2.1. Laplace transform of ultimate intensity integral

For convenience, the Laplace transform of intensity integral at the ultimate time $t\rightarrow \infty$ conditional on $X_0=x>0$ is denoted by

$$\phi(x) :=\mathbb{E}\left[ e^{{-}v \Lambda_{\infty }}\mid X_0=x \right],\quad v \geq 0.$$

Proposition 2.1. The Laplace transform of ultimate intensity integral $\Lambda _{\infty }$ conditional on $X_0=x>0$ satisfies

(2.3) \begin{equation} \phi(x)=\hat{p}\left( v h(x) \right) \left[\int_{0}^{x}\phi(y)g(y)\,\mathrm{d}y + \phi(x) \bar{G}(x) \right], \end{equation}

where

$$\hat{p}(v) := \int_{0}^{\infty }e^{{-}v u} p (u)\,\mathrm{d}u,\quad \bar{G}(u) := \int_{u}^{\infty } g(y)\,\mathrm{d}y.$$

Proof. By construction, the interarrival times of these jumps at $\{T_i\}_{i=1,2,\ldots }$ are independent of jump sizes $\{Y_i\}_{i=1,2,\ldots }$, therefore, $T_1$ and $X_{T_1}$ are independent, and we have

\begin{align*} \phi(x) & = \mathbb{E}\left[ \exp\left({-}v \int_0^{T_1}h(X_s)\,\mathrm{d}s \right) \times \exp\left({-}v \left.\int_{T_1}^{\infty}h(X_s)\,\mathrm{d}s \right) \right| X_0=x \right] \\ & = \mathbb{E}\left[ e^{{-}v h(x) T_1} \times \exp\left({-}v \left.\int_{T_1}^{\infty}h(X_s)\,\mathrm{d}s \right)\right| X_0=x \right] \\ & = \mathbb{E}\left[ \mathbb{E}\left[ e^{{-}v h(x) T_1} \times \exp\left({-}v \left.\int_{T_1}^{\infty}h(X_s)\,\mathrm{d}s \right) \right| X_0=x, T_1 \right] \right]\\ & = \mathbb{E}\left[ e^{{-}v h(x) T_1} \times \mathbb{E} \left[ \exp\left({-}v \left.\int_{T_1}^{\infty}h(X_s)\,\mathrm{d}s \right) \right| X_0=x, T_1 \right] \right] \\ & = \mathbb{E}\left[ e^{{-}v h(x) T_1} \times \phi(X_{T_1}) \right] \\ & = \hat{p}\left(v h(x)\right) \times \mathbb{E}\left[ \phi(X_{T_1}) \right]. \end{align*}

Note that, given the initial state level $X_0=x>0$ and the realization of the first jump size $Y_1=y$, the state process at the first-jump time, $X_{T_1}$, stays at the same level as $X_{T_1^{-}}$ if $y>X_{T_1^{-}}$, or moves down to a new level $y$ if $y< X_{T_1^{-}}$, so we have

$$\mathbb{E}\left[ \phi(X_{T_1}) \right]= \int_{0}^{x}\phi(y)g(y)\,\mathrm{d}y + \int_x^{\infty }\phi(x)g(y)\,\mathrm{d}y = \int_{0}^{x}\phi(y)g(y)\,\mathrm{d}y + \phi(x) \bar{G}(x).$$

Corollary 2.1. If jump sizes follow a uniform distribution on $[0,1]$, then, the Laplace transform of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.4) \begin{equation} \phi(x)=\hat{p}\left( v h(x) \right) \left[\int_{0}^{x}\phi(y)\,\mathrm{d}y + \left( 1-x \right) \phi(x) \right]. \end{equation}

Corollary 2.2. If jump sizes follow a uniform distribution on $[0,1]$ and interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p(t) =\nu \,e^{-\nu t}$, then the Laplace transform of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.5) \begin{equation} \phi(x)=\exp \left( - \int_{0}^{x}\frac{ v h^{\prime}(y)}{\nu y+v h(y)}\mathrm{d}y \right). \end{equation}

Proof. Set $p(t) =\nu e^{-\nu t}$ in (2.4), then we have

$$\phi(x)=\frac{\nu }{\nu +v h(x)} \left[\int_{0}^{x}\phi(y)\,\mathrm{d}y+\left( 1-x \right) \phi(x)\right],$$

or

$$\nu \int_{0}^{x}\phi(y)\,\mathrm{d}y=\left[ \nu x +v h(x)\right] \phi(x).$$

Differentiate it with respect to $x$ both sides, then we have

$$\left[ \nu x +v h(x) \right] \phi^{\prime}(x)+v h^{\prime}(x)\phi(x)=0,$$

or

$$\phi^{\prime}(x)={-}\frac{v h^{\prime }(x)}{\nu x +v h(x)}\phi(x).$$

Given the initial condition $\phi (0)=1$, we can solve this ODE by

$$\ln \phi(x) - \ln\phi(0)={-}\int_{0}^{x}\frac{v h^{\prime}(y)}{\nu y+v h(y)}\mathrm{d}y,$$

and the result follows.

By further specifying the functional form for $h(x)$, then, nicely, we can find the exact distribution of $\Lambda _{\infty }$.

Corollary 2.3. If jump sizes follow a uniform distribution on $[0,1]$, interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu \,e^{-\nu t}$ and $h(u)=u^{1+c},\ c> 0$, then the Laplace transform of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.6) \begin{equation} \phi(x)=\left(\frac{\frac{\nu }{x^{c}}}{\frac{\nu }{x^{c}}+v }\right)^{{(c+1)}/{c}}, \end{equation}

which implies

(2.7) \begin{equation} \Lambda_{\infty}\mid X_0=x \sim\textsf{Gamma}\left(\frac{c+1}{c},\frac{\nu }{x^{c}}\right). \end{equation}

Proof. Set $h(u)=u^{1+c},\ c>0$, then we have $h^{\prime }(u)= (c+1)u^{c}$. Hence, from (2.5), we have

\begin{align*} \phi(x) & = \exp \left( - (c+1) \int_{0}^{x}\frac{ v y^{c}}{\nu y+v y^{1+c}}\mathrm{d}y\right) \\ & = \exp \left( -\frac{c+1}{c} \int_{0}^{x}\frac{cy^{c-1}}{\frac{\nu }{v }+y^{c}}\mathrm{d}y \right) \\ & = \exp \left( -\frac{c+1}{c} \int_{0}^{x^{c} }\frac{1}{\frac{\nu }{v }+u}\mathrm{d}u \right) \\ & = \exp\left( - \frac{c+1}{c}\ln \left( \frac{\frac{\nu }{v }+x^{c}}{\frac{\nu }{v }} \right) \right)\\ & = \left( \frac{\frac{\nu }{x^{c}}}{\frac{\nu }{x^{c}}+v }\right)^{{(c+1)}/{c}}, \end{align*}

which is the Laplace transform of a gamma distribution with the shape parameter ${(c+1)}/{c}$ and rate parameter ${\nu }/{x^{c}}$.

2.2. Probability generating function of event numbers

Let us find the expression for the PGF of $N_\infty$.

Corollary 2.4. If jump sizes follow a uniform distribution on $[0,1]$ and interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu e^{-\nu t}$, then the PGF of $N_{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.8) \begin{equation} \mathbb{E}\left[ \theta ^{N_{\infty }}\mid X_0=x \right] = \exp \left( - \int_{0}^{x}\frac{( 1-\theta )h^{\prime }(y)}{\nu y+\left( 1-\theta \right) h(y)}\mathrm{d}y\right). \end{equation}

Corollary 2.5. If jump sizes follow a uniform distribution on $[0,1]$, interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu e^{-\nu t}$ and $h(u)=u^{1+c}, c>0$, then the PGF of $N_{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.9) \begin{equation} \mathbb{E}\left[ \theta ^{N_{\infty }}\mid X_0=x \right]= \left( \frac{1- \frac{x^{c}}{\nu +x^{c}} }{1- \frac{x^{c}}{\nu +x^{c}}\theta }\right)^{{(c+1)}/{c}}, \end{equation}

which implies

(2.10) \begin{equation} N_{\infty }\mid X_0=x \sim \textsf{Negative Binomial}\left( \frac{c+1}{c}, \frac{x^{c}}{\nu +x^{c}} \right). \end{equation}

Proof. Set $v =1-\theta$ in (2.8), then we have

$$\mathbb{E}\left[ \theta ^{N_{\infty }}\mid X_0=x \right]= \left( \frac{\frac{\nu }{x^{c}}}{\frac{\nu }{x^{c}}+ 1-\theta }\right)^{{(c+1)}/{c}}= \left( \frac{1- \frac{x^{c}}{\nu +x^{c}} }{1- \frac{x^{c}}{\nu +x^{c}}\theta }\right)^{{(c+1)}/{c}},$$

which is the PGF of a negative binomial distribution with parameters ${(c+1)}/{c}$ and ${x^{c}}/{(\nu +x^{c})}$.

2.3. Cumulants and moments of intensity integral

Let us find the expressions for the cumulants and the first two moments of $\Lambda _{\infty }$.

Theorem 2.1. The $m$th cumulant of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ is given by

(2.11) \begin{equation} \kappa_m = m! \int_{0}^{x} \frac{h^{\prime }(y)} {\nu y} \left( \frac{h(y)}{\nu y} \right)^{m-1} \mathrm{d}y,\quad m=1,2,\ldots, \end{equation}

Proof. By taking logarithm for the Laplace transform (2.5), we obtain the cumulant generating function

\begin{align*} \ln \phi(x) & = \ln \mathbb{E}\left[e^{{-}v \Lambda_{\infty }}\mid X_0=x \right] \nonumber \\ & ={-} \int_{0}^{x}\frac{ v h^{\prime}(y)}{\nu y+ v h(y)}\mathrm{d}y \nonumber \\ & ={-}\int_{0}^{x} \frac{ \frac{h^{\prime }(y)} {\nu y} v }{ 1+ \frac{h(y)}{\nu y} v} \mathrm{d}y \nonumber\\ & ={-}\int_{0}^{x} \frac{h^{\prime }(y)} {\nu y} v \sum_{k=0}^{\infty} \left( -\frac{h(y)}{\nu y} v \right)^{k} \mathrm{d}y \nonumber\\ & = \sum_{k=0}^{\infty} \left[ \int_{0}^{x} \frac{h^{\prime }(y)} {\nu y} ({-}1)^{k+1} \left( \frac{h(y)}{\nu y} \right)^{k} \mathrm{d}y \right] v^{k+1}\quad m=k+1 \nonumber\\ & = \sum_{m=1}^{\infty} \left[ \int_{0}^{x} \frac{h^{\prime }(y)} {\nu y} ({-}1)^{m} \left(\frac{h(y)}{\nu y} \right)^{m-1} \mathrm{d}y \right] v^{m} \nonumber\\ & = \sum_{m=1}^{\infty} ({-}1)^{m} \frac{\kappa_m}{m!} v^{m}. \end{align*}

Hence, we have

$$\int_{0}^{x} \frac{h^{\prime }(y)} {\nu y} ({-}1)^{m} \left( \frac{h(y)}{\nu y} \right)^{m-1} \mathrm{d}y = ({-}1)^{m}\frac{\kappa_m}{m!},$$

and the cumulants (2.11) for any $m=1,2,\ldots$.

The expression for any moment of $\Lambda _{\infty }$ given $X_0=x$ can easily be obtained using its cumulants (2.11). For example, we provide the mean and variance as below.

Corollary 2.6. The mean and variance of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ are, respectively, given by

(2.12) \begin{align} \mathbb{E}\left[ \Lambda_{\infty }\mid X_0=x \right] & = \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y, \end{align}

(2.13) \begin{align} \mathrm{Var}\left[ \Lambda_{\infty }\mid X_0=x \right] & = 2\int_{0}^{x} \frac{h^{\prime }(y) h(y)}{\nu ^{2}y^{2}} \mathrm{d}y. \end{align}

By specifying the form of $h(u),$ we may obtain the mean and variance explicitly.

Corollary 2.7. If $h(u)=u^{1+c},\ c>0$, then the mean and variance of $\Lambda _{\infty }$ conditional on $X_0=x \in (0,1)$ are respectively given by

\begin{align*} \mathbb{E}\left[ \Lambda_{\infty }\mid X_0=x \right] & = \frac{1}{\nu} \left( \frac{c+1}{c}\right) x^{c}, \\ \mathrm{Var}\left[ \Lambda_{\infty }\mid X_0=x \right] & = \frac{1}{\nu ^{2}}\left( \frac{c+1}{c}\right) x^{2c}. \end{align*}

2.4. Moments of event numbers

Let us find the expressions for the first two moments of $N_{\infty }$.

Corollary 2.8. The mean and variance of $N_{\infty }$ conditional on $X_0=x \in (0,1)$ are respectively given by

\begin{align*} \mathbb{E}\left[ N_{\infty }\mid X_0=x \right] & = \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y, \\ \mathrm{Var}\left[ N_{\infty }\mid X_0=x \right] & = 2\int_{0}^{x} \frac{h^{\prime }(y) h(y)}{\nu ^{2}y^{2}} \mathrm{d}y + \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y. \end{align*}

Proof. Since

$$\mathbb{E}\left[ N_{\infty} \mid X_0=x \right] = \left.\frac{\mathrm{d}}{\mathrm{d} \theta }\mathbb{E} \left[\theta^{N_{\infty}}\mid X_0=x \right]\right|_{\theta =1}= \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y,$$

and

\begin{align*} \mathbb{E}\left[ N_{\infty}\left( N_{\infty}-1\right) \mid X_0=x \right] & = \left.\frac{\mathrm{d}^{2}}{\mathrm{d}\theta^{2}}\mathbb{E} \left[\theta^{N_{\infty}} \mid X_0=x \right]\right|_{\theta =1} \\ & =\left(\int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y \right)^{2} + 2\int_{0}^{x} \frac{h^{\prime }(y) h(y)}{\nu ^{2}y^{2}}\mathrm{d}y, \end{align*}

the variance of $N_{\infty }$ can be obtained by

\begin{align*} \mathrm{Var}[N_{\infty} \mid X_0=x] & = \mathbb{E}\left[ N_{\infty}\left( N_{\infty}-1\right) \mid X_0=x \right] + \mathbb{E}\left[ N_{\infty} \mid X_0=x \right] - \left(\mathbb{E}\left[ N_{\infty} \mid X_0=x\right] \right)^{2} \\ & = \left( \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y \right)^{2} + 2\int_{0}^{x} \frac{h^{\prime }(y) h(y)}{\nu ^{2}y^{2}} \mathrm{d}y + \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y - \left( \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y \right)^{2} \\ & = 2\int_{0}^{x} \frac{h^{\prime }(y) h(y)}{\nu ^{2}y^{2}} \mathrm{d}y + \int_{0}^{x} \frac{h^{\prime }(y)}{\nu y} \mathrm{d}y. \end{align*}

Corollary 2.9. If $h(u)=u^{1+c}, c>0$, then the mean and variance of $N_{\infty }$ conditional on $X_0=x \in (0,1)$ are respectively given by

\begin{align*} \mathbb{E}\left[ N_{\infty }\mid X_0=x \right] & = \frac{1}{\nu}\left( \frac{c+1}{c}\right) x^{c}, \\ \mathrm{Var}\left[ N_{\infty }\mid X_0=x \right] & = \frac{1}{\nu ^{2}}\left( \frac{c+1}{c}\right) x^{2c} + \frac{1}{\nu }\left( \frac{c+1}{c}\right) x^{c}. \end{align*}

For example, by setting the initial state $x=0.9$ and varying the parameter $c$ for amplification effect, the conditional means and variances of $N_\infty$ for $\nu =0.5,1$ are plotted in Figure 3.

Figure 3. Conditional mean and variance of $N_\infty$ against the parameter $c$ for amplification effect.

2.5. Probability of last event

As the intensity process is decreasing and eventually approaching to zero, the resulting points (i.e., events) are gradually disappearing, and an interesting problem is the probability of no more event (e.g., insurance claim of a certain type) beyond a given time point $t$, that is,

(2.14) \begin{equation} \Pr \left\{ T^{*} < t \right\} = \Pr \left\{ N_{[t,\infty)} =0 \right\} = \mathbb{E}\left[ \exp \left(-\int_{t}^{\infty }h(X_{s})\,\mathrm{d}s\right)\right], \end{equation}

where $T^{*}$ is the time point at which the last event occurs. In order to find (2.14), let us first derive the CDF of $X_t$ in Lemma 2.1.

Lemma 2.1. If interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu \,e^{-\nu t}$, then the CDF of $X_{t}$ conditional on $X_0=x>0$ is given by

(2.15) \begin{equation} \Pr \left\{ X_{t}<\varsigma \right\} = 1-e^{-\nu t G(\varsigma )} \textbf{1}_{\left\{ \varsigma < x \right\}}. \end{equation}

Proof. By construction, $X_{t}$ is piecewise-constant and decreasing, which can be expressed by

$$X_{t}=\min\left\{ x,Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\},$$

where $M_{t}$ is a homogeneous Poisson process of constant rate $\nu$. Given $M_{t}=m$, the CDF of the minimum of $m$ i.i.d. random variables $\{Y_i\}_{i=1,\ldots,m}$ with the CDF $G(\varsigma )$ is $\left [1-\left (1-G(\varsigma )\right )^{m} \right ]$, that is,

$$\Pr \left\{\min\left\{Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\} < \varsigma \mid M_{t}=m\right\} = 1-\left( 1-G(\varsigma)\right)^{m}.$$

Then, we have

\begin{align*} \Pr \left\{\min\left\{Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\} <\varsigma\right\} & = \sum_{m=0}^{\infty}\left[ 1-\left( 1-G\left( \varsigma \right) \right)^{m} \right] \frac{e^{-\nu t}\left( \nu t\right) ^{m}}{m!} \\ & = 1 - \sum_{m=0}^{\infty}\left( 1-G\left( \varsigma \right) \right)^{m} \frac{e^{-\nu t}\left( \nu t\right) ^{m}}{m!}, \end{align*}

where

$$\sum_{m=0}^{\infty}\left(1-G\left( \varsigma \right) \right)^{m} \frac{e^{-\nu t}\left( \nu t\right) ^{m}}{m!} = e^{ -\nu t G\left( \varsigma \right) }.$$

Note that,

\begin{align*} \Pr \left\{ X_{t}<\varsigma\right\} & = \Pr \left\{\min \left\{\min\left\{ Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\}, x\right\} <\varsigma \right\} \\ & = 1- \Pr \left\{ \min \left\{ \min\left\{ Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\}, x\right\} >\varsigma \right\} \\ & = 1- \Pr \left\{ \min\left\{ Y_{1},Y_{2},\ldots, Y_{M_{t}}\right\} > \varsigma \right\} \times \textbf{1}_{\left\{ x > \varsigma \right\}} \\ & = 1- e^{-\nu t G(\varsigma )} \times \textbf{1}_{\left\{ \varsigma < x \right\}}, \end{align*}

and the result (2.11) follows.

Based on Lemma 2.1, let us now derive the PGF of $T^{*}$ in Theorem 2.2.

Theorem 2.2. If jump sizes follow a uniform distribution on $[0,1]$ and interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu \,e^{-\nu t}$, then the CDF of $T^{*}$ conditional on $X_0=x \in (0,1)$ is given by

(2.16) \begin{equation} \Pr \left\{ T^{*} < t \right\}=\nu t\int_{0}^{x} \exp \left( -\int_{0}^{\varsigma } \frac{h^{\prime }(y)}{\nu y+h(y)}\mathrm{d}y\right) e^{-\nu t\varsigma}\,\mathrm{d}\varsigma. \end{equation}

Proof. Set $v =1$ in (2.5), then we have

$$\mathbb{E}\left[ \exp \left( -\int_{t}^{\infty }h(X_s)\,\mathrm{d}s\right) \mid X_{t} \right] = \exp \left(-\int_{0}^{X_{t}}\frac{h^{\prime }(y)}{\nu y+h(y)}\mathrm{d}y\right).$$

Hence,

$$\Pr \left\{ T^{*} < t \right\}=\mathbb{E}\left[ \exp \left(-\int_{t}^{\infty }h(X_{s})\,\mathrm{d}s\right)\right] = \mathbb{E}\left[ \exp \left( -\int_{0}^{X_{t}}\frac{h^{\prime }(y)}{\nu y+h(y)}\mathrm{d}y\right) \right].$$

Based on the CDF (2.11) for $X_t$, we have the density of $X_t$ as

$$\Pr \left\{ X_{t} \in \mathrm{d} \varsigma \right\} = \nu t \,e^{-\nu t G(\varsigma )}g(\varsigma ) \textbf{1}_{\left\{ \varsigma < x \right\}} \mathrm{d} \varsigma,$$

and then,

$$\Pr \left\{ T^{*} < t\right\}=\nu t \int_{0}^{x} \exp \left(-\int_{0}^{\varsigma } \frac{h^{\prime }(y)}{\nu y+h(y)}\mathrm{d}y\right) e^{-\nu t G(\varsigma )}g(\varsigma)\,\mathrm{d}\varsigma.$$

With a uniform distribution on the interval $[0,1]$ for jump sizes, (2.12) follows immediately.

Corollary 2.10. If jump sizes follow a uniform distribution on $[0,1]$, interarrival times of jumps follow an exponential distribution with parameter $\nu >0$, that is, $p\left ( t\right ) =\nu \,e^{-\nu t}$ and $h(u)=u^{1+c},\ c>0$, then the CDF of $T^{*}$ conditional on $X_0=x \in (0,1)$ is given by

(2.17) \begin{equation} \Pr \left\{ T^{*} < t \right\} =\nu^{2+\frac{1}{c}} t \int_{0}^{x} \frac{1}{e^{\nu t \varsigma } \left( \nu + \varsigma^{c}\right)^{{(c+1)}/{c}}} \mathrm{d}\varsigma. \end{equation}

Proof. Based on Theorem 2.2 and the additional assumption of $h(u)=u^{1+c},\ c>0$, we have

\begin{align*} \Pr \left\{ T^{*} < t \right\} & = \nu t \int_{0}^{x} \exp \left( -\frac{c+1}{c}\int_{0}^{\varsigma } \frac{c y^{c-1}}{\nu +y^{c}}\mathrm{d}y\right) e^{-\nu t \varsigma}\,\mathrm{d}\varsigma \\ & = \nu t \int_{0}^{x} \exp \left( -\frac{c+1}{c} \ln\left( \frac{\nu + \varsigma^{c}}{\nu} \right) \right) e^{-\nu t \varsigma }\,\mathrm{d}\varsigma \\ & = \nu t \int_{0}^{x} \left( \frac{\nu}{\nu + \varsigma^{c}} \right)^{{(c+1)}/{c}} e^{-\nu t \varsigma}\,\mathrm{d}\varsigma\\ & = \nu^{2+\frac{1}{c}} t \int_{0}^{x} \frac{1}{e^{\nu t \varsigma } \left( \nu + \varsigma^{c}\right)^{{(c+1)}/{c}} } \mathrm{d}\varsigma. \end{align*}

We can calculate the probability of no event (e.g., insurance claim of a certain type) beyond time $t$ as given in (2.17) via numerical integration. For example, by setting parameters $\nu =2,\ c=1.2$ with the initial state $x=0.9$, the associated probabilities are reported in Table 1 and plotted in Figure 4. As time $t$ is getting larger, the intensity level is getting smaller and smaller. The intensity integral increases only marginally and events are still generated, but it will eventually arrive at the time point of no event anymore. Hence, the larger the time $t$ is, the higher the probability of no more event beyond time $t$ is.

Table 1. Probability of no event beyond time $t$.

Figure 4. Probability of no event beyond time $t$.

3.1. Life insurance

Medical innovation has made us live longer, healthier, and more prosperous lives. Additionally, due to artificial intelligence advances in health care, a much greater longevity would be possible in the long term as $t \rightarrow \infty$ or even within a couple of decades. Hence, we may apply our results to calculate survival probabilities which are the key inputs for life insurance. In general, by setting $v =1$ in (2.3) and solving the equation, we can obtain the ultimate survival probability (or survivor function) due to a particular event-type of interest as

(3.1) \begin{equation} \Pr \left\{ \tau^{*}_x = \infty \right\} = \mathbb{E}\left[ e^{-\Lambda_{\infty }}\mid X_0=x \right] = \hat{p}\left( h(x) \right) \times\left[ \int_{0}^{x}\phi(y)g(y)\,\mathrm{d}y + \phi(x) \bar{G}(x) \right], \end{equation}

where $\tau ^{*}$ is the first jump-arrival time in the point process $N_t$, that is,

$$\tau^{*}_x :=\inf\left\{ t>0: N_{t}=1\mid N_{0}=0,\ X_0=x \right\},$$

which is modeling the first event-arrival time of a particular type. More specifically, if jump sizes follow a uniform distribution on $[0,1]$, interarrival times of jumps follow an exponential distribution with its parameter $\nu$, that is, $p\left ( t\right ) =\nu \,e^{-\nu t}$ and $h(u)=u^{1+c},\ c>0$, then the survival probability $\Pr \left \{ \tau ^{*}_x = \infty \right \}$ can be immediately obtained by setting $v=1$ in (2.6). For example, by setting the initial state $x=0.9$ and varying the parameter $c$ for amplification effect, the survival probabilities for $\nu =0.5,1$, are plotted in Figure 5, and the associated detailed numerical output is reported in Table 2.

Figure 5. Survival probability against the parameter $c$ for amplification effect.

Table 2. Survival probability against the parameter $c$ for amplification effect.

Note that $\tau ^{*}_x$ is considered as a defective (improper) random variable with a point mass at time $t=\infty$. The survival probability (3.1) is corresponding to a particular type of gradually disappearing events (e.g., deaths caused by traffic accidents or deaths caused by the COVID-19) rather than overall events. Essentially, it is a marginal probability function, also well known as the subdistribution for a particular failure type of interest in the literature of survival analysis, see for example [Reference Fine and Gray19,Reference Gray22,Reference Kalbfleisch and Prentice29, Sect. 8]. The Cox process with piecewise-constant decreasing intensity as introduced in Section 2 provides us a key model component for modeling these types of gradually disappearing events rather than for all events within a general competing-risks framework. Of course, each life is not immortal in the reality, and we can simply add another model component acting as the force for all rest risks (e.g., the mortality force is increasing with respect to the age, year, or many other factors). This can be done by introducing another random time $\tau ^{*}_{b}$ acting as a competing risk that will eventually produce a death event at time

$$\tau^{*} := \min\left\{\tau^{*}_x,\tau^{*}_{b}\right\},$$

where $\tau ^{*}_{b}$ may be assumed to follow a simple Poisson process or more generally another Cox process with time-varying intensity $b(t)> 0$, and $b(t)$ could depends on years, ages, or many other factors. Therefore, the overall intensity is

$$\lambda_t + b(t),$$

where $b(t)$ is the baseline intensity for $\tau ^{*}$, and $\lambda _t$ is the type-specific intensity. Trivially, the ultimate overall survival probability (due to overall events) is zero, that is, $\Pr \left \{ \tau ^{*} = \infty \right \} = 0$. Therefore, here we are mainly interested in the nontrivial result of ultimate (marginal) survival probability (3.1) caused by events of a particular type, $\tau ^{*}_x$, driven by technological advances in a very long run. In fact, this research motivation is similar as the classical problem of ultimate ruin probability which are extensively studied in actuarial mathematics. In this paper, we are also dealing with the ultimate probability but within a different context considering the risk landscape change in the insurance industry.

3.2. Non-life insurance

The Advanced Driver Assistance Systems (ADAS) in highly automated vehicles may reduce or even eventually eliminate accidents by perceiving dangerous situations. Nearly perfect automated vehicles could be deployed as $t\rightarrow \infty$ or even within a couple of decades, which will minimize the number of accidents leading to fewer and fewer loss claims. Hence, we apply our results to the stop-loss reinsurance contract for a portfolio purely concentrated on the motor insurance business in the long run. Standard non-life insurance contracts are typically underwritten in the short term. However, they are often automatically renewed, and it is worth studying the long-term risk landscape in insurance similarly as the classical ruin problem in risk theory.

Let $\{Z_{i}\}_{i=1,2,\ldots }$ be the claim amounts, which are assumed to be i.i.d. with the CDF $H(z), z>0$. The total loss excess over the retention limit $K>0$ up to the time of $t\rightarrow \infty$ is given by

$$\left( C_{\infty }-K \right)^{+} :=\max\left\{ C_{\infty }-K,0\right\},$$

where $C_{\infty }=\sum _{i=1}^{N_{\infty }}Z_{i}$, and $N_{\infty }$ is the ultimate number of claims. Therefore, the stop-loss reinsurance premium is given by

$$\mathbb{E}\left[ \left( C_{\infty }-K \right)^{+} \mid X_0=x \right],$$

which is similar to a perpetual call option in finance.

Corollary 3.1. If jump sizes follow a uniform distribution on $[0,1]$, the interarrival times of jumps follow an exponential distribution with parameter $\nu$, that is, $p(t) =\nu \,e^{-\nu t}$ and $h(u)=u^{1+c},\ c>0$, and the claim sizes follow an Erlang distribution $\textsf {Erlang}(\varphi,\beta )$, then the expected ultimate total loss conditional on $X_0=x \in (0,1)$ is given by

$$\mathbb{E}\left[ C_{\infty} \mid X_0=x \right] = \frac{x^{c}}{\nu}\left( \frac{c+1}{c}\right) \frac{\varphi }{\beta },$$

and the stop-loss reinsurance premium with the retention limit $K$ is given by

(3.2) \begin{align} & \mathbb{E}\left[ \left( C_{\infty} - K\right)^{+} \mid X_0=x \right] \nonumber\\ & \quad =\left( \frac{x^{c}}{\nu +x^{c}} \right)^{{(c+1)}/{c}}\sum_{n=1}^{\infty } \frac{\Gamma(n+\frac{c+1}{c})}{n! \Gamma(n)}\left( \frac{\nu}{\nu +x^{c}} \right)^{n} \nonumber\\ & \qquad \times\left[\frac{n\varphi }{\beta } \left( 1-\frac{\gamma(n\varphi+1,\beta K)}{(n\varphi)!} \right) -K\left( 1-\frac{\gamma(n\varphi,\beta K)}{(n\varphi-1)!} \right)\right], \end{align}

where $\gamma (\cdot,\cdot )$ is the lower incomplete gamma function.

Proof. We assume that the claim sizes follow an Erlang distribution denoted by $\textsf {Erlang}(\varphi,\beta )$, that is,

$$h(z)=\frac{\beta^{\varphi }z^{\varphi -1}e^{-\beta z}}{(\varphi-1)!},\quad z,\beta >0,\enspace \varphi \geq 1,$$

where $\beta$ is the rate parameter and $\varphi$ is the shape parameter. Note that,

$$\sum_{i=1}^{N_{\infty}}Z_{i} \mid N_{\infty}\sim\textsf{Erlang}(N_{\infty}\varphi,\beta),$$

so, $C_{\infty }$ follows an Erlang-mixture distribution with the density function

$$\sum_{n=1}^{\infty }\Pr \left\{ N_{\infty }=n \mid X_0=x \right\}\times \frac{\beta ^{n\varphi }c^{n\varphi -1}e^{-\beta c}}{\left( n\varphi -1\right) !}.$$

Then, we have

$$\mathbb{E}\left[ C_{\infty} \mid X_0=x \right]= \mathbb{E}\left[ N_{\infty } \mid X_0=x \right] \times \mathbb{E}\left[ Z\right]= \frac{1}{\nu}\left( \frac{c+1}{c}\right) x^{c} \times\frac{\varphi }{\beta },$$

and

\begin{align*} & \mathbb{E}\left[ \left( C_{\infty} - K\right)^{+} \mid X_0=x \right] \\ & \quad =\mathbb{E}\left[ \left( C_{\infty }-K\right) \textbf{1}\{ C_{\infty }\geq K\} \mid X_0=x \right] \\ & \quad =\sum_{n=1}^{\infty }\Pr \left\{ N_{\infty }=n \mid X_0=x\right\} \times \left[\int_{K}^{\infty } c \frac{\beta ^{n\varphi }c^{n\varphi -1} e^{-\beta c}}{\left( n\varphi -1\right) !} \mathrm{d}c - K\int_{K}^{\infty } \frac{\beta ^{n\varphi }c^{n\varphi -1}e^{-\beta c}}{\left( n\varphi -1\right) !}\mathrm{d}c\right]\\ & \quad =\sum_{n=1}^{\infty }\Pr \left\{ N_{\infty }=n \mid X_0=x \right\} \times \left[\frac{n\varphi }{\beta }\int_{K}^{\infty } \frac{\beta ^{n\varphi+1}c^{n\varphi }e^{-\beta c}}{\left( n\varphi \right) !}\mathrm{d}c - K \int_{K}^{\infty }\frac{\beta ^{n\varphi }c^{n\varphi -1}e^{-\beta c}}{\left( n\varphi -1\right) !}\mathrm{d}c\right]\\ & \quad =\sum_{n=1}^{\infty }\Pr \left\{ N_{\infty }=n \mid X_0=x \right \} \times \left[\frac{n\varphi }{\beta } \left( 1-\frac{\gamma(n\varphi+1,\beta K)}{(n\varphi)!} \right) - K \left(1-\frac{\gamma(n\varphi,\beta K)}{(n\varphi-1)!} \right)\right] \\ & \quad =\sum_{n=1}^{\infty }\frac{\Gamma(n+\frac{c+1}{c})}{n! \Gamma(n)} \left(\frac{x^{c}}{\nu +x^{c}} \right)^{{(c+1)}/{c}}\left(\frac{\nu}{\nu +x^{c}} \right)^{n}\\ & \qquad \times\left[\frac{n\varphi }{\beta } \left( 1-\frac{\gamma(n\varphi+1,\beta K)}{(n\varphi)!} \right) - K \left(1-\frac{\gamma(n\varphi,\beta K)}{(n\varphi-1)!} \right)\right], \end{align*}

since

$$\int_{x}^{\infty} \frac{\beta ^{\varphi }z^{\varphi -1}e^{-\beta z}}{(\varphi-1)!} \mathrm{d}z = 1-\frac{\gamma(\varphi,\beta x)}{(\varphi-1)!},$$

and

$$N_{\infty} \mid X_0=x \sim \textsf{Negative Binomial}\left( \frac{c+1}{c}, \frac{x^{c}}{\nu +x^{c}} \right),$$

we have the probability mass function (PMF) of $N_{\infty }$ conditional on $X_0=x$ explicitly as

$$\Pr\left\{ N_{\infty }=n \mid X_0=x \right\}=\frac{\Gamma(n+\frac{c+1}{c})}{n! \Gamma(n)} \left(\frac{x^{c}}{\nu +x^{c}} \right)^{{(c+1)}/{c}} \left(\frac{\nu}{\nu +x^{c}} \right)^{n},\quad n=0,1,\ldots.$$

For example, by setting parameters $c=1.2,\varphi =1,\beta =1$ with the initial state $x=0.9$, we can calculate the expected total loss and stop-loss reinsurance premiums for $\nu =0.5,1$, which are plotted in Figures 6 and 7, respectively. The associated detailed numerical output is reported in Table 3.

Figure 6. Expected total loss against the parameter $c$ for amplification effect.

Figure 7. Stop-loss reinsurance premium.

Table 3. Stop-loss reinsurance premium.