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.

1 Introduction

Risk is a keyword in all definitions of actuarial sciences. Risk is present whenever the outcome is uncertain, whether favorable or unfavorable. Methods for transferring or distributing risk were practiced by Chinese and Babylonian traders as long ago as the 3rd and 2nd millennia BC, respectively. However actuarial science emerged significantly later (in the 17th century). It has an interesting history consisting of 4 periods, see, e.g., Bulinskaya (2017).

Not only insurance, but other applied probability research domains such as inventory and dams, finance, queueing theory, reliability and some others can be considered as special cases of decision making under uncertainty (or risk management) aimed at the systems performance optimization, thus eliminating or minimizing risk. For correct decision making one needs an appropriate mathematical model. Constructing an insurance company model one has to take into account its twofold nature. Originally all insurance societies were designed for risk sharing. Hence, their primary task is policyholders indemnification. Nowadays, for the most part they are joint stock companies. Thus, the secondary but very important task is dividend payments to shareholders.

The 20th century, belonging to the second (stochastic) period in actuarial sciences, is known for emergence of collective risk theory. The study of ruin probability for various modifications of (continuous-time) Cramér-Lundberg model has dominated. This allowed insurance companies to increase their solvency and thus provide a solution to the first task.

The modern period is characterized by investigation of complex systems, including dividends payment, reinsurance, tax, bank loans and investment. Interplay of actuarial and finance methods, in particular, unification of reliability and cost approaches is another feature of the last twenty years, see, e.g., Bulinskaya (2003). Furthermore, discrete-time models turned out to be more appropriate for description of some aspects of insurance company performance. On the other hand, discrete-time models can be used for approximation of continuous-time ones, see, e.g. Dickson and Waters (2004). Therefore we concentrate on such type of models although there exist a lot of simulation methods treating continuous-time insurance models (see, e.g., Dutang et al. (2008) and further modifications available at the site http://CRAN.R-project.org/). It is interesting to underline that discretization is widely used in many programs.

A first important discrete-time model giving start to investigation of dividends was introduced in a seminal paper Andersen (1957). The discrete model treating the ruin probability was firstly proposed in Gerber (1988). The paper Li et al. (2009) is a review of discrete-time models considered until 2009. Let us also mention three papers, not included in review: Chan and Zhang (2006) treating direct derivation of finite-time ruin probabilities, Alfa and Drekic (2007) presenting algorithmic analysis of the Sparre Andersen model in discrete time and Wei and Hu (2008) considering the models with stochastic rates of interest, see also Kordzakhia et al. (2012) and references therein. The model with dependence of finance and insurance risks and heavy tailed losses is considered in Tsitsiashvili (2010). Recurrent algorithms for ruin probability calculation are provided there.

The papers Blaževičius et al. (2010), Castañer et al. (2013) and Damarackas and Šiaulys (2014) investigate discrete-time models with different types of non-homogeneity. Inequalities for the ruin probability in a controlled discrete-time risk process are dealt with in Diasparra and Romera (2010). Sharp approximations of ruin probabilities in the discrete-time models are provided in Gajek and Rudz (2013). Quantitative analysis of ruin probability for a discrete-time insurance model with proportional reinsurance and surplus investment according to random interest rate is carried out in Jasulewicz and Kordecki (2015). Two cases of one period loss distributions were compared, namely, light-tailed (exponential) and heavy-tailed (Pareto).

Not only ruin probabilities were studied. Thus, discrete-time models with dividends and reinsurance are treated in Bulinskaya and Yartseva (2010). The discounted factorial moments of the deficit in discrete-time renewal risk model are dealt with in Bao and He (2012). Optimization of discrete-time insurance model with capital injections and reinsurance is carried out in Bulinskaya et al. (2015). Asymptotic analysis of insurance models with bank loans one can find in Bulinskaya (2014) and (2018). A substantial review of recent results on discrete-time models is supplied, as well, in Section 5 of Bulinskaya (2017). The author’s results concerning a discrete-time model with reinsurance and capital injections are given there in Section 6.2.

It is necessary to note that compound binomial models are still popular, see, e.g., Wat et al. (2018) and references therein.

In our paper we study the discrete-time insurance model under following assumptions. The premium obtained by the company each period is equal to a constant c > 0. The claims form a sequence of independent identically distributed nonnegative random variables. No assumptions are made about the claims distribution, except existence of density and finiteness of expectation. There exist investment possibility in a non-risky asset. More precisely, in contrast to above mentioned papers, we assume that only a certain part δ of surplus can be placed in a bank for $m\geqslant 1$ periods under a fixed interest rate β per period. At the end of the term the deposit is returned to the company along with interest and can be used for claims indemnification.

The paper is organized as follows. In Section 2 we give the model description. The ultimate and finite-time ruin probabilities are considered in Section 3. We obtain their expressions in terms of multi-dimensional integrals. For the particular case of exponentially distributed claims an explicit form of such integrals is calculated. The problem of dividends payment is dealt with in Section 4. Simulation results and sensitivity analysis of ruin probability for the case of exponential claims are presented in Section 5. In conclusion (Section 6) we discuss the obtained results and further research directions.

2 Model Description

We consider the following generalization of the model introduced in Bulinskaya and Kolesnik (2018). For certainty, we proceed in terms of insurance company performance. During the i th period (year, month, week or day) the company gets a fixed premium amount c and pays a random indemnity X_i, i = 1,2,…. It is supposed that $\{X_{i}\}_{i\geqslant 1}$ is a sequence of independent identically distributed nonnegative random variables (i.i.d. r.v.’s) with a known distribution function F(x), possessing a density p_X(x) and a finite mean. Let the initial capital S₀ = x be fixed and positive. It is possible to place a quota δ ∈ [0,1] of this amount in a bank for m periods ($m\geqslant 1$), the interest rate being β per period. Thus, the surplus at the end of the first period has the form

$$ S_{1} = (1-\delta)S_{0}+c-X_{1}. $$

The same procedure is repeated each period (investment of the part δ of available surplus, acquirement of premium and return of invested m periods earlier capital along with interest, payment of indemnity). For simplicity sake, we put u_m = (1 + β)^m getting the following recurrent formula

$$ S_{n} = \min[(1-\delta)S_{n-1},S_{n-1}]+c+u_{m}\delta S^{+}_{n-(m+1)}-X_{n}. $$

(1)

Here and further on it is assumed that S_k = 0 for k < 0. That means, nothing is returned if n < m + 1. The expression (1) is useful if it is permitted to delay the company insolvency, that is, the surplus can stay negative for some time. It takes into account the fact that negative surplus cannot be invested. Thus, for negative S_n− 1 we start from S_n− 1, whereas for positive values of S_n− 1 the amount (1 − δ)S_n− 1 is left after investment. In the same way, we write $u_{m}\delta S^{+}_{n-(m+1)}$ (where $S^{+}=\max \limits (S,0)$), because it was impossible to invest negative S_n−(m+ 1). Moreover, if the surplus at the previous step S_n− 1 was negative it is possible to obtain S_n > 0 due to arrival of previous investment (if any) and/or small indemnity. However in this case it is necessary to have $c-X_{n}+u_{m}\delta S^{+}_{n-(m+1)}>|S_{n-1}|$.

Hence, expression (1) is appropriate for the study of Parisian ruin τ^d which occurs if the process S_n stays below or at zero at least for a fixed time period d ∈{1,2,…}. More precisely, see, e.g., Czarna et al. (2017),

$$ \tau^{d}=\inf\{n>0 : S_{n}\leqslant 0, n-\sup\{k<n : S_{k}>0\}>d \}. $$

If we use the classical notion of ruin, then the ruin time τ is defined as follows

$$ \tau=\inf\{n>0 : S_{n}\leqslant 0\}, $$

(2)

that is, the delay d = 0. To calculate the ultimate ruin probability

$$ P(\tau<\infty)={\sum}_{n=1}^{\infty}P(S_{1}>0,\ldots, S_{n-1}>0,S_{n}\leqslant 0) $$

one can use instead of Eq. 1 the relation

$$ S_{n} = (1-\delta)S_{n-1}+c-X_{n}+u_{m}\delta S_{n-(m+1)}, $$

(3)

treated in Bulinskaya and Kolesnik (2018) only for m = 1. Clearly, expression (3) is also used for calculation of finite-time ruin probability.

There is no need to find the explicit form of S_n (although it is possible to obtain it). For further investigation we need only the following lemma.

Lemma 1

Put S₀ = x and

$$ S_{n}=f_{n}-\sum\limits_{i=1}^{n} g_{n,i}X_{i},\quad n\geqslant 1, $$

(4)

then f₀ = x, whereas

$$ f_{n}=(1-\delta)f_{n-1}+c, n=\overline{1,m},\quad f_{n}=(1-\delta)f_{n-1}+u_{m}\delta f_{n-(m+1)}+c, n>m, $$

(5)

and

$$ \begin{array}{@{}rcl@{}} g_{n,i}=(1-\delta)g_{n-1,i}+u_{m}\delta g_{n-(m+1),i},\quad i=\overline{1,n-(m+1)},\\ g_{n,i}=(1-\delta)g_{n-1,i},\quad i=\overline{n-m,n-1}, \quad g_{n,n}=1. \end{array} $$

(6)

Proof

Obviously, combination of Eqs. 3 and 4 provides the desired recursive relations (5) and (6). □

3 Ruin Probability: Theoretical Results

3.1 General Claim Distribution

Our next aim is investigation of ultimate and finite-time ruin probabilities. To this end, we reformulate Lemma 1 as follows. The company surplus S_n, $n\geqslant 1$, has the form S_n = f_n − Y_n with f_n defined by Eq. 5 and vector Y_n = (Y₁,…, Y_n) given by Y_n = G_n ⋅X_n. Here vector X_n = (X₁,…, X_n) and matrix G_n = (g_{k, i})_{k, i= 1,…, n}, where g_{k, i} = 0 for i > k, the others being specified by Eq. 6.

Now we prove the following result.

Theorem 1

The ultimate ruin probability is represented by

$$ \varphi(x) = \sum\limits_{n=1}^{\infty} \int\limits_{0}^{f_{1}} {\dots} \int\limits_{0}^{f_{n-1}} \int\limits_{f_{n}}^{+\infty} \prod\limits_{k=1}^{n} p_{X}(v_{k}(y_{1},\dots, y_{n})) dy_{1} \dots dy_{n}, $$

where x is the initial surplus, f_n, $n\geqslant 1$, and g_{n, i}, $n\geqslant 1$, $i=\overline {1,n}$, are given by Eqs. 5and 6, respectively. The function $v_{k}(y_{1},\dots , y_{n})$, $k=\overline {1,n}$, is the k th component of vector $\textbf {G}^{-1}_{n} \textbf {y}_{n}$ and y_n = (y₁,…, y_n).

Proof

The ruin time is defined by Eq. 2. So, we calculate the probability of ruin at the n th step obtaining the distribution of the ruin time. To this end, we use Lemma 1. Hence, the probability under consideration P(τ = n) is given by P(U_n) with

$$ U_{n}=\left\{g_{1,1} X_{1} < f_{1}, ..., \sum\limits_{i=1}^{n-1} g_{n-1,i} X_{i} < f_{n-1},\sum\limits_{i=1}^{n} g_{n,i} X_{i}\geqslant f_{n}\right\}. $$

In other words, P(τ = n) is a multiple integral over the domain

$$ U^{\prime}_{n}=\left\{g_{1,1} x_{1} < f_{1}, \dots, \sum\limits_{i=1}^{n-1} g_{n-1,i} x_{i} < f_{n-1}, \sum\limits_{i=1}^{n} g_{n,i}x_{i}\geqslant f_{n} \right\} $$

of $p_{X_{1}, \dots , X_{n}}(x_{1}, \dots , x_{n})$. Performing the change of variables y_n = G_n ⋅x_n, we see that the integral transforms into

$$ \int\limits_{0}^{f_{1}} {\dots} \int\limits_{0}^{f_{n-1}} \int\limits_{f_{n}}^{+\infty} \frac{p_{X_{1}, \dots, X_{n}}(v_{1}(y_{1}, \dots, y_{n}), {\dots} , v_{n}(y_{1}, \dots, y_{n}))} {|\det G_{n}(v_{1}(y_{1}, \dots, y_{n}), {\dots} , v_{n}(y_{1}, \dots, y_{n}))|} dy_{1} {\dots} dy_{n}. $$

Obviously, $\textbf {x}_{n} = \mathbf {G}^{-1}_{n} \cdot \mathbf {y}_{n}$, so $v_{i}(y_{1}, {\dots } , y_{n})=(\mathbf {G}^{-1}_{n}\cdot \mathbf {y}_{n})_{i}$.

Using the fact that the determinant is a product of diagonal elements g_{i, i} = 1 and the sequence X_i consists of i.i.d. r.v.’s we establish the desired form of probability

$$ \int\limits_{0}^{f_{1}} {\dots} \int\limits_{0}^{f_{n-1}} \int\limits_{f_{n}}^{+\infty} {\prod}_{k=1}^{n} p_{X}(v_{k}(y_{1},\dots, y_{n})) dy_{1} {\dots} dy_{n}. $$

(7)

Summing these expressions over all $n\geqslant 1$ we obtain the ultimate ruin probability. According to Lemma 1 integration limits f_k, $k\geqslant 1$, depend on x, so the ultimate ruin probability is also a function of the initial surplus x, that is, $\varphi (x)=P(\tau <\infty | S_{0}=x)$. □

Another interesting problem is calculation of finite-time ruin probability. In other words, we want to get the expression of probability

$$ \varphi_{N}(x)=P(\tau\leqslant N | S_{0}=x) $$

for a fixed number N.

Corollary 1

The following relation holds

$$ \varphi_{N} (x)=1 - \int\limits_{0}^{f_{1}} {\dots} \int\limits_{0}^{f_{N-1}} \int\limits_{0}^{f_{N}} {\prod}_{k=1}^{N} p_{X}(v_{k}(y_{1},\dots, y_{N})) dy_{1} {\dots} dy_{N}. $$

(8)

Proof

Clearly, using Eq. 7 which provides P(τ = n) we get φ_N(x) as a sum of corresponding summands for $n\leqslant N$. On the other hand, it can be written as 1 − P(τ > N) = 1 − P(S₁ > 0,…, S_N > 0). Thus, proceeding along the same lines as in the proof of Theorem 1 we immediately obtain expression (8). □

3.2 Exponential Claim Distribution

Theorem 1 gives an explicit formula for finding the probability of ultimate ruin. However, as it often happens in actuarial and financial mathematics, it contains multi-dimensional integrals that are difficult to calculate. As usual, we can apply the formula to the simplest particular case of the model and obtain analytical expressions. Specifically, in Theorem 2 we assume that X has an exponential distribution (which is a common assumption) and give an expression for ruin probability P(τ = n) containing no integrals. This result can be used for further theoretical analysis of the model.

Let E_n denote an n × n identity matrix. We first formulate a useful lemma.

Lemma 2

Let λ, $(a_{i})_{i = 1}^{n}$, $(f_{i})_{i = 1}^{n}$, $(l_{ij}, 1\leqslant j<i\leqslant n)$ be real numbers, λ > 0, f_i > 0 for all i. Let

$$ \begin{pmatrix} l_{1}\\ l_{2}\\ \vdots\\ l_{n -1}\\ l_{n} \end{pmatrix} = \begin{pmatrix} 0 & 0 & {\ldots} & 0 & 0\\ l_{21} & 0 & {\ldots} & 0 & 0\\ {\vdots} & {\vdots} & {\ddots} & {\vdots} & \vdots\\ l_{n-1 1} & l_{n-1 2} & {\ldots} & 0 & 0\\ l_{n 1} & l_{n 2} & {\ldots} & l_{n n-1} & 0 \end{pmatrix} \begin{pmatrix} y_{1}\\ y_{2}\\ \vdots\\ y_{n-1}\\ y_{n} \end{pmatrix} $$

be a vector of linear combinations of variables $(y_{i})_{i = 1}^{n}$ and denote

$$ I_{n} := {\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n}}^{f_{n}} \exp\left( - \lambda \sum\limits_{i = 1}^{n} a_{i} y_{i}\right) dy_{1} {\ldots} dy_{n}. $$

(9)

Then

$$ \begin{array}{@{}rcl@{}} I_{n} &=& \frac{1}{\lambda^{n}} \sum\limits_{k = 0}^{2^{n} -1} s_{k}, \end{array} $$

(10)

$$ \begin{array}{@{}rcl@{}} s_{k} &:=& (-1)^{k_{1} + {\ldots} k_{n}} \left( \prod\limits_{p = 1}^{n} {a_{p}^{k}}\right)^{-1} \exp\left( - \lambda \sum\limits_{p = 1}^{n - 1} k_{p} {a_{p}^{k}} f_{p}\right), \end{array} $$

(11)

where $\overline {k_{n} k_{n -1} {\ldots } k_{2} k_{1}}$ is the binary representation of k, $0\leqslant k\leqslant 2^{n} -1$, that is, k_i is either 0 or 1, and

$$ \begin{array}{@{}rcl@{}} & \begin{pmatrix} {a^{k}_{1}}\\ {a^{k}_{2}}\\ \vdots\\ a^{k}_{n - 1}\\ {a^{k}_{n}} \end{pmatrix} = (E_{n} + (1 - k_{1}) L_{1}) {\ldots} (E_{n} + (1 - k_{n}) L_{n}) \begin{pmatrix} a_{1}\\ a_{2}\\ \vdots\\ a_{n - 1}\\ a_{n} \end{pmatrix}, \end{array} $$

(12)

$$ \begin{array}{@{}rcl@{}} & (L_{i})_{st} = \begin{cases} l_{t s} & \text{if}~ t = i, s < i,\\ 0 & \text{otherwise}. \end{cases} \end{array} $$

(13)

Proof

We will argue by induction over n. If n = 1, the result is easily verified.

Assume the formula is true for all dimensions not higher than n − 1. Taking the integral with respect to y_n in Eq. 9 gives $I_{n} = I_{n - 1}^{\prime } + I_{n - 1}^{\prime \prime }$, where

$$ \begin{array}{@{}rcl@{}} I_{n-1}^{\prime} := \frac{1}{\lambda a_{n}} {\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n -1}}^{f_{n-1}} \exp\left( -\lambda \sum\limits_{i = 1}^{n - 1}(a_{i} + a_{n} l_{n i})y_{i}\right) dy_{1} {\ldots} dy_{n - 1},\\ I_{n-1}^{\prime\prime} := - \frac{1}{\lambda a_{n}} e^{-\lambda a_{n} f_{n}}{\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n -1}}^{f_{n-1}} \exp\left( - \lambda \sum\limits_{i = 1}^{n-1} a_{i} y_{i}\right) dy_{1} {\ldots} dy_{n-1}. \end{array} $$

After applying the formula for dimension n − 1 and getting

$$ \begin{array}{@{}rcl@{}} I_{n -1}^{\prime} &=& \frac{1}{\lambda^{n} a_{n}} \sum\limits_{k = 0}^{2^{n -1} -1} s_{k}^{\prime},\\ I_{n -1}^{\prime\prime} &=& -\frac{1}{\lambda^{n} a_{n}} e^{-\lambda a_{n} f_{n}} \sum\limits_{k = 0}^{2^{n -1} -1} s_{k}^{\prime\prime} \end{array} $$

for certain $s_{k}^{\prime }$ and $s_{k}^{\prime \prime }$, it can be easily seen that

$$ \begin{array}{@{}rcl@{}} \frac{1}{a_{n}} s_{k}^{\prime} &=& s_{k},\\ &&- \frac{1}{a_{n}} e^{-\lambda a_{n} f_{n}} s_{k}^{\prime\prime} = s_{k + 2^{n - 1}}, \end{array} $$

which completes the proof.

Theorem 2

If the density function of X is exponential, namely, p_X(y) = e^−λy1{y > 0}, where 1{⋅} is the indicator function, then the probability of ruin at a particular time is given by

$$ \begin{array}{@{}rcl@{}} P(\tau = n) &=& \frac{1}{\lambda^{n}} e^{- \lambda f_{n}} \sum\limits_{k = 0}^{2^{n - 1} -1} s_{k},\\ s_{k} &:=& (-1)^{k_{1} + {\ldots} k_{n-1}} \left( \prod\limits_{p = 1}^{n - 1} {a_{p}^{k}}\right)^{-1} \exp\left( - \lambda \sum\limits_{p = 1}^{n - 1} k_{p} {a_{p}^{k}} f_{p}\right), \end{array} $$

where $\overline {k_{n - 1} k_{n -2} {\ldots } k_{2} k_{1}}$ is the binary representation of k, $0\leqslant k\leqslant 2^{n-1}-1$, so k_i is either 0 or 1, and

$$ \begin{array}{@{}rcl@{}} \begin{pmatrix} {a^{k}_{1}}\\ {a^{k}_{2}}\\ \vdots\\ a^{k}_{n - 2}\\ a^{k}_{n - 1} \end{pmatrix} = (E + (1 - k_{1}) L_{1}) {\ldots} (E + (1 - k_{n -1}) L_{n - 1}) \begin{pmatrix} a_{1}\\ a_{2}\\ \vdots\\ a_{n - 2}\\ a_{n - 1} \end{pmatrix},\\ a_{i} = \begin{cases} (1 - u) \delta & \text{if} 1 \leqslant i \leqslant n - m - 1,\\ \delta & \text{if} n - m \leqslant i \leqslant n - 1, \end{cases}\\ (L_{i})_{st} = \begin{cases} u\delta & \text{if} (s, t) = (i - m - 1, i),\\ 1 - \delta & \text{if}(s, t) = (i - 1, i),\\ 0 & \text{otherwise}. \end{cases} \end{array} $$

Proof

Using the recurrence relations for g_{n, i} it is easy to show that

$$ \left( \mathbf{G}_{n}^{-1}\right)_{ij} = \begin{cases} 1 & \text{if $i = j$,}\\ \delta - 1 & \text{if $i - j = 1$,}\\ - u \delta & \text{if $i - j = m + 1$,}\\ 0 & \text{otherwise}. \end{cases} $$

Therefore,

$$ \begin{array}{@{}rcl@{}} \begin{array}{ll} v_{1}(y_{1}, \ldots, y_{n}) &= y_{1},\\ v_{2}(y_{1}, \ldots, y_{n}) &= y_{2} - (1 - \delta) y_{1},\\ v_{3}(y_{1}, \ldots, y_{n}) &= y_{3} - (1 - \delta) y_{2},\\ &\cdots\\ v_{m + 1}(y_{1}, \ldots, y_{n}) &= y_{m + 1} - (1 - \delta) y_{m},\\ v_{m + 2}(y_{1}, \ldots, y_{n}) &= y_{m + 2} - (1 - \delta) y_{m + 1} - u \delta y_{1},\\ v_{m + 3}(y_{1}, \ldots, y_{n}) &= y_{m + 3} - (1 - \delta) y_{m + 2} - u \delta y_{2},\\ &\cdots\\ v_{n}(y_{1}, \ldots, y_{n}) &= y_{n} - (1 - \delta) y_{n - 1} - u \delta y_{n - m - 1}. \end{array} \end{array} $$

It follows that

$$ \sum\limits_{k = 1}^{n} v_{k}(y_{1}, \ldots, y_{n}) = \sum\limits_{k = 1}^{n - m - 1} (1 - u) \delta y_{k} + \sum\limits_{k = n - m}^{n - 1} \delta y_{k} + y_{n}. $$

(14)

The expression for the ruin probability at time n

$$ P(\tau = n) = {\int}_{0}^{f_{1}} {\cdots} {\int}_{0}^{f_{n -1}} {\int}_{f_{n}}^{\infty} \prod\limits_{k = 1}^{\infty} p_{X}\left( v_{k}(y_{1}, \ldots, y_{n})\right) dy_{1} {\ldots} dy_{n} $$

can be rewritten in the form

$$ P(\tau = n) = {\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n - 1}}^{f_{n -1}} {\int}_{f_{n}}^{\infty} e^{- \lambda {\sum}_{k = 1}^{n} v_{k}(y_{1}, \ldots, y_{n})} dy_{1} {\ldots} dy_{n}. $$

(15)

The lower limits can be found by solving the inequalities v_k(y₁,…, y_n) ≥ 0:

$$ \begin{array}{@{}rcl@{}} \begin{array}{ll} l_{1} &= 0,\\ l_{2} &= (1 - \delta) y_{1},\\ l_{3} &= (1 - \delta) y_{2},\\ &\cdots\\ l_{m + 1} &= (1 - \delta) y_{m},\\ l_{m + 2} &= (1 - \delta) y_{m + 1} + u \delta y_{1},\\ l_{m + 3} &= (1 - \delta) y_{m + 2} + u \delta y_{2},\\ &\cdots\\ l_{n - 1} &= (1 - \delta) y_{n - 2} + u \delta y_{n - m - 2}. \end{array} \end{array} $$

Using (14) and taking the integral in (15) with respect to y_n, we obtain

$$ P(\tau = n) = \frac{1}{\lambda} e^{- \lambda f_{n}} {\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n - 1}}^{f_{n -1}} \exp\left( - \lambda \sum\limits_{p = 1}^{n - 1} a_{p} y_{p}\right) dy_{1} {\ldots} dy_{n - 1}. $$

(16)

Using Lemma 2, we get

$$ \begin{array}{@{}rcl@{}} {\int}_{l_{1}}^{f_{1}} {\cdots} {\int}_{l_{n - 1}}^{f_{n -1}} \exp\left( - \lambda \sum\limits_{p = 1}^{n - 1} a_{p} y_{p}\right) dy_{1} {\ldots} dy_{n - 1} \\= \frac{1}{\lambda^{n - 1}} {\sum}_{k = 0}^{2^{n - 1} -1} (-1)^{k_{1} + {\ldots} + k_{n-1}} \left( \prod\limits_{p = 1}^{n - 1} {a_{p}^{k}}\right)^{-1} \exp\left( - \lambda \sum\limits_{p = 1}^{n - 1} k_{p} {a_{p}^{k}} f_{p}\right). \end{array} $$

(17)

Combining (16) and (17) proves the desired result. □

Corollary 2

The finite-time ruin probability is given by

$$ \phi_{N} = 1 - \lambda^{N} I_{N} $$

for I_N as found in Eqs. 10–13.

4 Dividends

Another characteristic important to any insurance company is dividends payment to its shareholders. The simplest and widely used dividends strategy is a so-called barrier strategy. It is specified by some barrier level b > 0. If the company capital crosses upwards this level the excess is immediately paid out as dividends. The first problem is to find the probability that dividends will be paid at least once before the ruin. To solve it we define a new random variable

$$ t_{b}=\inf\{n>0: S_{n}\geqslant b\}. $$

Hence, we would like to calculate P(t_b < τ). It is not difficult to prove the following result.

Theorem 3

The relation

$$ P(t_{b}<\tau)= \sum\limits_{n=1}^{\infty} \int\limits_{f_{1} -b}^{f_{1}} {\dots} \int\limits_{f_{n-1}-b}^{f_{n-1}} \int\limits_{0}^{f_{n} -b} {\prod}_{k=1}^{n} p_{X}(v_{k}(y_{1},\dots, y_{n})) dy_{1} {\dots} dy_{n} $$

(18)

is valid for S₀ = x > 0. The expressions of f_k and v_k(y₁,…, y_n) are defined in Theorem 1.

Proof

We begin by establishing the form of P(t_b = n, t_b < τ). That means, we calculate the probability of crossing level b by surplus before ruin. Obviously, we can write

$$ P(t_{b}=n, t_{b}<\tau)=P(S_{1}\in (0,b),\ldots, S_{n-1}\in (0,b), S_{n}\geqslant b). $$

Recalling that S_k = f_k − Y_k, $k\geqslant 1$, we rewrite this probability as

$$ \begin{array}{@{}rcl@{}} P(f_{1}-b<Y_{1}<f_{1},\ldots,f_{n-1}-b<Y_{n-1}<f_{n-1},Y_{n}<f_{n}-b)\\ =\int\limits_{f_{1} -b}^{f_{1}} {\dots} \int\limits_{f_{n-1}-b}^{f_{n-1}} \int\limits_{0}^{f_{n} -b} \prod\limits_{k=1}^{n} p_{X}(v_{k}(y_{1},\dots, y_{n})) dy_{1} {\dots} dy_{n}. \end{array} $$

The last expression is obtained along the same lines as Eq. 7. Since $P(t_{b}<\tau )={\sum }_{n=1}^{\infty }P(t_{b}=n, t_{b}<\tau )$ we get immediately (18). □

Remark 1

Theorems 1, 3 and Corollary 1 can be extended to treat the case of independent non-identically distributed random variables X_i, $i\geqslant 1$. Instead of ${\prod }_{k=1}^{n} p_{X}(v_{k}(y_{1},\dots , y_{n}))$ one has to write ${\prod }_{k=1}^{n} p_{X_{k}}(v_{k}(y_{1},\dots , y_{n}))$, here $p_{X_{k}}(x)$ is the density of X_k.

5 Numerical Results

5.1 Simulation

Using the Python 3 programming language and the package numpy, we generated a large sample of pseudo-random variables to simulate the claim sizes X_i. The process S_n was modeled multiple times (100 000 times during this iteration) and the empirical probability of ruin was calculated. (It is equal to the number of attempts in which ruin occurred divided by the total number of attempts). In the same way, the empirical probability of the event that dividends are paid at least once before ruin was calculated as well. It is a very good numerical method for obtaining the values of these probabilities, and it is better than trying to calculate the multi-dimensional integrals found in explicit theoretical formulas above. Indeed, calculating those integrals by definition is unfeasible because the runtime grows exponentially as the number of dimensions increases. The simple Monte-Carlo method of calculating multi-dimensional integrals is also inappropriate in this case, because the measure of the domain grows very quickly, causing the standard error to grow so much that the answer contains no information (see Table 1).

Table 1 The probability of ruin calculated using the Monte-Carlo method

The reader may see the code of one simulation in Listing 1.

Listing 1

The code of one simulation

It is impossible to model the process S_n for all positive integers n, so we checked how quickly the probability of ruin over a finite horizon converges to the actual probability of ruin. Table 2 shows the results. It is clear from this table that it is enough to calculate S_n for only the first few values of n, the remaining steps do not change the probability much. We have verified this convergence quality for other values of parameters as well.

Table 2 The empirical probability of ruin and at least one payment of dividends before ruin: premium rate c and horizon are varied

Taking that into account, we will confine ourselves to only considering the horizon 1000. Table 3 shows how the two empirical probabilities depend on the distribution of X_i.

Table 3 The empirical probability of ruin and at least one payment of dividends before ruin: the distribution of X_i is varied

It is clear from this table that the result for X_i distributed exponentially differs from that for X_i distributed uniformly with the same mean. More precisely, if the distribution of claims is uniform, the situation is better for the insurance company: indeed, the claim amounts are bounded from above, so there is a smaller chance of losing too much in one period.

It is also very interesting how the performance of the company depends on the relative amount δ of the invested capital, and the number m of periods this fraction is kept in the bank. Table 4 shows our results. The data suggests that the probability of ruin is concave with respect to δ. Verifying this conjecture formally and finding the optimum is another theoretical problem.

Table 4 The empirical probability of ruin and at least one payment of dividends before ruin: m and δ are varied

A different problem is the complex behavior of the integrals in formulas (7), (8) and (18). In particular, it is impossible to give any simple expressions for the sequence f_n, because the characteristic polynomial x^m+ 1 − (1 − δ)x − uδ of the recurrence relation d_n = (1 − δ)d_n− 1 + uδd_n−(m+ 1) may not be solvable in radicals. Though it is easy to show (see Lemma 3) that the sequence f_n is monotonic if $c - \delta S_{0} \geqslant 0$, we cannot say much about the other case. For an example of such a sequence see Table 5. Numerical data suggests that f_n is always monotonic for sufficiently large n.

Table 5 The sequence f_n. Fixed parameters: δ = 0.5, β = 0.15, c = 2.0, S₀ = x = 10.0, m = 7

5.2 Sensitivity Analysis of the Ruin Probability

In the exponential case, the explicit formula for the probability of ruin with finite horizon has been obtained in Corollary 2. If the horizon is small enough, this formula is much more convenient to use than other methods, because the value is precise and takes little time to compute. We fix a small value of horizon (5 for speed and simplicity) and choose m to be 1. The probability of ruin is then a function of five parameters: δ, u, c, λ, x. We are interested in discovering how sensitive the probability is to changes in each parameter.

We consider the parameters to be independent uniformly distributed random variables in segments [0.0,1.0], [1.0,1.4], [2.0,10.0], [0.5,1.5], [0.0,16.0] respectively and simulate N = 20000 values of the target function. This way, we obtain five scatterplots shown in Fig. 1. It is immediately obvious that for the given segments the parameter c has the greatest influence, λ and x have a smaller effect on the function, δ still smaller but noticeable, and u does not seem to influence the probability of ruin at all (that may change if a much larger segment is chosen).

Fig. 1

The scatterplots versus δ, u, c, λ, x

We now prove this conclusion quantitatively. The target function Y is a function of five parameters (which are viewed as random variables): Y = h(Z₁, Z₂, Z₃, Z₄, Z₅). We are interested in the first-order sensitivity index of i th parameter S_i which represents the main effect contribution of this input factor:

$$ S_{i} =\frac{V[E(Y | Z_{i})]}{V(Y)}. $$

The total effect index accounts for the total contribution of the output variation due to factor Z_i, i.e. its first-order effect plus all higher-order effects due to interactions:

$$ S_{T_{i}} = 1 - \frac{V[E(Y |Z_{\sim i})]}{V(Y)}, $$

where by $Z_{\sim i}$ we mean all parameters other than the i th one. See Saltelli et al. (2008) or Bulinskaya and Gusak (2016) for background on these sensitivity indices and an algorithm for calculating them. In our case, we get the following values:

$$ \begin{aligned} S_{\delta} &= 0.031137, &\quad S_{T_{\delta}} &= 0.156902,\\ S_{u} &= 0.000145, &\quad S_{T_{u}} &= 0.001926,\\ S_{c} &= 0.174891, &\quad S_{T_{c}} &= 0.682256,\\ S_{\lambda} &= 0.096090, &\quad S_{T_{\lambda}} &= 0.481226,\\ S_{x} &= 0.066017, &\quad S_{T_{x}} &= 0.444585. \end{aligned} $$

These numbers unequivocally confirm our previous observation about the effects of the five parameters on the probability of ruin with finite horizon.

6 Conclusion and Further Research Directions

We have considered a periodic-review insurance model with investment in a non-risky asset. Although many researchers use investment in risky assets (see, e.g. Hussain and Parvez 2017), it may be dangerous for insurance companies (see, Kabanov and Pergamenshchikov 2016 and references therein). So, we treated the investment in a bank for a given time providing a fixed interest. The formulas for calculation of finite-time and ultimate ruin probabilities are obtained in terms of multi-dimensional integrals. For the case of exponential claims we computed these integrals getting the formulas useful for numerical investigations. Dividends payment under the barrier strategy is also considered.

We provided the following numerical results obtained by modeling the surplus dynamics using the programming language Python 3. The code of one simulation is given by Listing 1. Empirical ruin probability was calculated as well as probability of at least one dividends payment before the ruin. It turned out that such a method is better than calculation of ruin probability by Monte-Carlo method (see Table 1). The above mentioned empirical probabilities were calculated for exponential distribution if premium rate c and horizon are varied. As Table 2 shows, the ruin probability over a finite horizon quickly converges to the ultimate one. Results for exponential and uniform claim distributions are compared in Table 3 (for fixed investment parameters). Parameters m and δ are varied for exponential distribution in Table 4. The data suggests that the probability of ruin is concave with respect to δ.

Verification of this conjecture formally and finding the optimal investment policy (minimizing the ruin probability) is the theoretical problem we plan to solve.

Although it was proved in Lemma 3 that the sequence f_n monotonically grows for $c-\delta S_{0}\geqslant 0$, we cannot say much about the other case. An example of such a sequence is given in Table 5. Numerical data suggests that f_n is always monotonic for sufficiently large n. However it is necessary to prove this fact.

In Section 5.2 we carried out the sensitivity analysis of the model to small fluctuations of parameters for the case of exponential claim distribution using the methods presented in Saltelli et al. (2008), see Bulinskaya and Gusak (2016) as well. It turned out much more convenient to use the formula of Corollary 2 than other methods, since the result is precise and takes less time to compute. The methods of probability metrics, see, e.g., Rachev et al. (2013), will be useful for treatment of underlying processes perturbation.

Investigation of the asymptotic behavior of ruin probability for light- and heavy-tailed claim distributions is currently under development.

The results established for Parisian ruin and dividends payment with Parisian implementation delay will be published in a forthcoming paper.

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 $x_{1} < -\delta < \frac {c}{x} - \delta$. □