M12 Ruin and reinsurance

Topics in Insurance, Risk, and Finance

Author

Affiliation

Yuyu Chen

Department of Economics, University of Melbourne

Ruin and reinsurance

Ruin and reinsurance I

We have discussed how reinsurance affects insurers in a fixed period of time. Next, we study the case of classical risk process.

Given a compound Poisson aggregate claims process, the surplus claim process of an insurer without reinsurance is $$U(t)=u+ct-\sum _{i=1}^{N(t)}{X}_i,$$ where

• $u$ is the initial surplus
• $c$ is the premium income per unit of time
• $N(t)$ is the number of claims before or at time $t$
• $X_i$ is the $i$th claim without reinsurance.

Ruin and reinsurance II

With reinsurance, the surplus claim process will become $$U^{\ast }(t)=u+c^{\ast }t-\sum _{i=1}^{N(t)}{X}_i^{\ast },$$ where

• $u$ is the initial surplus
• $c^{\ast }$ is the premium income per unit of time, net of reinsurance
• $N(t)$ is the number of claims before time $t$
• $X_i^{\ast }$ is the $i$th claim, net of reinsurance.

Here, we make an implicit assumption that the premium is paid to reinsurer continuously.

Ruin and reinsurance III

• We are still in a classical risk model with reinsurance so the associated results can also be applied here.
• Since it is generally hard to obtain an analytic expression of the ultimate ruin probability, we focus on studying the adjustment coefficient. With reinsurance, the adjustment coefficient $R^{\ast }$ is given by $$\lambda +c^{\ast }{R}^{\ast }=\lambda \text{E}[\exp (R^{\ast }{X}_1^{\ast })].$$
• Then Lundberg’s inequality is used to approximate the ultimate ruin probability.
• We say a reinsurance arrangement is optimal in the sense that it maximizes the adjustment coefficient (hence minimise the approximated ultimate ruin probability).

Ruin and reinsurance IV

For the rest of the lecture, we make the assumptions below:

• Let $X$ be the individual claim with $X\sim F$, $F(0)=0$, and $f$ the density.
• The loading factors of insurer and reinsurer are $\theta$ and ${\theta }_R$.
• The Poisson parameter is $\lambda$.
• $h$ is a reinsurance arrangement, i.e., the insurer pays $h(X)$ under the reinsurance arrangement. For instance,
  • $h(x)=\alpha x$ gives the proportional reinsurance
  • $h(x)=min(x,M)$ gives the excess of loss reinsurance.
  • Note that we need $h(x)\le x$ (why?).

Then the insurer’s premium rate, net of reinsurance, is given by, $$c^{\ast }=(1+\theta )\lambda \text{E}[X]-(1+{\theta }_R)\lambda \text{E}[X-h(X)]$$ with $c^{\ast }>\lambda \text{E}[h(X)]$.

The optimal type of reinsurance

The optimal type of reinsurance

Consider the following question:

• An insurer can choose from different types of reinsurance contracts.

• It has a budget constraint on the reinsurance premium rate $c^{\ast }=C$, $C\in \text{R}$.

• What type of reinsurance contract should be considered?

The optimal type of reinsurance

Note that the cost of reinsurance is always the same with above assumption.

Proof I

Let $R_h$ and $R_e$ denote the adjustment coefficients of a reinsurance arrangement $h$ and the excess of loss reinsurance, respectively. We have the following two equations: $$\lambda +c^{\ast }{R}_h=\lambda \text{E}[\exp (R_{h}h(X))]=\lambda {\int }_0^{\mathrm{\infty }}\exp (R_{h}h(x))f(x)dx,$$ and $$\begin{array}{rl}\lambda +c^{\ast }{R}_e& =\lambda \text{E}[\exp (R_{e}min(X,M))]\\ & =\lambda ({\int }_0^M\exp (R_{e}x)f(x)dx+\exp (R_{e}M)(1-F(M))).\end{array}$$

Proof II

Define $$g_1(r)=\lambda {\int }_0^{\mathrm{\infty }}\exp (rh(x))f(x)dx-\lambda -c^{\ast }r,$$ and $$g_2(r)=\lambda ({\int }_0^M\exp (rx)f(x)dx+\exp (rM)(1-F(M)))-\lambda -c^{\ast }r.$$ Clearly, $R_h$ and $R_e$ are the roots of $g_1$ and $g_2$. We know that $g_1$ and $g_2$ are convex. Then, to show $R_e\ge R_h$, it is sufficient to show $g_1\ge g_2$.

Hence, we need to show $${\int }_0^{\mathrm{\infty }}\exp (rh(x))f(x)dx\ge {\int }_0^{\mathrm{\infty }}\exp (rϵ(x))f(x)dx,$$ where $ϵ$ is the excess of loss arrangement i.e., $ϵ(x)=min(x,M)$.

Proof III

To show the desired result, note that $$\exp (r(h(x)-ϵ(x)))\ge 1+r(h(x)-ϵ(x)).$$ Therefore, $$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\exp (rh(x))f(x)dx\ge & {\int }_0^{\mathrm{\infty }}\exp (rϵ(x))f(x)dx\\ & +r{\int }_0^{\mathrm{\infty }}\exp (rϵ(x))(h(x)-ϵ(x))f(x)dx.\end{array}$$ It suffices to see $$\begin{array}{rl}& {\int }_0^{\mathrm{\infty }}\exp (rϵ(x))(h(x)-ϵ(x))f(x)dx\\ & ={\int }_0^M\exp (rϵ(x))(h(x)-ϵ(x))f(x)dx+{\int }_M^{\mathrm{\infty }}\exp (rM)(h(x)-ϵ(x))f(x)dx\\ & \ge {\int }_0^M\exp (rM)(h(x)-ϵ(x))f(x)dx+{\int }_M^{\mathrm{\infty }}\exp (rM)(h(x)-ϵ(x))f(x)dx\\ & =\exp (rM){\int }_0^{\mathrm{\infty }}(h(x)-ϵ(x))f(x)dx=0.\end{array}$$

Example: Proportional reinsurance

Proportional reinsurance: premium income I

We next consider the optimal reinsurance with the same type.

• Assume that the retained proportion is $\alpha \in [0,1]$.
• Following our previous assumption, the premium rate, before reinsurance, is $$(1+\theta )\lambda m_1,$$ where $\theta$ is the loading factor for the insurer.
• Similarly, the premium rate for the reinsurer is$$(1+{\theta }_R)\lambda (1-\alpha )m_1,$$ where ${\theta }_R$ is the loading factor for the reinsurer.
• Consequently, the premium income for the insurer per unit of time, after reinsurance, is $$\begin{array}{rl}{c}^{\ast }& =(1+\theta )\lambda m_1-(1+{\theta }_R)\lambda (1-\alpha )m_1\\ & =((1+\theta )-(1+{\theta }_R)(1-\alpha ))\lambda m_1.\end{array}$$

Proportional reinsurance: premium income II

Constraint one: After reinsurance, the expected claim for the insurer is $\alpha m_1$. Therefore, we require the premium income per unit time, net of reinsurance, exceeds aggregate claims per unit time, i.e., $c^{\ast }\ge \lambda \alpha m_1$ and we get $$(1+\theta )-(1+{\theta }_R)(1-\alpha )\ge \alpha ,$$ which further gives $$\alpha \ge 1-\frac{\theta }{{\theta }_R}.$$

Proportional reinsurance: premium income IV

Constraint two: We assume that the insurer pays for reinsurance out of the premium income it receives, i.e., $c^{\ast }\ge 0$ $$(1+\theta )\lambda m_1\ge (1+{\theta }_R)\lambda (1-\alpha )m_1,$$ which gives $$\alpha \ge \frac{{\theta }_R-\theta }{1+{\theta }_R}.$$

Proportional reinsurance: premium income III

If not specified, we always assume ${\theta }_R\ge \theta$. With this condition, we can see that $$\alpha \ge \frac{{\theta }_R-\theta }{1+{\theta }_R},$$ is implied by $$\alpha \ge 1-\frac{\theta }{{\theta }_R}.$$

Hence, Constraint one is crucial and it is sufficient to only consider it with the assumption that ${\theta }_R\ge \theta$.

Propotional reinsurance: example

Suppose that the individual claim amount $X$ follows the exponential distribution $F(x)=1-e^{-x}.$ Recall that $M_X(r)=1/(1-r)$ for $r<1$. Hence, we have $$c^{\ast }=((1+\theta )-(1+{\theta }_R)(1-\alpha ))\lambda .$$ Then the adjustment coefficient can be calculated by $$\lambda M_{X^{\ast }}(r)-\lambda -c^{\ast }r=0,$$ where $M_{X^{\ast }}(r)=1/(1-\alpha r)$.

We will consider the following three cases:

• 1. $\theta ={\theta }_R=0.2$
• 2. $\theta =0.2$ and ${\theta }_R=0.25$
• 3. $\theta =0.05$ and ${\theta }_R=0.25$

Proportional reinsurance: example

1. $\theta ={\theta }_R=0.2$

In this case, $\alpha \in [0,1]$ and $$c^{\ast }=((1+\theta )-(1+{\theta }_R)(1-\alpha ))\lambda m_1=1.2\lambda \alpha .$$

Then $$\lambda M_{X^{\ast }}(r)-\lambda -c^{\ast }r=0$$ gives the adjustment coefficient $$R^{\ast }(\alpha )=\frac{1}{6\alpha }.$$ Therefore, $R^{\ast }$ is maximized when $\alpha =0$ which means that the insurer has no claim and no premium income. The surplus will be a constant $u$ and the ultimate ruin probability is 0.

Proportional reinsurance: example

2. $\theta =0.2$ and ${\theta }_R=0.25$

In this case, $\alpha \ge 1-\theta /{\theta }_R=1-0.8=0.2$ and hence $\alpha \in [0.2,1]$. We have $$c^{\ast }=((1+\theta )-(1+{\theta }_R)(1-\alpha ))\lambda m_1=(1.25\alpha -0.05)\lambda .$$ Then $$\lambda M_{X^{\ast }}(r)-\lambda -c^{\ast }r=0.$$ gives the adjustment coefficient $$R^{\ast }(\alpha )=\frac{0.25\alpha -0.05}{\alpha (1.25\alpha -0.05)}.$$ Taking the first derivative of $R^{\ast }$ gives $$\frac{d}{d\alpha }{R}^{\ast }(\alpha )=\frac{-0.3125{\alpha }^2+0.125\alpha -0.0025}{{\alpha }^2(1.25\alpha -0.05{)}^2}.$$ Therefore, $R^{\ast }$ is maximized when $\alpha =0.3789$ (The other root is 0.02 for the derivative being 0.).

Propotional reinsurance: example

3. $\theta =0.05$ and ${\theta }_R=0.25$

In this case, $\alpha \ge 1-\theta /{\theta }_R=1-0.2=0.8$ and hence $\alpha \in [0.8,1]$. We have $$c^{\ast }=((1+\theta )-(1+{\theta }_R)(1-\alpha ))\lambda m_1=(1.25\alpha -0.2)\lambda .$$ Then $$\lambda M_{X^{\ast }}(r)-\lambda -c^{\ast }r=0.$$ gives the adjustment coefficient $$R^{\ast }(\alpha )=\frac{1}{\alpha }-\frac{1}{1.25\alpha -0.2}.$$ Taking the first derivative of $R^{\ast }$ gives $$\frac{d}{d\alpha }{R}^{\ast }(\alpha )=\frac{0.8}{(\alpha -0.16{)}^2}-\frac{1}{{\alpha }^2}\ge 0,$$ for $\alpha \in [0.8,1]$. Hence $R^{\ast }$ is maximized when $\alpha =1$, i.e., no reinsurance is optimal.

Example: Excess of loss reinsurance

Excess of loss reinsurance: premium income

• Let $M$ be the retention level.
• The premium rate for the insurer, before reinsurance, is $$(1+\theta )\lambda m_1.$$
• The premium rate for the reinsurer is $$(1+{\theta }_R)\lambda \text{E}[max(X-M,0)].$$
• Then the premium rate, net of reinsurance, is $$c^{\ast }=(1+\theta )\lambda m_1-(1+{\theta }_R)\lambda \text{E}[max(X-M,0)].$$

Excess of loss reinsurance: example

• The adjustment coefficient can be solved by $$\lambda +c^{\ast }r=\lambda ({\int }_0^M\exp (rx)f(x)dx+\exp (rM)(1-F(M))).$$
• Assume that the individual claim follows $F(x)=1-e^{-x}$, $\theta =0.1$, ${\theta }_R=0.2$.

Excess of loss reinsurance: example

Constraint one: the premium income per unit time, net of reinsurance, exceeds aggregate claims per unit time: $$\begin{array}{rl}{c}^{\ast }& =(1+\theta )\lambda m_1-(1+{\theta }_R)\lambda \text{E}[max(X-M,0)]\\ & =1.1\lambda -1.2\lambda e^{-M}\\ & \ge \lambda \text{E}[min(X,M)]=\lambda (1-e^{-M}),\end{array}$$ which gives $M\ge \log 2=0.693$.

Excess of loss reinsurance: example

Constraint two: We assume that the insurer pays for reinsurance out of the premium income it receives: $$\begin{array}{r}1.1\lambda \ge 1.2\lambda {\int }_M^{\mathrm{\infty }}(x-M)e^{-x}dx=1.2\lambda e^{-M},\end{array}$$ which gives $M\ge \log (12/11)=0.0870$.

Hence, in this example it is sufficient to consider Constraint one, as in the case of proportional reinsurance.

Excess of loss reinsurance: example

The adjustment coefficient can be solved by $$1+(1+\theta -(1+{\theta }_R)e^{-M})r=\frac{1-r{e}^{-(1-r)M}}{1-r}.$$

This equation has to be solved numerically given a specific $M$.

We will not discuss how to solve the equation.

Excess of loss reinsurance: upper bound

• Remember an upper bound of adjustment coefficient without reinsurance is $$R\le \frac{2(c-\lambda m_1)}{\lambda m_2},$$ where $c$, $m_1$, $m_2$ are the premium income per unit of time, the first moment and the second moment of the individual claim.
• This formula can be still applied with reinsurance. However, $c$, $m_1$ and $m_2$ changes with reinsurance.
• We continue with the previous example with excess of loss reinsurance and compute an upper bound of $R$. That is, we need to compute $c^{\ast }$, $m_1^{\ast }$, and $m_2^{\ast }$ which denote the corresponding component with reinsurance.

Excess of loss reinsurance: upper bound

• The premium income per unit of time, net of reinsurance, is $$c^{\ast }=1.1\lambda -1.2\lambda e^{-M}.$$

• The first moment after reinsurance is $$m_1^{\ast }=1-e^{-M}.$$

• The second moment after reinsurance is $$\begin{array}{rl}{m}_2^{\ast }& ={\int }_0^M{x}^2{e}^{-x}dx+M^2{e}^{-M}\\ & =2(1-e^{-M}(1+M)).\end{array}$$

• Hence, we obtain the upper bound $$R\le \frac{e^M-2}{10(e^M-1-M)}.$$

• found an upper bound of $R$ in the case of excess of loss reinsurance. We can do similarly for proportional reinsurance.