Ruin Probability for Finite Erlang Mixture Claims Via Recurrence Sequences

Abstract

A new procedure to find the ultimate ruin probability in the Cramér-Lundberg risk model is presented for claims with a mixture of m Erlang distributions. The method requires to solve an m order linear recurrence sequence, which translates into finding the roots of an m-th degree polynomial and solving a system of m linear equations. We here study only the case when the roots of the polynomial are simple. A new approximation method for the ruin probability is also proposed based on this procedure and the simulation of a Poisson random variable. Several analytical expressions already known for the ruin probability in the case of Erlang claims, or mixtures of these, are recovered. Numerical results and plots from R programming are provided as examples.

Appendices

Appendix

General Solution to the Recurrence Sequence

Proposition 1

Suppose the roots $z_1,\ldots ,z_m$ of the characteristic polynomial (12) are simple. Then, for some constants $b_1,\ldots ,b_m$, the general solution to the linear recurrence sequence (11) can be written as

$$\begin{aligned} \overline{C}_{n}=\sum _{j=1}^mb_j\,z_j^n,\quad \text{ for }\ n\ge 0. \end{aligned}$$

Proof

For $n\ge 0$,

$$\begin{aligned} \overline{C}_{n+m} = \sum _{j=1}^{m} b_j\,z_j^{n+m} = \sum _{j=1}^m b_j\,z_j^{n} \sum _{k=0}^{m-1} \alpha _{m-k}\,z_j^{k} = \sum _{k=0}^{m-1} \alpha _{m-k}\,\sum _{j=1}^m b_j\,z_j^{n+k} = \sum _{k=0}^{m-1} \alpha _{m-k}\,\overline{C}_{n+k}. \end{aligned}$$

Sufficient Condition for Multiplicity One

Proposition 2

Suppose the equilibrium r.v. $N_e$, defined by (7), has a $\text{ unif }\,\{1,\ldots ,m\}$ distribution. Then the zeroes of the characteristic polynomial (12) are simple.

Proof

In this case, the values $\alpha _k$ are the constant $\alpha =\overline{C}_0/m$ and the characteristic polynomial takes the simpler form

$$\begin{aligned} p(y)=y^m-\alpha \sum _{j=0}^{m-1}y^j=y^m-\alpha \,\frac{1-y^m}{1-y},\quad \text{ for } \ y\ne 0,1. \end{aligned}$$

Define $H(y)=(1-y)p(y)=(1-y)y^m-\alpha (1-y^m)$. It follows that

• a) $H'(y)=(1-y)p'(y)-p(y)$, and

• b) $yH'(y)=mH(y)-y^{m+1}+\alpha m$.

Suppose that z is a zero of p(y). Then $z\,p(z)=0$, i.e. $z^{m+1}=\alpha \,\sum _{j=1}^mz^j$. Taking norms and using the fact that all roots have norm strictly less than one,

$$\begin{aligned} |z|^{m+1}\le \alpha \,\sum _{j=1}^m|z|^j<\alpha \,m. \end{aligned}$$

Then, $H(z)=0$ and by (b), $zH'(z)=-z^{m+1}+\alpha m\ne 0$. Hence, $H'(z)\ne 0$ and by (a), $p'(z)\ne 0$. This means z has multiplicity 1.

Constantinescu et al. Formula for $\text{ Erlang }(2,\alpha )$ Claims

For $\text{ Erlang }(2,\alpha )$ claims, the following ruin probability formula was obtained in Constantinescu et al. (2018),

$$\begin{aligned} \psi (u)=\frac{2\lambda }{\alpha c}\left( m_1\,e^{(s_1-\alpha )u} + m_2\,e^{(s_2-\alpha )u}\right) , \end{aligned}$$

(23)

for some values $m_1,m_2,s_1$ and $s_2$. We will show that (23) is equivalent to the ruin probability formula (18) obtained in this work. Substituting the value $c=(1+\theta )\lambda \mu =2(1+\theta )\lambda /\alpha$ in the formulas for $s_1$ and $s_2$ given in Constantinescu et al. (2018), one obtains

$$\begin{aligned} s_{1,2}=\frac{\lambda \pm \sqrt{\lambda ^2 + 4\lambda \alpha c} }{2c} =\alpha \,\frac{\overline{C}_0\pm \sqrt{\overline{C}_0^2 + 8\overline{C}_0}}{4} =\alpha \,z_{1,2}. \end{aligned}$$

(24)

Now, according to Constantinescu et al. (2018), the values of $m_1$ and $m_2$ are the solution to the partial fractions problem

$$\begin{aligned} \frac{s+\alpha /2}{s^2-\lambda (s+\alpha )/c}=\frac{m_1}{s-s_1} + \frac{m_2}{s-s_2}. \end{aligned}$$

This leads to

$$\begin{aligned} m_{1,2}=\frac{1}{2}\pm \frac{2+\overline{C}_0}{2\sqrt{\overline{C}_0^2+8\overline{C}_0}}.\end{aligned}$$

(25)

Thus, setting $b_{1,2}=m_{1,2}\,\overline{C}_0$ and $z_{1,2}=s_{1,2}/\alpha$, one can see that the formulas (18) and (23) are exactly the same.