2. Preliminaries

In this section we describe the risk models discussed in this note, and as the models are well-known, the descriptions are brief. The first is the compound binomial model introduced by Gerber (Reference Gerber1988). This is a discrete time model where the surplus process $\{ U_{d} (n)\} _{{n{\equals}0}}^{\infty} $ is given by $$U_{d} (n){\equals}u{\plus}n{\minus}\mathop{\sum}\limits_{i{\equals}1}^n {X_{i} } $$ with U _d (0)=u, where u is a non-negative integer. The premium income per unit time is 1 and $\{ X_{i} \} _{{i{\equals}1}}^{\infty} $ is a sequence of independent and identically distributed (i.i.d.) random variables, distributed on the non-negative integers, and representing aggregate claims per unit time. The ultimate ruin probability for this model can be defined in two different ways. The definition used in this note is that used by Li & Sendova (Reference Li and Sendova2013), namely

$$\psi _{d} (u){\equals}{\rm Pr}(U_{d} (n)\,\lt\,0\,\;{\rm for}\,{\rm some}\,n,\quad \,n{\equals}1,2,3,\,\ldots\,)$$

but we could have equally used the definition given by Gerber (Reference Gerber1988) and reworked the results we use below from Li & Sendova (Reference Li and Sendova2013). The net profit condition for this model is $E[X_{1} ]\,\lt\,1$ . When this condition holds, $\psi _{d} (u)\,\lt\,1$ , and when it does not, ψ _d (u)=1. The time of ruin is defined as $$T_{{d,u}} {\equals}{\rm min}\{ n\geq 1\,\colon\,\;U_{d} (n)\,\lt\,0\} $$ , and the probability function of the time of ruin is $$w_{d} \left( {u,n} \right){\equals}{\rm Pr}\left( {T_{d,u} _{\equals}n} \right)\,{\rm for}\;n{\equals}1,2,3,\,\ldots\,$$ .

The second model we consider is the model introduced by Sparre Andersen (Reference Sparre Andersen1957). Let $\{ W_{i} \} _{{i{\equals}1}}^{\infty} $ be a sequence of i.i.d. random variables representing the times between claims (with W ₁ being the time to the first claim), and let $\{ Y_{i} \} _{{i{\equals}1}}^{\infty} $ be a sequence of i.i.d. random variables representing the amounts of the individual claims. The premium income per unit time is c>0 and the surplus immediately following payment of the nth claim, n=1, 2, 3, … , is $U(n){\equals}u{\plus}\mathop{\sum}\limits_{i{\equals}1}^n {(cW_{i} {\minus}Y_{i} )} $ where u≥0 is the initial surplus. The ultimate ruin probability is defined as $\psi (u){\equals}{\rm Pr}(U(n)\,\lt\,0{\rm }\;{\rm for}\;{\rm some }\;n,{\rm }\quad n{\equals}1,2,3,\,\ldots\,)$ . The net profit condition for this model is c>E[Y ₁]/E[W ₁], and when this condition holds ψ(u)<1. If the net profit condition does not hold, we have ψ(u)=1. The time of ruin is defined as $$T_{u} {\equals}{\rm inf}\{ t\,\colon\,\;U(t)\,\lt\,0\} $$ , and the defective density of the time of ruin is denoted as w(u,t) for t>0, where $w(u,t){\equals}d[{\rm Pr}(T_{u} \leq t)]\,/\,dt$ .

The simplest version of the Sparre Andersen model is the classical risk model in which the distribution of W ₁ is exponential, meaning that claim arrivals are modelled as a Poisson process.

We now state two results that feature later in this note. First, in a Sparre Andersen model for which the net profit condition holds and the individual claim amount distribution is exponential with mean 1/α

(2.1) $$\psi (u){\equals}(1{\minus}R\,/\,\alpha ){\rm e}^{{{\minus}Ru}} $$

where R is the adjustment coefficient, and is the unique positive number satisfying $E[{\rm e}^{{{\minus}cRW_{1} }} ]{\rm }E[{\rm e}^{{RY_{1} }} ]{\equals}1$ (see, e.g., Grandell, Reference Grandell1991). When the distribution of W ₁ is exponential with parameter λ, R=α−λ/c. Second, in the case when the distribution of W ₁ is exponential with parameter λ and the net profit condition holds, $\psi (0){\equals}\lambda E[Y_{1} ]\,/\,c$ (see, e.g., Grandell, Reference Grandell1991).

Most of this note involves Erlang distributions. That is because few explicit results exist for finite time ruin problems, and most of these for Sparre Andersen models involve Erlang distributions, either as the distribution for inter-claim times or for individual claim amounts (see Landriault et al., Reference Landriault, Shi and Willmot2011, and references therein).

3. A Useful Identity

The generalised negative binomial distribution was introduced by Jain & Consul (Reference Jain and Consul1971). A generalised negative binomial random variable, N, has a probability function defined in terms of parameters k, p and β as

$${\rm Pr}(N{\equals}n){\equals}\left( {\matrix{ {\beta n{\plus}k} \hfill \cr \,\,\,\,\,\,n \hfill \cr } } \right){k \over {\beta n{\plus}k}}{\rm }p^{n} {\rm }(1{\minus}p)^{{n(\beta {\minus}1){\plus}k}} \quad {\rm for}\;n{\equals}0,1,2,\,\ldots\,$$

where k>0, 0<p<1 and $\,\mid\,\beta {\rm }p\,\mid\,\,\lt\,1$ . As the sum over n of this probability function is 1, it follows that

$$(1{\minus}p)^{{{\minus}k}} {\equals}\mathop \sum\limits_{n{\equals}0}^\infty \left( {\matrix{ {\beta n{\plus}k} \hfill \cr \,\,\,\,\,\,n \hfill \cr } } \right){k \over {\beta n{\plus}k}}{\rm }p^{n} {\rm }(1{\minus}p)^{{n(\beta {\minus}1)}} {\equals}\left( {{\cal B}_{\beta } \left( {p(1{\minus}p)^{{\beta {\minus}1}} } \right)} \right)^{k} $$

giving the useful identity

(3.1) $${\cal B}_{\beta } \left( {p(1{\minus}p)^{{\beta {\minus}1}} } \right){\equals}(1{\minus}p)^{{{\minus}1}} $$

In everything that follows, β is a positive integer at least equal to 2, and equation (3.1) offers a means of evaluating generalised binomial series. However, we need to be careful with the application of (3.1). Consider the function f(p)=p(1−p) ^β−1 over the interval [0, 1]. We have f(0)=f(1)=0, and the function has a unique maximum at p=1/β. So, if we can actually write z=p(1−p) ^β−1 in equation (3.1), then there will be two solutions for p unless we are writing z in terms of the value of p that maximises f(p).

In sections 4–6, we can make use of each solution. Only one solution is really of interest, and that solution corresponds to the case when the net profit condition holds for the risk model we consider. We will see that the condition

(3.2) $$\,\mid\,\beta {\rm }p\,\mid\,\,\lt\,1$$

on the parameters of the generalised negative binomial distribution tells us which of the two solutions to select.

When p=1/β then (3.1) still holds, even though the condition (3.2) does not, with

$${\cal B}_{\beta } \left( {{{(\beta {\minus}1)^{{\beta {\minus}1}} } \over {\beta ^{\beta } }}} \right){\equals}{\beta \over {\beta {\minus}1}}$$

This follows from the first identity in (5.59) of Graham et al. (Reference Graham, Knuth and Patashnik1994). This is not a particularly significant result in the context of ruin problems, but we state it because in all the applications of (3.1) that follow, the case p=1/β corresponds to the premium equalling expected claims. In the case of two solutions, one solution corresponds to the premium exceeding expected claims, the other to the premium being less than expected claims.

4. The Compound Binomial Model

In this section we give a simple illustration of the usefulness of equation (3.1). Li & Sendova (Reference Li and Sendova2013) show that the probability function of the time of ruin when aggregate claims are geometrically distributed with probability function (1−q)q ^x for x=0, 1, 2, …  is

$$w_{d} (u,n){\equals}q^{{n{\plus}u{\plus}1}} (1{\minus}q)^{{n{\minus}1}} {\rm }{{u{\plus}2} \over {2n{\plus}u}}{\rm }\left( {\matrix{ {2n{\plus}u} \hfill \cr \,\,\,{n\,{\minus}1} \hfill \cr } } \right),\quad {\rm for}\;n{\equals}1,2,3,\,\ldots\,$$

They also show that when the net profit condition holds, that is q<1/2, $\psi _{d} (u){\equals}(q\,/\,(1{\minus}q))^{{u{\plus}2}} \;{\rm for}\;u{\equals}0,1,2,\,\ldots\,$ . We can obtain this result from their expression for w _d (u, n) as

$$\eqalignno{ \psi _{d} (u) &#x0026; {\equals}\mathop \sum\limits_{n{\equals}1}^\infty w_{d} (u,n){\equals}\mathop \sum\limits_{m{\equals}0}^\infty q^{{m{\plus}u{\plus}2}} (1{\minus}q)^{m} {{u{\plus}2} \over {2m{\plus}u{\plus}2}}\left( {\matrix{ {2m{\plus}u{\plus}2} \cr m \cr } } \right) \cr &#x0026;{\equals}q^{{u{\plus}2}} {\rm }{\cal B}_{2} (q(1{\minus}q))^{{u{\plus}2}} $$

When q<1/2, the net profit condition holds, and condition (3.2) is satisfied (with β=2 and p=q), so ${\cal B}_{2} (q(1{\minus}q))^{{u{\plus}2}} {\equals}(1{\minus}q)^{{{\minus}(u{\plus}2)}} $ and hence $\psi _{d} (u){\equals}(q\,/\,(1{\minus}q))^{{u{\plus}2}} $ .

When q>1/2, the net profit condition does not hold, nor is condition (3.2) satisfied, so ${\cal B}_{2} (q(1{\minus}q))^{{u{\plus}2}} {\equals}q^{{{\minus}(u{\plus}2)}} $ and hence ψ _d (u)=1. The same conclusion holds when q=1/2.

Although formula (1.1) gives an expression for ${\cal B}_{2} (z)$ , we did not need it. Our approach gives the solution in the case of interest, that is the net profit condition holds, as well as giving the solution when it does not.

5. Erlang (n) Risk Model with Exponential Claims

In this section, we consider an Erlang risk model with exponential claims. Specifically, we assume that W ₁~Erl(n, λ) so that E(W ₁)=n/λ and Y ₁~Erl(1, α). We consider two formulae for the density of the time of ruin and explain how the ultimate ruin probability can be derived from each of them.

5.1. The density of the time of ruin

We consider first a formula which is implicit in the literature, namely

(5.1) $$w(u,t){\equals}e^{{{\minus}\alpha u{\minus}(\lambda {\plus}\alpha c)t}} \mathop \sum\limits_{j{\equals}1}^\infty {{(\alpha u)^{{j{\minus}1}} (\lambda t)^{{nj}} (nj)} \over {(j{\minus}1)\,!\,}}\mathop \sum\limits_{m{\equals}0}^\infty {{(\alpha c\lambda ^{n} )^{m} {\rm }t^{{m(n{\plus}1){\minus}1}} } \over {m\,!\,[n(m{\plus}j)]\,!\,}}$$

The above formula is obtained from results stated in Nie et al. (Reference Nie, Dickson and Li2015). Special cases of this formula have been derived previously by Drekic & Willmot (Reference Drekic and Willmot2003) in the case n=1, and by Dickson & Li (Reference Dickson and Li2010) in the case n=2. We start with this formula for w(u, t) as the manipulations are more straightforward than those in the next subsection. We now give the main result of this subsection.

Theorem 5.1 The ultimate ruin probability is

(5.2) $$\psi (u){\equals}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm exp}\left\{ {{\minus}\alpha u\left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)} \right\}$$

where $Q_{n} {\equals}\alpha c\lambda ^{n} \,/\,(\lambda {\plus}\alpha c)^{{n{\plus}1}} $ .

Proof 5.1 Integrating expression (5.1) with respect to t over (0, ∞) gives ψ(u), and thus

$$\eqalignno{ \psi (u) &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{j{\equals}1}^\infty {{(\alpha u)^{{j{\minus}1}} \lambda ^{{nj}} (nj)} \over {(j{\minus}1)\,!\,}}\mathop \sum\limits_{m{\equals}0}^\infty {{(\alpha c\lambda ^{n} )^{m} {\rm \Gamma }(nj{\plus}m(n{\plus}1))} \over {m\,!\,[n(m{\plus}j)]\,!\,(\lambda {\plus}\alpha c)^{{nj{\plus}m(n{\plus}1)}} }} \cr &#x0026;{\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{j{\equals}1}^\infty {{(\alpha u)^{{j{\minus}1}} \lambda ^{{nj}} } \over {(j{\minus}1)\,!\,{\rm }(\lambda {\plus}\alpha c)^{{nj}} }}\mathop \sum\limits_{m{\equals}0}^\infty \left( {\matrix{ {(n{\plus}1)m{\plus}nj} \cr m \cr } } \right){{nj{\rm }Q_{n}^{m} } \over {(n{\plus}1)m{\plus}nj}} $$

As

$$\mathop \sum\limits_{m{\equals}0}^\infty \left( {\matrix{ {(n{\plus}1)m{\plus}nj} \cr m \cr } } \right){{nj{\rm }Q_{n}^{m} } \over {(n{\plus}1)m{\plus}nj}}{\equals}{\cal B}_{{n{\plus}1}} (Q_{n} )^{{nj}} $$

it follows that

(5.3) $$\eqalignno { \psi (u) &#x0026; {{\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{j{\equals}1}^\infty {\rm \ztbnd }{{(\alpha u)^{{j{\minus}1}} \lambda ^{{nj}} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{nj}} } \over {(j{\minus}1)\,!\,{\rm }(\lambda {\plus}\alpha c)^{{nj}} }}} \hskip101pt $$

$$&#x0026;{\equals}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm exp}\left\{ {{\minus}\alpha u\left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)} \right\} $$

Before we explain how equation (3.1) can be used to evaluate the generalised binomial series in (5.2), we remark that from formula (2.1), when the net profit condition holds, the adjustment coefficient is given by

(5.4) $$R{\equals}\alpha \left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)$$

5.1.1. The case n=1

When n=1, our risk model is the classical risk model. If we first make the usual assumption that the net profit condition holds, that is c>λ/α, then from (3.1) with β=2, we see that the condition $\,\mid\,2p\,\mid\lt\,\,1$ leads to the choice p=λ/(λ+αc). Hence ${\cal B}_{2} (Q_{1} ){\equals}(\lambda {\plus}\alpha c)\,/\,(\alpha c)$ , formula (5.2) gives the familiar result

$$\psi (u){\equals}{\lambda \over {\alpha c}}{\rm exp}\{ {\minus}(\alpha {\minus}\lambda \,/\,c)u\} $$

and formula (5.4) gives R=α−λ/c. However, if we make the assumption that $0\,\lt\,c\leq \lambda \,/\,\alpha $ , then the net profit condition does not hold, and we see that the correct choice is p=αc/(λ+αc). In this case ${\cal B}_{2} (Q_{1} ){\equals}(\lambda {\plus}\alpha c)\,/\,\lambda $ , leading to ψ(u)=1.

5.1.2 The case n>1

When the net profit condition c>λ/(nα) holds, condition (3.1) is satisfied by $\tilde{p}\,\lt\,1\,/\,(n{\plus}1)$ such that

(5.5) $$\tilde{p}(1{\minus}\tilde{p})^{n} {\equals}\alpha c\lambda ^{n} \,/\,(\lambda {\plus}\alpha c)^{{n{\plus}1}} $$

Then

(5.6) $$\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} {\equals}\left( {{\lambda \over {(\lambda {\plus}\alpha c)(1{\minus}\tilde{p})}}} \right)^{n} {\equals}{{(\lambda {\plus}\alpha c){\rm }\tilde{p}} \over {\alpha c}}{\equals}\psi (0)$$

and we see that $\psi (u){\equals}\psi (0){\rm exp}\{ {\minus}Ru\} $ with $R{\equals}\alpha {\minus}\tilde{p}(\alpha {\plus}\lambda \,/\,c)$ .

In the case when $0\,\lt\,c\,\lt\,\lambda \,/\,(n\alpha )$ , condition (3.1) is satisfied by p=αc/(λ+αc), resulting in B _n+1(Q _n )=(λ+αc)/λ, which gives ψ(u)=1. The same conclusion is reached when c=λ/(nα).

5.2. A second identity

Dickson et al. (Reference Dickson, Hughes and Zhang2005) obtained a different formula for w(u, t), and they integrate this formula to obtain an expression for ψ(u). We now show that this expression leads to the same results for ψ(u) as in the previous subsection. To do this, we make use of identity (5.61) from Graham et al. (Reference Graham, Knuth and Patashnik1994), namely

(5.7) $$\mathop \sum\limits_{k{\equals}0}^\infty \left( {\matrix{ {\tau k{\plus}r} \cr k \cr } } \right)z^{k} {\equals}{{{\cal B}_{\tau } (z)^{r} } \over {1{\minus}\tau {\plus}\tau {\rm }{\cal B}_{\tau } (z)^{{{\minus}1}} }}$$

Starting from formula (3.12) of Dickson et al. (Reference Dickson, Hughes and Zhang2005), and with the notation (a) _n =Γ(n+a)/Γ(a) (Pochhammer’s symbol), we have

(5.8) $$\eqalignno{ \psi (u) &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{m{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm{\plus}n}} \mathop \sum\limits_{j{\equals}0}^m {\rm \ztbnd }{{(nm{\plus}n)_{j} } \over {j\,!\,{\rm }(m{\minus}j)\,!\,}}{\rm }\left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{j} {\rm }(\alpha u)^{{m{\minus}j}} \cr &#x0026; \quad {\minus}{{\alpha cne^{{{\minus}\alpha u}} } \over {\lambda {\plus}\alpha c}}\mathop \sum\limits_{m{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm{\plus}2n}} \mathop \sum\limits_{j{\equals}0}^m {{(nm{\plus}2n{\plus}1)_{j} } \over {j\,!\,{\rm }(m{\minus}j)\,!\,}}{\rm }\left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{j} {\rm }(\alpha u)^{{m{\minus}j}} \cr &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{m{\equals}0}^\infty \mathop \sum\limits_{j{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{n(m{\plus}j){\plus}n}} {{(n(m{\plus}j){\plus}n)_{j} } \over {j\,!\,{\rm }m\,!\,}}\left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{j} (\alpha u)^{m} \cr &#x0026; \quad {\minus}{{\alpha cne^{{{\minus}\alpha u}} } \over {\lambda {\plus}\alpha c}}\mathop \sum\limits_{m{\equals}0}^\infty \mathop \sum\limits_{j{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{n(m{\plus}j){\plus}2n}} {{(n(m{\plus}j){\plus}2n{\plus}1)_{j} } \over {j\,!\,{\rm }m\,!\,}}\left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{j} (\alpha u)^{m} \cr &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{m{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm{\plus}n}} {{(\alpha u)^{m} } \over {m\,!\,}}\mathop \sum\limits_{j{\equals}0}^\infty {{{\rm \Gamma }(n(m{\plus}j){\plus}n{\plus}j)} \over {j\,!\,{\rm \Gamma }(n(m{\plus}j){\plus}n)}}Q_{n}^{j}$$

(5.9) $${\minus}{{\alpha cne^{{{\minus}\alpha u}} } \over {\lambda {\plus}\alpha c}}\mathop \sum\limits_{m{\equals}0}^\infty \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm{\plus}2n}} \,{{(\alpha u)^{m} } \over {m\,!\,}}\mathop \sum\limits_{j{\equals}0}^\infty {{{\rm \Gamma }(n(m{\plus}j){\plus}2n{\plus}1{\plus}j)} \over {j\,!\,{\rm \Gamma }(n(m{\plus}j){\plus}2n{\plus}1)}}Q_{n}^{j} $$

where Q _n =αcλ ⁿ /(λ+αc) ⁿ⁺¹ as before. Note that we have used the identity $$\mathop{\sum}\limits_{m{\equals}0}^\infty {\mathop{\sum}\limits_{j{\equals}0}^m {g(m,j)} } {\equals}\mathop{\sum}\limits_{m{\equals}0}^\infty {\mathop{\sum}\limits_{j{\equals}0}^\infty {g(m{\plus}j,j)} } $$ in the first step. Now the sum over j in (5.8) can be written as

$$\eqalign{\mathop \sum\limits_{j{\equals}0}^\infty \left( {\matrix{ {(n{\plus}1)j{\plus}n(m{\plus}1){\minus}1} \cr j \cr } } \right)Q_{n}^{j} $$

and from (5.7) this can be written as

$${{{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n(m{\plus}1){\minus}1}} } \over {(n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n}}$$

and similarly the sum over j in (5.9) is

$$\mathop \sum\limits_{j{\equals}0}^\infty \left( {\matrix{ {(n{\plus}1)j{\plus}n(m{\plus}2)} \cr j \cr } } \right)Q_{n}^{j} {\equals}{{{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n(m{\plus}2)}} } \over {(n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n}}$$

So we obtain

$$\eqalignno{ \psi (u) &#x0026; {\equals}{{e^{{{\minus}\alpha u}} {\rm }\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\minus}1}} } \over {(n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n}}\mathop \sum\limits_{m{\equals}0}^\infty {{\left( {\alpha u\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)^{m} {\cal B}_{{n{\plus}1}} (Q_{n} )^{{nm}} } \over {m\,!\,}} \cr &#x0026; \quad {\minus}{\rm }{{\alpha cn\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{2n}} {\cal B}_{{n{\plus}1}} (Q_{n} )^{{2n}} e^{{{\minus}\alpha u}} } \over {\lambda {\plus}\alpha c}}\mathop \sum\limits_{m{\equals}0}^\infty {{\left( {\alpha u\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)^{m} {\cal B}_{{n{\plus}1}} (Q_{n} )^{{nm}} } \over {m\,!\,}} \cr &#x0026; {\equals}{{\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\minus}1}} } \over {(n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n}}{\rm exp}\left\{ {{\minus}\alpha u\left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)} \right\} \cr &#x0026; \quad {\minus}{\rm }{{\alpha cn\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{2n}} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{2n}} } \over {(\lambda {\plus}\alpha c)((n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n)}}{\rm exp}\left\{ {{\minus}\alpha u\left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)} \right\} $$

giving

(5.10) $$\psi (u){\equals}\psi (0){\rm exp}\left\{ {{\minus}\alpha u\left( {1{\minus}\left( {{{\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )} \over {\lambda {\plus}\alpha c}}} \right)^{n} } \right)} \right\}$$

with

$$\eqalignno{ \psi (0) &#x0026; {\equals}{{\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\minus}1}} } \over {(n{\plus}1){\cal B}_{{n{\plus}1}} (Q_{n} )^{{{\minus}1}} {\minus}n}}\left( {1{\minus}{{\alpha cn\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} {\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\plus}1}} } \over {\lambda {\plus}\alpha c}}} \right) \cr &#x0026; {\equals}{{\left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{n} } \over {n{\plus}1{\minus}n{\cal B}_{{n{\plus}1}} (Q_{n} )}}\left( {1{\minus}nQ_{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\plus}1}} } \right) $$

From the first identity in (5.59) of Graham et al. (Reference Graham, Knuth and Patashnik1994), we know that

$$Q_{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\plus}1}} {\equals}{\cal B}_{{n{\plus}1}} (Q_{n} ){\minus}1$$

giving

$$1{\minus}nQ_{n} {\rm }{\cal B}_{{n{\plus}1}} (Q_{n} )^{{n{\plus}1}} {\equals}n{\plus}1{\minus}n{\cal B}_{{n{\plus}1}} (Q_{n} )$$

and so, as before, $\psi (0){\equals}(\lambda {\cal B}_{{n{\plus}1}} (Q_{n} )\,/\,(\lambda {\plus}\alpha c))^{n} $ and formula (5.10) for ψ(u) is the same as (5.2).

6. The Number of Claims Until Ruin

In this section, we consider the number of claims until ruin from initial surplus u, denoted by N _u , with probability function p _k (u)=Pr(N _u =k). Our aim here is to illustrate how identity (3.1) allows us to show that $\mathop{\sum}\limits_{k{\equals}1}^\infty {p_{k} (u){\equals}\psi (u)} $ in particular cases. We start by considering the model of the previous section, then consider an identity under the classical risk model.

6.1. Erlang (n) risk model

For the model considered in section 5, Frostig et al. (Reference Frostig, Pitts and Politis2012) show that

(6.1) $$p_{k} (0){\equals}{{\lambda ^{{nk}} {\rm }(\alpha c)^{{k{\minus}1}} {\rm }[(n{\plus}1)k{\minus}2]\,!\,} \over {(\lambda {\plus}\alpha c)^{{(n{\plus}1)k{\minus}1}} {\rm }(nk{\minus}1)\,!\,{\rm }k\,!\,}},\quad {\rm for}\;k{\equals}1,2,3,\,\ldots\,$$

See also Landriault et al. (Reference Landriault, Shi and Willmot2011) and Zhao & Zhang (Reference Zhao and Zhang2013). We now generalise this result to the case u>0, and the key reason for taking the following approach is that our proof also shows why $\mathop{\sum}\limits_{k{\equals}1}^\infty {p_{k} (u){\equals}\psi (u)} $ . From arguments in Dickson & Qazvini (Reference Dickson and Qazvini2016) concerning the classical risk model with exponential claims, we can easily show that the probability generating function of N _u for this model is

(6.2) $$P_{{N_{u} }} (r){\equals}E\left[ {r^{{N_{u} }} I(T_{u} \,\lt\,\infty)} \right]{\equals}\mathop \sum\limits_{m{\equals}1}^\infty {{(\alpha u)^{{m{\minus}1}} e^{{{\minus}\alpha u}} } \over {(m{\minus}1)\,!\,}}P_{{N_{0} }} (r)^{m} $$

where I is the indicator function. So, by deriving $P_{{N_{0} }} (r)$ we can find $P_{{N_{u} }} (r)$ , and we do this in the next two theorems. Before doing so, we remark that Landriault et al. (Reference Landriault, Shi and Willmot2011) obtained an expression (their formula (23)) for the probability function of the number of claims until ruin in a Sparre Andersen model where the inter-claim time distribution is an infinite mixture of Erlang distributions with the same scale parameter and the claim size distribution is exponential. Formula (6.4) for p _m (u) is new, although it can also be obtained by manipulating formula (23) of Landriault et al. (Reference Landriault, Shi and Willmot2011). We do not consider the more general probability function obtained by Landriault et al. (Reference Landriault, Shi and Willmot2011) because, as far as we can tell, summation of it does not lead to generalised binomial series other than in the special case that a single mixing weight is 1, that is, the case we consider of the Erlang distribution.

Theorem 6.1 The probability generating function of N ₀ is

(6.3) $$P_{{N_{0} }} (r){\equals}{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}{\cal B}_{{n{\plus}1}} \left( {{{\alpha cr\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{{n{\plus}1}} }}} \right)^{n} $$

Proof 6.1 Using expression (6.1) we have

$$\displaylines{ P_{{N_{0} }} (r){\equals}\mathop \sum\limits_{k{\equals}1}^\infty r^{k} p_{k} (0){\equals}\mathop \sum\limits_{k{\equals}1}^\infty {{(r\lambda ^{n} )^{k} {\rm }(\alpha c)^{{k{\minus}1}} {\rm }[(n{\plus}1)k{\minus}2]\,!\,} \over {(\lambda {\plus}\alpha c)^{{(n{\plus}1)k{\minus}1}} {\rm }(nk{\minus}1)\,!\,{\rm }k\,!\,}} \cr \hskip19pt {\equals}\mathop \sum\limits_{t{\equals}0}^\infty {{(r\lambda ^{n} )^{{t{\plus}1}} {\rm }(\alpha c)^{t} {\rm }[(n{\plus}1)(t{\plus}1){\minus}2]\,!\,} \over {(\lambda {\plus}\alpha c)^{{(n{\plus}1)(t{\plus}1){\minus}1}} {\rm }[n(t{\plus}1){\minus}1]\,!\,{\rm }(t{\plus}1)\,!\,}} \cr \hskip40pt {\equals}{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}\mathop \sum\limits_{t{\equals}0}^\infty \left( {{{\alpha cr\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{{n{\plus}1}} }}} \right)^{t} {{(nt{\plus}t{\plus}n{\minus}1)\,!\,} \over {[n(t{\plus}1){\minus}1]\,!\,{\rm }(t{\plus}1)\,!\,}} \cr \hskip93pt {\equals}{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}\mathop \sum\limits_{t{\equals}0}^\infty {{[(n{\plus}1)t{\plus}n]\,!\,} \over {[n(t{\plus}1){\minus}1]\,!\,{\rm }(t{\plus}1)\,!\,}}{\rm }{1 \over {(n{\plus}1)t{\plus}n}}{\rm }\left( {{{\alpha cr\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{{n{\plus}1}} }}} \right)^{t} $$

Now the sum in the previous line is ${\cal B}_{{n{\plus}1}} (\alpha cr\lambda ^{n} \,/\,(\lambda {\plus}\alpha c)^{{n{\plus}1}} )^{n} $ if

$${{[(n{\plus}1)t{\plus}n]\,!\,} \over {[n(t{\plus}1){\minus}1]\,!\,{\rm }(t{\plus}1)\,!\,}}{\equals}n\left( {\matrix{ {(n{\plus}1)t{\plus}n} \cr t \cr } } \right)$$

and this is easily seen to be an equality, giving (6.3).

Theorem 6.2 For m=1, 2, 3, … ,

(6.4) $$p_{m} (u){\equals}e^{{{\minus}\alpha u}} \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm}} {\rm }\mathop \sum\limits_{k{\equals}0}^{m{\minus}1} \left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{k} {{(\alpha u)^{{m{\minus}k{\minus}1}} } \over {(m{\minus}k{\minus}1)\,!\,}}\left( {\matrix{ {nm{\plus}k} \cr k \cr } } \right){{n(m{\minus}k)} \over {nm{\plus}k}}$$

Proof 6.2 Starting from expression (6.2) and inserting expression (6.3) for $P_{{N_{0} }} (r)$ , we have

(6.5) $$\eqalignno{ P_{{N_{u} }} (r) &#x0026; {\equals}\mathop \sum\limits_{m{\equals}1}^\infty {{(\alpha u)^{{m{\minus}1}} e^{{{\minus}\alpha u}} } \over {(m{\minus}1)\,!\,}}\left( {{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}} \right)^{m} {\cal B}_{{n{\plus}1}} \left( {{{\alpha cr\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{{n{\plus}1}} }}} \right)^{{mn}}$$

$$\eqalignno {\quad \hskip15pt{\equals}\mathop \sum\limits_{m{\equals}1}^\infty \mathop \sum\limits_{t{\equals}0}^\infty {{(\alpha u)^{{m{\minus}1}} e^{{{\minus}\alpha u}} } \over {(m{\minus}1)\,!\,}}\left( {{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}} \right)^{m} \left( {\matrix{ {(n{\plus}1)t{\plus}mn} \cr t \cr } } \right) \cr {\times}\left( {{{\alpha cr\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{{n{\plus}1}} }}} \right)^{t} {{mn} \over {(n{\plus}1)t{\plus}mn}} \hskip 90pt $$

We now use the identity

(6.6) $$\mathop \sum\limits_{m{\equals}1}^\infty \mathop \sum\limits_{t{\equals}0}^\infty g(m,t){\equals}\mathop \sum\limits_{m{\equals}1}^\infty \mathop \sum\limits_{t{\equals}0}^{m{\minus}1} g(t{\plus}1,\,m{\minus}t{\minus}1)$$

for some function g to obtain

(6.7) $$\eqalignno{ P_{{N_{u} }} (r) &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{m{\equals}1}^\infty \left( {{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}} \right)^{m} {\rm }\mathop \sum\limits_{t{\equals}0}^{m{\minus}1} \left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{{m{\minus}t{\minus}1}} {\rm }{{(\alpha u)^{t} } \over {t\,!\,}}{\rm } \cr &#x0026; \quad {\rm }{\times}\left( {\matrix{ {(n{\plus}1)m{\minus}t{\minus}1} \cr \hfill{m{\minus}t{\minus}1} \hfill \cr } } \right){{n(t{\plus}1)} \over {(n{\plus}1)m{\minus}t{\minus}1}} \cr &#x0026; {\equals}e^{{{\minus}\alpha u}} \mathop \sum\limits_{m{\equals}1}^\infty \left( {{{r\lambda ^{n} } \over {(\lambda {\plus}\alpha c)^{n} }}} \right)^{m} {\rm }\mathop \sum\limits_{k{\equals}0}^{m{\minus}1} \left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{k} {\rm }{{(\alpha u)^{{m{\minus}k{\minus}1}} } \over {(m{\minus}k{\minus}1)\,!\,}} \cr &#x0026; \quad {\rm }{\times}\left( {\matrix{ {nm{\plus}k} \cr k \cr } } \right){{n(m{\minus}k)} \over {nm{\plus}k}} $$

As the coefficient of r ^m in (6.7) is p _m (u), we obtain

$$p_{m} (u){\equals}e^{{{\minus}\alpha u}} \left( {{\lambda \over {\lambda {\plus}\alpha c}}} \right)^{{nm}} {\rm }\mathop \sum\limits_{k{\equals}0}^{m{\minus}1} \left( {{{\alpha c} \over {\lambda {\plus}\alpha c}}} \right)^{k} {\rm }{{(\alpha u)^{{m{\minus}k{\minus}1}} } \over {(m{\minus}k{\minus}1)\,!\,}}{\rm }\left( {\matrix{ {nm{\plus}k} \cr k \cr } } \right){{n(m{\minus}k)} \over {nm{\plus}k}}$$

It is straightforward to show that $\mathop{\sum}\limits_{m{\equals}1}^\infty {p_{m} (u){\equals}\psi (u)} $ because all we have to do is set r=1 in (6.7) and then work backwards through the proof of Theorem 6.2 to (6.5). The only difficult step without knowing the derivation of the probability function is to apply identity (6.6). The equation we obtain for ψ(u) is the same as (5.2), because when r=1 expressions (6.5) and (5.3) are identical.

6.2. Classical risk model

For the classical risk model, a case other than that of exponential claims in which we can find an explicit form for p _n (u) is when u=0 and individual claims are Erl(m, μ) distributed. In this case we can show from formulae (6) and (7) of Dickson (Reference Dickson2012) that

$$p_{n} (0){\equals}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(m(n{\minus}1){\plus}n{\plus}i{\minus}1)\,!\,(i{\plus}1)} \over {(n{\minus}1)\,!\,(m(n{\minus}1){\plus}i{\plus}1)\,!\,}}{{\lambda ^{n} {\rm }(\mu c)^{{m(n{\minus}1){\plus}i}} } \over {(\lambda {\plus}\mu c)^{{m(n{\minus}1){\plus}n{\plus}i}} }}$$

We can use this to obtain ψ(0) since

$$\eqalignno{ \mathop \sum\limits_{n{\equals}1}^\infty p_{n} (0) &#x0026; {\equals}\mathop \sum\limits_{n{\equals}1}^\infty \mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(m(n{\minus}1){\plus}n{\plus}i{\minus}1)\,!\,(i{\plus}1)} \over {(n{\minus}1)\,!\,(m(n{\minus}1){\plus}i{\plus}1)\,!\,}}\,{{\lambda ^{n} {\rm }(\mu c)^{{m(n{\minus}1){\plus}i}} } \over {(\lambda {\plus}\mu c)^{{m(n{\minus}1){\plus}n{\plus}i}} }} \cr &#x0026; {\equals}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(i{\plus}1)(\mu c)^{i} } \over {(\lambda {\plus}\mu c)^{i} }}\mathop \sum\limits_{n{\equals}1}^\infty {{(m(n{\minus}1){\plus}n{\plus}i{\minus}1)\,!\,} \over {(n{\minus}1)\,!\,(m(n{\minus}1){\plus}i{\plus}1)\,!\,}}\,{{\lambda ^{n} {\rm }(\mu c)^{{m(n{\minus}1)}} } \over {(\lambda {\plus}\mu c)^{{m(n{\minus}1){\plus}n}} }} \cr &#x0026; {\equals}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(i{\plus}1)(\mu c)^{{i{\minus}m}} } \over {(\lambda {\plus}\mu c)^{{i{\minus}m}} }}\mathop \sum\limits_{n{\equals}1}^\infty {{(m(n{\minus}1){\plus}n{\plus}i)\,!\,} \over {(n{\minus}1)\,!\,(m(n{\minus}1){\plus}i{\plus}1)\,!\,}}\,{{\tilde{Q}_{m}^{n} } \over {m(n{\minus}1){\plus}n{\plus}i}} $$

where $\tilde{Q}_{m} {\equals}\lambda (\mu c)^{m} \,/\,(\lambda {\plus}\mu c)^{{m{\plus}1}} $ . Then

(6.8) $$\eqalignno{ \mathop \sum\limits_{n{\equals}1}^\infty p_{n} (0) &#x0026; {\equals}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(i{\plus}1)(\mu c)^{{i{\minus}m}} } \over {(\lambda {\plus}\mu c)^{{i{\minus}m}} }}\mathop \sum\limits_{r{\equals}0}^\infty \left( {\matrix{ {mr{\plus}r{\plus}i{\plus}1} \cr r \cr } } \right){{\tilde{Q}_{m}^{{r{\plus}1}} } \over {mr{\plus}r{\plus}i{\plus}1}} \cr &#x0026; {\equals}\tilde{Q}_{m} \mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(\mu c)^{{i{\minus}m}} } \over {(\lambda {\plus}\mu c)^{{i{\minus}m}} }}\mathop \sum\limits_{r{\equals}0}^\infty \left( {\matrix{ {r(m{\plus}1){\plus}i{\plus}1} \cr r \cr } } \right){{(i{\plus}1){\rm }\tilde{Q}_{m}^{r} } \over {r(m{\plus}1){\plus}i{\plus}1}} \cr &#x0026;{\equals}\tilde{Q}_{m} {\rm }\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} {{(\mu c)^{{i{\minus}m}} } \over {(\lambda {\plus}\mu c)^{{i{\minus}m}} }}{\cal B}_{{m{\plus}1}} (\tilde{Q}_{m} )^{{i{\plus}1}} \cr &#x0026;{\equals}{{\lambda {\cal B}_{{m{\plus}1}} (\tilde{Q}_{m} )} \over {\lambda {\plus}\mu c}}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} \left( {{{\mu c{\cal B}_{{m{\plus}1}} (\tilde{Q}_{m} )} \over {\mu c{\plus}\lambda }}} \right)^{i} $$

The net profit condition is c>λm/μ, and when this holds we see from (3.2) with β=m+1 that p=λ/(μc+λ) and so

(6.9) $${\cal B}_{{m{\plus}1}} (\tilde{Q}_{m} ){\equals}(\lambda {\plus}\mu c)\,/\,(\mu c).$$

Then from (6.8) we obtain the expected result

$$\mathop \sum\limits_{n{\equals}1}^\infty p_{n} (0){\equals}{{\lambda m} \over {\mu c}}{\equals}\psi (0)$$

When c<λm/μ, ${\cal B}_{{m{\plus}1}} (\tilde{Q}_{m} ){\equals}(1{\minus}\tilde{p})^{{{\minus}1}} $ where $\tilde{p}\,\gt\,1\,/\,(m{\plus}1)$ and

(6.9) $$\tilde{p}(1{\minus}\tilde{p})^{m} {\equals}{\lambda \over {\mu c{\plus}\lambda }}\left( {{{\mu c} \over {\mu c{\plus}\lambda }}} \right)^{m} $$

Then

$$\mathop \sum\limits_{n{\equals}1}^\infty p_{n} (0){\equals}{\lambda \over {(\lambda {\plus}\mu c)(1{\minus}\tilde{p})}}\mathop \sum\limits_{i{\equals}0}^{m{\minus}1} \left( {{{\mu c} \over {(\lambda {\plus}\mu c)(1{\minus}\tilde{p})}}} \right)^{i} $$

and straightforward manipulation using (6.9) gives $\mathop{\sum}\limits_{n{\equals}1}^\infty {p_{n} (0){\equals}1} $ . The same argument applies when c=λm/μ.

7. Concluding Remarks

As indicated earlier, there is only a small number of explicit formulae for the density function of the time of ruin or the probability function of the number of claims until ruin in Sparre Andersen models. Many of these have similar forms to the functions considered in this note. Consequently, we can apply the ideas in this note to other cases. We have not tried to give an exhaustive list in this note; rather, we have sought to show how seemingly complicated expressions can easily be simplified if we can evaluate generalised binomial series.