Parisian ruin of Brownian motion risk model over an infinite-time horizon

TL;DR

This paper derives the exact asymptotics for the probability and timing of Parisian ruin over an infinite horizon in a Brownian motion risk model with exponential discounting and interest, providing new insights into risk process behavior.

Contribution

It introduces the first precise asymptotic analysis of Parisian ruin probabilities and times for a Brownian motion-based risk model with interest and discounting.

Findings

Exact asymptotics of Parisian ruin probability derived

Asymptotic behavior of Parisian ruin time characterized

Results applicable to risk models with interest and volatility

Abstract

Let $B(t), t\in \mathbb{R}$ be a standard Brownian motion. In this paper, we derive the exact asymptotics of the probability of Parisian ruin on infinite time horizon for the following risk process \begin{align}\label{Rudef} R_u^{\delta}(t)=e^{\delta t}\left(u+c\int^{t}_{0}e^{-\delta v}d v-\sigma\int_{0}^{t}e^{-\delta v}d B(v)\right),\quad t\geq0, \end{align} where $u\geq 0$ is the initial reserve, $\delta\geq0$ is the force of interest, $c>0$ is the rate of premium and $\sigma>0$ is a volatility factor. Further, we show the asymptotics of the Parisian ruin time of this risk process.

Equations

$$R_u^{\delta }(t)=e^{\delta t}\left(u+c{\int }_0^t{e}^{-\delta v}dv-\sigma {\int }_0^t{e}^{-\delta v}dB(v)\right),t\ge 0,$$

$$R_u(t)=u+ct-X(t),t\ge 0,$$

$${\mathcal{P}}_S(u,T_u)=\mathbb{P}\left\{\underset{t\in [0,S]}{\inf }\underset{s\in [t,t+T_u]}{\sup }{R}_u(s)<0\right\},S\in (0,\infty ],$$

$${\mathcal{K}}_S^{\delta }(u,T_u):=\mathbb{P}\left\{\underset{t\in [0,S]}{\inf }\underset{s\in [t,t+T_u]}{\sup }{R}_u^{\delta }(s)<0\right\},S\in (0,\infty ),$$

$${\mathcal{K}}^{\delta }(u,T_u):=\mathbb{P}\left\{\underset{t\ge 0}{\inf }\underset{s\in [t,t+T_u]}{\sup }{R}_u^{\delta }(s)<0\right\},$$

$$\underset{u\to \infty }{\lim }{T}_u=T\in [0,\infty ].$$

$${\mathcal{K}}^0(u,T_u)=\mathbb{P}\left\{\underset{t\ge 0}{\inf }\underset{s\in [t,t+T_u]}{\sup }\left(u+cs-\sigma B(s)\right)<0\right\}\sim \mathcal{F}\left(\frac{2c^{2}T}{{\sigma }^2}\right)\exp \left(-\frac{2cu}{{\sigma }^2}\right),u\to \infty ,$$

$$\mathcal{F}\left(T\right)=\underset{\lambda \to \infty }{\lim }\frac{1}{\lambda }\mathbb{E}\left\{\underset{t\in [0,\lambda ]}{\sup }\underset{s\in [0,T]}{\inf }{e}^{\sqrt{2}B(t+s)-(t+s)}\right\}.$$

$$\tau (u)=\inf \{t>0:R_u^{\delta }(t)<0\}.$$

$$\eta (u)=\inf \{t\ge T_u:t-{\kappa }_{t,u}\ge T_u,R_u^{\delta }(t)<0\},\text{with}{\kappa }_{t,u}=\sup \{s\in [0,t]:R_u^{\delta }(s)\ge 0\},$$

$${\tilde{\mathcal{P}}}_a^f[0,\infty )=\underset{\lambda \to \infty }{\lim }{\tilde{\mathcal{P}}}_a^f[0,\lambda ]\in (0,\infty ),$$

$${\tilde{\mathcal{P}}}_a^f[0,\lambda ]=\mathbb{E}\left\{\underset{t\in [0,\lambda ]}{\sup }\underset{s\in [a,1]}{\inf }\exp \left(\sqrt{2}B(st)-st-f(st)\right)\right\}\in (0,\infty ),$$

$${\tilde{\mathcal{P}}}_1^f[0,\lambda ]=\mathbb{E}\left\{\underset{t\in [0,\lambda ]}{\sup }\exp \left(\sqrt{2}B(t)-\left|t\right|-f(t)\right)\right\},$$

$${\mathcal{K}}^{\delta }(u,T_u)$$

$$={\tilde{\mathcal{P}}}_a^f[0,\infty )\Psi \left(\frac{1}{\sigma }\sqrt{2\delta u^2+4cu}\right),u\to \infty ,$$

$${\mathcal{K}}^{\delta }(u,0)=\mathbb{P}\left\{\underset{t\ge 0}{\inf }{R}_u^{\delta }(s)<0\right\}\sim \mathbb{E}\left\{\underset{t\in [0,\infty )}{\sup }\exp \left(\sqrt{2}B(t)-t-{\left(\sqrt{t}-\frac{c}{\sigma \sqrt{\delta }}\right)}^2\right)\right\}\Psi \left(\frac{1}{\sigma }\sqrt{2\delta u^2+4cu}\right)$$

$$\displaystyle\mathcal{K}^{\delta}(u,0)=\Psi\left(\frac{\sqrt{2\delta}}{\sigma}\left(u+\frac{c}{\delta}\right)\right)\Big{/}\Psi\left(\frac{\sqrt{2}c}{\sigma\sqrt{\delta}}\right).$$

$$\displaystyle\mathbb{P}\left\{u^{2}\left(e^{-2\delta\eta(u)}-\left(\frac{c}{\delta u+c}\right)^{2}\right)\leq x\ \big{\lvert}\eta(u)<\infty\right\}\sim\frac{\widetilde{\mathcal{P}}^{f}_{a}[0,\frac{c^{2}}{\sigma^{2}\delta}+\frac{\delta x}{\sigma^{2}}]}{\widetilde{\mathcal{P}}^{f}_{a}[0,\infty)},\ u\rightarrow\infty.$$

$$\displaystyle\mathbb{P}\left\{u^{-\frac{1}{2}}\left(\eta(u)-\frac{u}{c}\right)\leq x\ \big{\lvert}\eta(u)<\infty\right\}\sim\Phi(cx),\ u\rightarrow\infty.$$

$$\displaystyle\mathbb{P}\left\{u^{2}\left(e^{-2\delta\tau(u)}-\left(\frac{c}{\delta u+c}\right)^{2}\right)\leq x\ \big{\lvert}\eta(u)<\infty\right\}\sim\frac{\widetilde{\mathcal{P}}^{f}_{1}[0,\frac{c^{2}}{\sigma^{2}\delta}+\frac{\delta x}{\sigma^{2}}]}{\widetilde{\mathcal{P}}^{f}_{1}[0,\infty)},\ u\rightarrow\infty,$$

$${\mathcal{K}}^{\delta }(u,T_u)=\mathbb{P}\left\{\underset{t\in [0,\infty )}{\inf }\underset{s\in [t,t+T_u]}{\sup }{R}_u^{\delta }(s)<0\right\}=\mathbb{P}\left\{\underset{t\in [0,\infty )}{\inf }\underset{s\in [t,t+T_u]}{\sup }{\tilde{R}}_u^{\delta }(s)<0\right\},$$

$${\tilde{R}}_u^{\delta }(s)=u+c{\int }_0^s{e}^{-\delta v}dv-\sigma {\int }_0^s{e}^{-\delta v}dB(v),t\ge 0.$$

$$\mathbb{E}\left\{{\left[\sigma {\int }_0^t{e}^{-\delta v}dB(v)\right]}^2\right\}=\frac{{\sigma }^2}{2\delta }\left(1-e^{-2\delta t}\right),$$

$$\underset{t\in [0,\infty )}{\sup }\mathbb{E}\left\{{\left[\sigma {\int }_0^t{e}^{-\delta v}dB(v)\right]}^2\right\}<\infty$$

$$\underset{t\in [0,\infty )}{\sup }\mathbb{E}\left\{\left|\sigma {\int }_0^t{e}^{-\delta v}dB(v)\right|\right\}<\infty ,$$

$$\psi (u):=$$

$$=$$

$$\psi (u)$$

$$=\mathbb{P}\left\{\underset{t^{\ast }\in [0,1]}{\sup }\underset{s^{\ast }\in [t^{\ast }{e}^{-2\delta T_u},t^{\ast }]}{\inf }\left(\sigma {\int }_0^{-\frac{1}{2\delta }\ln s^{\ast }}{e}^{-\delta v}dB(v)-\frac{c}{\delta }(1-{s^{\ast }}^{\frac{1}{2}})\right)>u\right\}.$$

$$V_Z^2(s)=Var\left(\sigma {\int }_0^{-\frac{1}{2\delta }\ln s}{e}^{-\delta v}dB(v)\right)=\frac{{\sigma }^2}{2\delta }(1-s),s\in [0,1].$$

$$M_u(t):=\frac{u{V}_Z(t)}{G_u(t)}=\frac{\frac{\sigma }{\sqrt{2\delta }}\sqrt{1-t}}{1+\frac{c}{\delta u}(1-t^{1/2})},0\le t\le 1,$$

$$t_u={\left(\frac{c}{\delta u+c}\right)}^2.$$

Full text

Parisian Ruin of Brownian Motion Risk Model over an Infinite-time Horizon

Long Bai

Long Bai, Department of Actuarial Science, University of Lausanne, UNIL-Dorigny, 1015 Lausanne, Switzerland

Long.Bai@unil.ch

Abstract: Let $B(t),t\in\mathbb{R}$ be a standard Brownian motion. In this paper, we derive the exact asymptotics of the probability of Parisian ruin on infinite time horizon for the following risk process

[TABLE]

where $u\geq 0$ is the initial reserve, $\delta\geq 0$ is the force of interest, $c>0$ is the rate of premium and $\sigma>0$ is a volatility factor. Further, we show the asymptotics of the Parisian ruin time of this risk process.

Key Words: Parisian ruin; ruin probability; ruin time; Brownian motion

AMS Classification: Primary 60G15; secondary 60G70

1. Introduction

In the risk theory, the surplus process of an insurance company can be modeled by

[TABLE]

see [11], where $u\geq 0$ is the initial reserve, $ct$ models the total premium received up to time $t$, and $X(t),t\geq 0$ denotes the aggregate claims process. In [7, 8], the Parisian ruin of $R_{u}(t)$ is defined by

[TABLE]

where $T_{u}$ models the pre-specified time which is a function of $u$. For $X(t),t\geq 0$ a Gaussian process, the asymptotics of $\mathcal{P}_{S}(u,T_{u})$ over finite-time horizon, i.e. $S\in(0,\infty)$, is investigated in [8]. Further, [7] showed the tail asymptotic results of $R_{u}(t)$ over infinite-time horizon, i.e. $S=\infty$ in (1.1), where $X(t)$ is a self-similar Gaussian process. In this paper considering the nature of the financial market, we introduce the force of interest $\delta$ into the model $R_{u}(t)$ as $R^{\delta}_{u}(t)$ in (0.1) when $X(t)=B(t)$. [4] gave an approximation of the Parisian ruin probability

[TABLE]

as $u\rightarrow\infty$. See [19, 6, 15] for more studies on risk models with force of interest. In the literature, no results are available for the approximation of Parisian ruin probability over infinite time horizon for $\delta>0$. In this contribution we shall investigate the asymptotics of the Parisian ruin probability

[TABLE]

as $u\rightarrow\infty$ where $T_{u}\geq 0$ models the pre-specified time satisfying

[TABLE]

When $\delta=0$ and $T\in[0,\infty)$, [7] showed that (hereafter $\sim$ means asymptotic equivalence)

[TABLE]

where

[TABLE]

Hereafter we make the convention that $\sup{\{\emptyset\}}=0$ and $\inf{\{\emptyset\}}=\infty$.

Complementary, we investigate the conditional distribution of the ruin time for the surplus process $R_{u}^{\delta}(t)$. The classical ruin time, e.g., [6, 13, 16], is defined as

[TABLE]

Here as in [7, 4] we define the Parisian ruin time of the risk process $R_{u}^{\delta}(t)$ by

[TABLE]

and $\tau(u)=\eta(u)$ when $T_{u}\equiv 0$.

Brief outline of the rest of the paper: In Section 2 we present our main results on the asymptotics of $\mathcal{K}^{\delta}(u,T_{u})$ as $u\rightarrow\infty$ and the approximation of the Parisian ruin time. All the proofs are relegated to Section 3.

2. Main results

Before giving the main results, we shall introduce a constant as

[TABLE]

with

[TABLE]

where $\lambda\geq 0,a\in[0,1]$ and $f(t)$ is a continuous function satisfying $\lim_{t\rightarrow\infty}\frac{f(t)}{t^{\epsilon}}=\infty$ for some $\epsilon>0$.

Note further that $\widetilde{\mathcal{P}}^{f}_{0}[0,\lambda]=e^{-f(0)}$ and

[TABLE]

see e.g. [9, 3, 14] for the bounds of $\widetilde{\mathcal{P}}^{f}_{a}[0,\infty)$ and more details.

Recall that $\Phi(\cdot),\Psi(\cdot)$ denote the distribution function and the survival function of an $\mathcal{N}(0,1)$ random variable, respectively, and $\Psi(u)\sim\frac{1}{\sqrt{2\pi}u}e^{-\frac{u^{2}}{2}},u\rightarrow\infty$.

Theorem 2.1**.**

For $\delta>0$ and $T_{u}$ satisfying (1.2), we have

[TABLE]

where $a=e^{-2\delta T}$ and $f(t)=\left(\sqrt{t}-\frac{c}{\sigma\sqrt{\delta}}\right)^{2}$.

Remark 2.2**.**

In Theorem 2.1, if $T=0$, $a=1$,we get the asymptotic result of the classical ruin probability, i.e., as $u\rightarrow\infty$

[TABLE]

*which corresponds to the results in [2].

Moreover, according to [12] (see also [10]) we have*

[TABLE]

Theorem 2.3**.**

Let $\eta(u)$ satisfy (1.4), under the assumptions and notation of Theorem 2.1, we have for $\delta>0$ and $x\in\left(-\frac{c^{2}}{\delta^{2}},\infty\right)$

[TABLE]

Remarks 2.4**.**

i) When $\delta=0$, [7] showed that for $x\in\mathbb{R}$

[TABLE]

ii) When $T_{u}\equiv 0$, $\eta(u)=\tau(u)$, by (2.3), we have

[TABLE]

which corresponds to the result in [2].

3. Proofs

Hereafter we assume that $\mathbb{C}_{i},i\in\mathbb{N}$ are positive constants.

Proof of Theorem 2.1 We have for $u>0$

[TABLE]

where

[TABLE]

Since for $t\in(0,\infty)$

[TABLE]

then

[TABLE]

implies that

[TABLE]

by the martingale convergence theorem, see [17], $\widetilde{R}_{u}^{\delta}(\infty):=\lim_{t\rightarrow\infty}\widetilde{R}_{u}^{\delta}(t)$ exists and is finite almost surely. Thus for any $u>0$

[TABLE]

Using a change of variable $s=-\frac{1}{2\delta}\ln s^{*},s^{*}\in[t^{*}e^{-2\delta T_{u}},t^{*}],\ t^{*}\in[0,1]$, we have

[TABLE]

For simplicity, we still use $s,t$ instead of $s^{*},t^{*}$.

Below, we set $Z(s)=\sigma\int_{0}^{-\frac{1}{2\delta}\ln s}e^{-\delta v}dB(v)$ with variance function given by

[TABLE]

We show next that for $u$ sufficiently large

[TABLE]

with $G_{u}(t):=u+\frac{c}{\delta}(1-t^{\frac{1}{2}})$ attains its maximum at the unique point

[TABLE]

In fact, we have for $t\in(0,1)$

[TABLE]

Letting $[M_{u}(t)]_{t}=0$, we get $t_{u}=\left(\frac{c}{\delta u+c}\right)^{2}$.

By (3.1), $[M_{u}(t)]_{t}>0$ for $t\in(0,t_{u})$ and $[M_{u}(t)]_{t}<0$ for $t\in(t_{u},1)$, so $t_{u}$ is the unique maximum point of $M_{u}(t)$ over $[0,1]$. Further

[TABLE]

Set $\delta(u)=\left(\frac{\ln u}{u}\right)^{2}$, $\Delta(u)=[0,t_{u}+\delta(u)]$ and for some positive constant $\lambda$

[TABLE]

We have for $u$ large enough

[TABLE]

where for $\theta\in(0,1)$

[TABLE]

First we show the asymptotic of $\Pi_{0}(u)$. For $u$ large enough

[TABLE]

where $\overline{Z}(t)=\frac{Z(t)}{V_{Z}(t)}$, $a=e^{-2\delta T}$, $\varepsilon_{1}\in(0,1)$ and $\varepsilon_{2}\in\left(0,(\frac{1}{a}-1)\wedge 1\right)$ if $T\in(0,\infty]$, $\varepsilon_{2}=0$ if $T=0$ . Similarly,

[TABLE]

We have

[TABLE]

and

[TABLE]

Since for any $t\in\Delta(u)$

[TABLE]

then for all large $u$

[TABLE]

and

[TABLE]

Consequently, we have

[TABLE]

For $0\leq t^{\prime}\leq t<1$, the correlation function of $Z(t)$ equals

[TABLE]

which implies that

[TABLE]

For $t,t^{\prime}\in\left[0,(1+\varepsilon_{1})\left(\frac{c^{2}}{\delta^{2}}+\lambda\right)\right]$ and $s,s^{\prime}\in(0,1]$

[TABLE]

where

[TABLE]

Since for $t,t^{\prime}\in\left[0,(1+\varepsilon_{1})\left(\frac{c^{2}}{\delta^{2}}+\lambda\right)\right]$ and $s,s^{\prime}\in(0,1]$

[TABLE]

we obtain

[TABLE]

For some small $\theta\in(0,1)$, by (3.5) we obtain that for $t,t^{\prime}\in[0,\theta]$

[TABLE]

holds. By (3.4), (3.6),(3.7), (3.8) and Lemma 5.1 in [8], as $u\rightarrow\infty$,

[TABLE]

and

[TABLE]

Letting $\varepsilon_{1},\varepsilon_{2}\rightarrow 0$, we have

[TABLE]

Next we show that

[TABLE]

Let $Y(t),t\in\mathbb{R}$ be a stationary Gaussian process with continuous trajectories, unit variance and correlation function satisfying for a constant $\varepsilon_{3}\in(0,\frac{1}{2})$

[TABLE]

By (3.4) and Slepian inequality in [18], we have

[TABLE]

where $\mathcal{A}_{u}(k):=\frac{u}{M_{u}}\left(1+\frac{1-\varepsilon_{4}}{2u^{2}}(\sqrt{u^{2}t_{u}+k\lambda}-c/\delta)^{2}-\frac{\varepsilon_{4}}{u^{2}}\right)$ and $\varepsilon_{4}\in(0,1)$ is a small constant. We observe that

[TABLE]

Further,

[TABLE]

and

[TABLE]

According to (3.10), (3), (3) and Lemma 5.3 of [5], we have as $u\rightarrow\infty,\varepsilon_{4}\rightarrow 0,\ \lambda\rightarrow\infty$

[TABLE]

Moreover, for all $u$ large

[TABLE]

holds for any $t\in\left[t_{u}+\delta(u),\theta\right]$, therefore

[TABLE]

Thus the above inequality combined with (3.8) and Theorem 8.1 in [18] derives that

[TABLE]

Finally, since

[TABLE]

by Borell inequality in [1]

[TABLE]

which combined with (3.2), (3.3), (3.9), (3.13) and (3.15) shows that

[TABLE]

Consequently, letting $\lambda\rightarrow\infty$, we have

[TABLE]

$\Box$

Proof of Theorem 2.3 We use the same notation as in the proof of Theorem 2.1. For $x\in\left(-\frac{c^{2}}{\delta^{2}},\infty\right)$ and $u>0$

[TABLE]

where $\tau^{*}_{u}=\{t\geq 0:\sigma\int_{0}^{-\frac{1}{2\delta}\ln t^{*}}e^{-\delta v}dB(v)-\frac{c}{\delta}(1-{t^{*}}^{\frac{1}{2}})>u\}$.

For $\psi_{x}(u):=\mathbb{P}\left\{\sup_{t^{*}\in[0,t_{u}+u^{-2}x]}\inf_{s^{*}\in[t^{*}e^{-2\delta T_{u}},t^{*}]}\left(\sigma\int_{0}^{-\frac{1}{2\delta}\ln s^{*}}e^{-\delta v}dB(v)-c\int_{0}^{-\frac{1}{2\delta}\ln s^{*}}e^{-\delta v}dv\right)>u\right\}$, using the similar argumentation about $\Pi_{0}(u)$ as in the proof of Theorem 2.1 with $\lambda=x$, we obtain

[TABLE]

Thus

[TABLE]

$\Box$

Acknowledgement: Thanks to Swiss National Science Foundation Grant no. 200021-166274.