2. Preliminaries

If there exist two positive functions $g(\cdot,\cdot)$ and $f(\cdot,\cdot)$ satisfying

\begin{align*} l_1 = \liminf_{x\rightarrow\infty} \frac{f(x,t)}{g(x,t)} \leq \limsup_{x\rightarrow\infty} \frac{f(x,t)}{g(x,t)} = l_2,\end{align*}

then we say that $f(x,t)\lesssim g(x,t)$ holds if $l_2\leq1$ ; $f(x,t)\gtrsim g(x,t)$ holds if $l_1\geq1$ ; $f(x,t)\sim g(x,t)$ holds if $l_1=l_2=1$ ; and $f(x,t)\asymp g(x,t)$ holds if $0<l_1 \leq l_2< \infty$ . All limit relations, unless otherwise stated, hold as $x\rightarrow\infty$ . Throughout this paper, C denotes a generic positive constant without relation to x, which may vary with the context. Moreover, for any $x,y\in\mathbb{R}$ , we write $x\wedge y=\min\left\{x,y\right\}$ , $x\vee y=\max\left\{x,y\right\}$ , and $x^{+}=x\vee 0$ .

In this paper, random and real vectors, supposed to be d-dimensional, are expressed by bold English letters. For two real vectors $\boldsymbol{a}=(a_1, \ldots,a_d)$ and $\boldsymbol{b}=(b_1,\ldots,b_d)$ , we assume that operations between vectors such as $\boldsymbol{a}>\boldsymbol{b}$ , $\boldsymbol{a}\pm\boldsymbol{b}$ , $\boldsymbol{a}\boldsymbol{b}$ , and $\boldsymbol{a}^{-1}$ should be interpreted componentwise, and we let $[\boldsymbol{a},\boldsymbol{b}]=[a_1,b_1]\times\cdots\times[a_d,b_d]$ , $[\boldsymbol{a},\boldsymbol{\infty})=[a_1,\infty)\times\cdots\times[a_d,\infty)$ . Additionally, for $y\in\mathbb{R}$ , write $y\boldsymbol{a}=\left(ya_1,\ldots,ya_d\right)$ as usual. We also write $\mathbf{0}=(0,\ldots,0)$ , $\mathbf{1}=(1,\ldots,1)$ , and $\mathbf{e}_{k}=(0,\ldots,1,\ldots,0)$ , where the value 1 occurs in the kth spot.

Definition 1. A distribution F on $[0,\infty)$ satisfying $\bar{F}(x)>0$ for all $x\geq0$ is said to be of regular variation, and we write $F\in\mathcal{R}_{-\alpha}$ , if

(1) \begin{align} \lim_{x\rightarrow\infty} \frac{ \bar{F}(xy) }{ \bar{F}(x) } =y^{-\alpha},\qquad y>0, \end{align}

for some $0<\alpha<\infty$ .

For some distribution function with $F\in\mathcal{R}_{-\alpha}$ , $0<\alpha<\infty$ , we get by [Reference Bingham, Goldie and Teugels5, Proposition 2.2.3] that there exist positive constants $C_F>1$ and $D_F$ such that, for all $x, xy\geq D_F$ and any $0<\varepsilon<\alpha$ ,

(2) \begin{align} \frac{1}{C_F}(y^{-\alpha+\varepsilon}\wedge y^{-\alpha-\varepsilon}) \leq \frac{\bar{F}(xy)}{\bar{F}(x)} \leq C_F(y^{-\alpha+\varepsilon}\vee y^{-\alpha-\varepsilon}).\end{align}

From the relation (2), it follows that for $p>\alpha$ ,

(3) \begin{align} x^{-p}=o(\bar{F}(x)),\qquad x\rightarrow \infty.\end{align}

Definition 2. An $\mathbb{R}_+^d$ -valued random vector $\boldsymbol{Z}$ is said to follow a multivariate regularly varying distribution if there exist a Radon measure $\nu$ not identically 0 and a positive normalizing function $a(\!\cdot\!)$ monotonically increasing to $\infty$ , such that when $t\rightarrow\infty$ ,

(4) \begin{align} t \, \mathbb{P}\left( \frac{\mathbf{Z}}{a(t)} \in \cdot \right) \stackrel{v}{\longrightarrow} \nu(\!\cdot\!) \quad\textrm{on } [0,\infty]^d\backslash \{\textbf{0}\}, \end{align}

where $\stackrel{v}{\longrightarrow}$ denotes vague convergence.

From the above definition, one can show that the Radon measure $\nu$ is homogeneous—namely, that there exists some $\alpha>0$ , denoting the MRV index, such that the equality

(5) \begin{align} \nu(sB)=s^{-\alpha}\nu(B)\end{align}

holds for every Borel set $B\subset[0,\infty]^d\backslash \{\textbf{0}\}$ . For details on this, see [Reference Resnick27, p. 178]. From [Reference Konstantinides17, p. 459], we obtain $a(t)\in\mathcal{R}_{1/\alpha}$ , and hence there exists some distribution $F\in\mathcal{R}_{-\alpha}$ such that

\begin{align*} \bar{F}(t) \sim \frac{1}{a^{\leftarrow}(t)}, \quad \textrm{as } t\rightarrow\infty,\end{align*}

where $a^{\leftarrow}(t)$ is the generalized inverse function of a(t). Consequently, the relation (4) can be expressed as follows:

(6) \begin{align} \frac{1}{\bar{F}(x)} \mathbb{P}\left(\frac{\boldsymbol{Z}}{x}\in\cdot\right) \stackrel{v}{\longrightarrow}\nu(\!\cdot\!) ,\quad\textrm{on } [0,\infty]^d\backslash \{\textbf{0}\}.\end{align}

Hence, we write $\boldsymbol{Z}\in \textrm{MRV}(\alpha,F,\nu)$ . For more details on MRV, the reader is referred to [Reference Resnick27, Chapter 6] or [Reference Konstantinides17, Chapter 13].

Next we present a lemma that is important in the following proofs.

Lemma 1. Let random vector $\boldsymbol{Z}\in \boldsymbol{MRV}(\alpha,F,\nu)$ for some $\alpha>0$ , let $\boldsymbol{\xi}$ be a positive random vector with arbitrarily dependent components satisfying $\mathbb{E}\boldsymbol{\xi}^{\beta}<\boldsymbol{\infty}$ for some $\beta>\alpha$ , and let $\left\{\Delta_t,t\in\mathcal{T}\right\}$ be a set of random events such that $\lim_{t\rightarrow t_0}\mathbb{P}(\Delta_t)=0$ for some $t_0$ in the closure of the index set $\mathcal{T}$ . Furthermore, assume that $\left\{\boldsymbol{\xi},\left\{\Delta_t,t\in\mathcal{T}\right\}\right\}$ and $\boldsymbol{Z}$ are independent. Then

\begin{align*} \lim_{t\rightarrow t_0}\limsup_{x\rightarrow\infty} \frac{\mathbb{P}\left(\boldsymbol{Z}\boldsymbol{\xi}>x\,\textbf{1},\Delta_t\right)} {\bar{F}(x)} =0. \end{align*}

Proof. Since the random vector $\boldsymbol{Z}\in \textrm{MRV}(\alpha,F,\nu)$ , the marginal tail of $\boldsymbol{Z}$ is regularly varying from [Reference Konstantinides17, p. 458]. Therefore, we have $Z_k\in\mathcal{R}_{-\alpha}$ , $1\leq k\leq d$ . Then

\begin{align*} \lim_{t\rightarrow t_0}\limsup_{x\rightarrow\infty} \frac{\mathbb{P}\left(\boldsymbol{Z}\boldsymbol{\xi}>x\,\textbf{1},\Delta_t\right)} {\bar{F}(x)} \leq \lim_{t\rightarrow t_0}\limsup_{x\rightarrow\infty} \frac{ \mathbb{P}\left( Z_k\xi_k>x, \Delta_t \right) } { \bar{F}(x) } =0, \end{align*}

where we used [Reference Tang and Yuan30, Lemma 6.2]. This completes the proof.

3. Main assumption

This paper proposes the following assumption to describe certain dependence structures between the random vectors $\mathbf{U}$ and $\mathbf{V}$ .

Remark 1. From Assumption 1, we can derive that

(7) \begin{align} \lim_{x\rightarrow\infty} \frac{\mathbb{P}\left(\bigcap_{k=1}^d\left\{U_k V_k>x\right\}\right)} {\bar{F}(x)} =\nu\left((\textbf{1},\boldsymbol{\infty})\right)>0 \end{align}

and

(8) \begin{align} \lim_{x\rightarrow\infty} \frac{ \mathbb{P}\left(U_k V_k>x\right) } { \bar{F}(x) } =\nu\left(\left(\mathbf{e}_{k},\boldsymbol{\infty}\right]\right) \;:\!=\;a_k >0, \qquad 1\leq k \leq d. \end{align}

The relation (7) indicates that $U_1V_1,\ldots,U_dV_d$ reveal large joint movements and accordingly are asymptotically dependent. The relation (8) indicates that the tails of the products $U_1V_1,\ldots,U_dV_d$ are regularly varying and that $U_iV_i$ and $U_jV_j$ have comparable tails. Furthermore, note that by (5) and (7), for any $(b_1, \ldots, b_d)\in[0,\infty]^d\backslash \{\textbf{0}\}$ ,

\begin{align*} \lim_{x\rightarrow\infty} \frac{\mathbb{P}\left(\bigcap_{k=1}^d\left\{U_k V_k>b_kx\right\}\right)} {\bar{F}(x)} =\nu\left((b_1,\infty]\times\cdots\times(b_d,\infty]\right)>0. \end{align*}

Remark 2. Set $\mathbf{U}=U\mathbf{1}$ ; then the product $\mathbf{UV}$ can be rewritten as

(9) \begin{align} \mathbf{UV}=(UV_1, \ldots, UV_d), \end{align}

which is called the scale mixture. This structure can be applied in various areas, including investments or insurance portfolios, and it has various interpretations in different applications. For example, [Reference Li and Sun18] developed a method to estimate the tail dependence of heavy-tailed scale mixtures of multivariate distributions. The paper [Reference Zhu and Li34] studied the asymptotic relations between the value-at-Risk and multivariate tail conditional expectation for heavy-tailed scale mixtures of multivariate distributions. Additionally, the class (9) of loss distributions is a subset of all multivariate regularly varying distributions considered in [Reference Joe and Li14], but it includes various loss distributions. In addition, the scale mixture can reflect both individual factors (via the $V_i$ ) and the common factor (via U) of a risk class. For instance, U is a common systemic risk factor associated with macroeconomic conditions, including regulatory bodies and economic conditions, while the quantities $V_k$ , $1\leq k\leq d$ , are individual risks explained as individual business risks and assets.

For Assumption 1, the following question arises naturally: what are the conditions under which the distribution of $\mathbf{UV}$ is MRV? This question has received an increasing amount of attention. The paper [Reference Basrak, Davis and Mikosch4] demonstrated the MRV of the product $\boldsymbol{UV}$ , when $\mathbf{U}$ is MRV and independent of $\mathbf{V}$ , as in Proposition 1 (see also [Reference Basrak, Davis and Mikosch4, Appendix], [Reference Fougeres and Mercadier11, Theorem 1], and [Reference Konstantinides and Li16, Lemma 3.1]).

Proposition 1. Let the random vector $\boldsymbol{U}\in {\boldsymbol{MRV}}(\alpha,F,\nu)$ for some $\alpha>0$ , and let $\boldsymbol{V}$ be a nonnegative random vector with dependent components satisfying $\mathbb{E}\boldsymbol{V}^p<\boldsymbol{\infty}$ for some $p>\alpha$ . Assume that $\boldsymbol{V}$ and $\boldsymbol{U}$ are independent. Then the following relation holds for every Borel set $K\subset[0,\infty]^d\backslash \{\textbf{0}\}$ :

\begin{align*} \lim_{x\rightarrow\infty}\frac{1}{\bar{F}(x)} \mathbb{P}\left(\frac{\boldsymbol{UV}}{x}\in K\right) =\mathbb{E}\left[\nu(\boldsymbol{V}^{-1}K)\right], \end{align*}

where

\begin{align*} \boldsymbol{V}^{-1}K &=\left\{\left(b_1,\ldots,b_d\right) |\left(V_1b_1,\ldots,V_db_d\right)\in K\right\}\\[5pt] &=\left\{\left(V^{-1}_1a_1,\ldots,V^{-1}_da_d\right), \, \left(a_1,\ldots,a_d\right)\in K\right\}. \end{align*}

The paper [Reference Chen, Wang and Zhang7] studied the tail asymptotics of the nonnegative random loss $\sum_{i=1}^d R_iS$ , where the stand-alone risk vector $\mathbf{R}=(R_1, \, \ldots, \, R_d)$ follows a multivariate regularly varying distributions with index $\alpha>0$ and is independent of S, representing the background risk factor. Furthermore, [Reference Chen, Wang and Zhang7] also assumed that $\mathbb{E}S^{\alpha+\delta}<\infty$ for some $\delta>0$ . Essentially, the conditions in [Reference Chen, Wang and Zhang7] imply that the random vector $(R_1S, \, \ldots, \, R_dS)$ is MRV by Proposition 1.

The hypothesis of independence between $\mathbf{U}$ and $\mathbf{V}$ in Proposition 1 may be too strong in certain settings. If this condition can be weakened, Assumption 1 will become broader and more meaningful. The paper [Reference Fougeres and Mercadier11] further improved Proposition 1 to the following result, Proposition 2, which seems meaningful in the context of actuarial risk theory.

Proposition 2. Let us assume that

(10) \begin{align} t \, \mathbb{P}\left( \left( \frac{\mathbf{U}}{a(t)}, \, \mathbf{V} \right) \in \cdot \right) \stackrel{v}{\longrightarrow} (\nu \times L) (\!\cdot\!) \end{align}

on the Borel sets of $([0,\infty]^d\backslash \{\textbf{0}\}) \times [0,\infty]^d$ , where L represents a probability measure on $[0,\infty]^d$ and $\nu$ denotes a Radon measure on $[0,\infty]^d\backslash \{\textbf{0}\}$ not concentrated at $\infty$ . Suppose that, for some $\delta>0$ and any $i=1, \, 2, \, \ldots, \, d$ ,

(11) \begin{align} \lim_{\varepsilon\rightarrow0} \, \limsup_{t\rightarrow\infty} t \mathbb{E}\left[ \left( \frac{|\mathbf{U}| \, |V_i|}{a(t)} \right)^{\delta} \mathbb{I}_{[|\mathbf{U}| / a(t) \leq \varepsilon ]} \right] =0, \end{align}

and also that

(12) \begin{align} \int_{[\mathbf{0}, \, \boldsymbol{\infty}]} || \mathbf{v} ||^{\alpha} L(\textrm{d}\mathbf{v}) <\infty, \end{align}

where $ \mathbb{I}_{[E]}$ is the indicator of the Borel set E and $|| \cdot ||$ denotes some norm on $\mathbb{R}^d$ . Then the random vector $\mathbf{UV}$ follows a multivariate regularly varying distribution under the relations (10), (11), and (12). More precisely,

\begin{align*} t \mathbb{P}(\mathbf{UV}\in a(t) \cdot ) \stackrel{v}{\longrightarrow} \nu_L(\!\cdot\!) \end{align*}

as $t\rightarrow\infty$ , where $\nu_L$ is the measure defined on $[0,\infty]^d\backslash \{\textbf{0}\}$ by

\begin{align*} \nu_L(A) =(\nu \times L) (\{(\mathbf{x},\mathbf{y}), \mathbf{x}\in \mathbf{y}^{-1}A\}) =\int_{[0, \infty]^d} \nu(\mathbf{y}^{-1}A)L(\textrm{d}y). \end{align*}

Assume that $(U_1, \ldots, U_d, V)$ , $d\geq1$ , is a positive random vector. Li [Reference Li19] introduced the following dependence structure among the components of $(U_1, \ldots, U_d, V)$ :

• • There are some d-variate function $f\;:\; \, [0,\infty]^d\backslash \{\textbf{0}\} \mapsto (0, \infty)$ , some univariate function $h\;:\; \, (0,\infty) \mapsto (0,\infty)$ satisfying

  (13) \begin{align} 0 < \inf_{0 < t < \infty} h(t) \leq \sup_{0 < t < \infty} h(t) <\infty, \end{align}
  and some distribution F supported on $(0,\infty)$ with an infinite upper endpoint, such that for every $\mathbf{b} \in [0,\infty]^d\backslash \{\textbf{0}\}$ the relation
  (14) \begin{align} \mathbb{P}(U_1>b_1u,\ldots,U_d>b_du | V = v) \sim f(\mathbf{b}) h(v) \bar{F}(u) \end{align}
  holds uniformly for $v\in(0,\infty)$ as $u\rightarrow\infty$ .

According to the relation (2.10) in [Reference Li19], taking $\mathbf{b}=\mathbf{e}_{k}$ , $1\leq k\leq d$ , in (14) yields that the relation

(15) \begin{align} \mathbb{P}(U_k>u|V=v) \sim \frac{1}{\mu}h(v)\bar{F}_k(u)\end{align}

holds uniformly for $v\in(0,\infty)$ as $u\rightarrow\infty$ , where $\bar{F}_k(u)$ is the distribution function of $U_k$ , and $\mu=\mathbb{E}[h(V)]\in(0,\infty)$ . Therefore, the dependence structure specified in (14) implies that the marginal vector $(U_k,V)$ follows a parallel two-dimensional dependence structure as shown in (15). The structure (15) is defined in [Reference Asimit and Jones2] and further developed in [Reference Asimit and Badescu1] and [Reference Li, Tang and Wu22]; it includes a wide range of commonly used bivariate copulas.

Now define a new positive random variable $V^*$ with distribution

\begin{align*} \mathbb{P}(V^*\in \textrm{d}v)=\frac{1}{\mu}h(v)\mathbb{P}(V\in \textrm{d}v).\end{align*}

For simplicity, we introduce a new random variable $\xi$ with distribution function $F\in\mathcal{R}_{-\alpha}$ for $0<\alpha<\infty$ , and let $\xi$ and $V^*$ be independent of all other sources of randomness. If

\begin{align*} \mathbb{E}(V^{\alpha+\varepsilon})<\infty\end{align*}

holds for $\varepsilon>0$ , then by (13),

\begin{align*}\mathbb{E}[(V^{*})^{\alpha+\varepsilon}]=\frac{1}{\mu}\int_{0}^{\infty}v^{\alpha+\varepsilon}h(v)\mathbb{P}(V\in \textrm{d}v)\leq C\mathbb{E}(V^{\alpha+\varepsilon})<\infty,\end{align*}

and for $\frac{\alpha}{\alpha+\varepsilon}<p<1$ and large $u>0$ ,

\begin{align*}\mathbb{P}(V>u^p)\leq \mathbb{E}(V^{\alpha+\varepsilon})u^{-p(\alpha+\varepsilon)}=o(1)\bar{F}(u)\end{align*}

and

\begin{align*}\mathbb{P}(V^*>u^p)\leq \mathbb{E}[(V^{*})^{\alpha+\varepsilon}]u^{-p(\alpha+\varepsilon)}=o(1)\bar{F}(u)\end{align*}

hold. Consequently, we have that by (14),

\begin{align*} \mathbb{P}(\mathbf{U} V>\mathbf{b}u) &=\mathbb{P}(\mathbf{U}V>\mathbf{b}u, 0<V\leq u^p) + \mathbb{P}(\mathbf{U} V>\mathbf{b}u, V> u^p)\\[5pt] &=\int_{0+}^{u^p}\mathbb{P} \left(\mathbf{U}>\mathbf{b}\frac{u}{v}|V=v\right) \mathbb{P}(V\in \textrm{d}v) +o(1)\bar{F}(u)\\[5pt] &\sim \mu f(\mathbf{b}) \int_{0+}^{u^p} \bar{F}\left(\frac{u}{v}\right)\mathbb{P}(V^*\in \textrm{d}v)\\[5pt] &=\mu f(\mathbf{b})\left[ \mathbb{P}(\xi V^*>u)-\mathbb{P}(\xi V^*>u, V^*>u^p) \right]\\[5pt] &\sim\mu f(\mathbf{b}) \mathbb{E}[(V^*)^{\alpha}]\bar{F}(u),\end{align*}

where $\mathbb{P}(\xi V^*>u)\sim \bar{F}(u)\mathbb{E}[(V^*)^{\alpha}]$ is due to the relation (4.4) in [Reference Tang, Wang and Yuen28]. Therefore, $(U_1V,\ldots,U_dV)$ satisfies Assumption 1.

4. The study of a d-dimensional discrete-time risk model under Assumption 1

This section considers an insurer who runs multiple lines of business and makes risky assets along individual lines. For every $i \in \mathbb{N}=\{1, \, 2, \, \ldots\}$ and integer $d\geq1$ , the real-valued random variable $X_{ki}$ , $1\leq k \leq d$ , denotes the net insurance loss (described by the aggregate claim amount minus the aggregate premium income) of the kth business of the insurer over the period i, and the positive random variable $\theta_{ki}$ denotes the discount factor, related to the return on the investment, of the kth business of the insurer over the same period. Let $\{(X_{1},\ldots,X_{d}), (X_{1i},\ldots,X_{di}), i\in \mathbb{N}\}$ and $\{(\theta_{1},\ldots,\theta_{d}), (\theta_{1i},\ldots,\theta_{di}), i\in \mathbb{N}\}$ be sequences of i.i.d. random vectors. The stochastic discounted value of total net insurance losses for the insurance company up to the time n can be described as

(16) \begin{align} \mathbf{S}_n =(S_{1n}, \, \ldots , \, S_{dn}) &=\left( \sum_{i=1}^n X_{1i} \prod_{j=1}^{i} \theta_{1j}, \, \ldots, \, \sum_{i=1}^n X_{di} \prod_{j=1}^{i} \theta_{dj} \right)\nonumber\\[5pt] &=\left( \sum_{i=1}^n X_{1i} Y_{1i}, \, \ldots, \, \sum_{i=1}^n X_{di} Y_{di} \right), \quad n\in \mathbb{N},\end{align}

where multiplication over the empty set is understood to be 1, and $Y_{ki}=\prod_{j=1}^{i} \theta_{kj}$ , $1\leq k\leq d$ . The product $\boldsymbol{\rho}x = (\rho_1x, \, \ldots, \rho_dx)$ denotes the vector of initial reserves assigned to different businesses, with positive $\rho_1, \, \ldots, \, \rho_d$ such that $\sum_{k=1}^d \rho_k =1$ . For $n\in\mathbb{N}$ , the ruin probability can be defined as

\begin{align*} \Psi(x, n) &\;:\!=\;\mathbb{P}\left( \bigcap_{k=1}^d \left\{\max_{1\leq m\leq n}{S}_{km} >\rho_k x \right\} \right) =\mathbb{P}\left(\max_{1\leq m\leq n}\mathbf{S}_m >\boldsymbol{\rho}x \right)\\[5pt] &=\mathbb{P}\left(\max_{1\leq m\leq n}\sum_{i=1}^m \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right),\end{align*}

which denotes the probability of the insurance company’s wealth process going below zero by time n. When $d=1$ , [Reference Fougeres and Mercadier11] considered the ruin probability of (16) if the random vector

\begin{equation*} (X_{11}Y_{11}, X_{12}Y_{12}, \,\ldots,\, X_{1n}Y_{1n})\end{equation*}

is MRV, and [Reference Tang and Yuan31] studied the ruin probability of (16) if the random vector $(X_{1i}, \theta_{1i})$ , $i\in\mathbb{N}$ , follows a bivariate regular variation structure. When $d=2$ , according to [Reference Chen and Yang8], we can obtain that the random vector

\begin{equation*} \left(\sum_{i=1}^mX_{1i}\prod_{j=1}^{i}\theta_{1j}, \sum_{i=1}^mX_{2i}\prod_{j=1}^{i}\theta_{2j}\right)\end{equation*}

still follows a bivariate-regular-variation structure if the pairs $(X_{1i}\theta_{1i}, X_{2i}\theta_{2i})$ , $i\in\mathbb{N}$ , follow some bivariate-regular-variation structure. In this section, we assume that $(X_{1i}^+\theta_{1i}, \ldots, X_{di}^+\theta_{di})$ , $i\in\mathbb{N}$ and $d\geq1$ , is a sequence of i.i.d. random vectors with generic vector $(X_{1}^+\theta_{1}, \ldots, X_{d}^+\theta_{d})\in \textrm{MRV}(\alpha, F,\nu)$ such that $\nu((\textbf{1},\boldsymbol{\infty}))>0$ ; then we study the asymptotic formula for the ruin probability for $n\in\mathbb{N}$ .

Theorem 1. Consider the d-dimensional discrete-time risk model (16). For each $i\in\mathbb{N}$ , assume that $\boldsymbol{X}_{i}^+ \boldsymbol{\theta}_{i}$ satisfies Assumption 1. If there is a constant $\beta>\alpha$ such that $\mathbb{E}\boldsymbol{\theta}^{\beta}<\textbf{1}$ , then it holds uniformly for $n\in\mathbb{N}$ that

(17) \begin{align} \Psi(x, n) &=\mathbb{P}\left( \max_{1\leq m\leq n}\sum_{i=1}^m \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{\rho}x \right)\nonumber\\[5pt] &\sim \mathbb{P}\left( \sum_{i=1}^n\boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) \sim \bar{F}(x)\sum_{i=1}^n \mathbb{E}\left[\nu(\boldsymbol{Y}_{i-1}^{-1}(\boldsymbol{\rho}, \boldsymbol{\infty}]) \right]. \end{align}

4.1. Numerical results

This subsection presents a two-dimensional numerical example to examine the accuracy of Theorem 1. All computations are conducted in R. For simplicity, we assume that $\theta_{1}=\theta_{2}=\theta$ , and let $\theta$ be exponential with parameter $\lambda=2$ .

We first construct a random pair $(X_1, X_2)$ . Let the distribution function F of the random variable Z follow a Pareto distribution, with scale parameter $\kappa=4$ and shape parameter $\alpha=2$ ( $\alpha>0$ ); that is, $F(x) = 1-\left( \frac{\kappa}{\kappa+x} \right)^{\alpha} \in \mathcal{R}_{-\alpha}$ , $x>0$ . Suppose that the dependence structure between Z and $\theta$ is established by a Farlie–Gumbel–Morgenstern (FGM) copula,

(18) \begin{align} C(u,v)=uv+\gamma uv(1-u)(1-v), \quad \gamma\in[-1,1], \, (u,v)\in[0,1]^2.\end{align}

Then, by [Reference Asimit and Badescu1, Example 2.2] or [Reference Li, Tang and Wu22, Example 3.2], the random pair $(Z, \theta)$ satisfies

\begin{align*} \mathbb{P}(Z>z\,\big|\, \theta=t) \sim \bar{F}(z) h(t), \quad\textrm{uniformly for all } t\in[0,\infty),\end{align*}

with $h(t)=1+\gamma(1-2e^{-\lambda t})$ . Let $(X_1, X_2)=(\zeta_1Z, \zeta_2Z)$ , where $\zeta_k$ , $k=1,2$ , are uniform on [0,1] and independent of $(Z, \theta)$ , and $\mathbb{E}\zeta_k^q<\infty$ for some $q>\alpha$ .

We next verify that $(X_1\theta, X_2\theta)$ satisfies Assumption 1. Using [Reference Li19, Example 3.1], we obtain

\begin{align*} \mathbb{P}(X_1>b_1x, X_2>b_2x|\theta=t) \sim V(b_1, b_2) h(t)\bar{F}(x), \quad\textrm{uniformly for all } t\in[0,\infty),\end{align*}

where $b_1,\, b_2>0$ are constants, and

\begin{align*} V(b_1, b_2) =\mathbb{E}\left( \frac{\zeta_1^{\alpha}}{b_1^{\alpha}}\bigwedge \frac{\zeta_2^{\alpha}}{b_2^{\alpha}} \right) =\frac{b_1+b_2}{(\alpha+1)(b_1\vee b_2)^{\alpha+1}} -\frac{2b_1b_2}{(\alpha+2)(b_1\vee b_2)^{\alpha+2}}.\end{align*}

Moreover, for $\beta=3>\alpha$ satisfying $\mathbb{E}\theta^{\beta}=\frac{6}{\lambda^3}<1$ and $\alpha/\beta<p<1$ , we obtain that

\begin{align*} &\mathbb{P}(X_1\theta>b_1x, X_2\theta>b_2x)\\[5pt] &=\mathbb{P}(X_1\theta>b_1x, X_2\theta>b_2x, 0<\theta\leq x^p) +\mathbb{P}(X_1\theta>b_1x, X_2\theta>b_2x, \theta> x^p)\\[5pt] &=\int_{0+}^{x^p} \mathbb{P}\left(X_1>b_1x/t, X_2>b_2x/t|\theta=t\right) \mathbb{P}(\theta\in \textrm{d}t) +o(1)\bar{F}(x)\\[5pt] &\sim V(b_1, b_2) \bar{F}(x) \int_{0+}^{x^p} t^{\alpha}h(t) \mathbb{P}(\theta\in \textrm{d}t)\\[5pt] &\sim V(b_1, b_2) \bar{F}(x) \left[ \mathbb{E}\theta^{\alpha}+\gamma \mathbb{E}\theta^{\alpha} -2\gamma \mathbb{E}\left(\theta^{\alpha}e^{-\lambda\theta}\right) \right] \;:\!=\;\mu V(b_1, b_2) \bar{F}(x),\end{align*}

where $\mu=\mathbb{E}\theta^{\alpha}+\gamma \mathbb{E}\theta^{\alpha}-2\gamma \mathbb{E}\left(\theta^{\alpha}e^{-\lambda\theta}\right)$ , and in the second and fourth steps we use the relation $\mathbb{P}(\theta>x^p)\leq x^{-p\beta} \mathbb{E}\theta^\beta=o(1)\bar{F}(x)$ from (3). This implies that $\mathbb{P}(X_1\theta>\cdot, X_2\theta>\cdot)$ possesses a bivariate regularly varying tail. Set $\gamma=0.5$ , so $\mu={11}/{16}$ and $\nu((1,\infty]\times(1,\infty])=\mu V(1,1)={11}/{96}>0$ ; then $(X_1\theta, X_2\theta)$ satisfies Assumption 1.

Finally, we estimate and compare the numerical results of the asymptotic formula (17) and the simulation of $\Psi(x,n)$ . Set $n=5$ ; then the asymptotic formula (17) becomes

\begin{align*} \Psi(x,n) \sim \mu V(\rho_1, \rho_2) \bar{F}(x) \sum_{i=1}^n \left(\mathbb{E}\theta^{\alpha}\right)^{i-1} \;:\!=\; \Psi_1(x,n).\end{align*}

Denote by $\Psi_2(x,n)$ the Monte Carlo simulation of $\Psi(x,n)$ . From [Reference Nelsen25, Exercise 3.23] we can generate an FGM random pair $(Z, \theta)$ , and then we get $(X_1, X_2, \theta)=(Z\zeta_1, Z\zeta_2, \theta)$ by generating two independent uniform (0,1) variates, $\zeta_1$ and $\zeta_2$ . We simulate $N=10^7$ samples of $\left((X_{11},X_{21}, \theta_{1}), \ldots,(X_{1n},X_{2n}, \theta_{n})\right)$ , and for each $k=1,\ldots,N$ , we denote by $\left( (X_{11}^{(k)},X_{21}^{(k)}, \theta_{1}^{(k)}), \ldots,(X_{1n}^{(k)},X_{2n}^{(k)}, \theta_{n}^{(k)}) \right)$ the independent copy of $\left((X_{11},X_{21}, \theta_{1}), \ldots,(X_{1n},X_{2n}, \theta_{n})\right)$ . Hence, $\Psi_2(x,n)$ is estimated by

\begin{align*} \frac{1}{N} \sum_{k=1}^{N} \mathbb{I}_{\left\{ \max_{1\leq m\leq n}\sum_{i=1}^m X_{1i}^{(k)}\prod_{j=1}^{i}\theta_j^{(k)} >\rho_1x, \quad \max_{1\leq m\leq n}\sum_{i=1}^m X_{2i}^{(k)}\prod_{j=1}^{i}\theta_j^{(k)} >\rho_2x \right\}}.\end{align*}

From Table 1, we observe that the ratio $\Psi_1(x,n)/\Psi_2(x,n)$ approaches 1 as x becomes large. Fixing x, we notice that the ruin probabilities decay when $\rho_1$ decreases from $0.5$ to $0.1$ , meaning that a larger value of $(1-\rho_1)$ leads to a smaller ruin probability of the corresponding business line from the insurance company.

Table 1.

Asymptotic versus simulated values

4.2. Some lemmas before the proof of Theorem 1

Clearly, one can derive the following relation for all $x>0$ :

(19) \begin{align} \mathbb{P}\left( \boldsymbol{X}_{i}^+ \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) =\mathbb{P}\left( \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x\right), \qquad \textrm{for any } \boldsymbol{b}\in[0,\infty]^d\backslash \{\textbf{0}\} \textrm{ and } i \in \mathbb{N}.\end{align}

Lemma 2. If the conditions of Theorem 1 hold, then for every fixed $\boldsymbol{b}\in[0,\infty]^d\backslash \{\textbf{0}\}$ and n we have

\begin{align*} \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \sim \sum_{i=1}^n \mathbb{P}\left( \boldsymbol{X}_{i}^+ \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \sim \sum_{i=1}^n\bar{F}(x) \mathbb{E}\left[\nu\left( \boldsymbol{Y}_{i-1}^{-1} (\boldsymbol{b},\boldsymbol{\infty}]\right)\right]. \end{align*}

Proof. For any $1\leq i\leq n$ , we have by the conditions of Theorem 1 that $\boldsymbol{X}_{i}^+ \boldsymbol{\theta}_i\in {\textrm{MRV}}(\alpha,F,\nu)$ and

(20) \begin{align} \mathbb{E}\boldsymbol{Y}_{i}^{\beta} =\left( \mathbb{E}Y_{1i}^{\beta}, \, \ldots, \, \mathbb{E}Y_{di}^{\beta} \right) =\left( \prod_{j=1}^i \mathbb{E}\theta_{1j}^{\beta}, \, \ldots, \, \prod_{j=1}^i \mathbb{E}\theta_{dj}^{\beta} \right) <\boldsymbol{\infty}. \end{align}

Since $\boldsymbol{X}_{i}^+ \boldsymbol{\theta}_i$ and $ \boldsymbol{Y}_{i-1}=\prod_{j=1}^{i-1}\boldsymbol{\theta}_{j} $ are independent, it follows from Proposition 1 that

(21) \begin{align} \mathbb{P}\left( \boldsymbol{X}_{i}^+ \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) &=\mathbb{P}\left(\frac{\boldsymbol{X}_{i}^+ \boldsymbol{\theta}_i \boldsymbol{Y}_{i-1}} {x} \in (\boldsymbol{b},\boldsymbol{\infty}]\right) \sim \bar{F}(x) \mathbb{E}\left[\nu\left( \boldsymbol{Y}_{i-1}^{-1} (\boldsymbol{b},\boldsymbol{\infty}]\right)\right]. \end{align}

Hence, it suffices to show that for every fixed $\boldsymbol{b}\in[0,\infty]^d\backslash \{\textbf{0}\}$ and n, the following asymptotic formula holds:

(22) \begin{align} \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \sim \sum_{i=1}^n \mathbb{P}\left( \boldsymbol{X}_{i}^+ \boldsymbol{Y}_{i} > \boldsymbol{b}x \right). \end{align}

Now we aim to show the upper bound of $\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right)$ . For an arbitrary $\varepsilon>0$ , choose small $v_1>0$ such that $(1-v_1)^{-\alpha}\leq1+\varepsilon$ and $(1+v_1)^{-\alpha}\geq 1-\varepsilon$ hold. Then

\begin{align*} &\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \leq \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+ \boldsymbol{Y}_{i} > \boldsymbol{b}x \right)\\[5pt] &= \mathbb{P}\left( \sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x, \bigcup_{m=1}^n \left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m} >\boldsymbol{b}(1-v_1)x\right) \right)\\[5pt] &\quad+\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x, \bigcap_{m=1}^n \left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m} >\boldsymbol{b}(1-v_1)x\right)^c\right)\\[5pt] & \;:\!=\; K_1(x,n)+K_2(x,n). \end{align*}

For $K_1(x,n)$ , we have by the relation (21) that

\begin{align*} K_1(x,n) &\leq \sum_{m=1}^n \mathbb{P}\left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m} >\boldsymbol{b}(1-v_1)x\right)\sim \sum_{m=1}^n\bar{F}\left((1-v_1)x\right) \mathbb{E}\left\{\nu(\boldsymbol{Y}_{m-1}^{-1}\boldsymbol{b}, \boldsymbol{\infty}]\right\}\\[4pt] &\sim (1-v_1)^{-\alpha}\bar{F}(x)\sum_{m=1}^n \mathbb{E}\left\{\nu \left( \boldsymbol{Y}_{m-1}^{-1}\boldsymbol{b}, \boldsymbol{\infty} \right] \right\}\\[4pt] &\leq (1+\varepsilon)\sum_{m=1}^n \bar{F}(x) \mathbb{E}\left\{\nu \left( \boldsymbol{Y}_{m-1}^{-1}\boldsymbol{b},\boldsymbol{\infty} \right] \right\} \sim (1+\varepsilon)\sum_{m=1}^n \mathbb{P}\left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m}>\boldsymbol{b}x\right). \end{align*}

For $K_2(x,n)$ we obtain, for $1\leq k\leq d$ and some $b_k>0$ ,

(23) \begin{align} &K_2(x,n) = \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x, \bigcap_{m=1}^n\bigcup_{l=1}^d \left(X_{lm}^+Y_{lm}\leq b_l(1-v_1)x\right) \right)\nonumber\\[4pt] & = \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x, \bigcup_{j=1}^n \left(X_{kj}^+Y_{kj}>\frac{b_kx}{n}\right), \bigcap_{m=1}^n\bigcup_{l=1}^d \left(X_{lm}^+Y_{lm}\leq b_l(1-v_1)x\right) \right)\nonumber\\[4pt] &\leq \sum_{j=1}^n \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x, X_{kj}^+Y_{kj}>\frac{b_kx}{n}, \bigcup_{l=1}^d \left(X_{lj}^+Y_{lj}\leq b_l(1-v_1)x\right) \right)\nonumber\\[4pt] &\leq \sum_{j=1}^n\sum_{l=1}^d \mathbb{P}\left(\sum_{i=1}^n X_{li}^+Y_{li}>b_lx, X_{kj}^+Y_{kj}>\frac{b_kx}{n}, X_{lj}^+Y_{lj}\leq b_l(1-v_1)x \right)\nonumber\\[4pt] &\leq \sum_{j=1}^n\sum_{l=1}^d\sum_{1\leq i\leq n, i\neq j} \mathbb{P}\left(X_{li}^+Y_{li}>\frac{b_lv_1x}{n-1}, X_{kj}^+Y_{kj}>\frac{b_kx}{n} \right)\nonumber\\[4pt] &=\left( \sum_{j=1}^n\sum_{l=1}^d\sum_{1\leq i<j\leq n} +\sum_{j=1}^n\sum_{l=1}^d\sum_{1\leq j<i\leq n} \right) \mathbb{P}\left(X_{li}^+\theta_{li}Y_{l,i-1}>\frac{b_lv_1x}{n-1}, X_{kj}^+\theta_{kj}Y_{k,j-1}>\frac{b_kx}{n} \right) \nonumber \\[4pt] &=o(1)\bar{F}(x)\,, \end{align}

where we applied (20), Lemma 1, and (8) in the last step, using the independence between

\begin{align*} X_{li}^+\theta_{li} \quad \textrm{and} \quad (Y_{l,i-1} , X_{kj}^+\theta_{kj}Y_{k,j-1}) \end{align*}

when $i>j$ and the independence between $X_{kj}^+\theta_{kj}$ and $(Y_{k,j-1} , X_{li}^+\theta_{li}Y_{l,i-1})$ when $i<j$ . Hence, for large x, we get

\begin{align*} \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \leq (1+C\varepsilon) \sum_{i=1}^n \mathbb{P}\left(\boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x\right). \end{align*}

Finally, we construct the lower bound of $\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right)$ . We have

\begin{align*} &\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{b}x \right) \geq \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i}>\boldsymbol{b}x, \bigcup_{m=1}^n \left(\boldsymbol{X}_{m}\boldsymbol{Y}_{m} >\boldsymbol{b}(1+v_1)x\right) \right)\\[5pt] &\geq \sum_{m=1}^n \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i}>\boldsymbol{b}x, \boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x \right)\\[5pt] &\qquad-\sum_{1\leq m<k\leq n} \mathbb{P}\left( \sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i}>\boldsymbol{b}x, \boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x, \boldsymbol{X}_{k}\boldsymbol{Y}_{k}>\boldsymbol{b}(1+v_1)x \right)\,, \end{align*}

so we find

\begin{align*} &\mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{b}x \right) \\[5pt] &\geq \sum_{m=1}^n \mathbb{P}\left(\boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x\right)\\[5pt] &\quad -\sum_{m=1}^n \mathbb{P}\left(\boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x, \bigcup_{k=1}^d \left( \sum_{i=1}^n X_{ki}Y_{ki} \leq b_kx \right) \right)\\[5pt] &\quad -\sum_{1\leq m<k\leq n} \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i}>\boldsymbol{b}x, \boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x, \boldsymbol{X}_{k}\boldsymbol{Y}_{k}>\boldsymbol{b}(1+v_1)x \right)\\[5pt] &\;:\!=\;K'_{\!\!1}(x,n)-K'_{\!\!2}(x,n)-K'_{\!\!3}(x,n). \end{align*}

For $K'_{\!\!1}(x,n)$ , we have by the relations (19) and (21) that

\begin{align*} &K'_{\!\!1}(x,n) \sim (1+v_1)^{-\alpha}\bar{F}(x) \sum_{m=1}^n \mathbb{E}\left(\nu \left( \boldsymbol{Y}_{m-1}^{-1}\boldsymbol{b},\boldsymbol{\infty} \right] \right)\\[5pt] &\geq (1-\varepsilon)\bar{F}(x) \sum_{m=1}^n \mathbb{E}\left(\nu \left( \boldsymbol{Y}_{m-1}^{-1}\boldsymbol{b},\boldsymbol{\infty} \right] \right)\\[5pt] &\geq (1-C\varepsilon)\sum_{m=1}^n \mathbb{P}\left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m}>\boldsymbol{b}x\right). \end{align*}

For $K'_{\!\!2}(x,n)$ , we obtain

\begin{align*} &K'_{\!\!2}(x,n) \leq \sum_{m=1}^n\sum_{k=1}^d \mathbb{P}\left(\boldsymbol{X}_{m}\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x, \sum_{i=1}^n X_{ki}Y_{ki}\leq b_kx \right)\\[5pt] &\leq \sum_{m=1}^n\sum_{k=1}^d \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, \sum_{1\leq i\neq m\leq n} X_{ki}Y_{ki}\leq-v_1b_kx \right)\\[5pt] &\leq \sum_{m=1}^n\sum_{k=1}^d\sum_{1\leq i< m\leq n} \mathbb{P}\left( X_{km}^+\theta_{km}Y_{k,m-1}>b_k(1+v_1)x, |X_{ki}|Y_{ki} \geq \frac{v_1b_kx}{n-1} \right) \\[5pt] &\quad + \sum_{m=1}^n\sum_{k=1}^d\sum_{1\leq m<i\leq n} \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, |X_{ki}|Y_{ki} \geq \frac{v_1b_kx}{n-1} \right) \\[5pt] & = o(1)\bar{F}(x) + \sum_{m=1}^n\sum_{k=1}^d\sum_{1\leq m<i\leq n} \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, |X_{ki}|Y_{ki} \geq \frac{v_1b_kx}{n-1} \right) , \end{align*}

where the last equality follows from Lemma 1, (20), and (8) by the independence between $X_{km}^+\theta_{km}$ and $(Y_{k,m-1} , |X_{ki}|Y_{ki})$ . For

\begin{align*} \frac{\alpha}{\beta} < p <1 \end{align*}

we find that, by Chebyshev’s inequality,

\begin{align*} &\sum_{m=1}^n\sum_{k=1}^d\sum_{1\leq m<i\leq n} \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, |X_{ki}|Y_{ki} \geq \frac{v_1b_kx}{n-1} \right)\leq \sum_{m=1}^n\sum_{k=1}^d\\[5pt] &\sum_{1\leq m<i\leq n} \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, |X_{ki}|Y_{ki} \geq \frac{v_1b_kx}{n-1}, \theta_{km} \leq x^{p} \right) +dn^3\mathbb{P}\left( \theta_{km}> x^{p} \right)\\[5pt] &\leq \sum_{m=1}^n\sum_{k=1}^d\sum_{1\leq m<i\leq n} \mathbb{P}\left( X_{km}^+Y_{km}>b_k(1+v_1)x, |X_{ki}| \prod_{j=1,j\neq m}^{i}\theta_{kj} \geq \frac{v_1b_kx^{1-p}}{n-1} \right) \\[5pt] &+dn^3x^{-p\beta}\mathbb{E}\theta_{km}^{\beta}=o(1) \bar{F}(x), \end{align*}

where the last step is due to (3), Lemma 1, (20), and (8), using the independence between the product $X_{km}^+\theta_{km}$ and $\left( Y_{k,m-1} , |X_{ki}| \prod_{j=1,j\neq m}^{i}\theta_{kj} \right)$ . For $K'_{\!\!3}(x,n)$ , by Lemma 1 we have

\begin{align*} K'_{\!\!3}(x,n) \leq \sum_{1\leq m<k\leq n} \mathbb{P}\left(\boldsymbol{X}_{m}^+\boldsymbol{Y}_{m}>\boldsymbol{b}(1+v_1)x, \boldsymbol{X}_{k}^+\boldsymbol{Y}_{k}>\boldsymbol{b}(1+v_1)x \right) =o(1)\bar{F}(x). \end{align*}

Therefore, for large x, we obtain

\begin{align*} \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i} \boldsymbol{Y}_{i} > \boldsymbol{b}x \right) \geq (1-C\varepsilon) \sum_{i=1}^n \mathbb{P}\left(\boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{b}x\right). \end{align*}

This completes the proof.

Lemma 3. If the conditions of Theorem 1 hold, then for every fixed $\boldsymbol{b}\in[0,\infty]^d\backslash \{\textbf{0}\}$ we get

\begin{align*} \lim_{N\rightarrow \infty}\limsup_{x\rightarrow\infty} \frac{\mathbb{P}\left(\sum_{i=N+1}^{\infty}\boldsymbol{X}_i^+\boldsymbol{Y}_i >\boldsymbol{b}x\right)} {\bar{F}(x)} = \lim_{N\rightarrow \infty}\limsup_{x\rightarrow\infty} \frac{\sum_{i=N+1}^{\infty} \mathbb{P}\left(\boldsymbol{X}_i^+\boldsymbol{Y}_i > \boldsymbol{b}x\right)} {\bar{F}(x)}=0. \end{align*}

Proof. For any fixed $0<p<\beta$ and $0<\varepsilon<\beta-\alpha$ , we choose some $0<q<1$ satisfying

\begin{align*} \left\{\left[\mathbb{E}\left(\frac{\theta_k}{q}\right)^{\alpha-\varepsilon} \right]\, \bigvee \, \left[\mathbb{E}\left(\frac{\theta_k}{q}\right)^{\alpha+\varepsilon}\right] \right\}<1, \quad 1\leq k \leq d. \end{align*}

Then there exists some $n_1>0$ such that the following relation holds:

\begin{align*} \sum_{i=n_1+1}^{\infty} \left\{ \left[\mathbb{E}\left( \frac{\theta_{1}}{q} \right)^{\alpha-\varepsilon}\right]^{i-1} \bigvee \left[\mathbb{E}\left( \frac{\theta_{1}}{q} \right)^{\alpha+\varepsilon}\right]^{i-1} \right\} \leq C\varepsilon. \end{align*}

Since $X_{1i}^+\theta_{1i}$ is of regular variation from (8), applying [Reference Li20, Lemma 1] for large x we obtain

\begin{align*} &\mathbb{P}\left(\sum_{i=n_1+1}^{\infty}\boldsymbol{X}_i^+\boldsymbol{Y}_i >\boldsymbol{b}x\right) \leq \mathbb{P}\left(\sum_{i=n_1+1}^{\infty}\boldsymbol{X}_i^+\boldsymbol{Y}_i >\sum_{i=n_1+1}^{\infty}\boldsymbol{b}(1-q)q^{i-1}x\right)\\[5pt] &\leq \sum_{i=n_1+1}^{\infty} \mathbb{P}\left(\boldsymbol{X}_i^+\boldsymbol{\theta}_i\frac{\boldsymbol{Y}_{i-1}}{q^{i-1}} >\boldsymbol{b}(1-q)x\right) \leq \sum_{i=n_1+1}^{\infty} \mathbb{P}\left( X_{1i}^+\theta_{1i}\frac{Y_{1,i-1}}{q^{i-1}}>b_1(1-q)x \right)\\[5pt] &\leq C\bar{F}(b_1(1-q)x)\sum_{i=n_1+1}^{\infty} \mathbb{E}\left[ \left( \frac{Y_{1,i-1}}{q^{i-1}} \right)^{\alpha-\varepsilon} \bigvee \left( \frac{Y_{1,i-1}}{q^{i-1}} \right)^{\alpha+\varepsilon} \right]\\[5pt] &\leq C\bar{F}(x)\sum_{i=n_1+1}^{\infty} \left\{ \left[\mathbb{E}\left( \frac{\theta_{1}}{q} \right)^{\alpha-\varepsilon}\right]^{i-1} \bigvee \left[\mathbb{E}\left( \frac{\theta_{1}}{q} \right)^{\alpha+\varepsilon}\right]^{i-1} \right\}\\[5pt] &\leq C\varepsilon \bar{F}(x). \end{align*}

This completes the proof.

4.3. Proof of Theorem 1

Proof. First we show that for any $n_2$ sufficiently large, the relation (17) holds uniformly for $1\leq n\leq n_2$ . By Lemma 2, the following relations hold uniformly for $1\leq n\leq n_2$ :

\begin{align*} \Psi(x,n)= \mathbb{P}\left(\max_{1\leq m\leq n}\sum_{i=1}^m \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) &\leq \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right)\\[5pt] &\sim \sum_{i=1}^n \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) \end{align*}

and

\begin{align*} \Psi(x,n)= &\mathbb{P}\left(\max_{1\leq m\leq n}\sum_{i=1}^m \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right)\\[5pt] &\geq \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) \sim \sum_{i=1}^n \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right). \end{align*}

Next, we study the uniformity of the relation (17) when $n>n_2$ . Clearly,

(24) \begin{align} 0&< \nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{i=1}^{\infty} \mathbb{E}\left( \bigwedge_{k=1}^d Y_{k, i-1}^{\alpha} \right) =\sum_{i=1}^{\infty}\mathbb{E}\left( \nu \left( \bigvee_{k=1}^d Y_{k, i-1}^{-1}\left( \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right)\right)\nonumber\\[5pt] &\leq \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left( \boldsymbol{Y}_{i-1}^{-1}\, \boldsymbol{\rho}, \boldsymbol{\infty} \right] \right) \leq \sum_{i=1}^{\infty}\mathbb{E}\left( \nu \left( \bigwedge_{k=1}^d Y_{k, i-1}^{-1}\left(\boldsymbol{\rho}, \boldsymbol{\infty}\right] \right)\right)\\[5pt] \nonumber &\leq \nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{i=1}^{\infty} \left(\mathbb{E}Y_{1, i-1}^{\alpha} +\cdots+ \mathbb{E}Y_{d, i-1}^{\alpha}\right) =\nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{k=1}^{d} \sum_{i=1}^{\infty} \left( \mathbb{E}\theta_k^{\alpha} \right) ^{i-1}\\[5pt] &<\infty\,\nonumber. \end{align}

On the one hand, it holds uniformly for $n>n_2$ that, by Lemmas 2 and 3 and the relation (24),

\begin{align*} &\mathbb{P}\left(\max_{1\leq k\leq n}\sum_{i=1}^k \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) \geq \mathbb{P}\left(\max_{1\leq k\leq n_2}\sum_{i=1}^k \boldsymbol{X}_{i}\boldsymbol{Y}_{i} > \boldsymbol{\rho}x \right)\\[5pt] &\geq \mathbb{P}\left(\sum_{i=1}^{n_2} \boldsymbol{X}_{i}\boldsymbol{Y}_{i} > \boldsymbol{\rho}x \right) \sim \sum_{i=1}^{n_2} \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i} > \boldsymbol{\rho}x \right)\\[5pt] &\geq \left(\sum_{i=1}^{n}-\sum_{i=n_2+1}^{\infty}\right) \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}x \right)\\[5pt] &\gtrsim \sum_{i=1}^{n} \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}x \right). \end{align*}

On the other hand, it holds uniformly for $n>n_2$ that, for $v_3$ satisfying $(1-v_3)^{-\alpha}\leq 1+\varepsilon$ ,

\begin{align*} &\mathbb{P}\left(\max_{1\leq k\leq n}\sum_{i=1}^k \boldsymbol{X}_{i}\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right) \leq \mathbb{P}\left(\sum_{i=1}^n \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i} >\boldsymbol{\rho}x \right)\\[5pt] &\leq \mathbb{P}\left( \sum_{i=1}^{n_2} \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}(1-v_3)x \right) +2\sum_{k=1}^d \mathbb{P}\left( \sum_{i=n_2+1}^{n} X_{ki}^+Y_{ki}>\rho_kv_3x \right)\\[5pt] & \;:\!=\; I_1(x,n_2)+I_2(x,n). \end{align*}

For $I_1(x,n_2)$ , we have by Lemma 2 that

\begin{align*} I_1(x,n_2) \sim (1-v_3)^{-\alpha}\bar{F}(x)\sum_{i=1}^{n_2} \mathbb{E}\left\{\nu(\boldsymbol{Y}_{i-1}^{-1}\boldsymbol{\rho},\boldsymbol{\infty}] \right\} &\sim (1-v_3)^{-\alpha} \sum_{i=1}^{n_2} \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}x \right)\\[5pt] &\leq (1+\varepsilon)\sum_{i=1}^{n} \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}x \right). \end{align*}

For $I_2(x,n)$ , we obtain by Lemmas 2 and 3 and the relation (24) that

\begin{align*} I_2(x,n) \leq 2\sum_{k=1}^d \mathbb{P}\left( \sum_{i=n_2+1}^{\infty}X_{ki}^+Y_{ki} >\rho_kv_3x \right) = o(1)\bar{F}(x) = o(1)\sum_{i=1}^{n} \mathbb{P}\left( \boldsymbol{X}_{i}^+\boldsymbol{Y}_{i}>\boldsymbol{\rho}x \right). \end{align*}

This completes the proof.

5. The study of d-dimensional continuous-time risk model under Assumption 1

In this section we consider an insurance company with d lines of business, for $d\geq1$ . Let x be the initial reserve and let $\boldsymbol{\rho}$ be the allocation vector. For each $1\leq k\leq d$ , the price process of the investment portfolio in the kth business is expressed by a geometric Lévy process $\{ e^{L_k(t)} , \, t\geq0 \}$ ; namely, $\{ L_k(t), \, t\geq0 \}$ is a Lévy process that has stationary and independent increments, is stochastically continuous, and starts from 0. Then the insurer’s discounted risk process, $\mathbf{R}(t)=(R_1(t), \, \ldots, \, R_d(t))$ , is given by

(25) \begin{align} \mathbf{R}(t) =\left( \sum_{i=1}^{N(t)} \!\!A_{1i} e^{-L_1(\tau_i)}-c_1\int_0^t\!\! e^{-L_1(s)} \textrm{d}s, \, \ldots, \, \sum_{i=1}^{N(t)}\!\! A_{di} e^{-L_d(\tau_i)}-c_d\int_0^t\!\! e^{-L_d(s)} \textrm{d}s \right),\end{align}

with $t\geq0$ , where $A_{ki}$ , $1\leq k \leq d$ and $i\in\mathbb{N}$ , denotes the ith claim amount from the kth business, and $(c_1, \, \ldots, \, c_d)$ denotes the vector of constant premium rates. The successive claim arrival times are denoted by $0<\tau_1<\tau_2<\cdots$ , and the claim arrival process $\{ N(t); \, t\geq0 \}$ is a renewal process with the following finite renewal function:

\begin{align*} \lambda(t) = \mathbb{E}N(t) = \sum_{i=1}^{\infty} \mathbb{P}(\tau_i \leq t)\,.\end{align*}

The vectors $ (A_{1i}, \, \ldots, \, A_{di})$ , $i\in\mathbb{N}$ , form a sequence of i.i.d. nonnegative random vectors with a generic random vector $(A_1,\, \ldots, \, A_d)$ . The inter-arrival times are denoted by $\chi_1=\tau_1$ and $\chi_i=\tau_i-\tau_{i-1}$ for $i=2, \, 3, \, \ldots$ . The sequence $\{\chi_i, \, i\geq1\}$ is i.i.d. with generic random variable $\chi$ . We define the Laplace exponent for the Lévy process { $L_k(t), \, t\geq0$ }, $1\leq k\leq d$ , by the formula

(26) \begin{align} \phi_k(s)=\log \mathbb{E} e^{-sL_k(1)}, \quad s\in(-\infty,\infty).\end{align}

If $\phi_k(s)$ is finite, then $\mathbb{E}e^{-sL_k(t)}=e^{t\phi_k(s)}<\infty$ for $t\geq0$ .

In the multivariate risk model case, ruin may appear in various situations. Thus, there are several versions of the probabilities of ruin based on various ruin sets. For the model (25) we adopt the following three ruin times:

(27) \begin{align} T_{\textrm{max}} =\inf\{t>0: \boldsymbol{\rho} x - \boldsymbol{R}(t) <\mathbf{0} \},\end{align}

which is the first instant when all line reserves become negative simultaneously,

(28) \begin{align} T_{\textrm{min}}=\inf\left\{t>0: \min_{1\leq k\leq d} \{\rho_kx-R_k(t)\}<0\right\},\end{align}

which is the first instant when at least one of the businesses is below zero, and

(29) \begin{align} T_{\textrm{ult}} =\inf\left\{t>0: \inf_{0\leq s\leq t} \{\rho_1x-R_1(s)\}<0,\ldots, \inf_{0\leq s\leq t} \{\rho_dx-R_d(s)\}<0 \right\},\end{align}

which is the first instant at which all lines have at some point run into deficit, though not necessarily simultaneously. Here, $\inf \emptyset$ is understood as $\infty$ by convention. In this section we study the following three types of ruin probabilities:

(30) \begin{align} \Psi_{\#} ( x, \infty) =\mathbb{P}\left(T_{\#}< \infty|\mathbf{R}(0)=\boldsymbol{\rho}x \right), \qquad 0<t\leq\infty,\end{align}

where ‘ $\#$ ’ denotes either ‘max’, ‘min’, or ‘ult’. The ruin probability represents a significant indicator of the functional quality of the insurance company and has been studied extensively by [Reference Asmussen and Albrecher3], [Reference Ji13], [Reference Konstantinides and Li16], and [Reference Li19], and others.

Theorem 2. Consider the d-dimensional continuous-time risk model (25). Assume that the vector $\mathbf{A}_i e^{ \mathbf{L}(\tau_{i-1})-\mathbf{L}(\tau_i) } =\left( A_{1i}e^{L_1(\tau_{i-1})-L_1(\tau_i)}, \, \ldots, \, A_{di}e^{L_d(\tau_{i-1})-L_d(\tau_i)}\right)$ , $i\in\mathbb{N}$ , satisfies Assumption 1. If $\mathbb{E}L_k(1)<\infty$ for $1\leq k\leq d$ and there is a constant $\beta>\alpha$ such that the Laplace exponent of the Lévy process $\{ L_k(t), \ t\geq0 \}$ satisfies $\phi_k(\beta)<0$ , then the following asymptotic formulas hold:

(31) \begin{align} \Psi_{\boldsymbol{max}}(x, \, \infty) \sim \Psi_{\boldsymbol{ult}}(x, \, \infty) \sim \bar{F}(x)\sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{ \left( e^{\mathbf{L}(\tau_{i-1})}\boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\} \right), \end{align}

(32) \begin{align} \Psi_{\boldsymbol{min}}(x, \, \infty) \notag \\[5pt] &\sim \bar{F}(x) \sum_{k=1}^{d}(\!-\!1)^{k-1}\sum_{1\leq m_1<\cdots< m_k\leq d} \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left( e^{\mathbf{L}(\tau_{i-1})}\sum_{l=1}^k\rho_{m_l}\boldsymbol{e}_{m_l}, \boldsymbol{\infty} \right] \right\} \right). \end{align}

Remark 4. Suppose that $L_k(t)$ , $1\leq k\leq d$ , follow the jump-diffusion process

(33) \begin{align} L_k(t) = r_k t + \sigma_k W_k(t) + \sum_{i=1}^{N(t)} B_{ki}, \end{align}

where $r_k\in \mathbb{R}$ stands for the log return rate, $\sigma_k>0$ denotes the volatility, and N(t) represents the aforementioned claim arrival process. Here, $\{ W_k(t), \, t\geq0 \}$ is a Wiener process and $B_{ki}$ describes the jump sizes. Assume that for $1\leq k \leq d$ all the random sources $\{(A_{ki}, B_{ki}), i\geq1\}$ , $\{N(t), t\geq0\}$ , and $\{W_k(t), t\geq0\}$ are mutually independent and $(A_{1i}e^{-B_{1i}}, \, \ldots, \, A_{di}e^{-B_{di}})$ satisfies Assumption 1. We observe that $(A_{1i}e^{-B_{1i}}, \, \ldots, \, A_{di}e^{-B_{di}})$ and $(e^{-r_1\chi_i-\sigma_1W_1(\chi_i)},\,\ldots,\,e^{-r_d\chi_i-\sigma_dW_d(\chi_i)})$ are independent and

\begin{equation*} \mathbb{E}[e^{-\beta r_1\chi_i-\beta\sigma_1W_1(\chi_i)}]<\infty\,, \end{equation*}

for $\beta>\alpha$ . Furthermore,

\begin{align*} &\mathbb{P}\left(\mathbf{A}_ie^{\mathbf{L}(\tau_{i-1})-\mathbf{L}(\tau_{i})} >\mathbf{b}x\right)\\[5pt] &=\mathbb{P}\left(A_{1i}e^{-B_{1i}}e^{-r_1\chi_i-\sigma_1W_1(\chi_i)}>b_1x, \ldots, A_{di}e^{-B_{di}}e^{-r_d\chi_i-\sigma_dW_d(\chi_i)}>b_dx \right) \end{align*}

for any $\mathbf{b}\in[0,\infty]^d\setminus \{\mathbf{0}\}$ , which implies that the random vector $\boldsymbol{A}_i e^{\mathbf{L}(\tau_{i-1}) - \mathbf{L}(\tau_{i})}$ is MRV by Proposition 1.

5.1. Some lemmas before the proof of Theorem 2

Lemma 4. Let the conditions of Theorem 2 hold. Then for any $\mathbf{b}\in[0,\infty]^d\setminus \{\mathbf{0}\}$ and n we get

\begin{align*} \mathbb{P}\left\{ \sum_{i=1}^{n} \boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_{i})} > \boldsymbol{b}x \right\} \sim \sum_{i=1}^{n} \bar{F}(x) \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{b}, \boldsymbol{\infty}\right]\right\}\right). \end{align*}

Proof. From the conditions of Theorem 2, we obtain by Hölder’s inequality that for any fixed $i\in\mathbb{N}$ ,

(34) \begin{align} \mathbb{E}e^{-p\mathbf{L}(\tau_{i-1})} \leq \left( \mathbb{E}e^{-\beta \mathbf{L}(\tau_{i-1})}\right) ^{p/\beta} = \left( \mathbb{E}e^{\tau_{i-1}\boldsymbol{\phi}(\beta)}\right) ^{p/\beta} < \boldsymbol{ \infty}, \qquad p\leq \beta. \end{align}

Hence by Proposition 1 we find

(35) \begin{align}\notag \mathbb{P}\left(\!\boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_{i})} \!>\! \boldsymbol{b}x \!\right) &= \mathbb{P}\!\left\{ \! \frac{ \boldsymbol{A}_{i} e^{[\mathbf{L}(\tau_{i-1})-\mathbf{L}(\tau_{i})]}\cdot e^{-\mathbf{L}(\tau_{i-1})} } {x} \in (\boldsymbol{b},\boldsymbol{\infty}] \!\right\}\!\\[5pt] &\sim\! \bar{F}(x) \mathbb{E}\left( \nu \left\{\!\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{b}, \boldsymbol{\infty}\right]\right\}\!\right). \end{align}

Indeed, in Lemma 2, for each $i=1, \ldots, n$ , take $\boldsymbol{X}_{i}=\boldsymbol{A}_{i}$ ,

\begin{align*} \boldsymbol{\theta}_{i}=e^{[\mathbf{L}(\tau_{i-1})-\mathbf{L}(\tau_{i})]}\,, \qquad \boldsymbol{Y}_{i-1}=e^{-\mathbf{L}(\tau_{i-1})}\,; \end{align*}

then, applying Lemma 2 together with (35), we obtain

\begin{align*} \mathbb{P}\left\{ \sum_{i=1}^{n} \boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_{i})} > \boldsymbol{b}x \right\} \sim \sum_{i=1}^{n} \bar{F}(x) \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{b}, \boldsymbol{\infty}\right]\right\}\right). \end{align*}

The proof is complete.

Lemma 5. Let the conditions of Theorem 2 hold. Then for any $\mathbf{b}\in[0,\infty]^d\setminus \{\mathbf{0}\}$ we get

\begin{align*} \mathbb{P}\left\{ \sum_{i=1}^{\infty} \boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_{i})} > \mathbf{b} x \right\} \sim \bar{F}(x) \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\,\boldsymbol{b}, \boldsymbol{\infty}\right]\right\} \right). \end{align*}

Proof. Take $\boldsymbol{X}_{i}=\boldsymbol{A}_{i}$ and

\begin{align*} \boldsymbol{\theta}_{i}=e^{[\mathbf{L}(\tau_{i-1})-\mathbf{L}(\tau_{i})]}\,, \qquad \boldsymbol{Y}_{i-1}=e^{-\mathbf{L}(\tau_{i-1})}\,, \end{align*}

for each $i=1, \ldots, n$ . Using Theorem 1 with $n=\infty$ and (35), we obtain

\begin{align*} \mathbb{P}\left\{ \sum_{i=1}^{\infty} \boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_{i})} > \mathbf{b} x \right\} \sim \bar{F}(x)\sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\,\boldsymbol{b}, \boldsymbol{\infty}\right]\right\} \right). \end{align*}

This completes the proof.

The following lemma is significant in the literature on the uniform estimates for the ruin probability of insurance risk models with an exponential Lévy process investment return (see for example [Reference Li19, Lemma 4.8] or [Reference Tang, Wang and Yuen28, Lemma 4.6]).

Lemma 6. Let $Z_k$ , $1\leq k\leq d$ , be an exponential functional of the Lévy process $\{ L_k(t), \, t \geq 0 \}$ defined as

\begin{align*} Z_k=\int_0^{\infty} e^{-L_k(t)} \textrm{d}t. \end{align*}

If $\mathbb{E}L_k(1)<\infty$ , then for every $\beta>0$ satisfying $\phi_k(\beta)<0$ , we have $\mathbb{E}Z_k^{\beta}<\infty$ .

Proof. From [Reference Tang, Wang and Yuen28], the condition $\phi_k(\beta)<0$ implies that $\mathbb{E}L_k(1)>0$ . Since $0<\mathbb{E}L_k(1)<\infty$ , we can apply [Reference Maulik and Zwart23, Lemma 2.1] to complete the proof.

5.2. Proof of Theorem 2

Proof. From the conditions in Theorem 2 we may conclude that $\phi_k(\alpha)<0$ for $\alpha<\beta$ and $1\leq k\leq d$ . Then

(36) \begin{align} 0&< \nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{i=1}^{\infty} \mathbb{E}\left(\bigwedge_{k=1}^de^{-\alpha L_k(\tau_{i-1})}\right) =\sum_{i=1}^{\infty}\mathbb{E}\left( \nu \left( \bigvee_{k=1}^d e^{{L}_k(\tau_{i-1})}\left(\boldsymbol{\rho}, \boldsymbol{\infty}\right]\right)\right)\nonumber\\[5pt] &\leq \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right) \leq \sum_{i=1}^{\infty}\mathbb{E}\left( \nu \left( \bigwedge_{k=1}^d e^{{L}_k(\tau_{i-1})}\left(\boldsymbol{\rho}, \boldsymbol{\infty}\right]\right)\right)\nonumber\\[5pt] &\leq \nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{i=1}^{\infty} \mathbb{E}\left\{ e^{-\alpha L_1(\tau_{i-1})}+ \cdots + e^{-\alpha L_d(\tau_{i-1})} \right\}\nonumber\\[5pt] &=\nu \left(\boldsymbol{\rho},\boldsymbol{\infty}\right] \sum_{k=1}^{d} \sum_{i=1}^{\infty} \left[ \mathbb{E}e^{\chi\phi_k(\alpha)}\right] ^{i-1} <\infty. \end{align}

Since

\begin{align*}\sum_{i=1}^{n} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right) \uparrow\sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right)\quad \textrm{as } n\rightarrow\infty,\end{align*}

there exists some large $n_3$ such that, for any $\varepsilon>0$ ,

(37) \begin{align} \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right) -\sum_{i=1}^{n_3} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right] \right) <\varepsilon. \end{align}

We first focus on the relation (31) for $\Psi_{\textrm{max}}(x, \, \infty)$ . By (25), (27), and (30), we have

\begin{align*} \Psi_{\textrm{max}}(x, \, \infty) =\mathbb{P}\left(\bigcup_{0< s< \infty} \left\{\boldsymbol{\rho}x +\boldsymbol{c}\int_0^s e^{-\mathbf{L}(u)} \textrm{d}u -\sum_{i=1}^{\infty}\boldsymbol{A}_{i} e^{-\mathbf{L}(\tau_i)}\mathbb{I}_{[\tau_i\leq s]}<\textbf{0} \right\} \right). \end{align*}

Then, by Lemma 5, it holds that

(38) \begin{align} \Psi_{\textrm{max}}(x, \, \infty) \leq \mathbb{P}\left(\sum_{i=1}^{\infty} \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} > \boldsymbol{\rho}x\right) \sim \sum_{i=1}^{\infty} \bar{F}(x) \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\}\right). \end{align}

Let

\begin{align*}\mathbf{Z}=(Z_1, \ldots, Z_d) =\left( \int_0^{\infty} e^{-L_1(s)} \textrm{d}s, \ldots, \int_0^{\infty} e^{-L_d(s)} \textrm{d}s\right).\end{align*}

By Lemma 1, we have

(39) \begin{align} &\Psi_{\textrm{max}}(x, \, \infty) \geq \mathbb{P}\left\{\sum_{i=1}^{n_3} \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho}x +\boldsymbol{c}\int_0^{\infty} e^{-\mathbf{L}(s)} \textrm{d}s\right\}\nonumber\\[5pt] &\geq \mathbb{P}\left\{\bigcup_{i=1}^{n_3} \left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho}x+\boldsymbol{c}\mathbf{Z} \right) \right\}\nonumber\\[5pt] & \geq \sum_{i=1}^{n_3} \mathbb{P}\left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho}x+\boldsymbol{c} \boldsymbol{Z} \right) \quad - \sum_{1\leq i<j\leq n_3} \mathbb{P}\left( \boldsymbol{A}_ie^{-\mathbf{L}(\tau_{i})}>\boldsymbol{\rho}x , \boldsymbol{A}_je^{-\mathbf{L}(\tau_{j})}>\boldsymbol{\rho}x \right) \nonumber\\[5pt] & \;:\!=\; I_1(x)+o(1)\bar{F}(x). \end{align}

For $I_1(x)$ , choose $v_4$ satisfying

\begin{align*} (1+v_4)^{-\alpha}\geq(1-\varepsilon)\,. \end{align*}

By Lemma 4, $F\in\mathcal{R}_{-\alpha}$ , and (37), we have

\begin{align*} &\sum_{i=1}^{n_3} \mathbb{P}\left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho} (1+v_4) x\right) \sim (1+v_4)^{-\alpha} \bar{F}(x) \sum_{i=1}^{n_3} \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\}\right) \\[5pt] &\geq (1-C\varepsilon)\bar{F}(x) \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right) -C\varepsilon\bar{F}(x). \end{align*}

From Markov’s inequality, Lemma 6, and the relation (3), we obtain

(40) \begin{align} \sum_{k=1}^{d} \mathbb{P}\left( c_k Z_k > \rho_k v_4 x\right) \leq C x^{ -\beta } \sum_{k=1}^{d} \mathbb{E}Z_k^{\beta} \leq C\varepsilon \bar{F} (x). \end{align}

Then we find

(41) \begin{align} I_1(x) &\geq \sum_{i=1}^{n_3} \mathbb{P}\left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho}x+\boldsymbol{c} \boldsymbol{Z}, \boldsymbol{c} \boldsymbol{Z}\leq \boldsymbol{\rho} v_4 x \right) \nonumber\\[5pt] &\geq \sum_{i=1}^{n_3} \mathbb{P}\left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho} (1+v_4) x, \boldsymbol{c} \boldsymbol{Z}\leq \boldsymbol{\rho} v_4 x \right) \geq \sum_{i=1}^{n_3} \mathbb{P}\left( \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} >\boldsymbol{\rho} (1+v_4) x\right) \nonumber\\[5pt] &-\sum_{i=1}^{n_3} \mathbb{P}\left\lbrace \boldsymbol{A}_i e^{-\mathbf{L}(\tau_{i})} > \boldsymbol{\rho}x, \bigcup_{k=1}^d \left( c_k Z_k > \rho_k v_4 x\right) \right\rbrace \nonumber\\[5pt] &\geq (1-C\varepsilon)\bar{F}(x) \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\}\right). \end{align}

Combining the relations (38), (39), and (41) yields the relation (31) for $\Psi_{\textrm{max}}(x, \, \infty)$ .

Now we establish the relation (32) for $\Psi_{\textrm{min}}(x,\infty)$ . We obtain by the inclusion–exclusion principle and Lemma 5 that

(42) \begin{align} &\Psi_{\textrm{min}}(x, \, \infty)\nonumber\\[5pt] & \leq \mathbb{P}\left(\bigcup_{k=1}^d \left\{ \sum_{i=1}^{\infty}A_{ki}e^{-L_k(\tau_i)}>\rho_kx\right\}\right)\nonumber\\[5pt] &= \sum_{k=1}^d (\!-\!1)^{k-1} \sum_{1\leq m_1<\cdots< m_k\leq d} \mathbb{P}\left(\sum_{i=1}^{\infty}\boldsymbol{A}_{i}e^{-\mathbf{L}(\tau_i)} >\left(\sum_{l=1}^k\rho_{m_l}\boldsymbol{e}_{m_l}\right)x\right)\nonumber\\[5pt] &\sim \bar{F}(x) \sum_{k=1}^{d}(\!-\!1)^{k-1}\sum_{1\leq m_1<\cdots< m_k\leq d} \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left( e^{\mathbf{L}(\tau_{i-1})}\sum_{l=1}^k\rho_{m_l}\boldsymbol{e}_{m_l}, \boldsymbol{\infty} \right] \right\} \right). \end{align}

By the inclusion–exclusion principle and (40), we get

(43)

Combining (42) and (43) gives the relation (32).

To establish the relation (31) for $\Psi_{\textrm{ult}}(x,\infty)$ , we note that by Lemma 5,

(44) \begin{align} \Psi_{\textrm{ult}}(x,\infty) &= \mathbb{P}\left\{ \inf_{0< s< \infty} \left( \rho_1x+c_1\int_0^s e^{-L_1(u)} \textrm{d}u -\sum_{i=1}^{N(s)}A_{1i}e^{-L_1{(\tau_i)}} \right)<0,\ldots,\right.\nonumber\\[5pt] &\left.\qquad\quad \inf_{0< s< \infty}\left( \rho_dx+c_d\int_0^s e^{-L_d(u)} \textrm{d}u -\sum_{i=1}^{N(s)}A_{di}e^{-L_d{(\tau_i)}} \right)<0 \right\}\nonumber\\[5pt] &\leq \mathbb{P}\left\{ \sum_{i=1}^{\infty}A_{1i}e^{-L_1{(\tau_i)}}>\rho_1x ,\ldots, \sum_{i=1}^{\infty}A_{di}e^{-L_d{(\tau_i)}}>\rho_dx \right\} \nonumber\\[5pt] & = \mathbb{P}\left\{ \sum_{i=1}^{\infty} \boldsymbol{A}_{i}e^{-\mathbf{L}{(\tau_i)}}> \boldsymbol{\rho} x \right\}\sim \sum_{i=1}^{\infty} \bar{F}(x) \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\}\right). \end{align}

Analogously to the reasoning for the lower bound of $\Psi_{\textrm{max}}(x, \, \infty)$ , it follows that

(45) \begin{align} &\Psi_{\textrm{ult}}(x,\infty) =\mathbb{P}\left\{ \sup_{0<s<\infty} \boldsymbol{R}(t)>\boldsymbol{\rho}x \right\}\nonumber\\[5pt] &\geq \mathbb{P}\left\{ \sum_{i=1}^{\infty}A_{1i}e^{-L_1{(\tau_i)}} -c_1Z_1 > \rho_1x, \, \ldots, \, \sum_{i=1}^{\infty}A_{di}e^{-L_d{(\tau_i)}} -c_dZ_d >\rho_dx \right\}\nonumber\\[5pt] &\geq \mathbb{P}\left\{ \sum_{i=1}^{n_3} \boldsymbol{A}_{i}e^{-\mathbf{L}{(\tau_i)}} >\boldsymbol{\rho} x + \boldsymbol{c}\mathbf{Z} \right\}\nonumber\\[5pt] & \geq (1-C\varepsilon)\bar{F}(x) \sum_{i=1}^{\infty} \mathbb{E}\left( \nu \left\{\left(e^{\mathbf{L}(\tau_{i-1})}\, \boldsymbol{\rho}, \boldsymbol{\infty}\right]\right\}\right). \end{align}

Combining the relations (44) and (45) gives the relation (31) for $\Psi_{\textrm{ult}}(x, \, \infty)$ . This completes the proof.