Discrete-Time Model of Company Capital Dynamics with Investment of a Certain Part of Surplus in a Non-Risky Asset for a Fixed Period

Abstract

A periodic-review insurance model is studied under the following assumptions. One-period insurance claims form a sequence of independent identically distributed nonnegative random variables with a finite mean. At the beginning of each period a quota δ of the company surplus is invested in a non-risky asset for m periods. Theoretical expressions for finite-time and ultimate ruin probabilities, in terms of multiple integrals, are presented and applied to the particular case where claims are exponential. Dividend problems are also considered. Numerical results obtained by virtue of simulation are provided and other algorithmic approaches are discussed. Sensitivity analysis of ruin probability is carried out for the case of exponential claims.

Appendix

For further investigation, the following results will be useful.

Lemma 3

The sequence $\{f_{n}\}_{n\geqslant 0}$ defined by Eq. 5is increasing if c > δx. If c = δx then the sequence $\{f_{n}\}_{n\geqslant m}$ is increasing whereas f_k = f₀ = x for $k=\overline {1,m}$.

Proof

Put h_n = f_n − f_n− 1, $n\geqslant 1$, then, according to Eq. 5, h₁ = f₁ − f₀ = c − δx, h_k = (1 − δ)^k− 1h₁, $k=\overline {2,m}$, h_m+ 1 = (1 − δ)^mh₁ + u_mδx and, for $k\geqslant 1$,

$$ h_{m+k+1}=(1-\delta)h_{m+k}+u_{m}\delta h_{k}. $$

(19)

Thus,

$$ h_{m+k}=[(1-\delta)^{m+k-1} +(k-1)u_{m}\delta(1-\delta)^{k-2}]h_{1}+u_{m}\delta(1-\delta)^{k-1} x,\quad k=\overline{2,m+1}, $$

whereas

$$ h_{2m+2}=[(1-\delta)^{2m+1} +(m+1)u_{m}\delta(1-\delta)^{m}]h_{1}+[u_{m}\delta(1-\delta)^{m+1}+(u_{m}\delta)^{2}] x. $$

Using Eq. 19 we conclude that h_n = ϰ_nh₁ + γ_nx with ϰ_n = (1 − δ)^n− 1, for $n=\overline {1,m+1}$, and γ_n = 0, $n=\overline {1,m}$, γ_m+ 1 = u_mδ. Moreover, ϰ_m+ 2 = (1 − δ)^m+ 1 + u_mδ, γ_m+ 2 = u_mδ(1 − δ) and ϰ_m+l+ 1 = (1 − δ)ϰ_m+l + u_mδϰ_l, for l > 1, while γ_m+l+ 1 = (1 − δ)γ_m+l + u_mδγ_l. It follows immediately that ϰ_n > 0 for $n\geqslant 1$, whereas all γ_n are non-negative. Thus, h_n > 0 for $n\geqslant 1$ if h₁ = c − δx > 0. If h₁ = 0 then h_n = 0 for $n=\overline {1,m}$ and h_n > 0 for n > m. Since f_n = f_n− 1 + h_n, it is clear that, for all n > m, f_n > f_n− 1 if $h_{1}=c-\delta x\geqslant 0$. □

It is possible to get an explicit form of coefficients ϰ_n and γ_n for any m and n.

Corollary 3

The following relations hold

$$ \varkappa_{n} = \sum\limits_{i=0}^{[\frac{n}{m+1}]} a_{i}^{(m,n)} (u_{m}\delta)^{i} (1-\delta)^{n-1-(m+1)i}, $$

with $a_{0}^{(m,n)}=1$, $a_{i}^{(m,n)}=a_{i}^{(m,n-1)}+a_{i-1}^{(m,n-m-1)}$ for i > 1.

Moreover,

$$ \gamma_{n}=\sum\limits_{k=1}^{[\frac{n}{m+1}]} a_{n,k}^{(m)}(u_{m}\delta)^{k} (1-\delta)^{n-(m+1)k} $$

with $a_{n,1}^{(m)}=1$ and $a_{n,k}^{(m)}=a_{n-1,k}^{(m)}+a_{n-1-m,k-1}^{(m)}$ for k > 1. The sum over empty set is equal to zero.

Proof

The results follow in a straightforward way from the expression h_n = h₁ϰ_n + xγ_n leading to the following recurrence relations:

$$ \begin{aligned} &\varkappa_{n}=(1-\delta)\varkappa_{n-1}+u_{m}\delta\varkappa_{n-1-m},\\ &\gamma_{n}=(1-\delta)\gamma_{n-1}+u_{m}\delta\gamma_{n-1-m}. \end{aligned} $$

□

Remark 2

It follows easily from definition of Y_n that it can be rewritten in a following form:

$$ Y_{n}={\sum}_{k=1}^{n} d_{n-k}X_{k}\quad \text{where}\quad d_{n-k}=g_{n,k}. $$

Hence, d_k = (1 − δ)^k, $k=\overline {0,m}$, and d_l = (1 − δ)d_l− 1 + u_mδd_l− 1−m for l > m. It is not difficult to obtain an explicit expression of d_n for any n :

$$ d_{n}={\sum}_{i=0}^{[\frac{n}{m+1}]} a_{i}^{(n)}(u_{m}\delta)^{i} (1-\delta)^{n-i(m+1)} $$

with $a_{0}^{(n)}=1$, $a_{1}^{(n)}=1+a_{1}^{(n-1)}$ for all n and $a_{i}^{(n)}=a_{i}^{(n-1)}+a_{i-1}^{(n-m-1)}$ for i > 1.

Lemma 4

Consider the sequence $\{f_{n}\}_{n \geqslant 0}$ for the case m = 1, given by the following recurrence relation:

$$ \begin{aligned} &f_{n} = (1 - \delta) f_{n - 1} + \delta u_{1} f_{n - 2} + c,\quad n \geq 2,\\ &f_{0} = x,\quad f_{1} = (1 - \delta) x + c. \end{aligned} $$

The solution can be written explicitly:

$$ f_{n} = c_{1} \frac{x_{1}^{n + 1} - 1}{x_{1} - 1} + c_{2} \frac{x_{2}^{n + 1} - 1}{x_{2} - 1}, \quad c_{1} = \frac{c - x \left( \delta + x_{2}\right)}{x_{1} - x_{2}},\quad c_{2} = \frac{c - x \left( \delta + x_{1}\right)}{x_{2} - x_{1}}, $$

where x₁ < x₂ are the roots of the quadratic equation y² − (1 − δ)y − u₁δ = 0.

Proof

For convenience, let f_− 1 = 0, and put h_n = f_n − f_n− 1 for $n \geqslant 0$. Then $\{h_{n}\}_{n \geqslant 0}$ satisfies:

$$ \begin{aligned} &h_{n} = (1 - \delta) h_{n - 1} + \delta u_{1} h_{n - 2},\quad n \geqslant 2,\\ &h_{0} = x,\quad h_{1} = c - \delta x. \end{aligned} $$

It is a simple homogeneous recurrence relation of order 2. The characteristic polynomial p(y) = y² − (1 − δ)y − u₁δ has two different roots x₁ < x₂, therefore, the solution is

$$ h_{n} = c_{1} x_{1}^ n + c_{2} x_{2}^ n. $$

The values of c₁ and c₂ are found from the initial conditions h₀ = c₁ + c₂ = x, h₁ = c₁x₁ + c₂x₂ = c − δx.

Recalling that

$$ f_{n} = {\sum}_{i = 0}^{n} h_{i} $$

completes the proof. □

Corollary 4

In the case m = 1, $f_{n} \sim d x_{2}^ n \to +\infty$ as $n \to + \infty$, where d is a positive constant. In particular, this sequence is strictly increasing for sufficiently large n.

Proof

Since the discriminant D = (1 − δ)² + 4u₁δ of p(y) is greater than (1 + δ)², the following three inequalities take place:

$$ \begin{aligned} &x_{1} = \frac{1 - \delta - \sqrt{D}}{2} < - \delta,\\ &x_{2} = \frac{1 - \delta + \sqrt{D}}{2} > 1,\\ &|x_{1}| < |x_{2}|. \end{aligned} $$

Then, obviously, the relation $f_{n} \sim d x_{2}^ n$ with $d = \frac {c - x \left (\delta + x_{1}\right )}{(x_{2} - x_{1}) (x_{2} - 1)} x_{2}$ follows from Lemma 4. We only need to show $c - x \left (\delta + x_{1}\right ) > 0$, or, equivalently, $x_{1} < \frac {c}{x} - \delta$. It is true because $x_{1} < -\delta < \frac {c}{x} - \delta$. □