Ruin Probability with Exponential Claims

The Exact Ruin Probability for Exponential Claims

When individual claim amounts $X$ follow an exponential distribution, the complex stochastic nature of the surplus process simplifies enough to yield an elegant, exact solution for the ruin probability.

Let's establish the context:

• The number of claims follows a Poisson process with rate $\lambda$.
• Individual claim amounts $X$ are independent and identically distributed, following an exponential distribution with parameter $\beta$, i.e., $X\sim \text{Exp}(\beta )$. The mean claim size is $\mu =1/\beta$.
• The premium is collected continuously at a rate $c$, which is set with a relative premium loading $\theta >0$, so $c=(1+\theta )\lambda \mu =(1+\theta )\lambda /\beta$.

For this specific model, the exact probability of ultimate ruin is given by a remarkably simple formula.

As the resource states, the formula is:

Connecting to the Adjustment Coefficient

This result has a powerful connection to our work in the previous lesson. Let's analyze the components of this formula.

First, consider the pre-factor $\frac{\lambda }{c\beta }$. By substituting the formula for the premium rate, $c=(1+\theta )\lambda /\beta$, we get:

So, the pre-factor is simply the reciprocal of one plus the premium loading.

Next, let's examine the exponent, $R=\beta -\lambda /c$. Let's see what this term is.

This expression is exactly the formula for the adjustment coefficient, $\gamma$, that we derived for exponentially distributed claims in our last lesson.

By substituting these findings back into the main formula, we arrive at a more insightful form of the exact ruin probability for exponential claims:

where $\gamma$ is the adjustment coefficient.

Compare this to Lundberg's inequality: $\psi (u)\le e^{-\gamma u}$. For the exponential case, the exact probability is simply the Lundberg bound scaled down by the factor $\frac{1}{1+\theta }$. Since the net profit condition requires $\theta >0$, this factor is always less than 1, confirming that the exact probability is indeed less than the bound.

Generalizing to Combinations of Exponentials

The single exponential distribution is mathematically convenient but may be too simple for real-world applications. A more flexible model can be constructed by assuming the claim size distribution is a combination of exponential distributions. This allows for more complex shapes, better fitting a wider range of empirical data. Given your background in probability, you can appreciate this as being analogous to how mixture models (like Gaussian Mixture Models) provide greater descriptive power.

A claim amount distribution that is a combination of $n$ exponentials has a probability density function of the form:

where $\sum A_i=1$. If all $A_i>0$, this is a hyperexponential distribution (a mixture of exponentials).

For this more general case, the ruin probability takes the form of a sum of exponential terms:

Here:

• The values $r_1,r_2,\dots ,r_n$ are the $n$ positive roots of the generalized adjustment coefficient equation, $c=\lambda {\sum }_{i=1}^n\frac{A_i{\beta }_i}{{\beta }_i-r}$.
• The coefficients $C_1,C_2,\dots ,C_n$ are constants determined by the parameters of the claim distribution.

Let's walk through a concrete example to see how this works in practice, which aligns with your preference for numerical application.

A Practical Example: Calculation for a Combination of Exponentials

We will use an example adapted from a classic actuarial paper by Dufresne and Gerber. This will demonstrate the full process of finding the roots $r_k$ and solving for the coefficients $C_k$.

Problem Setup:
Assume a claim density that is a combination of two exponentials ($n=2$) with the following parameters:

• ${\beta }_1=3$, ${\beta }_2=4$
• $A_1=4$, $A_2=-3$ (Note: since $A_2$ is negative, this is a combination rather than a mixture, which allows for more flexible distribution shapes).
• For simplicity, let the claim arrival rate $\lambda =1$ and the premium rate $c=1$.

Our goal is to find the exact expression for $\psi (u)=C_1{e}^{-r_{1}u}+C_2{e}^{-r_{2}u}$.

Step 1: Find the roots $r_1$ and $r_2$

The roots are the solutions to the equation $c=\lambda {\sum }_{i=1}^n\frac{A_i{\beta }_i}{{\beta }_i-r}$, but a simpler form for calculation is often used: $c=\lambda {\sum }_{i=1}^n\frac{A_i}{{\beta }_i-r}$.
Substituting our parameters:

To solve for $r$, we find a common denominator:

Factoring the quadratic equation gives $(r-1)(r-5)=0$. Thus, the two positive roots are $r_1=1$ and $r_2=5$.

Step 2: Find the coefficients $C_1$ and $C_2$

The coefficients are found by solving a system of linear equations derived from the structure of the problem.

The system of linear equations from formula (43) in the paper is:

For our two-component case ($n=2$), this gives two equations:

1. For $l=1$ (using ${\beta }_1=3$):

   $$\frac{C_1}{3-r_1}+\frac{C_2}{3-r_2}=\frac{1}{3}⟹\frac{C_1}{3-1}+\frac{C_2}{3-5}=\frac{1}{3}⟹\frac{C_1}{2}-\frac{C_2}{2}=\frac{1}{3}⟹C_1-C_2=\frac{2}{3}$$

2. For $l=2$ (using ${\beta }_2=4$):

   $$\frac{C_1}{4-r_1}+\frac{C_2}{4-r_2}=\frac{1}{4}⟹\frac{C_1}{4-1}+\frac{C_2}{4-5}=\frac{1}{4}⟹\frac{C_1}{3}-C_2=\frac{1}{4}$$

Now we solve this 2x2 system:
From the first equation, $C_1=C_2+2/3$. Substituting this into the second equation:

Now find $C_1$:

The coefficients are $C_1=5/8$ and $C_2=-1/24$.

Step 3: Write the final formula for $\psi (u)$

Substituting the roots and coefficients, the exact probability of ruin is:

This numerical example illustrates the complete process, which you can see summarized in the paper you will now read.

Conclusion

In this lesson, we made the significant leap from finding an upper bound to calculating the exact probability of ruin. This is a critical skill for the CM2 exam and a fundamental concept in actuarial science.

Key Takeaways:

• For exponentially distributed claims with parameter $\beta$, the exact ruin probability has a simple analytical form: $\psi (u)=\frac{1}{1+\theta }{e}^{-\gamma u}$, where $\gamma$ is the adjustment coefficient.
• This exact formula reveals that the true ruin probability is the Lundberg bound scaled by a factor of $\frac{1}{1+\theta }$.
• The method can be generalized to claim distributions that are combinations of exponentials, where the ruin probability becomes a weighted sum of exponential terms: $\psi (u)=\sum C_k{e}^{-r_{k}u}$.
• Calculating this requires a two-step process: first, find the roots ($r_k$) of the characteristic equation, and second, solve a system of linear equations to find the coefficients ($C_k$).

We have now covered the core of classical ruin theory. In our next lesson, we will shift our focus to another key area of actuarial work: claims reserving. We will introduce run-off triangles and the chain ladder method, a fundamental technique used to estimate an insurer's future claim liabilities.