2. Preliminaries

In this section, we obtain some results for Laplace transforms which we will apply in the proofs of Theorems 3.1 and 3.2, and it is convenient to deal with the ordinary and modified models separately. Throughout, we use the notation $\tilde a$ to denote the Laplace transform of a function a, with a second tilde when we consider a bivariate Laplace transform.

2.1. The ordinary SA model

By considering whether or not the surplus level ever falls below u, we have

\begin{equation*} \phi(u)=\phi(0)+\int_0^u g(0,y)\, \phi(u-y)\, dy\end{equation*}

(see, e.g. Willmot & Lin., Reference Willmot and Lin2001), so that

(1) \begin{equation} \tilde\phi(s) = \frac{\phi(0)}{s} + \tilde g(0,s)\, \tilde \phi(s)\end{equation}

giving

(2) \begin{equation} \frac{1}{1-\tilde g(0,s)}= \frac{s\tilde\phi(s)}{\phi(0)} = \frac{\tilde \phi_d(s)}{\phi(0)}+1\end{equation}

where $\phi_d(u)=\frac{d}{du}\,\phi(u)$ .

Second, the density of the time of ruin satisfies

(3) \begin{equation}w(u,t)= \int_0^t \int_0^u w(0,y,\tau)\, w(u-y,t-\tau)\, dy\, d\tau + \int_u^\infty w(0,y,t)\, dy \end{equation}

(see Dickson, Reference Dickson2008), and we define the bivariate Laplace transform of w(u, t) as

\begin{equation*} \tilde{\tilde w}(s,\delta)= \int_0^\infty \int_0^\infty e^{-su-\delta t}\, w(u,t)\, dt\,du \,.\end{equation*}

Further defining

\begin{equation*} \tilde{\tilde w}(0,s,\delta)= \int_0^\infty \int_0^\infty e^{-sy-\delta t}\, w(0,y,t)\, dy\,dt \end{equation*}

to be the bivariate Laplace transform of w(0, y, t) and defining

\begin{equation*} \tilde{\tilde{\mathscr{w}}}_1(s,\delta)= \int_0^\infty \int_0^\infty e^{-su-\delta t}\, \int_u^\infty w(0,y,t)\,dy\, dt\, du\,, \end{equation*}

we see that

(4) \begin{equation}\tilde{\tilde w}(s,\delta)= \tilde{\tilde w}(0,s,\delta)\,\tilde{\tilde w}(s,\delta) + \tilde{\tilde{\mathscr{w}}}_1(s,\delta) \,.\end{equation}

We next consider derivatives of these transforms with respect to $\delta$ as well as the transforms evaluated at $\delta=0$ . First,

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde w}(s,\delta)\bigg|_{\delta=0} = (\!-\!1)^n \int_0^\infty e^{-su}\, \int_0^\infty t^n\,w(u,t)\, dt\,du = (\!-\!1)^n \, \tilde\psi_n(s)\,.\end{equation*}

Next we have

\begin{equation*} \tilde{\tilde w}(0,s,\delta)\bigg|_{\delta=0}= \int_0^\infty e^{-sy}\,\int_0^\infty w(0,y,t)\, dt\,dy =\int_0^\infty e^{-sy}\,g(0,y)\,dy=\tilde g(0,s) \end{equation*}

and

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde w}(0,s,\delta)\bigg|_{\delta=0} = (\!-\!1)^n \int_0^\infty e^{-sy}\, \int_0^\infty t^n\,w(0,y,t)\, dt\,dy =(\!-\!1)^n\, \tilde\nu_n(s) \end{equation*}

where for $n=0,1,2,\ldots$ we define

(5) \begin{equation} \nu_n(y) = \int_0^\infty t^n\,w(0,y,t)\, dt\,.\end{equation}

We note that $\nu_0(y)=g(0,y)$ .

Finally,

\begin{equation*} \tilde{\tilde{\mathscr{w}}}_1(s,\delta)\bigg|_{\delta=0} = \int_0^\infty e^{-su}\, \int_u^\infty \int_0^\infty w(0,y,t)\,dt\, dy\, du = \tilde{\bar G}(0,s) \end{equation*}

where

\begin{equation*} \bar G(0,u)=\int_u^\infty g(0,y)\, dy \end{equation*}

and

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde{\mathscr{w}}}_1(s,\delta)=(\!-\!1)^n \, \int_0^\infty e^{-su}\, \int_u^\infty \int_0^\infty t^n\,w(0,y,t)\, dt\, dy\,du = (\!-\!1)^n \, \tilde{\bar N}_n(s) \end{equation*}

where for $n=1,2,3,\ldots$ we define

\begin{equation*} \bar N_n(u)=\int_u^\infty \nu_n(x)\, dx\,.\end{equation*}

2.2. The modified SA model

For the modified SA model, we need a smaller number of preliminary results. Using the arguments that underpin equation (3), we find that

(6) \begin{equation}w^m(u,t)= \int_0^t \int_0^u w^m(0,y,\tau)\, w(u-y,t-\tau)\, dy\, d\tau + \int_u^\infty w^m(0,y,t)\, dy \end{equation}

and we define

\begin{equation*} \tilde{\tilde w}^m(0,s,\delta)= \int_0^\infty \int_0^\infty e^{-sy-\delta t}\, w^m(0,y,t)\, dy\,dt \end{equation*}

to be the bivariate Laplace transform of $w^m(0,y,t)$ . Next define

\begin{equation*} \tilde{\tilde{\mathscr{w}}}_1^m(s,\delta)= \int_0^\infty \int_0^\infty e^{-su-\delta t}\, \int_u^\infty w^m(0,y,t)\,dy\, dt\, du\,, \end{equation*}

so that

(7) \begin{equation}\tilde{\tilde w}^m(s,\delta)= \tilde{\tilde w}^m(0,s,\delta)\,\tilde{\tilde w}(s,\delta) + \tilde{\tilde{\mathscr{w}}}_1^m(s,\delta) \,.\end{equation}

We then have

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde w}^m(s,\delta)\bigg|_{\delta=0} = (\!-\!1)^n \int_0^\infty e^{-su}\, \int_0^\infty t^n\,w^m(u,t)\, dt\,du = (\!-\!1)^n \, \tilde\psi^m_n(s)\,,\end{equation*}

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde w}^m(0,s,\delta)\bigg|_{\delta=0} = (\!-\!1)^n \int_0^\infty e^{-sy}\, \int_0^\infty t^n\,w^m(0,y,t)\, dt\,dy =(\!-\!1)^n\, \tilde\nu_n^m(s) \end{equation*}

where for $n=0,1,2,\ldots$ we define

\begin{equation*} \nu_n^m(y) = \int_0^\infty t^n\,w^m(0,y,t)\, dt\,,\end{equation*}

and

\begin{equation*} \frac{d^n}{d\delta^n}\, \tilde{\tilde{\mathscr{w}}}_1^m(s,\delta)=(\!-\!1)^n \, \int_0^\infty e^{-su}\, \int_u^\infty \int_0^\infty t^n\,w^m(0,y,t)\, dt\, dy\,du = (\!-\!1)^n \, \tilde{\bar N}_n^m(s) \end{equation*}

where for $n=0,1,2,\ldots$ we define

\begin{equation*} \bar N_n^m(u)=\int_u^\infty \nu_n^m(x)\, dx\,.\end{equation*}

3. The Main Results

We now derive expressions for moments of the time of ruin for ordinary and modified SA models in Theorems 3.1 and 3.2, respectively.

Theorem 3.1. For $n=1,2,3,\ldots$ , the nth moment of the time of ruin for the ordinary SA risk model is given by

(8) \begin{equation}\psi_n(u) = A_n(u)+\frac{1}{\phi(0)}\, \int_0^u A_n(x)\, \phi_d(u-x)\,dx \end{equation}

where

(9) \begin{equation} A_n(u)= \bar N_n(u)+ \sum_{j=1}^n \binom{n}{j} \int_0^u \nu_j(x)\, \psi_{n-j}(u-x)\,dx\,.\end{equation}

Proof. Differentiate equation (4) n times with respect to $\delta$ using Liebniz rule (e.g. Abramowitz & Stegun, Reference Abramowitz and Stegun1965) to get

\begin{align*}\frac{d^n}{d\delta^n}\,\tilde{\tilde w}(s,\delta)&= \tilde{\tilde w}(0,s,\delta)\,\frac{d^n}{d\delta^n}\,\tilde{\tilde w}(s,\delta)+ \sum_{j=1}^n \binom{n}{j} \frac{d^j}{d\delta^j}\, \tilde{\tilde w}(0,s,\delta)\, \frac{d^{n-j}}{d\delta^{n-j}}\,\tilde{\tilde w}(s,\delta)+ \frac{d^n}{d\delta^n}\,\tilde{\tilde{\mathscr{w}}}_1(s,\delta) \,,\end{align*}

and setting $\delta =0$ we obtain

\begin{align*}(\!-\!1)^n\, \tilde \psi_n(s) &= (\!-\!1)^n\,\tilde g(0,s) \,\tilde \psi_n(s) + \sum_{j=1}^n \binom{n}{j} (\!-\!1)^j\, \tilde \nu_j(s)\, (\!-\!1)^{n-j}\,\tilde\psi_{n-j}(s) + (\!-\!1)^n \, \tilde{\bar N}_n(s)\end{align*}

giving

(10) \begin{equation}\tilde \psi_n(s) = \frac{ \tilde A_{n}(s)}{1-\tilde g(0,s)} \end{equation}

where $A_n(u)$ is as defined in the statement of the theorem. Then, using (2) to replace $1/(1-\tilde g(0,s))$ in (10) we have for $n=1,2,3,\ldots$

\begin{align*}\psi_{n}(u) = A_{n}(u)+\frac{1}{\phi(0)}\, \int_{0}^{u} A_{n}(x)\, \phi_{d}(u-x)\,dx \,.\end{align*}

Corollary 3.1. When the initial surplus is 0, for $n=0,1,2,\ldots$ we have

\begin{equation*} \psi_n(0)=A_n(0)=\int_0^\infty\int_0^\infty t^n\, w(0,y,t)\, dt\,dy\,=\int_0^\infty t^n\, w(0,t)\,dt.\end{equation*}

Theorem 3.2. For $n=1,2,3,\ldots$ , the nth moment of the time of ruin for the modified SA risk model is given by

(11) \begin{equation} \psi_n^m(u)= \sum_{j=0}^n \binom{n}{j} \int_0^u \nu_j^m(x)\, \psi_{n-j}(u-x)\,dx + {\bar N}_n^m(u)\,.\end{equation}

Proof. Differentiate equation (7) n times with respect to $\delta$ using Liebniz rule to get

\begin{align*}\frac{d^n}{d\delta^n}\,\tilde{\tilde w}^m(s,\delta)&= \sum_{j=0}^n \binom{n}{j} \frac{d^j}{d\delta^j}\, \tilde{\tilde w}^m(0,s,\delta)\, \frac{d^{n-j}}{d\delta^{n-j}}\,\tilde{\tilde w}(s,\delta) + \frac{d^n}{d\delta^n}\,\tilde{\tilde{\mathscr{w}}}_1^m(s,\delta) \,,\end{align*}

and setting $\delta=0$ we get

\begin{equation*} (\!-\!1)^n\, \tilde \psi_n^m(s)= \sum_{j=0}^n \binom{n}{j} (\!-\!1)^j\, \tilde \nu_j^m(s)\, (\!-\!1)^{n-j}\,\tilde\psi_{n-j}(s) + (\!-\!1)^n \, \tilde{\bar N}_n^m(s)\end{equation*}

giving

\begin{equation*} \psi_n^m(u)= \sum_{j=0}^n \binom{n}{j} \int_0^u \nu_j^m(x)\, \psi_{n-j}(u-x)\,dx + {\bar N}_n^m(u)\,.\end{equation*}

To prove Theorems 3.1 and 3.2, we did not need to make any distributional assumptions. However, our third result applies only when the equilibrium distribution of K is of a particular form.

Theorem 3.3. Suppose that $k^e(t)=\sum_{i=1}^r p_i\, \kappa_i(t)$ where $\sum_{i=1}^r p_i = 1$ , and $\{\kappa_i\}_{i=1}^r$ are density functions. Let $\psi_n^{m_i}(u)$ denote the nth moment of the time of ruin for a modified SA process which has $\kappa_i$ as the density function of the time to the first claim for $i=1,2,3,\ldots,r$ . Then for $n=0,1,2,\ldots$

\begin{equation*} \psi_n^e(u)=\sum_{i=1}^r p_i \, \psi_n^{m_i}(u)\,.\end{equation*}

If there is value of i such that $\kappa_i(t)=k(t)$ , then we replace $\psi_n^{m_i}(u)$ with $\psi_n(u)$ .

We remark that we do not require positive values for each $p_i$ .

Proof. By conditioning on the time and the amount of the first claim, we have

\begin{align*}W^{m_i}(0,y,t)&=\int_0^t \kappa{_i}(s)\int_0^{cs} f(x)\, W(cs-x,y,t-s)\,dx\,ds + \int_0^t \kappa{_i}(s)\int_{cs}^{cs+y} f(x)\,\,dx\,ds\end{align*}

for $i=1,2,3,\ldots,r$ . Similarly,

\begin{align*}W^e(0,y,t)&=\int_0^t k^e(s)\int_0^{cs} f(x)\, W(cs-x,y,t-s)\,dx\,ds + \int_0^t k^e(s)\int_{cs}^{cs+y} f(x)\,\,dx\,ds \\[5pt]&= \sum_{i=1}^r p_i\, \int_0^t \kappa{_i}(s)\int_0^{cs} f(x)\, W(cs-x,y,t-s)\,dx\,ds\\[5pt]&\phantom{=}\,+ \sum_{i=1}^r p_i\,\int_0^t \kappa{_i}(s)\int_{cs}^{cs+y} f(x)\,\,dx\,ds \\[5pt]&= \sum_{i=1}^r p_i\, W^{m_i}(0,y,t)\end{align*}

so that

\begin{equation*} w^e(0,y,t)= \sum_{i=1}^r p_i\, w^{m_i}(0,y,t) \end{equation*}

and hence by (6)

\begin{align*}w^e(u,t) &= \sum_{i=1}^r p_i \,\int_0^t \int_0^u w^{m_i}(0,y,\tau)\, w(u-y,t-\tau)\, dy\, d\tau +\sum_{i=1}^r p_i \, \int_u^\infty w^{m_i}(0,y,t)\, dy\\[5pt]&= \sum_{i=1}^r p_i \, w^{m_i}(u,t)\,.\end{align*}

Thus, for $n=0,1,2,\ldots$

\begin{equation*} \psi_n^e(u)= \int_0^\infty t^n\, w^e(u,t) \,dt= \sum_{i=1}^r p_i \, \int_0^\infty t^n\,w^{m_i}(u,t)\,dt = \sum_{i=1}^r p_i \,\psi_n^{m_i}(u)\,.\end{equation*}

In order to obtain explicit solutions from these results for particular distributions for inter-claim times and individual claim amounts, it is clear that we need to know something about w(0, y, t) and $w^m(0,y,t)$ , but as we shall see in the following sections, we do not need explicit formulae.

4. SA Models with Exponential Claims

We first consider any ordinary SA model with exponential individual claims, that is, with $f(x)=\alpha e^{-\alpha x}$ , $x>0$ , since in this case we have the same general form of both the Laplace transform of $T_u$ and $\psi(u)$ regardless of the form of k whenever the negative solution of Lundberg’s fundamental equation exists. Specifically,

(12) \begin{equation} \mbox{E}\left[ e^{-\delta T_u}\,I(T_u<\infty)\right]=\left(1-\frac{R(\delta)}{\alpha}\right)e^{-R(\delta) u}\end{equation}

where $-R(\delta)$ is the negative solution of Lundberg’s fundamental equation. See Willmot & Woo (Reference Willmot and Woo2017). The special case $\delta=0$ gives $\psi(u)=\psi(0)e^{-Ru}$ where $R(=R(0))$ is the adjustment coefficient.

As $w(0,y,t)=w(0,t)\, \alpha e^{-\alpha y}$ by the memory-less property of the exponential distribution,

\begin{equation*} \nu_n(y)=\int_0^\infty t^n\, w(0,t)\, \alpha e^{-\alpha y}\, dt = \psi_n(0)\, \alpha e^{-\alpha y} \end{equation*}

and

\begin{equation*} \bar N_n(u)= \int_u^\infty \nu_n(y)\, dy = \psi_n(0)\, e^{-\alpha u} \,.\end{equation*}

Setting $n=1$ in equation (9) gives

\begin{align*}A_1(u)& =\bar N_1(u)+\int_0^u \nu_1(x)\, \psi(u-x)\, dx =\psi_1(0)\,e^{-Ru}\,,\end{align*}

and then equation (8) gives

\begin{align*}\psi_1(u)&=A_1(u)+\frac{1}{\phi(0)}\, \int_0^u A_1(x)\, \phi_d(u-x)\,dx \\[5pt]&= \psi_1(0) \, e^{-Ru} \left( 1+ \alpha u \psi(0)\right)\,.\end{align*}

One way to find the moments of $T_0$ is to set $u=0$ in (12) and to note that

\begin{equation*} \psi_n(0)=(\!-\!1)^{n+1}\, \frac{d^n}{d\delta^n} \frac{R(\delta)}{\alpha}\bigg|_{\delta=0}\,,\end{equation*}

and we can find the derivatives of $R(\delta)$ from Lundberg’s fundamental equation given the distribution of $V_1$ . For example, using formulae given by Dickson & Hipp (Reference Dickson and Hipp2001) for the first two derivatives of $R(\delta)$ evaluated when $\delta=0$ for the Erlang(2) risk model, we find that $\psi_1(0)=4.0744$ and $\psi_2(0)=187.4743$ when $c=1.2$ , $\alpha=1$ and the Erlang scale parameter is $\beta=2$ .

It is easily verified that when $n=2$ , equations (9) and (8) give

\begin{equation*} A_2(u)=\psi_2(0)\, e^{-Ru} + 2 \psi_1(0)^2 \, \alpha u\, e^{-Ru}\end{equation*}

and

\begin{equation*} \psi_2(u)=e^{-Ru}\bigg(\psi_2(0)+ \left(2\psi_1(0)^2+ \psi_2(0)\psi(0) \right)\alpha u + \psi(0) (\psi_1(0)\alpha u)^2\bigg)\,.\end{equation*}

We remark that Borovkov & Dickson (Reference Borovkov and Dickson2008) obtained a general expression for w(u, t) in terms of the convolutions of the density k, and that in some cases, for example Erlang inter-claim times, an explicit solution for w(u, t) exists. It is not an easy task to find the moments of the time of ruin directly from such solutions when $u>0$ ; this point is amply demonstrated by Drekic & Willmot (Reference Drekic and Willmot2003) who consider the case of the classical risk model.

For a modified SA model, we have $ w^m(0,y,t)=w^m(0,t)\, \alpha e^{-\alpha y}$ so that $ \nu^m_n(y)= \psi_n^m(0)\, \alpha e^{-\alpha y} $ and $ \bar N^m_n(u)= \psi_n^m(0)\, e^{-\alpha u}$ . Setting $n=1$ in equation (11) gives

\begin{equation*} \psi_1^m(u)=\bar N_1^m(u)+ \int_0^u \nu_0^m(x)\, \psi_{1}(u-x)\,dx +\int_0^u \nu_1^m(x)\, \psi_{0}(u-x)\,dx \end{equation*}

leading to

\begin{equation*} \psi_1^m(u)= \left(\psi_1^m(0)+ \psi_0^m(0)\psi_1(0) \alpha u\right)e^{-Ru}\,.\end{equation*}

Similarly, setting $n=2$ in equation (11) gives the final result as

\begin{equation*} \psi_2^m(u)= \left(\psi_2^m(0)+\left(\psi_0^m(0)\psi_2(0)+2\psi_1^m(0)\psi_1(0)\right)\alpha u+\psi_0^m(0)(\psi_1(0)\alpha u)^2\right)e^{-Ru}\,.\end{equation*}

We discuss an approach to finding $\psi_n^m(0)$ in section 7, and in section 8, we show how $\psi_n^m(0)$ can be obtained for some modified Erlang risk models.

5. The Classical Risk Model

We now consider the classical risk model, so that $\{N(t)\}_{t\ge0}$ is a Poisson process with parameter $\lambda$ . For this model, we can apply formula (8) for individual claim amount distributions for which the factorisation of f introduced by Willmot (Reference Willmot2007) holds, namely

(13) \begin{equation} f(x+y)=\sum_{j=1}^m \eta_j(x)\, \tau_j(y)\end{equation}

for functions $\{\eta_j\,, \tau_j\}_{j=1}^m$ . Willmot (Reference Willmot2007) shows that if the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter then this factorisation applies. See also Willmot & Woo (Reference Willmot and Woo2007) for further applications.

The bivariate Laplace transform of w(0, y, t) is

\begin{equation*}\int_0^\infty \int_0^\infty e^{-\delta t-sy}\, w(0,y,t)\, dy\, dt = \frac{\lambda}{c}\,\int_0^\infty \int_0^\infty e^{-\rho t-sy}\, f(t+y)\, dy\, dt\end{equation*}

where $\rho=\rho(\delta)$ is the unique positive solution of $\lambda + \delta - c s= \lambda \tilde f(s)$ , which is Lundberg’s fundamental equation for this model. See Gerber & Shiu (Reference Gerber and Shiu1998) for details. When (13) applies

\begin{align*}\int_0^\infty \int_0^\infty e^{-\delta t-sy}\, w(0,y,t)\, dy\, dt &= \frac{\lambda}{c}\,\int_0^\infty \int_0^\infty e^{-\rho t-sy}\, \sum_{j=1}^m \eta_j(t)\, \tau_j(y)\,dy\,dt\\[5pt]&= \frac{\lambda}{c}\, \sum_{j=1}^m \tilde \eta_j(\rho)\, \tilde\tau_j(s)\end{align*}

and so w(0, y, t) is of the form $ w(0,y,t)= \sum_{j=1}^m h_j(t)\,\tau_j(y)$ where $h_j(t)=\lambda \eta_j(t)/c$ .

To evaluate $\nu_n(y)$ from (5), we require quantities $H_j^n$ defined as

\begin{equation*} H_j^n=\int_0^\infty t^n\, h_j(t)\,dt\end{equation*}

for $n=1,2,3,\ldots$ and $j=1,2,3,\ldots,m,$ and we can find these quantities without identifying $\{h_j\}_{j=1}^m$ . We have

(14) \begin{equation} \int_0^\infty t^n\, h_j(t)\,dt = (\!-\!1)^n\, \frac{d^n}{d\delta^n} \,\tilde h_j(\rho)\bigg|_{\delta=0} \end{equation}

since $\rho=0$ when $\delta=0$ ; see Gerber & Shiu (Reference Gerber and Shiu1998). Further, the derivatives of $\rho$ when $\delta =0$ are easily found, and the first two are given in Dickson (Reference Dickson2016).

Example 5.1. Let $\lambda=1$ , $c=1.2$ and $f(x)=\frac{4}{3}\, e^{-2x}+\frac{1}{6}\, e^{-x/2}$ . Then

\begin{align*}f(x+y)&= \tfrac{2}{3}\, e^{-2x}\, 2e^{-2y} + \tfrac{1}{3}\, e^{-x/2}\, \tfrac{1}{2}\, e^{-y/2}\end{align*}

and

\begin{equation*} w(0,y,t)= h_1(t) \,2e^{-2y} + h_2(t)\, \tfrac{1}{2}\, e^{-y/2}\,,\end{equation*}

where

\begin{equation*} \tilde h_1(\delta)=\frac{5}{9(2+\rho)}\quad\mbox{and}\quad \tilde h_2(\delta)=\frac{5}{9(1+2\rho)}\,.\end{equation*}

Using (14) these lead to $H_1^1= \frac{25}{36}$ and $H_2^1=\frac{50}{9}$ . Further,

\begin{equation*} \psi(u)= 0.7990\, e^{-R_1 u}+0.0343\,e^{-R_2 u}\end{equation*}

where $R_1=0.1069$ and $R_2= 1.5598$ , leading to

\begin{equation*} A_1(u)= 6.2317\, e^{-R_1 u}+0.0183\, e^{-R_2 u}\end{equation*}

and

\begin{equation*} \psi_1(u)= (7.6152+3.1922u)\,e^{-R_1 u} - (1.3652-0.0059u)\,e^{-R_2 u}\,,\end{equation*}

in agreement with Example 6.1 of Lin & Willmot (Reference Lin and Willmot2000). Similarly we find that $H_1^2= 55\frac{5}{9}$ and $H_2^2=527\frac{7}{9}$ leading to

\begin{equation*} A_2(u) = (584.79+49.793u)\,e^{-R_1 u}-(1.4547-0.0063)\,e^{-R_2 u}\end{equation*}

and

\begin{align*}\psi_2(u)&= (705.93+360.36u+12.753u^2)\,e^{-R_1 u} -(122.60+0.4630u-0.0010u^2)\,e^{-R_2 u}\,.\end{align*}

6. Phase-Type(2) Risk Models

We can follow a process very similar to that in the previous section when the distribution of times between claims is phase-type(2) (also called Coxian(2)). However, we must now impose extra conditions on the individual claim amount distribution beyond the factorisation (13), but these conditions are not particularly restrictive. Following Dickson & Li (Reference Dickson and Li2010), we consider the situation when the functions $\{\eta_j\}_{j=1}^m$ satisfy the factorisation

(15) \begin{equation} \eta_j(x+y)=\sum_{i=1}^n \xi_{ij}(x)\, \zeta_{ij}(y)\,.\end{equation}

In the previous section, we made use of the unique positive solution $\rho$ of Lundberg’s fundamental equation for the classical risk model. We know from Ji & Zhang (Reference Ji and Zhang2012) that in the phase-type(2) risk model, there are two distinct positive solutions to Lundberg’s fundamental equation, and we can use these in a similar way.

If a density k belongs to the phase-type(2) class, its Laplace transform is of the form

(16) \begin{equation} \tilde k(s)=\frac{\lambda_1(1-p)s+\lambda_1 \lambda_2}{(s+\lambda_1)(s+\lambda_2)}\end{equation}

where $0<p\le1$ , $\lambda_i>0$ , for $i=1,2$ and $\lambda_2 \ne \lambda_1(1-p)$ . (See, e.g. Willmot & Woo, Reference Willmot and Woo2017). From Li & Garrido (Reference Li and Garrido2005), we know that

\begin{align*}\int_0^\infty\int_0^\infty e^{-\delta t- sy}\, w(0,y,t)\, dy\,dt &=\sum_{j=1}^2 b_j \int_0^\infty\int_0^\infty e^{-r_j t- sy}\,f(t+y)\,dy\,dt\end{align*}

where $r_1=r_1(\delta)$ and $r_2=r_2(\delta)$ are the positive solutions of Lundberg’s fundamental equation,

\begin{equation*} b_1 = \frac{\lambda^*+(\delta-cr_1)\chi}{c^2(r_2-r_1)}\quad\mbox{and}\quad b_2 = \frac{\lambda^*+(\delta-cr_2)\chi}{c^2(r_1-r_2)}\,,\end{equation*}

$\lambda^*=\lambda_1\lambda_2$ and $\chi=\lambda_1(1-p)$ .

Using the factorisation (13) of f, we get

\begin{align*}\int_0^\infty\int_0^\infty e^{-\delta t- sy}\, w(0,y,t)\, dy\,dt& =\sum_{j=1}^2 b_j \sum_{l=1}^m \tilde\eta_l(r_j)\,\tilde\tau_l(s) = \sum_{l=1}^m \sum_{j=1}^2 b_j \,\tilde\eta_l(r_j)\,\tilde\tau_l(s)\end{align*}

and so $w(0,y,t)= \sum_{l=1}^m \eta_l^*(t)\, \tau_l(y)$ where

\begin{align*}\tilde\eta_l^*(\delta)&= \sum_{j=1}^2 b_j \, \tilde\eta_l(r_j) \\[5pt]&= \frac{\lambda^*+(\delta-cr_1)\chi}{c^2(r_2-r_1)} \, \tilde\eta_l(r_1) - \frac{\lambda^*+(\delta-cr_2)\chi}{c^2(r_2-r_1)} \, \tilde\eta_l(r_2) \\[5pt]&= \frac{\lambda^*+\delta\chi}{c^2} \, \frac{\tilde\eta_l(r_1)-\tilde\eta_l(r_2)}{r_2 - r_1} - \frac{\chi}{c} \, \frac{r_1\,\tilde\eta_l(r_1)-r_2\,\tilde\eta_l(r_2)}{r_2 - r_1} \\[5pt]&=\frac{\lambda^*+\delta\chi}{c^2} \, \frac{\tilde\eta_l(r_1)-\tilde\eta_l(r_2)}{r_2 - r_1} - \frac{\chi}{c} \, \frac{\tilde\eta'_{\!\!l}(r_1)-\tilde\eta'_{\!\!l}(r_2)}{r_2 - r_1} \,.\end{align*}

We know that for a function $\alpha$ ,

\begin{equation*} \int_0^\infty\int_0^\infty e^{-rx-sy}\,\alpha(x+y)\, dy\,dx =\frac{\tilde\alpha(r)-\tilde\alpha(s)}{s-r}\end{equation*}

(e.g. Dickson & Hipp, Reference Dickson and Hipp2001) so when factorisation (15) applies

\begin{align*}\frac{\tilde\eta_l(r_1)-\tilde\eta_l(r_2)}{r_2 - r_1} &= \int_0^\infty\int_0^\infty e^{-r_1 x-r_2y}\,\eta_l(x+y)\, dy\,dx\\[5pt]&= \int_0^\infty\int_0^\infty e^{-r_1 x-r_2y}\,\sum_{i=1}^n \xi_{il}(x)\, \zeta_{il}(y)\, dy\,dx\\[5pt]&= \sum_{i=1}^n \tilde\xi_{il}(r_1)\, \tilde\zeta_{il}(r_2)\,.\end{align*}

Further, if a similar factorisation applies to $\eta'_{\!\!l}$ , say

\begin{equation*} \eta'_{\!\!l}(x+y)= \sum_{i=1}^k \vartheta_{il}(x)\, \varphi_{il}(y)\,,\end{equation*}

then

\begin{equation*} \tilde\eta_l^*(\delta) = \frac{\lambda^*+\delta\chi}{c^2} \, \sum_{i=1}^n \tilde\xi_{il}(r_1)\, \tilde\zeta_{il}(r_2) - \frac{\chi}{c} \, \sum_{i=1}^k \tilde\vartheta_{il}(r_1)\, \tilde\varphi_{il}(r_2)\, ,\end{equation*}

and we can differentiate this k times with respect to $\delta$ and then set $\delta=0$ to obtain quantities $H_l^n$ which are now defined as

\begin{equation*} H_l^n= \int_0^\infty t^n\, \eta_l^*(t)\,dt\,.\end{equation*}

We now illustrate ideas using the Erlang (2) distribution as our example. Although this is in some sense the simplest case (as we set $\lambda_1=\lambda_2=\beta$ and $p=1$ in (16), giving $\chi=0$ ), it is actually the case which yields most, as we can also obtain $\psi_n^e(u)$ in this case.

Example 6.1. We consider the same set-up as in Example 5.1, except that the claim inter-arrival times are Erlang(2) distributed and we set $\beta=2$ . The factorisation of f gave $\eta_1(x)= \tfrac{2}{3}\, e^{-2x}$ and $\eta_2(x)= \tfrac{1}{3}\, e^{-x/2}$ so we can set

\begin{equation*} \xi_{11}(x)= \tfrac{2}{3}\, e^{-2x},\quad \zeta_{11}(y)= e^{-2y},\quad \xi_{12}(x)= \tfrac{1}{3}\, e^{-x/2},\quad \zeta_{12}(y)= e^{-y/2}\,,\end{equation*}

resulting in $w(0,y,t)=\eta_1(t) \,2e^{-2y} + \eta_2(t)\, \tfrac{1}{2}\, e^{-y/2}$ where

\begin{equation*} \tilde \eta_1(\delta)=\tilde\xi_{11}(r_1)\, \tilde \zeta_{11}(r_2)= \frac{\beta^2}{c^2}\, \frac{2}{3(2+r_1)(2+r_2)}\end{equation*}

and

\begin{equation*} \tilde \eta_2(\delta)=\tilde\xi_{12}(r_1)\, \tilde \zeta_{12}(r_2)= \frac{\beta^2}{c^2}\,\frac{4}{3(1+2r_1)(1+2r_2)}\,. \end{equation*}

Differentiating $\tilde\eta_i(\delta)$ with respect to $\delta$ for $i=1,2$ , then setting $\delta=0$ , we find that $H_1^1= 0.5312$ and $H_2^1=6.0399$ . We also have

\begin{equation*} \psi(u)=0.7591\,e^{-R_1 u}+0.0304\,e^{-R_2 u} \end{equation*}

where $R_1=0.1253$ and $R_2 = 1.6806$ , leading to

\begin{equation*} A_1(u)= 6.5477\,e^{-R_1 u}+ 0.0233 \,e^{-R_2 u} \end{equation*}

and

\begin{equation*} \psi_1(u)=(7.5751+2.9576u)\,e^{-R_1 u} - (1.0040-0.0057u)\,e^{-R_2 u}\,.\end{equation*}

Taking the second derivatives of $\tilde\eta_i(\delta)$ for $i=1,2,$ we then find that $H_1^2 = 33.8510$ and $H_2^2=489.5552$ , giving

\begin{equation*} A_2(u)=(524.95+51.020u)\,e^{-R_1 u}-(1.5443-0.0087u)\,e^{-R_2 u}\end{equation*}

and

\begin{equation*} \psi_2(u)= (601.21 + 296.09 u + 11.523 u^2)\,e^{-R_1 u}-(77.807 + 0.3682 u - 0.0011 u^2)\,e^{-R_2 u}\,.\end{equation*}

To find $\psi_n^e(u),$ we apply results from Dickson & Li (Reference Dickson and Li2012) who consider modified Erlang( $\mathscr{n}$ ) risk models under which the distribution of the time to the first claim is Erlang(j) for $j=1,2,3,\ldots,\mathscr{n}-1$ . In the case of the modified Erlang(2) risk model under which the distribution of the time to the first claim is Erlang(1) with the same scale parameter $\beta$ , we find from formula (4.2) of Dickson & Li (Reference Dickson and Li2012) that $w^m(0,y,t)= \eta_1^m(t)\, 2e^{-2y} + \eta_2^m(t)\, \frac{1}{2} e^{-y/2}$ where

\begin{equation*} \tilde \eta_1^m(\delta)= \frac{2\beta}{3c}\left(\frac{1}{2+r_1}-\frac{\beta}{c}\, \frac{\sqrt{\frac{4}{3(2+r_1)}+\frac{1}{3(1+2r_1)}}}{(2+r_1)(2+r_2)}\right) \end{equation*}

and

\begin{equation*} \tilde \eta_2^m(\delta)= \frac{\beta}{3c}\left(\frac{2}{1+2r_1}-\frac{4\beta}{c}\, \frac{\sqrt{\frac{4}{3(2+r_1)}+\frac{1}{3(1+2r_1)}}}{(1+2r_1)(1+2r_2)}\right)\,. \end{equation*}

Extending the previous notation we find that $H_1^{0,m}= \tilde \eta_1^m(0)= 0.3560$ and $H_2^{0,m}= \tilde \eta_2^m(0)= 0.5212$ giving

\begin{equation*} \nu_0^m(x)=0.7119 \, e^{-2x}+1.0424\, e^{-x/2}\,.\end{equation*}

We next find that $H_1^{1,m}=0.3587$ and $H_2^{1,m}=3.5965$ , and from (11) we then find that

\begin{equation*} \psi_1^m(u)= (5.9902 + 3.1799 u)\,e^{-R_1 u}-(2.0350-0.0114u)\,e^{-R_2 u}\,.\end{equation*}

As the equilibrium distribution of the Erlang(2, $\beta$ ) distribution is an equally weighted average of the Erlang (1, $\beta$ ) and Erlang(2, $\beta$ ) distributions, Theorem 3.3 gives

\begin{align*} \psi_1^e(u)&= \tfrac{1}{2}\left( \psi_1(u)+\psi_1^m(u)\right) \\[5pt] &= (6.7826 + 3.0687 u)\,e^{-R_1 u} -(1.5195 - 0.0085 u) \,e^{-R_2 u}\,. \end{align*}

We then find $H_1^{2,m}=19.8204$ and $H_2^{2,m}=286.3038$ leading to

\begin{align*}\psi_2^m(u)&=(461.17+301.56u+12.389u^2)\,e^{-R_1 u} -(155.04+0.7464u-0.0021u^2)\,e^{-R_2 u}\end{align*}

and

\begin{align*}\psi_2^e(u)&=(531.19+298.83u+11.956u^2)\,e^{-R_1 u} -(116.42+0.5573 u-0.0016 u^2)\,e^{-R_2 u}.\end{align*}

We remark that the differentiation of $\tilde\eta_i(\delta)$ and $\tilde\eta_i^m(\delta)$ is easily done with mathematical software, and the derivatives of $r_i(\delta)$ when $\delta=0$ are easily computed – see Dickson & Hipp (Reference Dickson and Hipp2001).

The arguments that allow us to obtain the functions $\tilde\eta_i^m(\delta)$ for $i=1,2$ in the above example are given in section 4 of Dickson & Li (Reference Dickson and Li2012). Unfortunately, similar arguments do not seem to apply for other phase-type(2) risk models. However, in the next section we show how the negative solutions of Lundberg’s fundamental equation can be used to obtain the bivariate Laplace transform of $w^m(0,y,t)$ , giving us what we need to find $\psi_n^m(u)$ .

7. Modified SA Models

We now illustrate how we can find the Laplace transform of $w^m(0,y,t)$ (leading to the functions $\nu_n^m(u)$ ) using the negative solutions of Lundberg’s fundamental equation. We can do this by considering the Gerber-Shiu function with initial surplus 0 for a modified SA model of the form

\begin{align*} \varphi^m(0)&=\mbox{E}[\exp\{-\delta T_0^m-s Y_0^m\}\,I(T_0^m<\infty)]\\[5pt] &=\int_0^\infty e^{-\delta t}\, k_1(t)\int_0^{ct}f(x)\,\varphi(ct-x)\, dx\,dt + \int_0^\infty e^{-\delta t}\, k_1(t)\int_{ct}^\infty f(x)\, e^{-s(x-ct)}\, dx\,dt \end{align*}

where $ \varphi(u)=\mbox{E}[\exp\{-\delta T_u-s Y_u\}\,I(T_u<\infty)]$ is the equivalent Gerber-Shiu function for the ordinary SA model. The objective is to first solve for $\varphi(u)$ and then use this to express $ \varphi^m(0)$ as a sum of products of Laplace transforms in $\delta$ and s. We illustrate the idea in the next example. It seems difficult to give general results as we require an assumption for $k_1$ in order to solve for $ \varphi^m(0)$ .

Example 7.1. Suppose that in an ordinary SA model, the claim inter-arrival times are distributed as the sum of two independent exponential random variables with parameters $\lambda_1$ and $\lambda_2$ respectively where $\lambda_1<\lambda_2$ , and $f(x)=\sum_{i=1}^2 q_i\, \alpha_i \, e^{-\alpha_i x}$ where $q_1+q_2=1$ . Gerber & Shiu (Reference Gerber and Shiu2005) show that for this model

\begin{equation*} \varphi(u)=C_1\, e^{-R_1 u}+C_2\, e^{-R_2 u} \end{equation*}

where $-R_i (=-R_i(\delta))$ for $i=1,2$ are the negative solutions of Lundberg’s fundamental equation

(17) \begin{equation} (\lambda_1+\delta-cs)(\lambda_2+\delta-cs)=\lambda_1\, \lambda_2\, \tilde f(s)\,,\end{equation}

and

(18) \begin{equation} \sum_{k=1}^2 \frac{C_k}{\alpha_i-R_k}=\frac{1}{\alpha_i+s} \end{equation}

for $i=1,2$ . Equations (18) are exactly the same equations that are solved in the context of the classical risk model by Dickson & Drekic (Reference Dickson and Drekic2006), so we have

\begin{align*} C_j=\sum_{k=1}^2 \gamma_{j,k}(\delta)\,\frac{\alpha_k}{\alpha_k+s} \end{align*}

where

\begin{align*} \gamma_{1,1}(\delta)&=\frac{(\alpha_1-R_1)(\alpha_1-R_2)(\alpha_2-R_1)}{\alpha_1(R_2-R_1)(\alpha_1-\alpha_2)} \,,\quad \gamma_{1,2}(\delta)=\frac{-(\alpha_1-R_1)(\alpha_2-R_1)(\alpha_2-R_2)}{\alpha_1(R_2-R_1)(\alpha_1-\alpha_2)} \,,\\[5pt] \gamma_{2,1}(\delta)&=\frac{-(\alpha_1-R_1)(\alpha_1-R_2)(\alpha_2-R_2)}{\alpha_2(R_2-R_1)(\alpha_1-\alpha_2)} \,,\quad \gamma_{2,2}(\delta)=\frac{(\alpha_1-R_2)(\alpha_2-R_1)(\alpha_2-R_2)}{\alpha_2(R_2-R_1)(\alpha_1-\alpha_2)} \,. \end{align*}

Suppose that we want to find the moments $\psi_n^e(u)$ . As both k and $k^e$ are combinations of exponentials, it is sufficient to consider the case when $k_1(t)=\lambda e^{-\lambda t}$ for $t>0$ , as we can then find the required quantities using Theorems 3.2 and 3.3. We get

\begin{align*} \varphi^m(0)&= \sum_{i=1}^2 \sum_{j=1}^2 \int_0^\infty \lambda\,e^{-(\lambda+\delta) t}\int_0^{ct} q_i\, \alpha_i\, e^{-\alpha_i x}\,C_j\, e^{-R_j(ct-x)}\, dx\,dt \\[5pt] &\phantom{=}\, + \sum_{i=1}^2 \int_0^\infty \lambda\,e^{-(\lambda+\delta) t}\int_{ct}^\infty q_i\, \alpha_i\, e^{-\alpha_i x}\, e^{-s(x-ct)}\, dx\,dt \\[5pt] &= \lambda \sum_{i=1}^2 \sum_{j=1}^2 q_i\, \alpha_i\,C_j\,\int_0^\infty e^{-(\lambda+\delta+cR_j) t} \,\frac{1-e^{-(\alpha_i - R_j)ct}}{\alpha_i-R_j}\,dt \\[5pt] &\phantom{=}\, + \lambda \sum_{i=1}^2 q_i\, \alpha_i\,\int_0^\infty \frac{e^{-(\lambda+\delta+\alpha_i c) t}}{\alpha_i+s}\, dt \\[5pt] &= \lambda \sum_{i=1}^2 \sum_{j=1}^2 \frac{q_i\, \alpha_i\,C_j}{(\lambda+\delta+cR_j)(\alpha_i-R_j)} - \lambda \sum_{i=1}^2 \sum_{j=1}^2 \frac{q_i\, \alpha_i\, C_j}{(\lambda+\delta+\alpha_i c)(\alpha_i-R_j)} \\[5pt] &\phantom{=}\, + \lambda \sum_{i=1}^2 \frac{q_i\, \alpha_i}{(\lambda+\delta+\alpha_i c)(\alpha_i +s)} \\[5pt] &= \lambda \sum_{i=1}^2 \sum_{j=1}^2 \frac{q_i\, \alpha_i\,C_j \,c}{(\lambda+\delta+cR_j)(\lambda+\delta+\alpha_i c)} + \lambda \sum_{i=1}^2 \frac{q_i\, \alpha_i}{(\lambda+\delta+\alpha_i c)(\alpha_i +s)} \\[5pt] & = \lambda \sum_{i=1}^2 \sum_{j=1}^2 \frac{q_i\, \alpha_i\,c}{(\lambda+\delta+cR_j)(\lambda+\delta+\alpha_i c)} \sum_{k=1}^2 \gamma_{j,k}(\delta)\,\frac{\alpha_k}{\alpha_k+s}\\[5pt] &\phantom{=}\, + \lambda \sum_{i=1}^2 \frac{q_i\, \alpha_i}{(\lambda+\delta+\alpha_i c)(\alpha_i +s)} \,. \end{align*}

So

\begin{equation*} w^m(0,y,t)=\sum_{l=1}^2\eta_l(t)\, \alpha_l e^{-\alpha_l y}\end{equation*}

where

\begin{align*} \tilde\eta_l(\delta)&= \lambda \sum_{i=1}^2 \sum_{j=1}^2 \frac{q_i\, \alpha_i\,c\, \gamma_{j,l}(\delta)}{(\lambda+\delta+cR_j)(\lambda+\delta+\alpha_i c)} + \frac{\lambda\,q_l}{(\lambda+\delta+\alpha_l c)} \,. \end{align*}

We can evaluate $\tilde\eta_l(0)$ and by differentiating Lundberg’s fundamental equation (17) we can find the derivatives of $R_i(\delta)$ when $\delta =0$ , and we can insert these into the derivatives of $ \tilde\eta_l(\delta)$ to evaluate the quantities $H_l^{i,m}$ for $l=1,2$ and $i=1,2,3,\ldots$ . All other details required to find $\psi_n(u)$ for this example can be found in Gerber & Shiu (Reference Gerber and Shiu2005).

We remark that this approach can also be used to find the bivariate transform of w(0, y, t). For this, we simply require to find $\varphi(u)$ . However, as in the above example, the transforms with respect to $\delta$ may be complicated expressions in terms of the negative solutions of Lundberg’s fundamental equation.

It is also possible to use this approach to find an expression for the Laplace transform with respect to $\delta$ of the time of ruin for the modified process, and to then find the moments of the time of ruin for the modified process by differentiating this Laplace transform. However, this can be a very cumbersome process; interested readers can try to reproduce the results for $\psi_n^m(u)$ in Example 6.1 using this approach.

8. The Erlang( $\mathscr{n}$ ) Risk Model

In this section, we consider the Erlang( $\mathscr{n}$ ) risk model and take a different approach to finding the quantities for this model that are equivalent to $H_j^n$ in section 5 and $H_j^{n,m}$ in section 6. We extend ideas from Dickson & Li (Reference Dickson and Li2012), and we start with some definitions from that paper. For $m=0,1,2,\ldots$ let

\begin{equation*} p_m(t)=e^{-\beta t}\, \frac{(\beta t)^m}{m!}\,,\end{equation*}

where $\beta$ is the scale parameter of the Erlang( $\mathscr{n}$ ) distribution, and for $j=0,1,2,\ldots,\mathscr{n}-1$ define

\begin{equation*} g_{\mathscr{n},j}(x,t)=\sum_{r=1}^\infty p_{r\mathscr{n}+j}(t) f^{r*}(x) \end{equation*}

where f is still the density function for individual claims and $f^{r*}$ is the r-fold convolution of f with itself. Dickson & Li (Reference Dickson and Li2012) show that when the factorisation (13) holds, for a modified Erlang( $\mathscr{n}$ ) risk model under which the distribution of the time to the first claim is Erlang(j) with scale parameter $\beta$ , then, with $m_j$ denoting this modification, the joint density of the time of ruin and deficit at ruin when the initial surplus is zero is of the form

(19) \begin{equation} w^{m_j}(0,y,t)= \sum_{l=1}^m h_{j,l}(t)\, \tau_l(y)\,, \end{equation}

and this holds for $j=1,2,3,\ldots,\mathscr{n}-1$ . They give an expression for $\tilde h_{j,l}(\delta)$ (the final formula in their section 3.1), but this expression does not appear helpful for our purpose. It is convenient to adopt the notation $w^{m_\mathscr{n}}(0,y,t)$ for the joint density of $T_0$ and $Y_0$ for the ordinary Erlang( $\mathscr{n}$ ) risk model.

Using the arguments in Dickson & Li (Reference Dickson and Li2012), for this modified Erlang( $\mathscr{n}$ ) risk model claims occur at occurrences $j, \mathscr{n}+j,2\mathscr{n}+j,\ldots$ of an underlying Poisson process, and we find that

\begin{align*}w^{m_j}(0,y,t)&= \beta \, p_{j-1}(t)\, f(ct+y)+\beta \sum_{r=1}^\infty p_{r\mathscr{n}+j-1}(t)\,\int_0^{ct} f^{r*}(x)\,f(ct-x+y)\,dx \\[5pt]&\phantom{=}\, - c\, \sum_{i=0}^{\mathscr{n}-1} \int_0^t g_{\mathscr{n},j+i-\mathscr{n}}(cs,s)\, w^{m_{\mathscr{n}-i}}(0,y,t-s)\,ds\,.\end{align*}

The first two terms in the above consider the situation when a claim occurs in the interval ( $t,t+dt)$ , while the third term adjusts for upcrossings of the surplus process through level 0 prior to time t. Suppose the last upcrossing through 0 occurs with $\mathscr{n}-i$ Poisson events until the next claim, for $i=0,1,2,\ldots,\mathscr{n}-1$ . Then, we have a contribution $w^{m_{\mathscr{n}-i}}(0,y,t-s)$ . This means there have been i Poisson events from the time of the previous claim to time s, so the number of Poisson events to time s is $r\mathscr{n}+j+i$ for some r, $r=0,1,2,\ldots$ . Using (19) and (13) we get

\begin{align*}w^{m_j}(0,y,t)&= \beta \, p_{j-1}(t)\,\sum_{l=1}^m \eta_l(ct)\, \tau_l(y)+ \beta \sum_{r=1}^\infty p_{r\mathscr{n}+j-1}(t)\,\int_0^{ct} f^{r*}(x)\,\sum_{l=1}^m \eta_l(ct-x)\, \tau_l(y) \,dx\\[5pt]&\phantom{=}\, - c\, \sum_{i=0}^{\mathscr{n}-1} \int_0^t g_{\mathscr{n},j+i-\mathscr{n}}(cs,s)\, \sum_{l=1}^m h_{\mathscr{n}-i,l}(t-s)\, \tau_l(y)\,ds\,,\end{align*}

so that

(20) \begin{align}h_{j,l}(t) &= \beta \, p_{j-1}(t)\, \eta_l(ct) + \beta \sum_{r=1}^\infty p_{r\mathscr{n}+j-1}(t)\,\int_0^{ct} f^{r*}(x)\, \eta_l(ct-x) \,dx \nonumber\\[5pt]&\phantom{=}\, - c\, \sum_{i=0}^{\mathscr{n}-1} \int_0^t g_{\mathscr{n},j+i-\mathscr{n}}(cs,s)\, h_{\mathscr{n}-i,l}(t-s)\, ds \,.\end{align}

In order to find moments of the time of ruin for the modified processes, we require quantities we denote by $H_{j,l}^r$ given by

\begin{equation*} H_{j,l}^r = \int_0^\infty t^r \, h_{j,l}(t) \, dt\end{equation*}

for $r=0,1,2,\ldots$ . For brevity, we now write (20) as

(21) \begin{equation}h_{j,l}(t) = a_{j,l}(t) -\sum_{i=0}^{\mathscr{n}-1}\int_0^t \gamma_{j,i}(s)\, h_{\mathscr{n}-i,l}(t-s)\,ds \end{equation}

where

\begin{equation*} a_{j,l}(t) = \beta \, p_{j-1}(t)\, \eta_l(ct) + \beta \sum_{r=1}^\infty p_{r\mathscr{n}+j-1}(t)\,\int_0^{ct} f^{r*}(x)\, \eta_l(ct-x) \,dx \end{equation*}

and $ \gamma_{j,i}(s)=c g_{\mathscr{n},j+i-\mathscr{n}}(cs,s)$ . Further, for $r=0,1,2,\ldots$ let

\begin{equation*} A_{j,l}^r = \int_0^\infty t^r\, a_{j,l}(t)\, dt\qquad\mbox{ and }\qquad\Gamma^r_{j,i}=\int_0^\infty t^r\, \gamma_{j,i}(t)\,dt \,.\end{equation*}

Then from (21), we get

(22) \begin{align}H_{j,l}^r & = A_{j,l}^r -\sum_{i=0}^{\mathscr{n}-1}\int_0^\infty t^r\,\int_0^t \gamma_{j,i}(s)\, h_{\mathscr{n}-i,l}(t-s)\,ds\,dt \nonumber \\[5pt]&= A_{j,l}^r -\sum_{i=0}^{\mathscr{n}-1}\int_0^\infty \gamma_{j,i}(s)\, \int_s^\infty t^r\,h_{\mathscr{n}-i,l}(t-s)\,dt\,ds \nonumber \\[5pt]&= A_{j,i}^r -\sum_{i=0}^{\mathscr{n}-1}\int_0^\infty \gamma_{j,i}(s)\, \int_0^\infty (y+s)^r\,h_{\mathscr{n}-i,l}(y)\,dy\,ds \nonumber \\[5pt]&= A_{j,l}^r -\sum_{i=0}^{\mathscr{n}-1}\sum_{k=0}^r \binom{r}{k}\, \Gamma^{r-k}_{j,i}\, H^k_{\mathscr{n}-i,l} \,. \end{align}

Thus, we have a system of equations that we can use to find the quantities $H_{j,l}^r$ . We note that for $n=0,1,2,\ldots$

\begin{equation*} \psi^{m_j}_n(0)=\int_0^\infty t^n \int_0^\infty w^{m_j}(0,y,t)\,dy\,dt=\sum_{l=1}^m H_{j,l}^n\,.\end{equation*}

Table 1.

Values of $H_{j,i}^r$ .

Example 8.1. We consider the situation when the claim inter-arrival time distribution is Erlang(3, $\beta$ ), and the individual claim amount distribution is Erlang(2, $\alpha$ ), and we find the moments $\psi_n(u)$ and $\psi_n^e(u)$ for $n=1,2$ . As

\begin{equation*} k^e(t)= \frac{\beta}{3}\, e^{-\beta x}\sum_{j=0}^2 \frac {(\beta x)^j}{j!}= \frac{1}{3}\sum_{i=1}^3 \kappa_i(t)\end{equation*}

where $\kappa_i(t)$ is the Erlang(i, $\beta$ ) density, we can find the moments $\psi_n^e(u)$ using Theorem 3.3 by finding $\psi_n^{m_j}(u)$ , $j=1,2,3$ . The factorisation (13) gives

\begin{equation*} \eta_1(x)=\alpha x e^{-\alpha x}, \quad \eta_2(x)= e^{-\alpha x}, \quad \tau_1(y)=\alpha e^{-\alpha y}, \quad \tau_2(y)=\alpha^2 y e^{-\alpha y}\,,\end{equation*}

so that for $j=1,2,3$

\begin{equation*} w^{m_j}(0,y,t)=h_{j,1}(t)\, \tau_1(y) + h_{j,2}(t)\, \tau_2(y)\end{equation*}

and for the ordinary Erlang(3) risk model

(23) \begin{equation}g(0,y)=H_{3,1}^0\, \tau_1(y)+H_{3,2}^0\, \tau_2(y) \,.\end{equation}

For $j=1,2$ we find that

\begin{align*}A_{j,1}^r&= \sum_{m=0}^\infty \frac{\beta^{3m+j}}{(3m+j-1)!}\, \frac{(\alpha c)^{2m+1}}{(2m+1)!} \, \frac{(r+5m+j)!}{(\beta+\alpha c)^{r+5m+j+1}} \, \\[5pt]A_{j,2}^r&= \sum_{m=0}^\infty \frac{\beta^{3m+j}}{(3m+j-1)!}\, \frac{(\alpha c)^{2m}}{(2m)!} \, \frac{(r+5m+j-1)!}{(\beta+\alpha c)^{r+5m+j}} \,,\end{align*}

and for $i=0,1,2$

\begin{equation*} \Gamma_{j,i}^r =\sum_{m=1}^\infty \frac{\beta ^{3m+j+i-3}}{(3m+j+i-3)!}\, \frac{ (\alpha c)^{2m} }{(2m-1)!}\, \frac{(5m+j+i+r-4)!}{(\beta+\alpha c)^{5m+j+i+r-3}} \,.\end{equation*}

We now set the parameter values as $\alpha=2$ , $\beta=3$ and $c=1.2$ . Solving the system of equations (22) for $r=0,1,2$ yields the values in Table 1.

Noting that $\phi(0)=1-H_{3,1}^0-H_{3,2}^0=0.2640,$ we can use (1) and (23) to obtain $\phi(u)$ , from which

\begin{equation*} \psi(u)=0.7556 \, e^{-R_1 u} - 0.0196 \, e^{-R_2 u} \end{equation*}

where $R_1=0.3952$ and $R_2=2.6721$ . Applying the theorems from section 3, we obtain the first moments as

\begin{align*}\psi_1(u)&=(2.5005+3.1066u)\,e^{-R_1 u} +(0.2325+0.0027u)\,e^{-R_2 u} \,,\\[5pt]\psi_1^{m_2}(u)&=(1.4012+3.5976u)\,e^{-R_1 u} +(0.4866+0.0055u)\,e^{-R_2 u} \, ,\\[5pt]\psi_1^{m_1}(u)&=({-}0.1080+4.1663u)\,e^{-R_1 u} +(1.0179+0.0114u)\,e^{-R_2 u} \, ,\\[5pt]\psi_1^{e}(u)&=(1.2646+3.6235u)\,e^{-R_1 u} +(0.5790+0.0065u)\,e^{-R_2 u} \, ,\end{align*}

and the second moments as

\begin{align*}\psi_2(u)&=(64.500+99.264u+12.772u^2)\,e^{-R_1 u} + (5.3566-0.0625u-0.0004u^2)\,e^{-R_2 u} \, , \\[5pt]\psi_2^{m_2}(u)&=(33.322+102.67u+14.791 u^2)\,e^{-R_1 u} + (10.952-0.1307u-0.0007u^2)\,e^{-R_2 u} \, ,\\[5pt]\psi_2^{m_1}(u)&=({-}3.4108+104.66u+17.129u^2)\,e^{-R_1 u} + (22.386-0.2735u-0.0016u^2)\,e^{-R_2 u} \, , \\[5pt]\psi_2^{e}(u)&=(31.470+102.20u+14.897u^2)\,e^{-R_1 u} + (12.898-0.1556u-0.0009u^2)\,e^{-R_2 u} \,.\end{align*}

Figures 1 and 2 respectively show the mean and standard deviation of the time of ruin, given that ruin occurs, for both the ordinary and the equilibrium models. It is not surprising that the plots are close in each figure, given that the only difference for these risk processes is the distribution of the time to the first claim.

Figure 1

Mean time of ruin.

Figure 2

Standard deviation of time of ruin.

9. Concluding Remarks

We have presented general formulae for moments of the time of ruin in ordinary and modified Sparre Andersen risk models, and we have shown how they can be applied to find exact results. Our approach does not offer any advantage over the existing result by Lin & Willmot (Reference Lin and Willmot2000) for the classical risk model. However, it does provide a fairly simple way of finding solutions for $\psi_n(u)$ in phase-type(2) risk models. In the case of Erlang risk models, we are able to find moments in the ordinary, equilibrium, and some modified cases.

Application of our approach in sections 5 and 6 hinges on some knowledge of the functions w(0, y, t) and $w^m(0,y,t)$ , as well as the factorisations we have imposed. In practice, these factorisations are not serious impositions as many individual claim amount distributions satisfy them. The approach in section 7 is more generally applicable, but is likely to require more tedious differentiation than in sections 5 and 6. To apply the approach in section 8, we must be able to compute convolutions of the individual claim amount distribution. This is not a great restriction as we can do this when the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter, and Willmot & Woo (Reference Willmot and Woo2007) show that a wide variety of distributions are of this type.

An open question is how to find the quantity $\nu_n(y)$ given by (5), either analytically or numerically (which is perhaps more likely), for forms of f for which the factorisation (13) does not apply. If we can do this, then it should be possible to use numerical approaches to find $\psi_n(u)$ for $u>0$ . An approach for the case $u=0$ is presented in Dickson & Hipp (Reference Dickson and Hipp2001) where the inter-claim times are Erlang(2) distributed and the claim amount distribution is Pareto.