2. Our setting and results

We consider the following generalization of the classical Cramér–Lundberg model of the risk theory of non-life insurance.

• • The claim number processes of the n insured people, $N_i=\{ N_i(t) \}_{t \in [0,\infty)}$ , $ i=1,\ldots, n$ , are independent mixed Poisson processes, and their aggregation is denoted by $N(t)\;:\!=\; \sum_{i=1}^n N_i(t)$ , $t \geq 0$ . For each $ i \in \{ 1,\ldots, n \}$ , the mixing random variable $\Lambda_i$ of $N_i$ , obtained by $\Lambda_i=\lim_{t \to \infty} {N_i(t)} / {t}$ almost surely (a.s.), is assumed to satisfy $\mathbb{E}[ \Lambda_i] <\infty$ . We do not assume that the $\Lambda_i$ s are identically distributed. The distribution function of $\Lambda_i$ is denoted by $F_i(\cdot)$ . Moreover, the filtration generated by process $N_i$ is denoted by $\mathcal{F}_i=\{ \mathcal{F}_i(t) \}_{t \in [0,\infty)}$ .

• • Let the claim sizes $X_1, X_2, \ldots$ be a sequence of independent and identically distributed (i.i.d.) positive random variables that are independent of the processes $N_i$ . We assume that $\mu\;:\!=\; \mathbb{E}[X_k] < \infty$ .

• • The instantaneous premium rate $c=\{ c(t) \}_{t \in [0,\infty)} $ is defined as

  \begin{equation*}c(t) \;:\!=\; (1+\theta) \mu \sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ],\qquad t \geq 0,\end{equation*}
  where the safety loading $\theta$ is a positive constant.

• • The initial surplus $u_0$ is a positive constant.

Furthermore, we define the following four concepts:

• • The filtration $\mathcal{F}=\{ \mathcal{F}(t)\}_{t \in [0,\infty)}$ is defined as $ \mathcal{F}(t) \;:\!=\; \mathcal{F}_1(t) \vee \cdots \vee \mathcal{F}_n(t). $

• • Let $\tau_k$ be the kth arrival time of N for each positive integer k. We set $\tau_0\;:\!=\;0$ .

• • Let $\tau$ be the ruin time defined as $\tau\;:\!=\; \inf\{t \mid U(t)<0\}$ , where $\inf \emptyset\;:\!=\; \infty $ by convention.

• • Finally, for each $k\in \{ 1,2,\ldots \}$ , the random variable $I_k$ is defined as

  \begin{equation*}I_k\;:\!=\; \int_{\tau_{k-1}}^{\tau_k} \sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ] \, \mathrm{d} t.\end{equation*}

For comparison with our model, we introduce the surplus of the classical Cramér–Lundberg model: $U^0(t) \;:\!=\; u_0 + c t - \sum_{k=1}^{N^0(t)} X_k$ , $t \geq 0$ . The intensity of the Poisson process $N^0$ is the constant $\lambda$ , and the premium rate c is defined as $c \;:\!=\; (1+\theta) \mu \lambda$ . The setting is the same as in Definition 1 except for the claim number process and premium rate. The ruin probability of the classical Cramér–Lundberg model and the ruin time are defined as

\begin{align*} \psi^0(u) & \;:\!=\; \mathbb{P}\Big( \inf_{t \in [0,\infty)} U^0(t) < 0 \Big), \\[5pt] \tau^0 & \;:\!=\; \inf\!\big\{t \mid U^0(t)<0\big\},\end{align*}

where $\inf \emptyset\;:\!=\; \infty$ . Additionally, $\tau^0_k$ is defined as the kth arrival time of $N^0$ for each positive integer k. Note that the ruin probability $\psi^0(u)$ does not depend on the constant intensity $\lambda$ .

Remark 2. Using Bayes’ theorem, Dubey (1977) showed that the expression $\mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ]$ equals

\begin{equation*}\frac{\int_0^\infty \lambda^{1+N_i(t)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(t)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}.\end{equation*}

See also Proposition 4.1 of Grandell (1997).

For our generalized model, we have the following theorem; the second assertion addresses the ruin probability.

Figure 1. Typical sample path of the surplus process and premium rate.

If the $X_k$ have a light-tailed distribution, that is, there is some $\delta>0$ that satisfies $\mathbb{E}[\!\exp\!(\delta X_k)] < \infty$ , then we also have the following.

Corollary 1. If the $X_k$ have a light-tailed distribution and a positive R exists that satisfies $\mathbb{E}[\!\exp\!(RX_k)]-1-(1+\theta)\mu R=0$ , then this is the Lundberg adjustment coefficient of our generalized Cramér–Lundberg model, and the ruin probability is written as

\begin{equation*}\mathbb{P}\Big( \inf_{t \in [0,\infty)} U_t < 0 \Big)=\frac{\exp\!(\!-\!Ru)}{\mathbb{E}[\exp\!(\!-\!RU_\tau) \mid \tau<\infty]}.\end{equation*}

The following lemma is the key to the proof of our theorem.

The next proposition addresses the dependence of $\{ U({\tau_k}) \}_{k\in\{1,2,\ldots \}} $ and the $\Lambda_i$ .

• • Suppose that $\Lambda_i \sim \Gamma(\alpha_i, \beta)$ for each i, where the second parameter $\beta$ is common for all $\Lambda_i$ . It then follows from Remark 2 that

  \begin{equation*}\sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ] =\sum_{i=1}^n \frac{N_i(t)+\alpha_i }{\beta+t} =\frac{N(t) + \sum_{i=1}^n \alpha_i }{\beta+t} \quad \mbox{a.s.}\end{equation*}
  Moreover, because $\sum_{i=1}^n \Lambda_i \sim \Gamma\big(\!\sum_{i=1}^n \alpha_i, \beta\big)$ ,
  \begin{equation*}\mathbb{E} \Bigg[ \sum_{i=1}^n\Lambda_i \mid \mathcal{F}^N(t) \Bigg] =\frac{N(t) + \sum_{i=1}^n \alpha_i }{\beta+t} \quad \mbox{a.s.}\end{equation*}
  Therefore, for this example, $\sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ] =\mathbb{E} \big[ \sum_{i=1}^n\Lambda_i \mid \mathcal{F}^N(t) \big]$ a.s.

• • Suppose that each $\Lambda_i$ is uniformly distributed in the interval $(0,a_i)$ . Because

  \begin{equation*}\int_0^{a_i} \lambda^y \mathrm{e}^{-\lambda t} \,\mathrm{d}\lambda = \frac{y! }{t^{y+1}}\Bigg\{1- \mathrm{e}^{-a_it} \sum_{k=0}^y \frac{(a_it)^k}{k!}\Bigg\} = \frac{y! \mathrm{e}^{-a_it}}{t^{y+1}} \sum_{k=y+1}^\infty \frac{(a_it)^k}{k!}\end{equation*}
  for each nonnegative integer y, we see from Remark 2 that
  \begin{equation*}\sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(t) ] =\sum_{i=1}^n \frac{\{ N_i(t)+1 \} \sum_{k=N_i(t)+2}^\infty \frac{(a_it)^k}{k!}}{t \sum_{k=N_i(t)+1}^\infty \frac{(a_it)^k}{k!}} \quad \mbox{a.s.}\end{equation*}
  Because this is not $\mathcal{F}^N(t)$ -measurable, for this example
  \begin{equation*}\mathbb{P} \Bigg[\sum_{i=1}^n \mathbb{E} [\Lambda_i \mid \mathcal{F}_i(t) ] \neq\mathbb{E} \Bigg[ \sum_{i=1}^n\Lambda_i \mid \mathcal{F}^N(t) \Bigg]\Bigg] > 0.\end{equation*}

We now present a numerical example of the ruin probability in finite time using a Monte Carlo simulation. We assume the following: the claim size $X_k$ is exponentially distributed with parameter 1, $\Lambda_i$ is exponentially distributed with parameter 1, the initial surplus $u_0=5$ , and $\theta=0.2$ .

We consider the three cases of $n=1$ , 10, and 100 insureds. We generate 100 000 random scenarios and calculate the ruin probability numerically at time $t=1$ , 5, 10, 50, 100, and 1000.

In all cases, the ruin probability for infinite time equals $\frac{1}{1.2} \exp\!\big({-}\frac{0.2 \cdot 5}{1.2}\big) \simeq 0.362$ . The results for finite time are shown in Table 1. The following two points are observed:

1. 1. For finite time, the smaller the number of insureds, the lower the probability of ruin. This result is easy to understand because the smaller the number of insureds, the larger the variance of the surplus in finite time.

2. 2. In all cases, the ruin probability approaches a certain level as t approaches infinity. As expected from Theorem 1, the ruin probability for infinite time is independent of the number of insureds.

Table 1. Ruin probability for finite time

3. Proofs

We first provide a lemma that is a consequence of Bayes’ theorem.

Lemma 2. For each nonnegative integer k, the posterior distribution of $(\Lambda_1, \ldots, \Lambda_n) \mid \mathcal{F}(\tau_k) $ satisfies

\begin{equation*}\mathbb{P} \big( \Lambda_i \leq \lambda_i, i=1,\ldots, n \mid \mathcal{F}(\tau_k) \big) =\prod_{i=1}^n\frac{\int_0^{\lambda_i} \nu_i^{N_i(\tau_k)} \mathrm{e}^{-\nu_i \tau_k} \, \mathrm{d} F_i(\nu_i)}{\int_0^\infty \nu_i^{N_i(\tau_k)} \mathrm{e}^{-\nu_i \tau_k } \, \mathrm{d} F_i(\nu_i)}\end{equation*}

for $\lambda_i>0$ , $i=1,\ldots, n$ .

Proof of Lemma 2. Let $\{ t_\ell \}_{\ell \in \{0,1, \ldots, k\}}$ be a strictly increasing sequence of nonnegative real numbers with $t_0=0$ , and let $\{ y_{i, \ell} \}$ , where $i\in \{1, \ldots, n\}$ and $\ell \in \{0,1, \ldots, k\}$ , be a double sequence of nonnegative integers that satisfy the following two properties:

• • for all i, the sequence $\{ y_{i, \ell} \}_{\ell \in \{0,1, \ldots, k\}}$ is non-decreasing;

• • for all $\ell$ , $\sum_{i=1}^n y_{i, \ell}=\ell$ .

Conditional on $\sigma(\Lambda_1, \ldots, \Lambda_n)$ , the processes $N_i$ are independent Poisson processes, and

\begin{multline*} \mathbb{P}\big( \tau_\ell \in [t_\ell, t_\ell+dt_\ell), \, \ell=1,\ldots, k \mid \Lambda_i =\lambda_i, \, i=1,\ldots, n \big) \\[5pt] = \Bigg\{ \prod_{\ell=1}^k \Bigg( \sum_{i=1}^n \lambda_i \Bigg) \mathrm{e}^{-\left( \sum_{i=1}^n \lambda_i \right) (t_\ell-t_{\ell-1})} \Bigg\} \, \mathrm{d} t_1 \ldots \mathrm{d} t_k = \Bigg( \sum_{i=1}^n \lambda_i \Bigg)^k \mathrm{e}^{-\left( \sum_{i=1}^n \lambda_i \right) t_k } \, \mathrm{d} t_1 \ldots \mathrm{d} t_k.\end{multline*}

Moreover,

\begin{multline*} \mathbb{P}\big( N_i(\tau_\ell)=y_{i,\ell}, \, i=1,\ldots, n, \, \ell=1,\ldots, k \mid \Lambda_i =\lambda_i, \, \tau_\ell =t_\ell, \, i=1,\ldots, n, \, \ell=1,\ldots, k \big) \\[5pt] = \prod_{\ell=1}^k \frac{ \prod_{i=1}^n \lambda_i^{ y_{i,\ell} - y_{i,\ell-1} }}{ \sum_{i=1}^n \lambda_i}=\frac{ \prod_{i=1}^n \lambda_i^{ y_{i,k} }}{ \big(\!\sum_{i=1}^n \lambda_i \big)^k}.\end{multline*}

It thus follows that

\begin{multline*} \mathbb{P}\big( \tau_\ell \in [t_\ell, t_\ell+dt_\ell), \, N_i(\tau_\ell)=y_{i,\ell}, \, i=1,\ldots, n, \, \ell=1,\ldots, k \mid \Lambda_i =\lambda_i, \, i=1,\ldots, n \big) \\[5pt] = \Bigg( \prod_{i=1}^n \lambda_i^{y_{i,k} } \Bigg) \mathrm{e}^{-\left( \sum_{i=1}^n \lambda_i\right) t_k } \, \mathrm{d} t_1 \ldots \mathrm{d} t_k = \Bigg( \prod_{i=1}^n \lambda_i^{y_{i,k} } \mathrm{e}^{-\lambda_i t_k} \Bigg) \, \mathrm{d} t_1 \ldots \mathrm{d} t_k,\end{multline*}

which, together with Bayes’ theorem, completes the proof.

Proof of Lemma 1. It suffices to show that

(1) \begin{equation} \mathbb{E} \big[ \mathrm{e}^{-a I_{k+1} } \mid \mathcal{F}({\tau_k}) \big] = \frac{1}{a+1} \qquad \mathrm{for\ all\ } a \in [0,\infty).\end{equation}

It follows from Remark 2 that

\begin{equation*}I_{k+1}=\sum_{i=1}^n\int_{\tau_k}^{\tau_{k+1}}\frac{\int_0^\infty \lambda^{1+N_i(t)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(t)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t=\sum_{i=1}^n\int_{\tau_k}^{\tau_{k+1}}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t,\end{equation*}

which implies that the left-hand side of (1) is equal to

(2) \begin{align}& \mathbb{E}\Bigg[\!\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_{k+1}}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\, \big| \, \mathcal{F}(\tau_k)\Bigg] \nonumber \\ & = \mathbb{E}\Bigg[\!\mathbb{E}\Bigg[\!\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_{k+1}}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\, \big| \,\sigma(\Lambda_1, \ldots, \Lambda_n) \vee \mathcal{F}(\tau_k)\Bigg]\mid \mathcal{F}(\tau_k)\!\Bigg]\end{align}

according to the tower property of conditional expectations. Because the conditional distribution of the inter-arrival time $\tau_{k+1}-\tau_k$ given $\sigma(\Lambda_1, \ldots, \Lambda_n) \vee \mathcal{F}(\tau_k)$ is exponential with mean $(\!\sum_{i=1}^n \Lambda_i)^{-1}$ , expression (2) is further equal to

(3) \begin{align}&\mathbb{E}\Bigg[\int_0^\infty\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_k+x}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\Bigg(\sum_{i=1}^n \Lambda_i \Bigg) \mathrm{e}^{-x \sum_{i=1}^n \Lambda_i}\, \mathrm{d} x\, \big| \,\mathcal{F}(\tau_k)\Bigg]\nonumber \\[5pt]&=\int_0^\infty\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_k+x}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\mathbb{E} \Bigg[ \Bigg(\sum_{i=1}^n \Lambda_i \Bigg) \mathrm{e}^{-x \sum_{i=1}^n \Lambda_i} \, \big| \, \mathcal{F}(\tau_k) \Bigg]\, \mathrm{d} x\nonumber \\[5pt]&=- \int_0^\infty\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_k+x}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\bigg(\frac{\mathrm{d}}{\mathrm{d} x}\mathbb{E} \big[ \mathrm{e}^{-x \sum_{i=1}^n \Lambda_i} \, \big| \, \mathcal{F}(\tau_k) \big]\bigg)\, \mathrm{d} x\nonumber \\[5pt]&=- \int_0^\infty\exp\!\Bigg\{{-}a \sum_{i=1}^n\int_{\tau_k}^{\tau_k+x}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t\Bigg\}\nonumber \\[5pt]& \qquad \qquad\times \Bigg(\frac{\mathrm{d}}{\mathrm{d} x}\prod_{i=1}^n\frac{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k } \, \mathrm{d} F_i(\lambda)}\Bigg)\, \mathrm{d} x\end{align}

by Lemma 2.

In the following, we show that (3) equals ${1}/{(a+1)}$ . The idea is to observe that

\begin{equation*}\frac{\mathrm{d}}{\mathrm{d} t} \int_0^\infty \lambda^y \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)=\int_0^\infty \lambda^y \frac{\mathrm{d}\mathrm{e}^{-\lambda t}}{\mathrm{d} t} \, \mathrm{d} F_i(\lambda)=-\int_0^\infty \lambda^{1+y} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)\end{equation*}

for each i and $y\geq 0$ , where the interchangeability of integration and differentiation is guaranteed by Lebesgue’s dominated convergence theorem and our assumption that $\mathbb{E}[\Lambda_i]<\infty$ . Consequently, the integral with respect to t in (3) is, via integration by substitution, rewritten as follows:

\begin{align*}\int_{\tau_k}^{\tau_k+x}\frac{\int_0^\infty \lambda^{1+N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} t&=\bigg[{-} \log\!\left(\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda t} \, \mathrm{d} F_i(\lambda)\right)\bigg]_{t=\tau_k}^{t=\tau_k +x}\\[5pt]&=-\log\!\Bigg(\frac{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k} \, \mathrm{d} F_i(\lambda)}\Bigg).\end{align*}

Expression (3) is therefore equal to

\begin{align*}& - \int_0^\infty\Bigg\{\prod_{i=1}^n\frac{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k} \, \mathrm{d} F_i(\lambda)}\Bigg\}^a\Bigg(\frac{\mathrm{d}}{\mathrm{d} x}\prod_{i=1}^n\frac{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k } \, \mathrm{d} F_i(\lambda)}\Bigg)\, \mathrm{d} x\\[5pt]&=\frac{- \int_0^\infty\big\{\!\prod_{i=1}^n\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)\big\}^a\big\{\frac{\mathrm{d}}{\mathrm{d} x}\prod_{i=1}^n\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)\big\}\, \mathrm{d} x}{\big\{\!\prod_{i=1}^n\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k} \, \mathrm{d} F_i(\lambda)\big\}^{a+1}}\\[5pt]&=\frac{-\frac{1}{a+1}\left[\left\{\prod_{i=1}^n\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda (\tau_k+x)} \, \mathrm{d} F_i(\lambda)\right\}^{a+1}\right]_{x=0}^{x=\infty}}{\big\{\!\prod_{i=1}^n\int_0^\infty \lambda^{N_i(\tau_k)} \mathrm{e}^{-\lambda \tau_k} \, \mathrm{d} F_i(\lambda)\big\}^{a+1}}= \frac{1}{a+1},\end{align*}

which verifies our desired equality (1).

Proof of Theorem 1(i). Each increment of the discrete-time process $\{ U({\tau_k}) \}_{k\in\{1,2,\ldots \}} $ satisfies

(4) \begin{equation}U(\tau_k)- U(\tau_{k-1}) = (1+\theta) \mu I_k -X_k\end{equation}

for $k= 1, 2,\ldots$ . By Lemma 1, it also follows that $\mathcal{F}(\tau_{k-1}) \vee \sigma(X_1, \ldots, X_{k-1})$ , $I_k$ , and $X_k$ are independent, and thus the right-hand side of (4) is independent of $\mathcal{F}(\tau_{k-1})\vee \sigma(X_1, \ldots, X_{k-1})$ . This implies that the discrete-time process $\{ U(\tau_k) \}_{k \in \{0,1, \ldots \}}$ is a random walk. Moreover, each increment of the discrete-time process, that is, the right-hand side of (4), has a distribution that does not depend on either n or the distributions of the $\Lambda_i$ , because $I_k$ is independent of $X_k$ and is exponentially distributed with mean 1. This completes the proof of the first assertion in Theorem 1.

Proof of Proposition 1. The conditional distribution of $\tau_1$ given $\sigma(\Lambda_1, \ldots, \Lambda_n) $ is exponential with mean $\big(\!\sum_{i=1}^n \Lambda_i\big)^{-1}$ , and thus

\begin{align*}\mathbb{E} [ I_1 \mid \sigma(\Lambda_1, \ldots, \Lambda_n ) ]& = \int_0^\infty\Bigg\{\sum_{i=1}^n\int_0^t\frac{\int_0^\infty \lambda \mathrm{e}^{-\lambda u} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \mathrm{e}^{-\lambda u} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} u\Bigg\}\Bigg(\sum_{i=1}^n \Lambda_i \Bigg) \mathrm{e}^{-t \sum_{i=1}^n \Lambda_i}\, \mathrm{d} t\\[5pt]& = \int_0^\infty\Bigg\{\sum_{i=1}^n\int_0^{s \left(\sum_{i=1}^n \Lambda_i\right)^{-1}}\frac{\int_0^\infty \lambda \mathrm{e}^{-\lambda u} \, \mathrm{d} F_i(\lambda)}{\int_0^\infty \mathrm{e}^{-\lambda u} \, \mathrm{d} F_i(\lambda)}\, \mathrm{d} u\Bigg\}\mathrm{e}^{-s} \, \mathrm{d} s \quad \mbox{a.s.,}\end{align*}

where $s\;:\!=\;t \sum_{i=1}^n \Lambda_i$ . Moreover, it follows readily by definition that

\begin{equation*} \mathbb{E} [ U({\tau_1}) \mid \sigma(\Lambda_1, \ldots, \Lambda_n ) ]= u_0 + (1+\theta) \mu \mathbb{E} [ I_1 \mid \sigma(\Lambda_1, \ldots, \Lambda_n ) ]- \mathbb{E} [ X_1 ],\end{equation*}

and $\mathbb{E} [U({\tau_1}) \mid \sigma(\Lambda_1, \ldots, \Lambda_n ) ]$ is therefore almost surely equal to a strictly decreasing function of $\sum_{i=1}^n \Lambda_i$ . Because $\sum_{i=1}^n \Lambda_i$ is non-constant by our assumption, the random variable $U({\tau_1})$ is not independent of the $\Lambda_i$ .

4. Second proof via stochastic calculus

In this section we use stochastic calculus to provide another proof of our key lemma.

Proof of Lemma 1. For each $i\in \{ 1,\ldots, n \}$ , the process

(5) \begin{equation}\bigg\{ N_i(t)- \int_0^t \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s \bigg\}_{t \in [0,\infty)}\end{equation}

is an $\mathcal{F}_i$ -martingale. Indeed, for $0\leq u<t$ ,

\begin{align*} \mathbb{E} \bigg[ N_i(t) & - \int_0^t \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s \mid \mathcal{F}_i(u) \bigg] - \bigg\{ N_i(u)- \int_0^u \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s \bigg\} \\[5pt] & \quad= \mathbb{E} [ N_i(t)- N_i(u) \mid \mathcal{F}_i(u) ] - \mathbb{E} \bigg[ \int_u^t \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s \mid \mathcal{F}_i(u) \bigg] \\[5pt] & \quad= \mathbb{E} \big[ \mathbb{E} [ N_i(t)- N_i(u) \mid \sigma(\Lambda_i) \vee \mathcal{F}_i(u) ] \mid \mathcal{F}_i(u) \big] - \int_u^t \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(u) ] \, \mathrm{d} s \\[5pt] & \quad = \mathbb{E} [ (t-u)\Lambda_i \mid \mathcal{F}_i(u) ] - (t-u) \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(u) ] = 0 \quad \mbox{a.s.}\end{align*}

Because every $\mathcal{F}_i$ -martingale is also an $\mathcal{F}$ -martingale, the process (5) is an $\mathcal{F}$ -martingale. By aggregation, we see that the process $\big\{ N(t)- \int_0^t \sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s \big\}_{t \in [0,\infty)}$ is also an $\mathcal{F}$ -martingale, and hence the $\mathcal{F}$ -compensator of the point process N is $A(t) \;:\!=\; \int_0^t \sum_{i=1}^n \mathbb{E} [ \Lambda_i \mid \mathcal{F}_i(s) ] \, \mathrm{d} s$ . Because this compensator has continuous paths, it follows from the Grigelionis (Reference Grigelionis1977) Poisson reduction result (see also Kallenberg Reference Kallenberg2017, Corollary 9.30) that the time-changed process $N(A^{-1}(\cdot))$ , where $A^{-1}(s) \;:\!=\; \inf\{ t \geq 0 \mid A(t) > s \}$ , $s\in [0,\infty)$ , is an $\mathcal{F}(A^{-1}$ $(\cdot))$ -Poisson process with intensity parameter 1. Consequently, for each $k\geq 0$ , the random variable $I_{k+1}=A(\tau_{k+1})-A(\tau_k)$ is independent of $\mathcal{F}(\tau_k)$ and is exponentially distributed with mean 1.