The thresholds can be viewed as a criterion for classifying claims as large or small. In this section, we suppose that the random thresholds \(\{Q_{i},i=1,2,\dots\}\) follow a geometric distribution with p.f. \((1-p_{3})p_{3}^{n-1}\), \(n\in\mathbb{N}^{+}\), \(0< p_{3}<1\). We derive an explicit expression for the Gerber-Shiu function when the claim sizes belong to the discrete \(K_{m}\)-family and present some special cases.

### 4.1 The Gerber-Shiu function with \(K_{m}\)-family claim sizes

We assume that the claim sizes \(\{X_{i},i=1,2,\dots\}\) follow a distribution from discrete \(K_{m}\)-family, that is, the generating function of the density function \(b(\cdot)\) satisfies the following form:

$$ \hat{b}(z) =\frac{z[\prod_{i=1}^{m}(1-q_{i})+\sum_{j=1}^{m-1}\beta_{j}(z-1)^{j}]}{ \prod_{i=1}^{m}(1-q_{i}z)},\quad \mathbf{R}(z)< \min\biggl\{ \frac{1}{q_{i}},1\leq i \leq m\biggr\} , $$

(4.1)

where \(0< q_{i}<1\), \(i=1,2,\dots,m\), and the coefficients \(\beta_{1},\beta _{2},\dots,\beta_{m-1}\) are such that \(b(\cdot)\) is a p.f. This class of distributions includes, as particular cases, the shifted geometric, shifted or truncated negative binomial, and linear combinations of these; see Li [2] for more examples. Let \(p_{m}(z)=\prod_{i=1}^{m}(1-q_{i}z)\) and \(p_{m-1}^{*}(z)=\prod_{i=1}^{m}(1-q_{i})+\sum_{j=1}^{m}\beta_{j}(z-1)^{j}\). Then (4.1) becomes \(\hat{b}(z)=\frac{zp_{m-1}^{*}(z)}{p_{m}(z)}\).

Applying definition (2.2), it follows that

$$ \hat{\chi}(z)=\sum_{y=1}^{\infty}(zp_{3})^{y}b(y)= \hat{b}(p_{3}z). $$

On the other hand, we can rewrite generalized Lundberg’s equation (2.12) as

$$\begin{aligned} \hat{h}_{1,v}(z)-\hat {h}_{2,v}(z)={}& \bigl(z^{c}-vp_{1}\bigr) \bigl(z^{c}-vp_{2} \bigr)-v(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr) \hat{b}(z) \\ &{} -v\bigl[(1-p_{2}) \bigl(z^{c}-vp_{1} \bigr)-(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr)\bigr] \hat{\chi}(z) \\ ={}&0. \end{aligned}$$

(4.2)

We further denote \(q_{m}(z)=p_{m}(p_{3}z)\) and \(q_{m-1}^{*}(z)=p_{m-1}^{*}(p_{3}z)\). Then (4.2) can be rewritten as

$$\begin{aligned}& \bigl(z^{c}-vp_{1}\bigr) \bigl(z^{c}-vp_{2} \bigr)-v(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr) \frac{zp_{m-1}^{*}(z)}{p_{m}(z)} \\& \quad{} -v\bigl[(1-p_{2}) \bigl(z^{c}-vp_{1} \bigr)-(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr)\bigr] \frac{p_{3}zq_{m-1}^{*}(z)}{q_{m}(z)}=0 \end{aligned}$$

or, equivalently,

$$\begin{aligned} \bigl(z^{c}-vp_{1}\bigr) \bigl(z^{c}-vp_{2}\bigr)p_{m}(z)q_{m}(z)-v(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr)zp_{m-1}^{*}(z)q_{m}(z) \\ \quad{} -v p_{3}\bigl[(1-p_{2}) \bigl(z^{c}-vp_{1} \bigr)-(1-p_{1}) \bigl(z^{c}-vp_{2}\bigr) \bigr]zq_{m-1}^{*}(z)p_{m}(z)=0. \end{aligned}$$

(4.3)

It is obvious that the left side of equation (4.3) is a polynomial of degree \(2c+2m\) in *z* with leading coefficient \(\alpha=p_{3}^{m}\prod_{i=1}^{m}q_{i}^{2}\), which implies that there are \(2c+2m\) roots in total. According to Lemmas 1 and 2, exactly 2*c* roots are located in the unit circle \(\mathcal {C}\), denoted by \(z_{1}, z_{2}, \dots, z_{2c}\) as before. Therefore, we conclude that the remaining 2*m* roots must lie in the region \(\{z:\vert z\vert >1\}\), denoted by \(\rho_{1}, \rho_{2}, \dots, \rho_{2m}\). Since the left side of (4.3) is equivalent to \(p_{m}(z)q_{m}(z)[\hat{h}_{1,v}(z)-\hat{h}_{2,v}(z)]\), equation (4.3) can be reexpressed as

$$\begin{aligned} p_{m}(z)q_{m}(z)\bigl[\hat{h}_{1,v}(z)- \hat{h}_{2,v}(z)\bigr]&=\alpha p_{m}^{o}(z)q_{m}^{o}(z) \bigl[\hat{h}_{1,v}(z)-\hat{h}_{2,v}(z)\bigr] \\ &=\alpha\prod_{i=1}^{2c}(z-z_{i}) \prod_{l=1}^{2m}(z-\rho_{l}), \end{aligned}$$

(4.4)

where \(p_{m}^{o}(z)=\prod_{i=1}^{m}(z-\frac{1}{q_{i}})\) and \(q_{m}^{o}(z)=\prod_{i=1}^{m}(z-\frac{1}{p_{3}q_{i}})\) are both polynomials of degree 2*m* in *z* with the leading coefficient 1. Substituting (4.4) into (3.1), we have

$$ \hat{m}_{1}(z)=v\frac{p_{m}^{o}(z)q_{m}^{o}(z)}{\prod_{l=1}^{2m}(z-\rho _{l})}\hat{ \eta}_{1}(z)=v \biggl[1+\frac{S(z)}{D(z)} \biggr]\hat{ \eta}_{1}(z), $$

(4.5)

where \(S(z)=p_{m}^{o}(z)q_{m}^{o}(z)-\prod_{l=1}^{2m}(z-\rho_{l})\) is a polynomial of degree \(2m-1\) or less, and \(D(z)=\prod_{l=1}^{2m}(z-\rho _{l})\) is a polynomial of degree *m*. Further, if \(\rho_{1}, \rho_{2}, \dots, \rho_{2m}\) are distinct, then \(\hat{g}(z)\doteq\frac{S(z)}{D(z)}\) can be obtained by partial fractions as

$$ \hat{g}(z)=\sum_{i=1}^{2m} \frac{\sigma_{i}}{\rho_{i}-z}, $$

(4.6)

where

$$ \sigma_{i}=\frac{\prod_{j=1}^{m}(\frac{1}{q_{j}}-\rho_{i})\prod_{j=1}^{m}(\frac{1}{p_{3}q_{j}}-\rho_{i})}{\prod_{j=1,j\neq i}^{2m}(\rho _{j}-\rho_{i})},\quad i=1, 2, \ldots, 2m. $$

Thus, inverting the generating function of (4.6) gives

$$ g(y)=\sum_{i=1}^{2m} \sigma_{i}\rho_{i}^{-(y+1)},\quad y\in\mathbb{N}. $$

(4.7)

Utilizing (4.7), we can invert the generating function in (4.5) to

$$ m_{1}(u)=v \Biggl[\eta_{1}(u)+\sum _{i=0}^{u}g(u-i)\eta_{1}(i) \Biggr]. $$

Similarly, from (3.9) we obtain

$$ m_{2}(u)=v \Biggl[\eta_{2}(u)+\sum _{i=0}^{u}g(u-i)\eta_{2}(i) \Biggr]. $$

### 4.2 The generating function of the time to ruin with geometric distribution

In this section, we consider the generating function of the time to ruin, which is one of important quantities in risk analysis. Let the penalty function \(\omega(x_{1},x_{2})=1\) for all \(x_{1}\in\mathbb{N}\) and \(x_{2}\in\mathbb{N}^{+}\). Then the Gerber-Shiu function (1.3) reduces to

$$ \varphi_{i}(u)=E\bigl\{ v^{T_{i}}I(T_{i}< \infty) \vert U(0)=u\bigr\} ,\quad u\in\mathbb{N}, i=1,2. $$

We assume that the claim sizes \(\{X_{i}, i=1, 2, \ldots\}\) follow geometric distribution with p.f. \(b(y)=(1-q)q^{y-1}\), \(y\in\mathbb {N}^{+}\), and the generating function \(\hat{b}(z)=\frac{z(1-q)}{1-zq}\). By (2.1) and (2.2) some simple algebras lead to

$$ \hat{\chi}(z)=\frac{z p_{3}(1-q)}{1-p_{3}qz},\qquad \hat{\xi}(z)=\frac {z(1-q)(1-p_{3})}{(1-qz)(1-p_{3}qz)}. $$

(4.8)

Implementing the expressions into (4.8), generalized Lundberg’s equation (2.12) becomes

$$\begin{aligned}& \bigl(z^{c}-vp_{1}\bigr) \bigl(z^{c}-vp_{2} \bigr)-v(1-p_{2}) \bigl(z^{c}-vp_{1}\bigr) \frac {zp_{3}(1-q)}{1-p_{3}qz}\\& \quad{}-v(1-p_{1}) \bigl(z^{c}-vp_{2} \bigr) \frac{z(1-q)(1-p_{3})}{(1-qz)(1-p_{3}qz)}=0, \end{aligned}$$

which can be further rearranged to the following equation without changing the roots:

$$\begin{aligned}& \bigl(z^{c}-vp_{1}\bigr) \bigl(z^{c}-vp_{2}\bigr) \biggl(z-\frac{1}{p_{3}q}\biggr) \biggl(z-\frac{1}{q}\biggr)+\frac {v(1-q)(1-p_{2})}{q}\biggl(z-\frac{1}{q} \biggr) \bigl(z^{c}-vp_{1}\bigr) \\& \quad{} +\frac{v(1-p_{1})(1-q)(1-p_{3})z(z^{c}-vp_{2})}{q^{2}p_{3}}=0. \end{aligned}$$

(4.9)

As discussed in Section 4.1, there are two roots lying in the complex plane \(\{z:\vert z\vert >1\}\), denoted by \(\rho_{1}\), \(\rho_{2}\). Henceforth, we assume that \(\rho_{1}\neq\rho_{2}\). The other 2*c* roots are in the unit circle \(\mathcal {C}\), still denoted by \(z_{i}\), \(i=1,2,\dots,2c\). Moreover, we have

$$ \biggl(z-\frac{1}{p_{3}q}\biggr) \biggl(z-\frac{1}{q} \biggr)\bigl[\hat{h}_{1,v}(z)-\hat {h}_{2,v}(z)\bigr]=\prod _{i=1}^{2c}(z-z_{i}) (z- \rho_{1}) (z-\rho_{2}). $$

(4.10)

On the other hand, substituting (4.8) into the expression of \(\hat{\eta}_{1}(z)\) finally proves that

$$ \hat{\eta}_{1}(z)=- \biggl[\frac{\kappa_{1}}{z-\frac{1}{q}}+ \frac{\kappa _{2}}{z-\frac{1}{p_{3}q}} \biggr], $$

(4.11)

where the constants \(\kappa_{1}\) and \(\kappa_{2}\) are determined respectively by

$$\begin{aligned}& \kappa_{1}=(1-p_{1})\sum_{j=1}^{2c} \frac{z_{j}^{c}-vp_{2}}{\pi^{\prime}(z_{j})(qz_{j}-1)}, \\& \kappa_{2}=\frac{v(1-q)}{q}\sum_{u=0}^{c-1} \delta_{1}(u)\sum_{j=1}^{2c} \frac{z_{j}^{u}}{ \pi^{\prime}(z_{j})(p_{3}qz_{j}-1)}. \end{aligned}$$

Inserting (4.10) and (4.11) into (3.6), we obtain

$$ \hat{\varphi}_{1}(z)=-\frac{v [\kappa_{1}(z-\frac{1}{p_{3}q})+\kappa _{2}(z-\frac{1}{q}) ]}{(z-\rho_{1})(z-\rho_{2})}. $$

(4.12)

Using partial fractions, we rewrite (4.12) as

$$ \hat{\varphi}_{1}(z)=\sum_{i=1}^{2} \frac{\lambda_{i}}{\rho_{i}-z}, $$

(4.13)

where

$$ \lambda_{i}=\frac{v [\kappa_{1}(\rho_{i}-\frac{1}{p_{3}q})+\kappa_{2}(\rho _{i}-\frac{1}{q}) ]}{\rho_{i}-\rho_{j}},\quad i, j=1, 2, i\neq j. $$

Finally, the inversion of (4.11) yields

$$ \varphi_{1}(u)=\sum_{i=1}^{2} \lambda_{i}\rho_{i}^{-(u+1)}. $$

(4.14)

As for \(\varphi_{2}(u)\), direct calculations yield

$$\hat{\varphi}_{2}(z)=-\frac{v [\iota_{1}(z-\frac{1}{p_{3}q})-\iota _{2}z+\iota_{3} ]}{(z-\rho_{1})(z-\rho_{2})}, $$

where the constants \(\iota_{i}\), \(i=1, 2, 3\), are given by

$$\begin{aligned}& \iota_{1}=(1-p_{2})\sum_{j=1}^{2c} \frac{z_{j}^{c}-vp_{1}}{\pi^{\prime}(z_{j})(qz_{j}-1)}, \\& \iota_{2}=v(1-q) (1-p_{3})\sum _{u=0}^{c-1}\delta_{2}(u) \sum _{j=0}^{2c}\frac{z_{j}^{u+1}}{\pi^{\prime}(z_{j})(qz_{j}-1)(p_{3}qz_{j}-1)}, \\& \iota_{3}=\frac{v(1-q)(1-p_{3})}{q^{2}p_{3}}\sum_{u=0}^{c-1} \delta_{2}(u) \sum_{j=0}^{2c} \frac{z_{j}^{u}}{\pi^{\prime}(z_{j})(qz_{j}-1)(p_{3}qz_{j}-1)}. \end{aligned}$$

Consequently, \(\varphi_{2}(u)\) can be obtained by partial fractions as

$$ \varphi_{2}(u)=\sum_{i=1}^{2} \sigma_{i}\rho_{i}^{-(u+1)}, $$

(4.15)

where

$$ \sigma_{i}=\frac{v [\iota_{1}(\rho_{i}-\frac{1}{p_{3}q})-\iota_{2}\rho_{i}+\iota _{3} ]}{\rho_{i}-\rho_{j}},\quad i, j=1, 2, i\neq j. $$

### Example

In this example, we assume that the thresholds \(\{Q_{i}\}\) and claim sizes \(\{X_{i}\}\) both follow geometric distributions with parameters \(p_{3}=0.2\) and \(q=0.6\). Let \(c=1\), \(p_{1}=0.7\), \(p_{2}=0.8\), \(v=0.85\). It is easy to check that the positive loading condition is fulfilled. Generalized Lundberg’s equation has four roots: \(z_{1}=0.7410\), \(z_{2}=0.6601\), \(\rho _{1}=1.3692\), \(\rho_{2}=8.3914\). Furthermore, we use (4.14) and (4.15) to calculate the exact values for \(\varphi_{i}(u)\):

$$\begin{aligned}& \varphi_{1}(u)=0.3491\times1.3692^{-(u+1)}-0.0016 \times8.3914^{-(u+1)}, \\& \varphi_{2}(u)=0.5969\times1.3692^{-(u+1)}-0.0228 \times8.3914^{-(u+1)}. \end{aligned}$$

Figure 1 describes the behavior of \(\varphi_{i}(u)\), \(i=1, 2\), with respect to initial surplus *u*. As expected, \(\varphi _{i}(u)\) decreases as the initial surplus *u* increases. Moreover, \(\varphi_{1}(u)\) is always smaller than \(\varphi_{2}(u)\).