Approximations of the Ruin Probability
in a Discrete Time Risk Model

Several models have been proposed for a discrete time¹¹1We will reserve the use of letter $n$ for the approximation procedures proposed later on. risk process $\{U(t):t=0,1,\ldots\}$. The following model is known as a compound binomial process and was first considered in [7],

$$U(t)=u+t-\sum_{i=1}^{N(t)}X_{i},$$

(1)

where $U(0)=u\geq 0$ is an integer representing the initial capital and the counting process $\{N(t):t=0,1,\ldots\}$ has a $\mbox{Binomial}(t,p)$ distribution, where $p$ stands for the probability of a claim in each period. The discrete random variables $X_{1},X_{2},\ldots$ are i.i.d. with probability function $f_{X}(x)=P(X_{i}=x)$ for $x=1,2,\ldots$ and mean $\mu_{X}$ such that $\mu_{X}\cdot p<1$. This restriction comes from the net profit condition. Each $X_{i}$ represents the total amount of claims in the $i$-th period where claims existed. In each period, one unit of currency from premiums is gained. The top-left plot of Figure 1 shows a realization of this risk process. The ultimate ruin time is defined as

$$\tau=\min\,\{t\geq 1:U(t)\leq 0\},$$

as long as the indicated set is not empty, otherwise $\tau:=\infty$. Hence, the probability of ultimate ruin is

$$\psi(u)=P(\tau<\infty\mid U(0)=u).$$

One central problem in the theory of ruin is to find $\psi(u)$. For the above model this probability can be calculated using the following relation known as Gerber’s formula [7],

	$\displaystyle\psi(0)$	$\displaystyle=$	$\displaystyle p\cdot\mu_{X},$		(2)
	$\displaystyle\psi(u)$	$\displaystyle=$	$\displaystyle(1-p)\psi(u+1)+p\sum_{x=1}^{u}\psi(u+1-x)\,f_{X}(x)+p\,\overline{F}_{X}(u),$		(3)

for $u=1,2,\ldots$ where $\overline{F}_{X}(u)=P(X_{i}>u)=\sum_{x=u+1}^{\infty}f_{X}(x)$.

An apparently simpler risk model is defined as follows.

In this case, at each unit of time there is always a claim of size $Y$. If $\mu_{Y}$ denotes the expectation of this claim, the net profit condition now reads $\mu_{Y}<1$. It can be shown [4, pp. 467] that this condition implies that $\psi(u)<1$, where the time of ruin $\tau$ and the ultimate ruin probability $\psi(u)$ are defined as before. Under a conditioning argument it is easy to show that the probability of ruin satisfies the recursive relation

	$\displaystyle\psi(0)$	$\displaystyle=$	$\displaystyle\mu_{Y},$		(5)
	$\displaystyle\psi(u)$	$\displaystyle=$	$\displaystyle\sum\limits_{y=0}^{u}{f_{Y}(y)\,\psi(u+1-y)}+\overline{F}_{Y}(u),\quad u\geq 1.$		(6)

Now, given a compound binomial model (1) we can construct a Gerber-Dickson model (4) as follows. Let $R_{1},R_{2},\ldots$ be i.i.d. $\mbox{Bernoulli}(p)$ random variables and define $Y_{i}=R_{i}\cdot X_{i}$, $i\geq 1$. The distribution of these claims is $f_{Y}(0)=1-p$ and $f_{Y}(y)=p\cdot f_{X}(y)$ for $y\geq 1$.

Conversely, given model (4) and defining $p=1-f_{Y}(0)$, we can construct a model (1) by letting claims $X_{i}$ have distribution $f_{X}(x)=f_{Y}(x)/p$, for $x\geq 1$. It can be readily checked that $\mu_{Y}=p\cdot\mu_{X}$ and that the probability generating function of $U(t)$ in both models coincide. This shows models (1) and (4) are equivalent in the sense that $U(t)$ has the same distribution in both models. As expected, the recursive relations (3) and (6) can be easily obtained one from the other.

In this work we will use the notation in the Gerber-Dickson risk model (4) and drop the subindex in the distribution of claims. Also, as time an other auxiliary variables are considered discrete, we will write, for example, $t\geq 0$ instead of $t=0,1,\ldots$ Our main objective es to provide some methods to approximate the ultimate ruin probability in the discrete risk model of Gerber and Dickson.

A survey of results and models for discrete time risk models can be found in [12].

The continuous version of this formula plays a major role in the theory of ruin for the Cramér-Lundberg model. On the contrary, the discrete version is seldom mentioned in the literature on discrete time risk models. In this section we develop this formula and apply it later to find a general method to calculate ultimate ruin probabilities for claims with particular distributions. The construction procedure resembles closely that for the continuous case.

Assuming $\tau<\infty$, the non-negative random variable $W=|U(\tau)|$ is known as the severity of ruin. It indicates how large the capital drops below zero at the time of ruin. See the top-right plot of Figure 1. The joint probability of ruin and severity not greater than $w=0,1,\ldots$ is denoted by

$$\varphi(u,w)=P(\tau<\infty,W\leq w\mid U(0)=u).$$

(7)

In [5] it is shown that, in particular,

$$\varphi(0,w)=\sum_{x=0}^{w}\overline{F}(x),\quad w\geq 0.$$

(8)

Hence,

$$P(\tau<\infty,W=w\mid U(0)=0)=\varphi(0,w)-\varphi(0,w-1)=\overline{F}(w).$$

(9)

This probability will be useful in finding the distribution of the size of the first drop of the risk process below its initial capital $u$, see Proposition 2.3 below, which will ultimately lead us to the Pollaczeck–Khinchine formula. For every claim distribution, there is an associated distribution which often appears in the calculation of ruin probabilities. This is defined next.

The probability function defined by (10) is also known as the integrated-tail distribution, although this name is best suited to continuous distributions. For example, the equilibrium distribution associated to a $\mbox{Geometric}(p)$ claim distribution with mean $\mu=1/(1-p)$ is the same geometric since

$$f_{e}(y)=\overline{F}(y)/\mu=(1-p)^{y+1}\,p/(1-p)=p\,(1-p)^{y},\quad y\geq 0.$$

(11)

As in the continuous time risk models, let us define the surplus process $\{Z(t):t\geq 0\}$ by

$$Z(t)=u-U(t)=\sum_{i=1}^{t}(Y_{i}-1).$$

(12)

This is a random walk that starts at zero, it has stationary and independent increments and $Z(t)\rightarrow-\infty\ a.s.$ as $t\to\infty$ under the net profit condition $\mu<1$. See bottom-right plot of Figure 1. In terms of this surplus process, ruin occurs when $Z(t)$ reaches level $u$ or above. Thus, the ruin probability can be written as

$$\psi(u)=P(Z(t)\geq u\mbox{ for some }t\geq 1)=P\left(\max\limits_{t\geq 1}\left\{Z(t)\right\}\geq u\right),\quad u\geq 1.$$

(13)

As $u\geq 1$ and $Z(0)=0$, we can also write

$$\psi(u)=P\left(\max\limits_{t\geq 0}\left\{Z(t)\right\}\geq u\right).$$

(14)

We next define the times of records and the severities for the surplus process.

The random variables $\tau_{0}^{*}<\tau_{1}^{*}<\cdots$ represent the stopping times when the surplus process $\left\{Z(t):t\geq 0\right\}$ arrives at a new or the previous maximum, and the severity $Y_{i}^{*}$ is the difference between the maxima at $\tau_{i}^{*}$ and $\tau_{i-1}^{*}$. A graphical example of these record times are shown in the bottom-right plot of Figure 1. In particular, observe $\tau_{1}^{*}$ is the first positive time the risk process is less than or equal to its initial capital $u$, that is,

$$\tau_{1}^{*}=\min\,\{t>0:u-U(t)\geq 0\},$$

(16)

and the severity is $Y_{1}^{*}=Z(\tau_{1}^{*})=u-U(\tau_{1}^{*})$ and this is the size of this first drop below level $u$. Also, since the surplus process has stationary increments, all severities share the same distribution, that is,

$$Y_{i}^{*}=Z(\tau_{i}^{*})-Z(\tau_{i-1}^{*})\sim Z(\tau_{1}^{*})-Z(0)=Y_{1}^{*},\quad i\geq 1,$$

(17)

assuming $\tau_{i}^{*}<\infty$. We will next find out that distribution.

Proof. By (17), it is enough to find the distribution of $Y_{1}^{*}$. Observe that $\tau_{1}^{*}=\tau$ when $U(0)=0$. By (9) and (LABEL:dickson-ruin-0), for $x\geq 0$,

	$\displaystyle P(Y_{1}^{*}=x\mid\tau_{1}^{*}<\infty))$	$\displaystyle=$	$\displaystyle P(u-U(\tau_{1}^{*})=x\mid\tau_{1}^{*}<\infty)$
		$\displaystyle=$	$\displaystyle P(|U(\tau)|=x\mid\tau<\infty,U(0)=0)$
		$\displaystyle=$	$\displaystyle P(\tau<\infty,Y=x\mid U(0)=0)/P(\tau<\infty\mid U(0)=0)$
		$\displaystyle=$	$\displaystyle\overline{F}(x)/\mu.$

The independence property follows from the independence of the claims. Indeed, the severity of the $i$-th record time is

$$Y_{i}^{*}=Z(\tau_{i}^{*})-Z(\tau_{i-1}^{*})=\sum_{j=\tau_{i-1}^{*}+1}^{\tau_{i}^{*}}(Y_{j}-1),\quad i\geq 1.$$

Therefore,

$$P\left(\bigcap_{i=1}^{k}\left(Y_{i}^{*}=y_{i}\right)\right)=P\left(\bigcap_{i=1}^{k}\left(\sum_{j=\tau_{i-1}^{*}+1}^{\tau_{i}^{*}}(Y_{j}-1)=y_{i}\right)\right)=\prod_{i=1}^{k}\,P\left(Y_{i}^{*}=y_{i}\right).$$

∎

Since the surplus process is a Markov process, the following properties hold: For $i\geq 2$ and assuming $\tau_{i}^{*}<\infty$, for $0<s<x$,

$$P(\tau_{i}^{*}=x\mid\tau_{i-1}^{*}=s)=P(\tau_{i}^{*}-\tau_{i-1}^{*}=x-s\mid\tau_{i-1}^{*}=s)=P(\tau_{1}^{*}=x-s).$$

(19)

Also, for $k\geq 1$,

	$\displaystyle P(\tau_{k}^{*}<\infty\mid\tau_{k-1}^{*}<\infty)$	$\displaystyle=$	$\displaystyle P(\tau_{1}^{*}<\infty),$		(20)
	$\displaystyle P(\tau_{k}^{*}=\infty\mid\tau_{k-1}^{*}<\infty)$	$\displaystyle=$	$\displaystyle P(\tau_{1}^{*}=\infty).$		(21)

The total number of records of the surplus process $\{Z(t):t\geq 0\}$ is defined by the non-negative random variable

$$K=\max\,\{k\geq 1:\tau_{k}^{*}<\infty\},$$

(22)

when the indicated set is not empty, otherwise $K:=0$. Note that $0\leq K<\infty$ a.s. since $Z(t)\rightarrow-\infty$ a.s. under the net profit condition. The distribution of this random variable is established next.

Proof. The case $k=0$ can be related to the ruin probability with $u=0$ as follows,

$$f_{K}(0)=P(\tau_{1}^{*}=\infty)=P(\tau=\infty\mid U(0)=0)=1-\psi(0)=1-\mu.$$

Hence, $P(K>0)=\psi(0)=\mu$. Let us see the case $k=1$,

$$f_{K}(1)=P(\tau_{1}^{*}<\infty,\tau_{2}^{*}=\infty)=P(\tau_{2}^{*}=\infty\mid\tau_{1}^{*}<\infty)P(\tau_{1}^{*}<\infty).$$

By (20),

$$f_{K}(1)=P(\tau_{1}^{*}=\infty)P(\tau_{1}^{*}<\infty)=P(K>0)f_{K}(0)=\mu(1-\mu).$$

Now consider the case $k\geq 2$ and let $A_{k}=(\tau_{k}^{*}<\infty)$. Conditioning on $A_{k-1}$ and its complement,

	$\displaystyle P(A_{k})$	$\displaystyle=$	$\displaystyle P(\tau_{k}^{*}<\infty\mid A_{k-1})P(A_{k-1})$
		$\displaystyle=$	$\displaystyle P(\tau_{k}^{*}<\infty\mid\tau_{k-1}^{*}<\infty)P(A_{k-1})$
		$\displaystyle=$	$\displaystyle P(\tau_{1}^{*}<\infty)P(A_{k-1})$
		$\displaystyle=$	$\displaystyle\psi(0)P(A_{k-1}).$

An iterative argument shows that $P(A_{k})=(\psi(0))^{k}$, $k\geq 2$. Therefore,

$$f_{K}(k)=P(\tau_{k+1}^{*}=\infty,A_{k})=P(\tau_{k+1}^{*}=\infty\mid A_{k})P(A_{k})=P(\tau_{1}^{*}=\infty)(\psi(0))^{k}=(1-\mu)\mu^{k}.$$

∎

Proof.

$$\sum_{i=1}^{K}Y_{i}^{*}=\sum_{i=1}^{K}\left(Z(\tau_{i}^{*})-Z(\tau_{i-1}^{*})\right)=Z(\tau_{K}^{*})=\max_{t\geq 0}\,\{Z(t)\}\quad a.s.$$

(26)

Thus, for $u\geq 1$,

$$\psi(u)=P\left(\max_{t\geq 0}\,\{Z(t)\}\geq u\right)=P\left(\sum_{i=1}^{K}Y_{i}^{*}\geq u\right).$$

∎

Proof. For $u=0$, the sum in (27) reduces to $\mu$ which we know is $\psi(0)$. For $u\geq 1$, by (23) and (25),

$$\psi(u)=P\left(\sum_{i=1}^{K}Y^{*}_{i}\geq u\right)=\sum_{k=0}^{\infty}P\left(\sum_{i=1}^{K}Y^{*}_{i}\geq u\mid K=k\right)f_{K}(k)=(1-\mu)\sum_{k=1}^{\infty}P(S_{k}^{*}\geq u)\mu^{k}.$$

∎

For example, suppose claims have a $\mbox{Geometric}(p)$ distribution with mean $\mu=(1-p)/p$. The net profit condition $\mu<1$ implies $p>1/2$. We have seen that the associated equilibrium distribution is again $\mbox{Geometric}(p)$, and hence the $k$-th convolution is $\mbox{Negative Binomial}(k,p)$, $k\geq 0$. Straightforward calculations show that the Pollaczeck–Khinchine formula gives the known solution for the probability of ruin,

$$\psi(u)=\left(\frac{1-p}{p}\right)^{u+1},\quad u\geq 0.$$

(28)

This includes in the same formula the case $u=0$. In the following section we will consider claims that have a mixture of some distributions.

Negative binomial mixture (NBM) distributions will be used to approximate the ruin probability when claims have a mixed Poisson (MP) distribution. Although NBM distributions are the analogue of Erlang mixture distributions, they cannot be used to approximate any discrete distribution with non-negative support. However, it turns out that they can approximate mixed Poisson distributions. This is stated in [17, Theorem 1], where the authors define NBM distributions those with probability generating function

$$G(z)=\lim_{m\rightarrow\infty}\sum_{k=1}^{m}q_{k,m}\left(\frac{1-p_{k,m}}{1-p_{k,m}\,z}\right)^{r_{k,m}},\quad z<1,$$

where $q_{k,m}$ are positive numbers and sum $1$ over index $k$. This is a rather general definition for a NBM distribution. In this work we will consider a particular case of it.

We will denote by $\mbox{nb}(k,p)(x)$ the probability function of a negative binomial distribution with parameters $k$ and $p$, and by $\mbox{NB}(k,p)(x)$ its distribution function, namely, for $x\geq 0$,

$$\mbox{nb}(k,p)(x)=\binom{k+x-1}{x}p^{k}(1-p)^{x},\,\mbox{ and }\,\mbox{NB}(k,p)(x)=1-\sum_{i=0}^{k-1}\mbox{nb}(x+1,1-p)(i).$$

It is useful to observe that any NBM distribution can be written as a compound sum of geometric random variables. Indeed, let $N$ be a discrete random variable with probability function $q_{k}=f_{N}(k)$, $k\geq 1$, and define $S_{N}=\sum_{i=1}^{N}X_{i}$, where $X_{1},X_{2},\ldots$ are i.i.d. r.v.s $\mbox{Geometric}(p)$ distributed and independent of $N$. Then

$$\sum_{k=1}^{\infty}q_{k}\cdot\mbox{nb}(k,p)(x)=\sum_{k=1}^{\infty}q_{k}\cdot P\left(\sum_{i=1}^{k}X_{i}=x\right)=P(S_{N}=x),\quad x\geq 0.$$

Thus, given any $\mbox{NBM}(\boldsymbol{\pi},p)$ distribution with $\boldsymbol{\pi}=(f_{N}(1),f_{N}(2),\ldots)$, we have the representation

$$S_{N}=\sum_{i=1}^{N}X_{i}\sim\mbox{NBM}(\boldsymbol{\pi},p).$$

(29)

In particular,

	$\displaystyle E(S_{N})$	$\displaystyle=$	$\displaystyle E(N)\left(\frac{1-p}{p}\right),$		(30)
	$\displaystyle F_{S_{N}}(x)$	$\displaystyle=$	$\displaystyle\sum_{k=1}^{\infty}f_{N}(k)\cdot\mbox{NB}(k,p)(x),\quad x\geq 0,$		(31)

and the p.g.f. has the form $G_{S_{N}}(r)=G_{N}(G_{X}(r))$. The following is a particular way to write the distribution function of a NBM distribution.

Proof.

	$\displaystyle F_{S_{N}}(x)$	$\displaystyle=$	$\displaystyle\sum_{k=1}^{\infty}f_{N}(k)\cdot\mbox{NB}(k,p)(x)$
		$\displaystyle=$	$\displaystyle\sum_{k=1}^{\infty}f_{N}(k)\,\left[1-\sum_{i=0}^{k-1}\mbox{nb}(x+1,1-p)(i)\right]$
		$\displaystyle=$	$\displaystyle\sum_{i=0}^{\infty}\left[\sum_{k=1}^{i}f_{N}(k)\right]\,\mbox{nb}(x+1,1-p)(i)$
		$\displaystyle=$	$\displaystyle E(F_{N}(Z)).$

∎

We will show next that the equilibrium distribution associated to a NBM distribution is again NBM. For a distribution function $F(x)$, $\overline{F}(x)$ denotes $1-F(x)$.

Proof.

$$f_{e}(x)=\frac{\overline{F}_{S_{N}}(x)}{E(S_{N})}=\frac{p\sum_{i=0}^{\infty}\overline{F}_{N}(i)\binom{x+i}{i}p^{i}(1-p)^{x+1}}{(1-p)E(N)}=\sum_{i=0}^{\infty}\frac{\overline{F}_{N}(i)}{E(N)}\binom{x+i}{i}p^{i+1}(1-p)^{x}.$$

Naming $j=i+1$,

$$f_{e}(x)=\sum_{j=1}^{\infty}\frac{\overline{F}_{N}(j-1)}{E(N)}\binom{j+x-1}{x}p^{j}(1-p)^{x}=\sum_{j=1}^{\infty}f_{Ne}(j)\cdot\mbox{nb}(j,p)(x).$$

∎

It can be checked that (33) is a probability function. It is the equilibrium distribution associated to $N$. In what follows, a truncated geometric distribution will be used. This is denoted by $\mbox{TGeometric}(\rho)$, where $0<\rho<1$, and defined by the probability function $f(k)=\rho(1-\rho)^{k-1}$, for $k\geq 1$.

The following proposition states that a compound geometric NBM distribution is again NBM. This result is essential to calculate the ruin probability when claims have NBM distribution.

Proof. For $x\geq 1$ and $m\geq 1$,

$$P(S=x\mid M=m)=P\left(\sum_{j=1}^{m}S_{N_{j}}=x\right)=P\left(\sum_{j=1}^{m}\sum_{i=1}^{N_{j}}X_{i\,j}=x\right)=P\left(\sum_{\ell=1}^{N_{m}}X_{\ell}=x\right),$$

(37)

where $N_{m}=\sum_{i=1}^{m}N_{i}$ and $X_{\ell}\sim\mbox{Geometric}(p)$ for $\ell\geq 1$. Therefore,

$$P(S=x)=\sum_{m=1}^{\infty}P(S=x\mid M=m)f_{M}(m)=\sum_{m=1}^{\infty}P\left(\sum_{\ell=1}^{N_{m}}X_{l}=x\right)f_{M}(m)=P\left(\sum_{\ell=1}^{N^{*}}X_{\ell}=x\right),$$

where $N^{*}=\sum_{j=1}^{M}N_{j}$. Using Panjer’s formula it can be shown that $N^{*}$ has distribution $\boldsymbol{\pi}^{*}$ given by (35) and (36). Since $X_{\ell}\sim\mbox{Geometric}(p)$, $\sum_{\ell=1}^{N^{*}}X_{\ell}\sim\mbox{NBM}(\boldsymbol{\pi}^{*},p)$. Lastly, we consider the probability of the event $(S=0)$.

$$P(S=0)=\sum_{k=1}^{\infty}f_{N^{*}}(k)\,\mbox{nb}(k,p)(0)=\sum_{k=1}^{\infty}f_{N^{*}}(k)\,p^{k}=f_{N^{*}}(1)\,p+\sum_{k=2}^{\infty}f_{N^{*}}(k)\,p^{k}.$$

Substituting $f_{N^{*}}(k)$ from (35) and (36), one obtains

$\displaystyle P(S=0)$

$\displaystyle=$

$\displaystyle\rho\,G_{N}(p)+(1-\rho)\,G_{N}(p)\,P(S=0).$

Therefore,

$$P(S=0)=\frac{\rho\,G_{N}(p)}{1-(1-\rho)\,G_{N}(p)}=G_{M}(G_{N}(p))=G_{M}(G_{N}(G_{X_{i\,j}}(0))).$$

(38)

The last term is the p.g.f. of a $\mbox{NBM}(\boldsymbol{\pi}^{*},p)$ distribution evaluated at zero. ∎