Ruin probabilities for a regenerative Poisson gap generated risk process

Abstract

A risk process with constant premium rate c and Poisson arrivals of claims is considered. A threshold r is defined for claim interarrival times, such that if k consecutive interarrival times are larger than r, then the next claim has distribution G. Otherwise, the claim size distribution is F. Asymptotic expressions for the infinite horizon ruin probabilities are given for both light- and the heavy-tailed cases. A basic observation is that the process regenerates at each G-claim. Also an approach via Markov additive processes is outlined, and heuristics are given for the distribution of the time to ruin.

Appendix

1.1 1. Regenerative calculations for k > 2

Now, let us prove that for k ≥ 2,

$$ \mu^{(k)} = \frac{{\hbox{e}}^{\lambda r}}{\lambda} \frac{{\hbox{e}}^{\lambda k r}-1}{{\hbox{e}}^{\lambda r}-1}, $$

(36)

where μ^(k) is the mean of the cycle length in the k-case, which we denote a k-cycle. For k = 2, we can easily check that

$$ \frac{1}{\lambda} {\hbox{e}}^{\lambda r}(1+{\hbox{e}}^{\lambda r}) = \frac{{ \hbox{e}}^{\lambda r}}{\lambda} \frac{{\hbox{e}}^{\lambda 2 r}-1}{{\hbox{e}}^{\lambda r}-1}. $$

We assume now the result holds for some k ≥ 2. After a k-cycle, we have two possibilities. The next interarrival time is long and we get a (k + 1)-cycle. This gives an additional contribution to the mean of a (k + 1)-cycle length of p ₊ m ₊. The second possibility is that the next interarrival time is short and the (k + 1)-cycle starts afresh after that. Putting things together gives

$$ \mu^{(k+1)} = \mu^{(k)} + p_+m_+ + p_-(m_-+\mu^{(k+1)}) $$

so that

$$ \begin{aligned} \mu^{(k+1)} &\; =\; \frac{1}{p_+} \left[ \mu^{(k)} + p_+m_+ + p_-m_- \right] \; =\; {\hbox{e}}^{\lambda r} \left[ \frac{{\hbox{e}}^{\lambda r}}{\lambda} \frac{{\hbox{e}}^{\lambda k r}-1}{{\hbox{e}}^{\lambda r}-1} + \frac{1}{\lambda} \right],\\ &\; =\; \frac{{\hbox{e}}^{\lambda r}}{\lambda} \frac{{\hbox{e}}^{\lambda (k+1) r}-1}{{\hbox{e}}^{\lambda r}-1}, \end{aligned} $$

which ends the proof.

1.2 2. Light-tailed calculations for k > 2

Here, let us prove that, for k > 2,

$$ \varphi^{(k)}(\alpha,\beta) = \frac{g(\alpha,\beta)^k\widehat{F} [\alpha]^{k-1}\widehat{G}[\alpha]}{1-k(\alpha,\beta)\widehat{F}[\alpha] [1-(g(\alpha,\beta)\widehat{F}[\alpha])^k ]/[1-g(\alpha,\beta)\widehat{F}[\alpha]]}, $$

where

$$ \begin{aligned} g(\alpha,\beta) &\; =\; \frac{\lambda}{\lambda+c\alpha-\beta} {\hbox{e}}^{(\beta-c\alpha-\lambda)r},\\ k(\alpha,\beta) &\; =\; \frac{\lambda}{\lambda+c\alpha-\beta} (1-{\hbox{e}}^{(\beta-c\alpha-\lambda)r}), \end{aligned} $$

and ${\varphi^{(k)}(\alpha,\beta) = {\mathbb E}\left[{\hbox{e}}^{\alpha S(\omega)}{\hbox{e}}^{\beta \omega}\right]}$ in the k-case. We use the same notation as in Sect. 5 A (k > 2)-cycle begins as a (k = 2)-cycle with some short interarrival times, each followed by an F-claim with contribution to $\varphi^{(k)}(\alpha,\beta)$ equals to

$$ \frac{p_+}{1-p_-\widehat F[\alpha]\widehat A^-[\beta-c\alpha]}, $$

followed by a long interarrival time and an F-claim which gives a contribution of

$$ \widehat F[\alpha] \widehat A^+[\beta-c\alpha]. $$

After that, there are k possible paths. One path is constituted with k − 1 long interarrival times. The first k − 2 times are followed by an F-claim and the last one by a G-claim and so the cycle terminates. This path has a probability p ^k–1₊ to occur and gives a contribution of

$$ \left(p_+A^+[\beta-c\alpha]\right)^{k-1}\widehat{F} [\alpha]^{k-2}\widehat{G}[\alpha]. $$

For 0 ≤ i ≤ k − 2, the i-path is constituted with i long interarrival times and one short time, each followed by an F-claim, then the cycle starts afresh after that. The i-path occurs with probability p ⁱ₊ p ₋ and gives a contribution of

$$ {p_+}^ip_- \widehat{F}[\alpha]^{i+1} A^+[\beta-c\alpha]^i A^-[\beta-c\alpha] \varphi^{(k)}(\alpha,\beta). $$

Putting things together gives

$$ \begin{aligned} \varphi^{(k)}(\alpha,\beta)&=\frac{p_+}{1-p_-\widehat F [ \alpha ] \widehat A^- [ \beta-c\alpha ]} F [ \alpha ] \widehat A^+ [ \beta-c\alpha ] \times \left(\vphantom{\sum_{i=0}^{k-2} {p_+}^ip_- \widehat{F}}\left(p_+A^+ [ \beta-c\alpha ] \right)^{k-1} \widehat{F} [ \alpha ] ^{k-2}\widehat{G} [ \alpha ]\right.\\ &\qquad\left.+\sum_{i=0}^{k-2} {p_+}^ip_- \widehat{F}[\alpha]^{i+1} A^+ [ \beta-c\alpha ] ^i A^- [ \beta-c\alpha ] \varphi^{(k)}(\alpha,\beta) \right), \end{aligned} $$

which ends the proof.

1.3 3. Proof of Eqs. 23–27

The expression Eq. 23 for h ^θ(−, 0) is just Proposition 5.1.

To compute h ^θ(i) for all i, we first introduce some notation, with V an exponential random variable with rate λ:

$$ \begin{aligned} p_{(a)+}&={\mathbb P}\left( V > r-a \right)={\hbox{e}}^{-\lambda (r-a)} \;,\\ p_{(a)-}&={\mathbb P}\left( V \le r-a \right)= 1- {\hbox{e}}^{-\lambda (r-a)} \;,\\ \widehat A [\alpha] &={\mathbb E}\left[ {\hbox{e}}^{\alpha V} \right] = \frac{\lambda}{\lambda+\alpha},\\ \widehat A^{(a)+}[\alpha] &={\mathbb E}\left[ {\hbox{e}}^{\alpha V} \big| V > r-a \right]=\frac{\lambda}{\lambda-\alpha}{\hbox{e}}^{\alpha (r-a)} \;, \\ \widehat A^{(a)-}[\alpha]&={\mathbb E}\left[ {\hbox{e}}^{\alpha V} \big| V \le r-a \right] = \frac{\lambda}{\lambda-\alpha}\left(\frac{1-{\hbox{e}}^{-(\lambda-\alpha) (r-a)}}{1-{\hbox{e}}^{-\lambda (r-a)}}\right)\;, \end{aligned} $$

and we are using p ₊, p ₋, A ⁺[α] and A ⁻[α] defined in Sects. 3, 5.

The easiest case is the (+,  +)-one. Actually, after an exponential time, a G-claim occurs which ends the cycle, so

$$ \begin{aligned} h^{\theta}(+,+) &={\mathbb E}_{(+,+)} {\hbox{e}}^{\theta S(\omega_1) - \kappa(\theta)\omega_1} ={\mathbb E} {\hbox{e}}^{-(c\theta+\kappa(\theta))V}{\hbox{e}}^{U_G}\\ &=\widehat{A}[-c\theta - \kappa(\theta)] \widehat{G}[\theta]= \frac{\lambda}{\lambda+c\theta+\kappa(\theta)} \widehat{G}[\theta]. \end{aligned} $$

For the (+, a)-case, two cases have to be considered. If the exponential time before the first claim is larger than r − a then a G-claim occurs, otherwise, an F-claim occurs and the state becomes (−, 0):

$$ \begin{aligned} h^{\theta}(+,a) &={\mathbb E}_{(+,a)} {\hbox{e}}^{\theta S(\omega_1) - \kappa(\theta)\omega_1},\\ &=p_{(a)-} \widehat A^{(a)-}[-c\theta-\kappa(\theta)] \widehat{F}[\theta] h^{\theta}(-,0) \\ & \quad + p_{(a)+} \widehat A^{(a)+}[-c\theta-\kappa(\theta)] \widehat{G}[\theta]\\ &=\frac{k^{(a)}(\theta,-\kappa(\theta)) g(\theta,-\kappa(\theta))^2 \widehat{F}[\theta]^2\widehat{G}[\theta]}{1-k(\theta,-\kappa(\theta)) \widehat{F}[\theta]\left(1+g(\theta,-\kappa(\theta)) \widehat{F}[\theta]\right)}\\ & \quad + g^{(a)}(\theta,-\kappa(\theta))\widehat{G}[\theta]. \end{aligned} $$

The particular case (+, 0) gives

$$ \begin{aligned} h^{\theta}(+,0) &=k^{(0)}(\theta,-\kappa(\theta)) h^{\theta}(-,0) + g^{(0)}(\theta,-\kappa(\theta))\\ &=\frac{g(\theta,-\kappa(\theta)) \widehat{G}[\theta]\left(1-k(\theta,-\kappa(\theta)) \widehat{F}[\theta]\right)}{1-k(\theta,-\kappa(\theta)) \widehat{F}[\theta]\left(1+g(\theta,-\kappa(\theta)) \widehat{F}[\theta]\right)}. \end{aligned} $$

The (−,  +)-case can be easily solved using the (+, 0) one. Actually, after an exponential time, an F-claim occurs and the state becomes (+, 0):

$$ \begin{aligned} h^{\theta}(-,+) &={\mathbb E}_{(-,+)} {\hbox{e}}^{\theta S(\omega_1) - \kappa(\theta)\omega} = \widehat{A}[-c\theta - \kappa(\theta)] \widehat{F}[\theta] h^{\theta}(+,0)\\ & =\frac{\lambda}{\lambda+c\theta+\kappa(\theta)} \frac{g(\theta,-\kappa(\theta))\widehat{F} [\theta]\widehat{G}[\theta]\left(1-k(\theta,-\kappa(\theta)) \widehat{F}[\theta]\right)}{1-k(\theta,-\kappa(\theta)) \widehat{F}[\theta]\left(1+g(\theta,-\kappa(\theta)) \widehat{F}[\theta]\right)}. \end{aligned} $$

We solve the (−, a)-case using both (+, 0) and (−, 0) ones. If the exponential time before the first claim is larger than r − a, then an F-claim occurs and the state becomes (+, 0), otherwise, an F-claim occurs and the state becomes (−, 0):

$$ \begin{aligned} h^{\theta}(-,a) &={\mathbb E}_{(-,a)} {\hbox{e}}^{\theta S(\omega_1) - \kappa(\theta)\omega_1},\\ &=p_{(a)-} \widehat A^{(a)-}[-c\theta-\kappa(\theta)] \widehat{F}[\theta] h^{\theta}(-,0)\\ & \quad + p_{(a)+} \widehat A^{(a)+} [ -c\theta-\kappa(\theta) ] \widehat{F} [ \theta ] h^{\theta}(+,0),\\ &=\frac{g(\theta,-\kappa(\theta))\widehat{F} [ \theta ] \widehat{G} [ \theta ] }{1-k\left(\theta,-\kappa(\theta)\right) \widehat{F} [ \theta ] \left(1+g\left(\theta,-\kappa(\theta)\right) \widehat{F} [ \theta ] \right)} \quad \times \left\{k^{(a)}(\theta,-\kappa(\theta))g(\theta,-\kappa(\theta)) \widehat{F} [ \theta ]\right.\\ &\quad\left.+g^{(a)}\left(\theta,-\kappa(\theta)\right)\widehat{G} [ \theta ] \left(1-k\left(\theta,-\kappa(\theta)\right) \widehat{F} [ \theta ] \right)\right\}. \end{aligned} $$

1.4 4. A counterexample

We give here an example of random variables X, Y such that X is subexponential and Y light-tailed, but the tail of X − Y is lighter than the tail of X (of course, such r.v.’s have to be dependent).

We take X with tail $\overline F(x) = { \hbox{e}}^{-x/\log x}, x>{\hbox{e}}.$ Then the density and failure rates are

$$ f(x) = \left[\frac{1}{\log x}-\frac{1}{\log^2 x}\right]{\hbox{e}}^{-x/\log x},\ \hbox{resp.} \,\lambda(x)=\frac{f(x)}{\overline F(x)}=\frac{1}{\log x}-\frac{1}{\log^2 x}. $$

The integral to be inspected for Pitman’s criterion [p. 299, 5] is

$$ \int_{\hbox{e}}^\infty {\hbox{e}}^{x\lambda (x)}f(x) {\hbox{d}} x = \int_{\hbox{e}}^\infty{\hbox{e}}^{-x/\log^2 x}\left[\frac{1}{\log x}-\frac{1}{\log^2 x}\right] {\hbox{d}} x, $$

which is finite so that X is subexponential.

Let $Y=\varphi(X)=X/\log X$ and note that whenever $z<\varphi^{-1}(x),$ i.e. $\varphi(z)<x,$ then

$$ {\mathbb P}(Y>x)={\mathbb P}\left(X>\varphi^{-1}(x)\right) \le {\mathbb P}(X>z). $$

Taking z = z(x) = x log x, we have

$$ \varphi(z)=\frac{x\log x}{\log x+\log\log x} < x . $$

and thus

$$ {\mathbb P}(Y>x) \le {\mathbb P}(X>z)=\exp\left\{-x\log x/[\log x+\log\log x]\right\} < {\hbox{e}}^{-x/2} $$

for all large x. Thus Y is light-tailed.

Let 0 < b < a < 1 and note that X − Y = ψ(X) = X(1 − 1/log X). Taking z = z(x) = x + ax/log x, we get

$$ \psi(z) = (x+ax/\log x)\left[1-\frac{1}{\log x\left[1+\log(1+a/\log x)/\log x\right]}\right] < x $$

for large x (using a < 1), and thus as above,

$$ {\mathbb P}(X-Y>x) \le {\mathbb P}(X>z)={\hbox{e}}^{-z/\log z}. $$

Now

$$ \begin{aligned} \frac{z}{\log z}=&\frac{x+ax/\log x}{\log x\left[1+\log(1+a/\log x)/\log x\right]}\\ \ge& \frac{x+ax/\log x}{\log x}\left[1-\left[1+\log(1+a/\log x)/\log x\right]\right] \\ \ge&\frac{x}{\log x}+\frac{bx}{\log^2 x} \end{aligned} $$

for all large x (using (1 + v)⁻¹ > 1 − v in the second step). Thus

$$ {\mathbb P}(X-Y>x) \le {\mathbb P}(X>x){\hbox{o}}(1), $$

and the counterexample is complete.