\star_{{r}+1,{N(t)}}} ) = ({r} + 1){L_{{r}}(t)} - {r} {L_{{r}+1}(t)}.\end{equation*}

Thus, in the ECOMOR treaty, the functional ${\phi}$ in (6) is the mapping ${\varphi_{r}}\,:\,{\mathds{D}} \to {\mathds{R}}$ defined for every ${\xi} \in {\mathds{D}}$ as

(10) \begin{equation} {{\varphi_{r}}(\xi)} = \sup_{t \in [0,1]} \Big\{ {{\xi}(t)} - {c} t - ({r} + 1){{\mathfrak J}^{{r}}_{{\xi}}(t)} + {r} {{\mathfrak J}^{{r}+1}_{{\xi}}(t)} \Big\},\end{equation}

while the pre-image of $[{a},\infty)$ under ${\varphi_{r}}$ , i.e. ${{\mathcal{A}}^{r}_{{c},{a}}} = {\varphi^{-1}_{r}}( [{a},\infty) )$ , is defined as

(11) \begin{equation} {{\mathcal{A}}^{r}_{{c},{a}}} = \bigg\{ {\xi} \in {\mathds{D}}: \sup_{t \in [0,1]} \Big\{ {{\xi}(t)} - {c} t - ({r} + 1){{\mathfrak J}^{{r}}_{{\xi}}(t)} + {r} {{\mathfrak J}^{{r}+1}_{{\xi}}(t)} \Big\} \geqslant {a} \bigg\}.\end{equation}

2.2. Preliminaries on the Skorokhod topology and notation

Consider the complete metric space $({\mathds{D}},{d(,)})$ . The functional ${{\mathfrak J}^{m}_{{\xi}}(t)}$ defined in (7) will play a significant role in the forthcoming analysis. Thus, it is important to confirm that it is well defined. For this reason, let ${\mathcal{D}({\xi})}$ be the set of discontinuities of ${\xi} \in {\mathds{D}}$ , i.e.

\begin{equation*} {\mathcal{D}({\xi})} = \{ t \in [0,1]\,:\,{{\xi}(t^-)} \neq {{\xi}(t)}\},\end{equation*}

and let ${\mathcal{D}({\xi},\varepsilon)}$ be the set of discontinuities of magnitude at least $\varepsilon$ , i.e.

(12) \begin{equation} {\mathcal{D}({\xi},\varepsilon)} = \{ t \in [0,1]\,:\,{\left| {{{{\xi}(t^-)} - {{\xi}(t)}}} \right|} \geqslant \varepsilon\}.\end{equation}

The following result is standard.

Consequently, the supremum in (7) is attained because only finitely many jumps can exceed a given positive number. As a result, ${{\mathfrak J}^{m}_{{\xi}}(t)}$ is well defined.

Some important subspaces of ${\mathds{D}}$ for our analysis are those restricted to step functions. We let $\smash{{\mathds{D}^{\uparrow}_{\mathcal{S}}}}$ be the set of all non-decreasing step functions vanishing at the origin. Furthermore, ${ {\mathds{D}}_{j}}$ is the subspace of ${\mathds{D}}$ consisting of non-decreasing step functions, vanishing at the origin, with exactly j steps, and similarly, ${ {\mathds{D}}_{ \leqslant j}} = \bigcup_{0 \leqslant i \leqslant j} { {\mathds{D}}_{i}}$ consists of non-decreasing step functions, vanishing at the origin, with at most j steps. Finally, if ${\mathcal{D}_+({\xi})}$ denotes the number of discontinuities of ${\xi} \in {\mathds{D}}$ , we can then formally define the integer-valued set function ${\mathcal{J}(A) }$ appearing in (2) by

(13) \begin{equation} {\mathcal{J}(A) } = \inf_{{\xi} \in {A} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}}} {\mathcal{D}_+({\xi})},\end{equation}

which we call the rate function. Observe that every ${\xi} \in { {\mathds{D}}_{j}}$ is determined by the pair of jump sizes and jump times $({\textbf{\textit{x}}},{\textbf{\textit{u}}}) \in {\mathds{R}}^{j}_+ \times [0,1]^{\,j}$ , i.e. ${{\xi}(t)} = \sum_{i = 1}^{j} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t)$ , where ${\textbf{1}_{{ B }}}$ is the indicator function on the set ${ B }$ . For $ {\textbf{\textit{x}}}= ({{x} _{1}} ,\dots,{{x} _{j}} )$ and $ {\textbf{\textit{u}}} = ({{u} _{1}} ,\dots,{{u} _{j}} )$ , we define the sets

\begin{align*} { {\mathds{R}}^{j\downarrow}_+ } &= \big\{ {\textbf{\textit{x}}} \in {\mathds{R}}^{j}_+ \,:\, {{x} _{1}} \geqslant {{x} _{2}} \geqslant \dots \geqslant {{x} _{j}} >0\big\} \intertext{and} { S_{j}} &= \big\{ ({\textbf{\textit{x}}}, {\textbf{\textit{u}}}) \in { {\mathds{R}}^{j\downarrow}_+ } \times (0,1)^{\,j}\,:\, {{u} _{1}} ,\dots,{{u} _{j}} \text{ are all distinct} \big\},\end{align*}

where the ${{u} _{j}} $ do not follow the ordering of the ${{x} _{j}} $ , i.e. ${{x} _{k}} \geqslant {{x} _{l}} \not \Rightarrow {{u} _{k}} \geqslant {{u} _{l}} $ . Thus, we can formally define the mapping ${T_{j}}\,:\, { S_{j}} \to { {\mathds{D}}_{j}}$ by ${{T_{j}}({\textbf{\textit{x}}}, {\textbf{\textit{u}}})} = \sum_{i = 1}^{j} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ .

Furthermore, let ${{\nu}_{\alpha}}(x,\infty) = x^{-{\alpha}}$ (i.e. the pure power decay part of the regularly varying claim sizes), and let ${{\nu}^{j}_{\alpha}}$ denote the restriction to ${ {\mathds{R}}^{j\downarrow}_+ }$ of the j-fold product measure of ${{\nu}_{\alpha}}$ . We define, for each $j \geqslant 1$ , the measure ${ C_{j} }$ concentrated on ${ {\mathds{D}}_{j}}$ as

(14) \begin{equation} { C_{j} }(\bullet) = \mathbb{E} \Bigg[ {{\nu}^{j}_{\alpha}} \Bigg\{{\textbf{\textit{y}}} \in {\mathds{R}}^{j}_+ : \sum_{i=1}^j y_i {\textbf{1}_{\{{U_{i,}1}\}}} \in \bullet \Bigg\} \Bigg],\end{equation}

where the random variables ${U_{i}}$ , $i=1,\dots,j$ , are i.i.d. uniform on [0, 1].

Finally, we say that a set ${A} \subseteq {\mathds{D}}$ is bounded away from another set ${ B } \subseteq {\mathds{D}}$ if $\inf_{x \in {A}, y \in { B }} {d(x,y)}>0$ . Additionally, we let $_{\delta}{{A}} = \{ {\xi} \in {\mathds{D}}: {d({\xi},{A})} \leqslant {\delta} \}$ for any ${\delta}>0$ . We conclude this section with a large deviations result from [Reference Rhee18]. Let ${Y(t)}$ , $t \geqslant 0$ , be a centered Lévy process with a Lévy measure ${\nu}$ that is concentrated on $(0,\infty)$ and satisfies ${\nu}(x,\infty) = x^{-\alpha} {L(x)}$ for some $\alpha>1$ (in our case, $\nu$ is the claim arrival rate times the measure of the claim size). Moreover, let ${\bar{Y}_{n}} = \{ {\bar{Y}_{n}(t)},\, t \in [0,1] \}$ , where ${\bar{Y}_{n}(t)} = {Y(n t)}/n$ , $t \geqslant 0$ .

Theorem 1. (Theorem 3.2 of [Reference Rhee18].) Suppose that A is a measurable set. If ${\mathcal{J}(A) }$ as in (13) equals $j^*$ and ${d( A,{{ {\mathds{D}}_{ \leqslant j^*-1}}} )} > 0$ , then

\begin{equation*}{ C_{j^*} }({{A}^\circ})\leqslant \varliminf_{n\to\infty} \frac{{\mathbb{P}} ( {\bar{Y}_{n}} \in A ) }{(n{\nu}[n,\infty))^{\,j^*}}\leqslant \varlimsup_{n\to\infty} \frac{{\mathbb{P}} ( {\bar{Y}_{n}} \in A ) }{(n{\nu}[n,\infty))^{\,j^*}}\leqslant { C_{j^*} }({\bar{A}}),\end{equation*}

where ${ C_{j} }$ is defined in (14).

3.1. Proof of Theorem 2

The first step is to show that both mappings ${\phi_{r}}, {\varphi_{r}}\,:\, {\mathds{D}} \to {\mathds{R}}$ from (8) and (10), respectively, are Lipschitz continuous. Due to their continuity, (6) will hold and, consequently, we will be able to write ${\mathbb{P}} ( {\phi_{r}} ({\bar{S}_{n}}) \geqslant {a} )= {\mathbb{P}} ( {\bar{S}_{n}} \in {{A}^{r}_{{c},{a}}} )$ and ${\mathbb{P}} ( {\varphi_{r}} ({\bar{S}_{n}}) \geqslant {a} )= \allowbreak {\mathbb{P}} ( {\bar{S}_{n}} \in {{\mathcal{A}}^{r}_{{c},{a}}} )$ . For this, we need the following intermediate result.

Lemma 2. For every $({\xi}, {\zeta}) \in {\mathds{D}^{2}}$ , $m \in {\mathds{N}}$ , and ${h} \in {\Lambda}$ , we have

\begin{equation*} {\left| {{{{\mathfrak J}^{m}_{{\zeta} \circ {h}}(t)} - {{\mathfrak J}^{m}_{{\xi}}(t)} }} \right|} \leqslant 2 m { \| {\xi} - {\zeta} \circ {h} \| }, \qquad {for\ all}\ t \in [0,1].\end{equation*}

Proof. By the definition of ${{\mathfrak J}^{m}_{{\zeta} \circ {h}}(t)}$ , there exists $(\sigma_1,\dots,\sigma_m) \in [0,t]^m$ , with $\sigma_i \neq \sigma_j$ for all $i \neq j$ , such that

(16) \begin{equation} {{\mathfrak J}^{m}_{{\zeta} \circ {h}}(t)} = \sum_{i=1}^m ( {\zeta} \circ {h}(\sigma_i)-{\zeta} \circ {h}(\sigma_i^-) ).\end{equation}

In addition, we have that

(17) \begin{equation} {{\mathfrak J}^{m}_{{\xi}}(t)} = \max_{\substack{ (s_1,\dots,s_m)\in[0,t]^m\\s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^m ( {{\xi}(s_i)}-{{\xi}(s_i^-)}) \geqslant \sum_{i=1}^m ( {{\xi}(\sigma_i)}-{{\xi}(\sigma_i^-)} ).\end{equation}

Subtracting (16) and (17), we obtain

\begin{align*} {{\mathfrak J}^{m}_{{\zeta} \circ {h}}(t)} - {{\mathfrak J}^{m}_{{\xi}}(t)} &\leqslant \sum_{i=1}^m ( {\zeta} \circ {h}(\sigma_i)-{\zeta} \circ {h}(\sigma_i^-) - {{\xi}(\sigma_i)}+{{\xi}(\sigma_i^-)} )\\ &\leqslant \sum_{i=1}^m \big( {\left| {{{\zeta} \circ {h}(\sigma_i) - {{\xi}(\sigma_i)}}} \right|} + {\left| {{{\zeta} \circ {h}(\sigma_i^-) - {{\xi}(\sigma_i^-)}}} \right|} \big)\\ &\leqslant 2 m { \| {\xi} - {\zeta} \circ {h} \| }.\end{align*}

Following similar arguments, we can also show that ${{\mathfrak J}^{m}_{{\xi}}(t)} - {{\mathfrak J}^{m}_{{\zeta} \circ {h}}(t)} \leqslant 2 m { \| {\xi} - {\zeta} \circ {h} \| }$ , which completes the proof. □

We are now ready to establish the desired continuity.

Proof. Without loss of generality, we assume that ${{\phi_{r}}(\xi)} \geqslant {{\phi_{r}}({\zeta})}$ , otherwise we switch the roles of ${\xi}$ and ${\zeta}$ . For every $\varepsilon>0$ , there exists $t_* \in [0,1]$ such that

(18) \begin{equation} {{\xi}(t_*)} - {c} t_* - {{\mathfrak J}^{{r}}_{{\xi}}(t_*)} > {{\phi_{r}}(\xi)} - \varepsilon.\end{equation}

On the other hand, by the definition of ${J_1}$ , there exists ${h} = {h}({\xi},{\zeta},\varepsilon) \in {\Lambda}$ so that

(19) \begin{equation} {d({\xi},{\zeta})} + \varepsilon = { \| {h} - {id} \| } \vee { \| {\xi} - {\zeta} \circ {h} \| } \geqslant ({h}(t_*) - t_* ) \vee ( {{\xi}(t_*)} - {\zeta} \circ {h}(t_*) ).\end{equation}

Furthermore, using the fact that ${h}$ is a homeomorphism on [0, 1], we obtain

(20) \begin{align} {\zeta} \circ {h}(t_*) - & {c} {h}(t_*) - {{\mathfrak J}^{{r}}_{{\zeta} \circ {h}}(t_*)} \notag \\[3pt] &= {\zeta} \circ {h}(t_*) - {c} {h}(t_*) - \max_{\substack{ (s_1,\dots,s_{r})\in[0,t_*]^{r}\\ s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^{r} ( {\zeta} \circ {h}(s_i)- {\zeta} \circ {h}(s_i^-) ) \notag \\ &= {\zeta} ( {h}(t_*)) - {c} {h}(t_*) - \max_{\substack{ (s_1,\dots,s_{r})\in[0,{h}(t_*)]^{r}\\s_i\neq s_j,\forall i\neq j}} \sum_{i=1}^{r} ( {\zeta}(s_i)- {\zeta}(s_i^-) ) \notag \\ &= {\zeta} \big( {h}(t_*) \big) - {c} {h}(t_*) - {{\mathfrak J}^{{r}}_{{\zeta}}({h}(t_*))} \leqslant {{\phi_{r}}({\zeta})}.\end{align}

Subtracting (20) from (18) yields

\begin{align*} {{\phi_{r}}(\xi)} - {{\phi_{r}}({\zeta})} &< \varepsilon + ( {{\xi}(t_*)} - {\zeta} \circ {h}(t_*) ) + {c} ({h}(t_*) - t_* ) + ( {{\mathfrak J}^{{r}}_{{\zeta} \circ {h}}(t_*)} - {{\mathfrak J}^{{r}}_{{\xi}}(t_*)} ) \notag \\ & < \varepsilon + ( {d({\xi},{\zeta})} + \varepsilon ) + {\left| {{c}} \right|} ( {d({\xi},{\zeta})} + \varepsilon ) + 2 {r} ( {d({\xi},{\zeta})} + \varepsilon ) \notag \\ &= (2 + {\left| {{c}} \right|} + 2 {r}) \varepsilon + (1 + {\left| {{c}} \right|} + 2 {r}) {d({\xi},{\zeta})},\end{align*}

where we have also used (19) and ${{\mathfrak J}^{{r}}_{{\zeta} \circ {h}}(t_*)}- {{\mathfrak J}^{{r}}_{{\xi}}(t_*)} \leqslant 2 {r} { \| {\xi} - {\zeta} \circ {h} \| }$ by applying Lemma 2 with $t=t_*$ and $m={r}$ . Letting $\varepsilon \to 0$ , we conclude that ${{\phi_{r}}(\xi)} - {{\phi_{r}}({\zeta})} \leqslant (1 + {\left| {{c}} \right|} + 2 {r}) {d({\xi},{\zeta})}$ , i.e. ${\phi_{r}}$ is Lipschitz continuous. The Lipschitz continuity for the ${\varphi_{r}}$ mapping can be shown in an analogous manner. More precisely, for every $\varepsilon>0$ , there exists $t_* \in [0,1]$ such that

(21) \begin{equation} {{\xi}(t_*)} - {c} t_* - ({r}+1) {{\mathfrak J}^{{r}}_{{\xi}}(t_*)} + {r} {{\mathfrak J}^{{r}+1}_{{\xi}}(t_*)} > {{\varphi_{r}}(\xi)} - \varepsilon.\end{equation}

For a homeomorphism ${h}$ on [0, 1] satisfying (19), we have

(22) \begin{align} & {\zeta} \circ {h}(t_*) - {c} {h}(t_*) - ({r}+1) {{\mathfrak J}^{{r}}_{{\zeta} \circ {h}}(t_*)} + {r} {{\mathfrak J}^{{r}+1}_{{\zeta} \circ {h}}(t_*)}\nonumber\\ & \quad = {\zeta} ( {h}(t_*) ) - {c} {h}(t_*) - ({r}+1) {{\mathfrak J}^{{r}}_{{\zeta}}({h}(t_*))} + {r} {{\mathfrak J}^{{r}+1}_{{\zeta}}({h}(t_*))} \leqslant {{\varphi_{r}}({\zeta})}.\end{align}

We assume now, without loss of generality, that ${{\varphi_{r}}(\xi)} \geqslant {{\varphi_{r}}({\zeta})}$ and we subtract (22) from (21) to attain

\begin{align*} {{\varphi_{r}}(\xi)} - {{\varphi_{r}}({\zeta})} <& \ \varepsilon + ( {{\xi}(t_*)} - {\zeta} \circ {h}(t_*) ) + {c} ({h}(t_*) - t_* ) + ({r} +1) ( {{\mathfrak J}^{{r}}_{{\zeta} \circ {h}}(t_*)} - {{\mathfrak J}^{{r}}_{{\xi}}(t_*)} ) \\ &+ {r} ( {{\mathfrak J}^{{r}+1}_{{\xi}}(t_*)} - {{\mathfrak J}^{{r}+1}_{{\zeta} \circ {h}}(t_*)}) \notag \\ <& ( 2 + {\left| {{c}} \right|} + 4 {r} ({r} +1) ) \varepsilon + ( 1 + {\left| {{c}} \right|} + 4 {r} ({r} +1) ) {d({\xi},{\zeta})},\end{align*}

where we have also used (19) and twice Lemma 2 with $m={r},{r}+1$ and $t=t_*$ . Letting $\varepsilon \to 0$ , the result is immediate. □

As a next step, we calculate the rate functions ${\mathcal{J}({{A}^{r}_{{c},{a}}}) }$ and ${\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) }$ that appear in (2) and are formally defined in (13). For simplicity, we set ${c}_+=\max \{0,{c}\}$ and ${c}_-=\max \{0,-{c}\}$ .

Lemma 4. (Evaluation of the rate function.) The rate function defined by (13) is equal to ${r} +1$ in both treaties, i.e.

\begin{equation*} {\mathcal{J}({{A}^{r}_{{c},{a}}}) } = {\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) } = {r} +1.\end{equation*}

Proof. We first need to show that ${\mathcal{J}({{A}^{r}_{{c},{a}}}) }$ cannot take any value smaller than or equal to ${r}$ . Let us assume, on the contrary, that $\smash{{\xi} \in {{A}^{r}_{{c},{a}}} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}}}$ such that ${\mathcal{D}_+({\xi})} =k \leqslant {r}$ . This means that ${\xi} = \sum_{i \leqslant k} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ , with ${{x} _{1}} \geqslant {{x} _{2}} \geqslant \cdots \geqslant {{x} _{k}} >0$ and $\{ 0,{{u} _{1}} ,{{u} _{2}} ,\dots,{{u} _{k}} ,1 \}$ all distinct. By taking into account Assumptions 1 and 2, we calculate

\begin{align*} {{\phi_{r}}(\xi)} &= \sup_{t \in [0,1]} \big\{ {{\xi}(t)} - {c} t - {{\mathfrak J}^{{r}}_{{\xi}}(t)}\big\} = \sup_{t \in [0,1]} \Bigg\{ \cancel{{\sum_{i=1}^k {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t)}} - {c} t - \cancel{{\sum_{i=1}^k {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t)}} \Bigg\}\\ & = {c}_-,\end{align*}

which states that ${\xi} \not\in {{A}^{r}_{{c},{a}}}$ because ${{\phi_{r}}(\xi)} = {c}_- \not \geqslant {a}$ . As a result, ${\mathcal{J}({{A}^{r}_{{c},{a}}}) } \not = k$ , $k \leqslant r$ .

Let us assume now that ${\xi} \in {{A}^{r}_{{c},{a}}} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}}$ such that ${\mathcal{D}_+({\xi})} = {r}+1$ , i.e. ${\xi} = \sum_{i = 1}^{{r}+1} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ , with ${{x} _{1}} \geqslant {{x} _{2}} \geqslant \cdots \geqslant {{x} _{{r}+1}} >0$ and $\{ 0,{{u} _{1}} ,{{u} _{2}} ,\dots,{{u} _{{r}+1}} ,1 \}$ all distinct. To calculate ${{\phi_{r}}(\xi)}$ , observe first that

(23) \begin{equation} {{\xi}(t)} - {{\mathfrak J}^{{r}}_{{\xi}}(t)} = \sum_{i=1}^{{r}+1} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t) - {{\mathfrak J}^{{r}}_{{\xi}}(t)} = \begin{cases} 0, & t < \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} , \\[4pt] {{x} _{{r}+1}} , & t \geqslant \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} , \end{cases} \end{equation}

because all the claims are ‘absorbed’ according to Assumption 2 before the arrival of the ( ${r}+1$ )st claim, which happens at time $t^* = \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \}$ . Thus, we can write

\begin{align*} {{\phi_{r}}(\xi)} &= \sup_{t \in [0,1]} \big\{ {{\xi}(t)} - {c} t - {{\mathfrak J}^{{r}}_{{\xi}}(t)} \big\} = \sup_{t \in [0,1]} \Bigg\{ {{x} _{{r}+1}} \prod_{i=1}^{{r}+1} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t) - {c} t \Bigg\}\\ &= {{x} _{{r}+1}} - {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} +{c}_- ,\end{align*}

since ${{x} _{{r}+1}} \prod_{i=1}^{{r}+1} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t)$ remains fixed at the value ${{x} _{{r}+1}} $ from $t^*= \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \}$ onward, while $-{c} t$ decreases or increases depending on the value of ${c}$ . Due to the fact that ${\xi} \in {{A}^{r}_{{c},{a}}}$ , we get

\begin{equation*}{{\phi_{r}}(\xi)} \geqslant {a} \Rightarrow {{x} _{{r}+1}} \geqslant a + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- \geqslant a - {c}_- > 0,\end{equation*}

i.e. ${{A}^{r}_{{c},{a}}} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}} \not = \emptyset$ but contains all step functions with ${r}+1$ steps such that the ( ${r}+1$ )st largest step satisfies ${{x} _{{r}+1}} \geqslant a + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_-$ . Thus, ${\mathcal{J}({{A}^{r}_{{c},{a}}}) } = {r} +1$ .

The proof for ${\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) } = {r} +1$ in the ECOMOR treaty is similar. More precisely, it can easily be shown that there is no $\smash{{\xi} \in {{\mathcal{A}}^{r}_{{c},{a}}} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}}}$ with ${\mathcal{D}_+({\xi})} =k \leqslant {r}$ . Consequently, ${\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) } \not = k$ , $k \leqslant r$ . Let us assume next that ${\xi} \in {{\mathcal{A}}^{r}_{{c},{a}}} \cap {\mathds{D}^{\uparrow}_{\mathcal{S}}}$ such that ${\mathcal{D}_+({\xi})} = {r}+1$ , i.e. ${\xi} = \sum_{i = 1}^{{r}+1} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ , with ${{x} _{1}} \geqslant {{x} _{2}} \geqslant \cdots \geqslant {{x} _{{r}+1}} >0$ and $\{ 0,{{u} _{1}} ,{{u} _{2}} ,\dots,{{u} _{{r}+1}} ,1 \}$ all distinct. We have

(24) \begin{equation} {r} {{\mathfrak J}^{{r}+1}_{{\xi}}(t)} - ({r} + 1){{\mathfrak J}^{{r}}_{{\xi}}(t)} = -{{\mathfrak J}^{{r}}_{{\xi}}(t)} + \begin{cases} 0, & t < \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} , \\[4pt] {r} {{x} _{{r}+1}} , & t \geqslant \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} , \end{cases} \end{equation}

due to Assumption 1. By combining (23) and (24), we calculate

\begin{align*} {{\varphi_{r}}(\xi)} &= \sup_{t \in [0,1]} \big\{ {{\xi}(t)} - {c} t - ({r} + 1){{\mathfrak J}^{{r}}_{{\xi}}(t)} + {r} {{\mathfrak J}^{{r}+1}_{{\xi}}(t)} \big\} \\ &= \sup_{t \in [0,1]} \Bigg\{ ({r}+1){{x} _{{r}+1}} \prod_{i=1}^{{r}+1} {\textbf{1}_{\{{{u} _{i}} ,1\}}}(t) - {c} t \Bigg\}\\[4pt] &=({r}+1){{x} _{{r}+1}} - {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} +{c}_- .\end{align*}

Since ${\xi} \in {{\mathcal{A}}^{r}_{{c},{a}}}$ , we get ${{\varphi_{r}}(\xi)} \geqslant {a} \Rightarrow ({r} +1){{x} _{{r}+1}} \geqslant {a} + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_-$ , i.e. ${{\mathcal{A}}^{r}_{{c},{a}}} \cap \smash{{\mathds{D}^{\uparrow}_{\mathcal{S}}} \not = \emptyset}$ but contains all step functions with ${r}+1$ steps such that the ( ${r}+1$ )st largest step satisfies ${{x} _{{r}+1}} \geqslant ( {a} + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- )/({r}+1) $ . Thus, ${\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) } = {r} +1$ , and the proof is complete. □

An essential condition of Theorem 3.2 in [Reference Rhee18] is that the sets $_{\delta}{{{A}^{r}_{{c},{a}}}} \cap { {\mathds{D}}_{ \leqslant {{\mathcal{J}({{A}^{r}_{{c},{a}}}) }}}}$ and $_{\delta}{{{\mathcal{A}}^{r}_{{c},{a}}}} \cap { {\mathds{D}}_{ \leqslant {{\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) }}}}$ are bounded away from ${ {\mathds{D}}_{ \leqslant {{\mathcal{J}({{A}^{r}_{{c},{a}}}) } - 1}}}$ and ${ {\mathds{D}}_{ \leqslant {{\mathcal{J}({{\mathcal{A}}^{r}_{{c},{a}}}) } - 1}}}$ , respectively. Verifying this condition allows us then to derive the result (2) for both treaties. We can directly use the value of the rate function in the following result due to Lemma 4.

Proof. To simplify the notation in the proof, we write ${A}$ instead of ${{A}^{r}_{{c},{a}}}$ and ${\mathcal{A}}$ instead of ${{\mathcal{A}}^{r}_{{c},{a}}}$ ; the notation $_{\delta}{{A}}$ , $_{\delta}{{\mathcal{A}}}$ follows naturally.

We start by showing that $_{\delta}{{A}} \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ is bounded away from ${ {\mathds{D}}_{ \leqslant {r}}}$ for some ${\delta}>0$ . Thanks to Lemma 2, we have that $_{\delta}{{A}} \subset {A}({\delta})$ , where ${A}({\delta}) = {\phi_{r}}^{-1}( [{a} - ({\left| {{c}} \right|} + 2 {r} + 1){\delta} ,\infty) )$ . Hence, it suffices to show that ${A}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ is bounded away from ${ {\mathds{D}}_{ \leqslant {r}}}$ . Let ${\xi} \in {A}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ . Since ${\xi} \in { {\mathds{D}}_{ \leqslant {r}+1}}$ , we can write ${\xi} = \sum_{i = 1}^{{r}+1} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ with ${{x} _{1}} \geqslant {{x} _{2}} \geqslant \dots \geqslant {{x} _{{r}+1}} \geqslant 0$ , for which we have ${{\phi_{r}}(\xi)} \leqslant {{x} _{{r}+1}} - {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \}+{c}_- \leqslant {{x} _{{r}+1}} + {c}_-$ according to the proof of Lemma 4. Furthermore, ${\xi} \in {A}({\delta}) \Leftrightarrow {{\phi_{r}}(\xi)} \geqslant {a} - ({\left| {{c}} \right|} + 2 {r} + 1){\delta}$ . Combining the two inequalities, we obtain that ${{x} _{{r}+1}} \geqslant ({a} - {c}_-) - ({\left| {{c}} \right|} + 2 {r} + 1){\delta} \geqslant ({a}-{c}_-)/2$ for ${\delta} \leqslant ({a}-{c}_-)/2 ({\left| {{c}} \right|} + 2 {r} + 1)$ . In other words, for ${\delta} \leqslant ({a}-{c}_-)/2 ({\left| {{c}} \right|} + 2 {r} + 1)$ , ${\xi} \in {A}({\delta}) \cap { {\mathds{D}}_{{r}+1}} \subset {A}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ with jump sizes bounded from below by $({a}-{c}_-)/2$ , which implies that ${A}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ is bounded away from ${ {\mathds{D}}_{ \leqslant {r}}}$ .

In a similar manner, it suffices to show that ${\mathcal{A}}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ is bounded away from ${ {\mathds{D}}_{ \leqslant {r}}}$ , where ${\mathcal{A}}({\delta}) = {\varphi_{r}}^{-1}( [{a} - ({\left| {{c}} \right|} + 4 {r}^2 + 4 {r} + 1){\delta} ,\infty) )$ . Let ${\xi} \in {\mathcal{A}}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ . Since ${\xi} \in { {\mathds{D}}_{ \leqslant {r}+1}}$ , we can write ${\xi} = \sum_{i = 1}^{{r}+1} {{x} _{i}} {\textbf{1}_{\{{{u} _{i}} ,1\}}}$ with ${{x} _{1}} \geqslant {{x} _{2}} \geqslant \dots \geqslant {{x} _{{r}+1}} \geqslant 0$ , for which it holds that ${{\varphi_{r}}(\xi)} \leqslant ({r}+1){{x} _{{r}+1}} - {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} + {c}_- \leqslant ({r}+1){{x} _{{r}+1}} + {c}_-$ . Furthermore, ${\xi} \in {\mathcal{A}}({\delta}) \Leftrightarrow {{\varphi_{r}}(\xi)} \geqslant {a} - ({\left| {{c}} \right|} + 4 {r}^2 + 4 {r} + 1){\delta}$ . Combining the two inequalities, we obtain that $({r}+1){{x} _{{r}+1}} \geqslant ({a}-{c}_-) - ({\left| {{c}} \right|} + 4 {r}^2 + 4 {r} + 1){\delta} \geqslant ({a}-{c}_-)/2$ for ${\delta} \leqslant ({a}-{c}_-)/2 ({\left| {{c}} \right|} + 4 {r}^2 + 4 {r} + 1)$ . In other words, the jump sizes of ${\xi}$ are bounded from below by $({a}-{c}_-)/2({r}+1)$ , which implies that ${\mathcal{A}}({\delta}) \cap { {\mathds{D}}_{ \leqslant {r}+1}}$ is bounded away from ${ {\mathds{D}}_{ \leqslant {r}}}$ for ${\delta} \leqslant ({a}-{c}_-)/2 ({\left| {{c}} \right|} + 4 {r}^2 + 4 {r} + 1)$ , and the proof is complete. □

Let ${ \mathcal{C}_{{r}+1}^{\text L} }\coloneqq { C_{{r}+1} ({{A}^{r}_{{c},{a}}}) }$ and ${ \mathcal{C}_{{r}+1}^{\text E} }\coloneqq { C_{{r}+1} ({{\mathcal{A}}^{r}_{{c},{a}}}) }$ . According to Section 3.1 in [Reference Rhee18], the $\liminf$ and $\limsup$ in (2) yield the same result when

\begin{equation*} { C_{{\mathcal{J}(A) }} ({{A}^\circ}) } = { C_{{\mathcal{J}(A) }} (A) }= { C_{{\mathcal{J}(A) }} ({\bar{A}}) }.\end{equation*}

However, the above equality holds when the set ${A}$ is ${ C_{{\mathcal{J}(A) }} }$ -continuous, i.e. ${ C_{{\mathcal{J}(A) }} }({\partial{A}})=0$ , where the boundary ${\partial{A}} = {\bar{A}} \setminus {{A}^\circ}$ of a set ${A}$ is the closure of ${A}$ without its interior. We prove in the next lemma that the sets ${{A}^{r}_{{c},{a}}}$ and ${{\mathcal{A}}^{r}_{{c},{a}}}$ are both ${ C_{{r}+1} }$ -continuous.

Proof. To simplify the notation in the proof, we again write ${A}$ instead of ${{A}^{r}_{{c},{a}}}$ and ${\mathcal{A}}$ instead of ${{\mathcal{A}}^{r}_{{c},{a}}}$ ; the notation ${{{A}}^\circ}$ , ${{{\mathcal{A}}}^\circ}$ , ${\bar{A}}$ , ${\bar{A}}$ follows naturally.

We start by showing the ${ C_{{r}+1} }$ -continuity of ${A}$ . In compliance with the notation introduced in Subsection 2.2, we consider the function ${T^{-1}_{{r}+1}}\,:\, { {\mathds{D}}_{r+1}} \to { S_{r+1}}$ such that

\begin{align*} {{T^{-1}_{{r}+1}}({{{A}}^\circ})} &= {{T^{-1}_{{r}+1}}({\phi_{r}}^{-1}( ({a},\infty) ))} \\[4pt] &= \big\{ ({\textbf{\textit{x}}}, {\textbf{\textit{u}}}) \in { S_{r+1}}\, :\, {{x} _{{r}+1}} > {a} + {c}_+ \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- \big\}, \\[4pt] {{T^{-1}_{{r}+1}}({\bar{A}})} &= {{T^{-1}_{{r}+1}}({\phi_{r}}^{-1}( [{a},\infty) ))}\\[4pt] &= \big\{ ({\textbf{\textit{x}}}, {\textbf{\textit{u}}}) \in { S_{r+1}}\, :\, {{x} _{{r}+1}} \geqslant {a} + {c}_+ \max \{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- \big\}.\end{align*}

Obviously, the set ${{T^{-1}_{{r}+1}}({\bar{A}})} \setminus {{T^{-1}_{{r}+1}}\left({{{A}}^\circ}\right)}$ has zero Lebesgue measure. Combining this with ${{{A}}^\circ} \subseteq {A} \subseteq {\bar{A}}$ and ${\phi_{r}}$ being a continuous function, we conclude that ${ C_{{r}+1} ({{\partial{A}}}) } = 0$ , i.e. ${A}$ is ${ C_{{r}+1} }$ -continuous. To prove the ${ C_{{r}+1} }$ -continuity of ${\mathcal{A}}$ , it suffices to observe that the set ${{T^{-1}_{{r}+1}}({\bar{\mathcal{A}}})} \setminus {{T^{-1}_{{r}+1}}\left({{{\mathcal{A}}}^\circ}\right)}$ has zero Lebesgue measure, where

\begin{align*} {{T^{-1}_{{r}+1}}({{{\mathcal{A}}}^\circ})} &= {{T^{-1}_{{r}+1}}({\varphi_{r}}^{-1}( ({a},\infty) ))}\\ &= \big\{ ({\textbf{\textit{x}}}, {\textbf{\textit{u}}}) \in { S_{r+1}}\, :\, {{x} _{{r}+1}} > ({a} + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- )/({r}+1) \big\}, \\ {{T^{-1}_{{r}+1}}({\bar{\mathcal{A}}})} &= {{T^{-1}_{{r}+1}}({\varphi_{r}}^{-1}( [{a},\infty)))}\\ &= \big\{ ({\textbf{\textit{x}}}, {\textbf{\textit{u}}}) \in { S_{r+1}}\, :\, {{x} _{{r}+1}} \geqslant ({a} + {c}_+ \max\{ {{u} _{1}} ,\dots,{{u} _{{r}+1}} \} - {c}_- )/({r}+1) \big\},\end{align*}

which follows by the same reasoning. □

We now calculate the pre-constants ${ C_{{\mathcal{J}({{A}^{r}_{{c},{a}}}) }}}({{A}^{r}_{{c},{a}}})$ and ${ C_{{\mathcal{J} }({{\mathcal{A}}^{r}_{{c},{a}}})} }({{\mathcal{A}}^{r}_{{c},{a}}})$ .

Lemma 7. (Calculation of the pre-constant.) The constants ${ \mathcal{C}_{{r}+1}^{\text L} }$ and ${ \mathcal{C}_{{r}+1}^{\text E} }$ are given by

\begin{alignat*}{2} { \mathcal{C}_{{r}+1}^{\text L} } =& \quad \frac{1}{({r}+1)!} &&\times \begin{cases} {a}^{-({r} + 1) {\alpha}} \cdot \prescript{}{2}{F}_1[{r}+1, ({r} + 1) {\alpha};\ {r} + 2; -{c}/{a}], & {c}>0, \\[3pt] ({a}+{c})^{-({r}+1){\alpha}}, & {c}<0 ; \end{cases} \\[5pt] { \mathcal{C}_{{r}+1}^{\text E} } =& \frac{({r}+1)^{({r}+1){\alpha}}}{({r}+1)!} &&\times \begin{cases} {a}^{-({r} + 1) {\alpha}} \cdot \prescript{}{2}{F}_1[{r}+1, ({r} + 1) {\alpha};\ {r} + 2; -{c}/{a}], & {c}>0, \\[3pt] ({a}+{c})^{-({r}+1){\alpha}}, & {c}<0. \end{cases}\end{alignat*}

Proof. Recall that ${ \mathcal{C}_{{r}+1}^{\text L} }\coloneqq { C_{{r}+1} ({{A}^{r}_{{c},{a}}}) }$ and ${ \mathcal{C}_{{r}+1}^{\text E} }\coloneqq { C_{{r}+1} ({{\mathcal{A}}^{r}_{{c},{a}}}) }$ . To calculate these constants, we use the definition of the measure ${ C_{{r}+1} (\bullet) }$ in (14). We start with ${ \mathcal{C}_{{r}+1}^{\text L} }$ . It is known that for ${U_{1}},\dots,{U_{{r}+1}} \sim {\mathcal{U}}(0,1)$ , the distribution of the random variable $\max \{ {U_{1}},\dots,{U_{{r}+1}} \}$ is given by the formula ${\mathbb{P}}(\max \{ {U_{1}},\dots,{U_{{r}+1}} \leqslant t) = t^{{r}+1}$ . Furthermore, by using that $\int_b^{+\infty} {\alpha} y^{-n {\alpha} - 1} \, {\text d} y = b^{-n {\alpha}}/n$ with $b>0$ , we calculate recursively the following multiple integrals for $n \in {\mathds{N}}$ and positive $y_i$ :

\begin{align*} {\mathcal{I}_{n}} &= \int\limits_{y_1 \geqslant \dots \geqslant y_{n+1}} \prod_{i=1}^n {\alpha} y_i^{-{\alpha}-1} \, {\text d} y_1 \cdots {\text d} y_n \\ &= \int\limits_{y_n = y_{n+1}}^{+\infty} \int\limits_{y_{n-1} = y_n}^{+\infty} \dots \int\limits_{y_2 = y_3}^{+\infty} \prod_{i=2}^{n} {\alpha} y_i^{-{\alpha}-1} \underbrace{\left(\;\int\limits_{y_1 = y_2}^{+\infty} {\alpha} y_1^{-{\alpha}-1} \, {\text d} y_1 \;\right)}_{= y_2^{-{\alpha}}} \, {\text d} y_2 \cdots {\text d} y_n \\ &= \int\limits_{y_n = y_{n+1}}^{+\infty} \dots \int\limits_{y_3 = y_4}^{+\infty} \prod_{i=3}^{n} {\alpha} y_i^{-{\alpha}-1} \underbrace{\left(\; \int\limits_{y_2 = y_3}^{+\infty} {\alpha} y_2^{-2{\alpha}-1} \, {\text d} y_2\;\right)}_{= y_3^{-2 {\alpha}}/2} \, {\text d} y_3 \cdots {\text d} y_n = \dots \\ &= \frac{1}{n!}(y_{n+1})^{- n {\alpha}}.\end{align*}

Consequently, in the case ${c}>0$ , we obtain, by virtue of Remark 1,

\begin{align*} { \mathcal{C}_{{r}+1}^{\text L} } &= \mathbb{E} \Bigg[ {{\nu}^{{r}+1}_{\alpha}} \Bigg\{ {\textbf{\textit{y}}} \in {\mathds{R}}^{{r}+1}_+ : \sum_{i=1}^{{r}+1} y_i {\textbf{1}_{\{{U_{i}},1\}}} \in {{A}^{r}_{{c},{a}}} \Bigg\} \Bigg] \\ &= \mathbb{E} \left[ \int\limits_{y_1 \geqslant \dots \geqslant y_{{r}+1}>0} \prod_{i=1}^{{r}+1} {\alpha} y_i^{-{\alpha}-1} {\textbf{1}_{{ \{ y_{{r}+1} \geqslant {a} + {c} \max \{ {U_{1}},\dots,{U_{{r}+1}}\} \} }}}\, {\text d} y_1 \cdots {\text d} y_{{r}+1} \right] \\ &= \int\limits_{t \in [0,1]} \int\limits_{y_1 \geqslant \dots \geqslant y_{{r}+1}>0} \prod_{i=1}^{{r}+1} {\alpha} y_i^{-{\alpha}-1} {\textbf{1}_{\{y_{{r}+1} \geqslant {a} + {c} t \}\}}} ({r}+1) t^{r} \, {\text d} y_1 \cdots {\text d} y_{{r}+1} {\text d} t\\ &= \int\limits_{t \in [0,1]} \int\limits_{ y_{{r}+1}>0} {\mathcal{I}_{{r}}} {\alpha} y_{{r}+1}^{-{\alpha}-1} {\textbf{1}_{\{y_{{r}+1} \geqslant {a} + {c} t \} \}}} ({r}+1) t^{r} \, {\text d} y_{{r}+1} {\text d} t\displaybreak\\ &= \int\limits_{t \in [0,1]} \int\limits_{ y_{{r}+1}={a} + {c} t }^{+\infty} \frac{1}{{r}!}(y_{{r}+1})^{- {r} {\alpha}} {\alpha} y_{{r}+1}^{-{\alpha}-1} ({r}+1) t^{r} \, {\text d} y_{{r}+1} {\text d} t\\[3pt] &= \frac{{r}+1}{{r}!} \int\limits_0^1 t^{r} \left(\; \int\limits_{{a} + {c} t }^{+\infty} {\alpha} (y_{{r}+1})^{-({r}+1){\alpha}-1} \, {\text d} y_{{r}+1} \;\right) {\text d} t = \frac{1}{{r}!} \int\limits_0^1 t^{r} ({a} + {c} t)^{-({r}+1){\alpha}} \, {\text d} t\\[3pt] &=\frac{{a}^{-({r} + 1) {\alpha}}}{({r}+1)!} \cdot \prescript{}{2}{F}_1[{r}+1, ({r} + 1) {\alpha};\ {r} + 2; -{c}/{a}].\end{align*}

Analogously, we find

\begin{align*} { \mathcal{C}_{{r}+1}^{\text E} } &= \mathbb{E} \Bigg[{{\nu}^{{r}+1}_{\alpha}} \Bigg\{ {\textbf{\textit{y}}} \in {\mathds{R}}^{{r}+1}_+ : \sum_{i=1}^{{r}+1} y_i {\textbf{1}_{\{{U_{i}},1\}}} \in {{\mathcal{A}}^{r}_{{c},{a}}} \Bigg\} \Bigg] \\ &=\int\limits_{y_1 \geqslant \dots \geqslant y_{{r}+1}>0} \prod_{i=1}^{{r}+1} {\alpha} y_i^{-{\alpha}-1} {\textbf{1}_{{ \{y_{{r}+1} \geqslant ({a} + {c} \max \{ {U_{1}},\dots,{U_{{r}+1}}\} )/({r}+1) \} }}}\, {\text d} y_1 \cdots {\text d} y_{{r}+1} \\[3pt] &= \int\limits_{t \in [0,1]} \int\limits_{ y_{{r}+1}>0} {\mathcal{I}_{{r}}} {\alpha} y_{{r}+1}^{-{\alpha}-1} {\textbf{1}_{\{y_{{r}+1} \geqslant ({a} + {c} t)/({r}+1) \} \}}} ({r}+1) t^{r} \, {\text d} y_{{r}+1} {\text d} t\\[3pt] &= \frac{1}{{r}!} \int\limits_0^1 t^{r} \left( \frac{{a} + {c} t}{{r}+1} \right)^{-({r}+1){\alpha}} \, {\text d} t = ({r}+1)^{({r}+1){\alpha}} \frac{1}{{r}!} \int\limits_0^1 t^{r} ( {a} + {c} t )^{-({r}+1){\alpha}} \, {\text d} t\\ &= ({r}+1)^{({r}+1){\alpha}} \frac{{a}^{-({r} + 1) {\alpha}}}{({r}+1)!} \cdot \prescript{}{2}{F}_1[{r}+1, ({r} + 1) {\alpha};\ {r} + 2; -{c}/{a}].\end{align*}

In the case ${c}<0$ , the coefficients simplify to

\begin{align*} \hspace*{15pt} { \mathcal{C}_{{r}+1}^{\text L} } &= \mathbb{E} \left[ \int\limits_{y_1 \geqslant \dots \geqslant y_{{r}+1}>0} \prod_{i=1}^{{r}+1} {\alpha} y_i^{-{\alpha}-1} {\textbf{1}_{\{y_{{r}+1} \geqslant {a} + {c} \}}}\, {\text d} y_1 \cdots {\text d} y_{{r}+1} \right] = \int\limits_{ {a} + {c}}^{+\infty} {\mathcal{I}_{{r}}} {\alpha} y_{{r}+1}^{-{\alpha}-1} \, {\text d} y_{{r}+1}\\[4pt] &= \cdots = \frac{1}{({r}+1)!} ({a}+{c})^{-({r}+1){\alpha}}, \quad \text{and}\\[4pt] { \mathcal{C}_{{r}+1}^{\text E} } &= ({r}+1)^{({r}+1){\alpha}} \frac{1}{({r}+1)!} ({a}+{c})^{-({r}+1){\alpha}}.\hspace*{175pt}\Box\end{align*}

Remark 2. When ${c}>0$ , the coefficients ${ \mathcal{C}_{{r}+1}^{\text L} }$ and ${ \mathcal{C}_{{r}+1}^{\text E} }$ can be equivalently expressed in terms of finite sums involving the Gamma function. More precisely, by applying integration by parts ${r}$ times, we calculate for $k>{r}+1$ that

\begin{align*} \frac{1}{{r}!} \int t^{r} ({a} + {c} t)^{-k} \, {\text d} t =& \sum_{m=1}^{{r}+1} \frac{(-1)^{m+1} t^{{r}+1-m}}{({r}+1-m)!} \frac{({a} + {c} t)^{m-k}}{{c}^m \prod_{j=1}^m(j-k)}\\ =& \sum_{m=1}^{{r}+1} \frac{(-1)^{m+1} t^{{r}+1-m}}{({r}+1-m)!} \frac{({a} + {c} t)^{m-k}}{{c}^m (1-k)_m}\end{align*}

\begin{align*} =& -\sum_{m=1}^{{r}+1} \frac{ t^{{r}+1-m}}{({r}+1-m)!} \frac{({a} + {c} t)^{m-k}}{{c}^m (k-m)_m} \Rightarrow \\ \frac{1}{{r}!} \int\limits_0^1 t^{r} ({a} + {c} t)^{-k} \, {\text d} t =& \frac{ {a}^{{r}+1-k} }{{c}^{{r}+1} (k-{r}-1)_{{r}+1}} -\sum_{m=1}^{{r}+1} \frac{({a} + {c})^{m-k}}{({r}+1-m)! {c}^m (k-m)_m},\end{align*}

where $(b)_k = \Gamma(b+k)/\Gamma(b)$ is again the Pochhammer symbol. Thus,

\begin{align*} { \mathcal{C}_{{r}+1}^{\text L} } =& \frac{ {a}^{-({r}+1)({\alpha}-1)} \Gamma ( ({r}+1) {\alpha} ) }{{c}^{{r}+1} \Gamma ( ({r}+1)({\alpha}-1) )} -\sum_{m=1}^{{r}+1} \frac{({a} + {c})^{m-({r}+1){\alpha}} \Gamma ( ({r}+1) {\alpha} ) }{({r}+1-m)! {c}^m \Gamma ( ( {r}+1){\alpha} - m )}, \\[3pt] { \mathcal{C}_{{r}+1}^{\text E} } =& ({r}+1)^{({r}+1){\alpha}} \\[3pt] &\times \left( \frac{ {a}^{-({r}+1)({\alpha}-1)} \Gamma ( ({r}+1) {\alpha} ) }{{c}^{{r}+1} \Gamma ( ({r}+1)({\alpha}-1) )} -\sum_{m=1}^{{r}+1} \frac{({a} + {c})^{m-({r}+1){\alpha}} \Gamma ( ({r}+1) {\alpha} ) }{({r}+1-m)! {c}^m \Gamma ( ( {r}+1){\alpha} - m )} \right).\end{align*}

Remark 3. In the absence of reinsurance ( ${r} = 0$ ), the pre-constant simplifies to

\begin{equation*} \frac{{a}^{-{\alpha} +1} - ({a} + {c})^{-{\alpha} +1}}{{c} ({\alpha} - 1)},\end{equation*}

which can also be derived from existing results; see, e.g., [Reference Asmussen and Klüppelberg6,Reference Embrechts, Goldie and Veraverbeke10].

Finally, we know from [Reference Kyprianou14] that the compound Poisson aggregate claim process ${S(t)} = \sum_{i=1}^{{N(t)}} {{X}_{i}}$ is a special Lévy process with Lévy measure ${\nu}({\text d} x) = {\lambda} {{F}({\text d} x)}$ , which means that $n \cdot {\nu}[n,\infty) = {\lambda} n {\bar{{F}}(n)} = {\lambda} {L(n)} n^{-{\alpha}+1}$ , $n \in {\mathds{N}}$ . We conclude the proof by combining this result with Lemma 7 to obtain the expression (15).