Precise large deviations of aggregate claims in a size-dependent renewal risk model with stopping time claim-number process

Abstract

In this paper, we consider a size-dependent renewal risk model with stopping time claim-number process. In this model, we do not make any assumption on the dependence structure of claim sizes and inter-arrival times. We study large deviations of the aggregate amount of claims. For the subexponential heavy-tailed case, we obtain a precise large-deviation formula; our method substantially relies on a martingale for the structure of our models.

1 Introduction

Consider the following renewal risk model. Let $\{X_{k}, k\in \mathbb{N}\}$ and $\{\theta_{k}, k\in \mathbb{N}\}$ be claim sizes and inter-arrival times, respectively. Assume that $(X_{k},\theta_{k})$, $k\in \mathbb{N}$, form a sequence of independent and identically distributed (i.i.d.) copies of a generic random pair $(X,\theta)$ with marginal distribution functions $F=1-\overline{F}$ on $[0,\infty)$ and G on $[0,\infty)$, with dependent components X and θ. The claim arrival times are $\tau_{k}=\sum_{i=1}^{k}\theta_{i}, k\in \mathbb{N}$, with $\tau_{0}=0$. The number of claims is defined by

$$\begin{aligned} N_{t}^{*}=\inf_{k}\{k\in \mathbb{N}: \tau_{k}\geq t\}, \end{aligned}$$

(1.1)

then $N_{t}^{*}$ is a stopping time. In this way, the aggregate amount of claims over the $[0,t]$ is of the form

$$\begin{aligned} S_{t}^{*}=\sum_{k=1}^{N_{k}^{*}}X_{k}, \qquad S_{0}^{*}=X_{0}=0,\quad t\geq0. \end{aligned}$$

(1.2)

Note that $N_{t}^{*}=\sup_{k}\{k\in \mathbb{N}:\tau_{k}\leq t\}\triangleq N_{t}$ if $\tau_{k}=t$, whereas $N_{t}^{*}=N_{t}+1$ if $\tau_{k}\neq t$.

We study large deviations of $S_{t}^{*}$ in (1.2). We only consider the case of heavy-tailed claim-size distributions. One of the most important classes of heavy-tailed distributions is the class $\mathcal{S}$ of subexponential distributions. By definition, a distribution F on $[0,\infty)$ is subexponential if $\overline{F}(x)=1-F(x)$ for all $x\geq0$ and the relation

$$\begin{aligned} \lim_{x\rightarrow\infty}\frac{\overline {F^{*n}}(x)}{\overline{F}(x)}=n \end{aligned}$$

(1.3)

holds for all $n\geq2$, where $F^{*n}$ denotes the n-fold convolution of F. Clearly, (1.3) implies

$$\begin{aligned} \lim_{x\rightarrow\infty}\frac{P(X_{1}+X_{2}+\cdots +X_{n}>x)}{P(\max\{ X_{1},\ldots,X_{n}\}>x)}=1, \end{aligned}$$

(1.4)

where $X_{1},X_{2},\ldots$ is a sequence of i.i.d. r.v.’s with the common distribution function (d.f.) F. See Embrechts et al. [1] for a nice review of subexponential distributions in the context of insurance and finance.

Our main result is given now.

Theorem 1.1

Consider the aggregate amount of claims $S_{t}^{*}$ in (1.2), assume that $F\in\mathcal{S}$, $E[X]=\mu\in(0,\infty)$ and $E[\theta]=1/\lambda\in(0,\infty)$. Then, for arbitrarily given $\gamma>0$, we have uniformly for all $x\geq\gamma t$

$$\begin{aligned} P\bigl(S_{t}^{*}-\mu\lambda t >x\bigr)\thicksim\lambda t \overline{F}(x), \quad t\rightarrow\infty. \end{aligned}$$

(1.5)

A non-standard renewal risk model with dependent components X and θ, which was firstly proposed by Albrecher and Teugels [2] and further studied by Boudreault et al. [3], Cossette et al. [4], Badescu et al. [5], and among many others. Recently, Asimit and Badescu [6] introduced a general dependence structure for $(X,\theta)$, via the conditional tail probability of X given θ; see also Li et al. [7]. In particular, Chen and Yuen [8] considered that there is a random variable θ̃ such that

$$\begin{aligned} \operatorname{Pr}(\theta>t|X>x)\leq \operatorname{Pr}(\tilde{\theta}>t) \end{aligned}$$

(1.6)

holds for $t\geq0$ and large enough x, they studied the large deviations of the aggregate amount of $\mathcal{C}$ heavy-tailed claims, where $\mathcal{C}\subset\mathcal{S}$ (see Embrechts et al. [1]).

We now comment on the approaches used in this work. First, in Theorem 1.1, we do not make an assumption on the dependence structure of $(X,\theta)$. The existing results usually require a conditional tail probability of X given θ, e.g., Chen and Yuen [8] made the assumption (1.6), to say the least. Second, we extend the asymptotic behavior of the large deviations of $S_{t}^{*}$ to the case of $\mathcal{S}$ heavy-tailed claims. Finally, we construct a martingale to prove our result.

The rest of the paper is organized as follows. Section 2 recalls various preliminaries and prepares a few lemmas. Section 3 presents the proof of the main result. We end the paper with conclusions in Section 4.

2 Preliminaries

Throughout this paper, for two positive functions $a(\cdot)$ and $b(\cdot)$, we write $a(x)\lesssim b(x)$ if $\limsup_{x\rightarrow \infty}a(x)/b(x)\leq1$, write $a(x)\gtrsim b(x)$ if $\liminf_{x\rightarrow\infty}a(x)/b(x)\geq1$, and write $a(x)\sim b(x)$ if both. Very often we equip limit relationships with certain uniformity, which is crucial for our purpose. For instance, for two positive bivariate functions $a(\cdot,\cdot)$ and $b(\cdot,\cdot)$, we say that $a(\cdot,\cdot)\sim b(\cdot,\cdot)$ holds uniformly for $x\in\Delta\neq\emptyset$ if

$$\lim_{t\rightarrow\infty}\sup_{x\in\Delta } \biggl\vert \frac{a(x;t)}{b(x;t)}-1 \biggr\vert =0. $$

Clearly, the asymptotic relation $a(\cdot,\cdot)\sim b(\cdot,\cdot)$ holds uniformly for $x\in\Delta$ if and only if

$$\limsup_{t\rightarrow\infty}\sup_{x\in\Delta } \frac{a(x;t)}{b(x;t)} \leq1\quad \mbox{and}\quad \liminf_{t\rightarrow\infty}\inf _{x\in\Delta } \frac{a(x;t)}{b(x;t)}\geq1. $$

To obtain our desired results, we need to mention the following useful lemma.

Lemma 2.1

Consider the renewal counting process $N_{t}^{*}$ in (1.1). Under the assumption $E[\theta]=1/\lambda<\infty$. Then, for any $p\geq1$, we have

$$\begin{aligned} E\bigl[N_{t}^{*}\bigr]^{p}\sim(\lambda t)^{p}, \quad \textit{as }t\rightarrow\infty. \end{aligned}$$

(2.1)

Proof

First, for arbitrarily $\varepsilon>0$, by definition of $N^{*}$ and $E[\theta]=1/\lambda$, we have

$$\begin{aligned} \mathbb{P}\bigl\{ N^{*}_{t}>(\lambda+\varepsilon)t\bigr\} =&\mathbb{P}\Biggl\{ \sum_{j=1}^{(\lambda +\varepsilon)t}\theta_{j} \leq t\Biggr\} \\ =&\mathbb{P}\Biggl\{ \frac{1}{(\lambda+\varepsilon)t}\sum_{j=1}^{(\lambda +\varepsilon)t} \theta_{j}\leq\frac{1}{\lambda+\varepsilon}\Biggr\} \\ \rightarrow&0. \end{aligned}$$

(2.2)

Similarly,

$$\begin{aligned} \mathbb{P}\bigl\{ N^{*}_{t}< (\lambda-\varepsilon)t\bigr\} \rightarrow0. \end{aligned}$$

(2.3)

Combining (2.2) and (2.3), we have

$$\begin{aligned} \frac{N_{t}^{*}}{t}\stackrel{P}{\rightarrow}\lambda. \end{aligned}$$

(2.4)

It remains to show

$$\begin{aligned} \sup_{t\geq1}\mathbb{E}\biggl(\frac{N_{t}^{*}}{t} \biggr)^{p}< \infty,\quad \mbox{for } \forall p\geq1. \end{aligned}$$

Indeed, for any $\delta>0$,

$$\begin{aligned} \mathbb{E}\biggl(\frac{N_{t}^{*}}{t} \biggr)^{p} =&p \int_{0}^{\infty}u^{p-1}\mathbb{P}\bigl\{ N_{t}^{*}>ut\bigr\} \, du \\ =&p(\lambda+\delta)^{p} \int_{0}^{\infty}u^{p-1}\mathbb{P}\bigl\{ N_{t}^{*}>(\lambda +\delta )ut\bigr\} \,du \\ \leq&p(\lambda+\delta)^{p} \int_{0}^{\infty}u^{p-1}\mathbb{P}\Biggl\{ \sum _{j=1}^{(\lambda+\delta)ut}\theta_{j}\leq t\Biggr\} \,du. \end{aligned}$$

(2.5)

For the case when $u\geq1$ is an integer

$$\begin{aligned} \mathbb{P}\Biggl\{ \sum_{j=1}^{(\lambda+\delta)ut} \theta_{j}\leq t\Biggr\} \leq&\mathbb{P}\Biggl\{ \bigcap _{k=1}^{u} \Biggl\{ \sum_{j=(k-1)(\lambda+\delta)t+1}^{k(\lambda+\delta)t} \theta_{j}\leq t \Biggr\} \Biggr\} = \Biggl(\mathbb{P}\Biggl\{ \sum_{j=1}^{(\lambda+\delta)t}\theta _{j}\leq t \Biggr\} \Biggr)^{u}. \end{aligned}$$

By the law of large numbers

$$\mathbb{P}\Biggl\{ \sum_{j=1}^{(\lambda+\delta)t} \theta_{j}\leq t \Biggr\} \rightarrow0\quad (t\rightarrow\infty). $$

So we have the bound, for any constant $c>0$,

$$\begin{aligned} \mathbb{P}\Biggl\{ \sum_{j=1}^{(\lambda+\delta)ut} \theta_{j}\leq t\Biggr\} \leq e^{-cu}. \end{aligned}$$

(2.6)

Combining (2.5) and (2.6), uniformly for large u and t,

$$\begin{aligned} \mathbb{E}\biggl(\frac{N_{t}^{*}}{t} \biggr)^{p} \leq&p(\lambda+ \delta)^{p} \int _{0}^{\infty }u^{p-1}e^{-cu}\,du< \infty. \end{aligned}$$

(2.7)

By (2.4) and (2.7), we obtain Lemma 2.1. □

3 Proof of Theorem 1.1

By $F\in\mathcal{S}$ and (1.4), we need only to prove

$$\mathbb{P}\Bigl\{ \max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x \Bigr\} \sim \lambda t\overline{F}(x)\quad \mbox{for }t,x\rightarrow\infty. $$

Write $\xi_{k}=I_{\{X_{k}-\mu>x\}}$. First,

$$I_{\{\max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x\}}\leq\sum_{k=1}^{N_{t}^{*}} \xi_{k}. $$

By Wald’s equation

$$\mathbb{P}\Bigl\{ \max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x\Bigr\} \leq\bigl(\mathbb{E}N_{t}^{*}\bigr)\mathbb{P}\{X>x+\mu\}. $$

From Lemma 2.1 and $\overline{F}(x+\mu)\sim\overline{F}(x)$ (see class $\mathcal {S}\subset\mathcal{L}$ and definition of $\mathcal{L}$, Embrechts et al. [1]), we have

$$\begin{aligned} \mathbb{P}\Bigl\{ \max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x \Bigr\} \lesssim \lambda t\overline{F}(x)\quad \mbox{for }t,x\rightarrow\infty. \end{aligned}$$

(3.1)

On the other hand,

$$\begin{aligned} I_{\{\max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x\}}\geq\sum_{k=1}^{N_{t}^{*}} \xi_{k}-\sum_{1\leq j< k\leq N_{t}^{*}}\xi_{j} \xi_{k}. \end{aligned}$$

(3.2)

All we need is to bound the second term. Notice that

$$M_{n}\equiv\sum_{1\leq j< k\leq n}\xi_{j}( \xi_{k}-\mathbb{E}\xi_{k}),\quad n=1,2,\ldots $$

is a martingale. By Doob’s stopping rule,

$$\mathbb{E}M_{N_{t}^{*}}=\mathbb{E}M_{1}=0. $$

Or

$$\mathbb{E}\biggl( \sum_{1\leq j< k\leq N_{t}^{*}}\xi_{j}( \xi_{k}-\mathbb{E}\xi_{k}) \biggr)=0. $$

Hence,

$$\mathbb{E}\biggl( \sum_{1\leq j< k\leq N_{t}^{*}}\xi_{j} \xi_{k} \biggr)=\mathbb{E}\xi\cdot \mathbb{E}\biggl( \sum_{1\leq j< k\leq N_{t}^{*}} \xi_{j} \biggr)=\mathbb{E}\xi\cdot \mathbb{E}\Biggl( \sum_{j=1}^{ N_{t}^{*}-1} \xi_{j}\bigl(N_{t}^{*}-j\bigr) \Biggr). $$

Write $Z_{n}=\sum_{j=1}^{n}\xi_{j}$. Notice that

$$\sum_{j=1}^{ N_{t}^{*}-1}\xi_{j} \bigl(N_{t}^{*}-j\bigr)=\sum_{n=1}^{N_{t}^{*}-1}Z_{n}. $$

Therefore,

$$\begin{aligned} \mathbb{E}\Biggl(\sum_{j=1}^{ N_{t}^{*}-1} \xi_{j}\bigl(N_{t}^{*}-j\bigr) \Biggr) =&\mathbb{E}\Biggl(\sum _{n=1}^{N_{t}^{*}-1}Z_{n} \Biggr) \\ =&\sum_{n=1}^{\infty} \mathbb{E}(Z_{n} I_{\{N_{t}^{*}\geq n+1\}}) \\ =&\sum_{n=1}^{\infty} \mathbb{E}(Z_{n} I_{\{\tau_{n+1}\leq t\}}). \end{aligned}$$

(3.3)

For each n,

$$\begin{aligned} \mathbb{E}(Z_{n} I_{\{\tau_{n+1}\leq t\}}) =&\sum _{j=1}^{n}\mathbb{E}(\xi_{j} \cdot I_{\{\tau_{n+1}\leq t\}}) \\ \leq&\sum_{j=1}^{n}\mathbb{E}(\xi_{j} \cdot I_{\{\sum_{k\neq j}^{n+1}\theta_{k}\leq t\} } ) \\ =&\sum_{j=1}^{n}\mathbb{E}\xi \cdot \mathbb{P}\Biggl\{ \sum _{k=1}^{n}\theta_{k}\leq t\Biggr\} =n\mathbb{E}\xi\cdot \mathbb{P}\bigl\{ N_{t}^{*}\geq n\bigr\} . \end{aligned}$$

(3.4)

By (3.3) and (3.4), $\exists c>0$ such that

$$\mathbb{E}\Biggl(\sum_{j=1}^{ N_{t}^{*}-1}\xi_{j} \bigl(N_{t}^{*}-j\bigr) \Biggr)\leq \mathbb{E}\xi\cdot\sum _{n=1}^{\infty}n \mathbb{P}\bigl\{ N_{t}^{*}\geq n\bigr\} \leq c\cdot \mathbb{E}\xi\cdot \mathbb{E}\bigl(N_{t}^{*}\bigr)^{2}. $$

Combining our computation and Lemma 2.1,

$$\mathbb{E}\biggl( \sum_{1\leq j< k\leq N_{t}^{*}}\xi_{j} \xi_{k} \biggr)\leq c\cdot(\mathbb{E}\xi)^{2}\cdot \mathbb{E}\bigl(N_{t}^{*}\bigr)^{2}\leq c\bigl(\lambda t\cdot \overline{F}(x)\bigr)^{2}=o\bigl(\lambda t\overline{F}(x)\bigr). $$

By (3.2), therefore,

$$\begin{aligned} \mathbb{P}\Bigl\{ \max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x \Bigr\} \gtrsim\lambda t\overline{F}(x)\quad \mbox{for }t,x\rightarrow\infty. \end{aligned}$$

(3.5)

Hence, by (3.1) and (3.5), we have

$$\mathbb{P}\Bigl\{ \max_{k\leq N_{t}^{*}}(X_{k}-\mu)>x \Bigr\} \sim \lambda t\overline{F}(x)\quad \mbox{for }t,x\rightarrow\infty. $$

4 Conclusions

As was remarked by a few researchers in the area, precise large-deviation results of size-dependent renewal risk models are particularly useful for evaluating some risk measures such as the conditional tail expectation of the aggregate amount of claims from a large insurance portfolio. Finally, we would like to point out that equation (1.5) agrees with existing ones in the literature. This indicates that the aggregate amount of $S_{t}^{*}$ defined by (1.1) does not affect the asymptotic behavior of the large deviations.