Discrete Ruin Probabilities: Recursive Methods

The Discrete-Time Risk Model

First, let's set up the framework. We consider a simplified model of an insurer's surplus that evolves in discrete time steps (e.g., year by year).

The key components are:

• Initial Surplus, $u$: The capital the insurer holds at the beginning, which we assume to be a non-negative integer.
• Premium Income: To keep the model simple, we assume the insurer receives a net premium of 1 unit at the start of each time period.
• Aggregate Claims, $X_n$: The total claims paid out during period $n$. This is a random variable, and we assume it takes integer values $k=0,1,2,\dots$ with probabilities $f_k=P(X_n=k)$. The claim amounts are independent and identically distributed across periods.

The surplus at the end of period $n$, denoted $U_n$, evolves according to the following relationship:
$U_n=U_{n-1}+1-X_n$
with the starting condition $U_0=u$.

Ruin occurs if the surplus ever becomes non-positive. Note that in this discrete model, we define ruin as hitting or crossing zero ($U_n\le 0$). This is because we are approximating a continuous process; a surplus of zero in discrete time implies the continuous-time surplus must have been negative just before the premium for the next period was received.

Our goal is to calculate $\psi (u)=P(T<\infty |U_0=u)$, the probability of ultimate ruin given an initial surplus $u$.

Developing the Recursive Formula

The core of this method lies in relating the ruin probability at one surplus level to the ruin probabilities at others. We can achieve this by conditioning on the outcome of the very first time period.

Consider an insurer with initial surplus $u$. After one period, the surplus will be $u+1-X_1$, where $X_1$ is the claim amount in that period.

1. If $X_1>u$: The new surplus $u+1-X_1$ will be less than or equal to 0. Ruin occurs immediately and with certainty in this scenario.
2. If $X_1\le u$: The new surplus $u+1-X_1$ is positive. The process continues from this new surplus level. The probability of eventual ruin from this point is simply $\psi (u+1-X_1)$.

Using the law of total probability, we can write an expression for $\psi (u)$ by summing over all possible claim amounts $k$:
$\psi (u)={\sum }_{k=0}^{\infty }P(\text{ruin}|X_1=k)P(X_1=k)$
Breaking this down into the two cases:
$\psi (u)={\sum }_{k=0}^u\psi (u+1-k)f_k+{\sum }_{k=u+1}^{\infty }1\cdot f_k$
The second term is simply the probability that the first claim causes ruin, which is $P(X_1>u)$. Let $F_u={\sum }_{k=0}^u{f}_k$ be the cumulative distribution function of the claim amount. Then $P(X_1>u)=1-F_u$.

So, the fundamental equation, sometimes called De Pril's recursion, is:
$\psi (u)={\sum }_{k=0}^u\psi (u+1-k)f_k+(1-F_u)$
This equation connects the ruin probabilities for different surplus levels. It forms the basis of our recursive calculation.

Calculating Ruin Probabilities

To use the formula, we can rearrange it to solve for the ruin probability at the next highest surplus level. Let's isolate the term involving the highest surplus, which occurs when $k=0$ (no claims).
$\psi (u)=f_0\psi (u+1)+{\sum }_{k=1}^u\psi (u+1-k)f_k+(1-F_u)$
Solving for $\psi (u+1)$, we get:
$\psi (u+1)=\frac{1}{f_0}\left[\psi (u)-{\sum }_{k=1}^u\psi (u+1-k)f_k-(1-F_u)\right]$
This formula is the engine of our method. It shows that if we know the ruin probabilities $\psi (0),\psi (1),\dots ,\psi (u)$, we can calculate $\psi (u+1)$. This suggests a step-by-step process, but it requires a starting point: we need to know $\psi (0)$.

For this discrete risk model, there's a neat result that connects back to our previous lesson. The ruin probability with zero initial surplus, $\psi (0)$, is given by:
$\psi (0)=\frac{\mu -1}{\mu }$
where $\mu =E[X]$ is the expected claim amount per period. Since the premium is 1 per period, the denominator $\mu$ represents the risk premium, and $(\mu -1)$ is the net cost. The ratio is effectively the proportion of the risk premium that isn't covered by the safety loading, leading to ruin from a starting point of zero. For this to be a valid probability, we need $\mu >1$, which is equivalent to a negative safety loading—a situation guaranteed to lead to eventual ruin. If premium $c>\mu$, $\psi (0)=1$. The general formula for $\psi (0)$ is $\psi (0)=\mu /c$. With our convention of $c=1$, this is just $\psi (0)=\mu$.

Let's work through an example to see how this comes together.

Example: Calculating $\psi (1)$ and $\psi (2)$

Suppose an insurer's aggregate claims per period, $X$, follow a geometric distribution with $P(X=k)=(1/3)(2/3{)}^k$ for $k=0,1,2,\dots$. Premiums are 1 unit per period.

1. Claim Probabilities ($f_k$):

   • $f_0=1/3$
   • $f_1=(1/3)(2/3)=2/9$
   • $f_2=(1/3)(2/3{)}^2=4/27$, etc.

2. Expected Claim Amount ($\mu$):
   For a geometric distribution starting at 0, $E[X]=\frac{p}{1-p}$ where $p$ is the success probability. Here the formula is $f_k=p(1-p{)}^k$. Our $p=1/3$, so $\mu =E[X]=\frac{1-p}{p}=\frac{2/3}{1/3}=2$. The mean of a geometric distribution of the form $p(1-p{)}^k$ for $k=0,1,2,...$ is $(1-p)/p$. Here, $p=1/3$, so $\mu =(1-1/3)/(1/3)=2$.
   Note: The safety loading is $\theta =\frac{c-\mu }{\mu }=\frac{1-2}{2}=-0.5$. Since it's negative, eventual ruin is certain, i.e., $\psi (\infty )=1$.

3. Calculate $\psi (0)$:
   Using the formula $\psi (0)=\mu /c$, with $c=1$:
   $\psi (0)=\mu =2$
   Wait, a probability cannot be greater than 1. This formula $\psi (0)=\mu /c$ applies when there is a positive safety loading ($c>\mu$). When there is a negative safety loading ($c\le \mu$), ruin is certain from any starting surplus if the process runs forever. So, $\psi (u)=1$ for all $u$.

Let's use a more realistic example where survival is possible.

Corrected Example with Positive Safety Loading

Suppose claims $X$ have the distribution $f_0=0.6,f_1=0.3,f_2=0.1$, and $f_k=0$ for $k>2$. The premium per period is $c=1$.

1. Expected Claim Amount ($\mu$):
   $\mu =E[X]=(0\cdot 0.6)+(1\cdot 0.3)+(2\cdot 0.1)=0.5$
   The safety loading is $\theta =\frac{1-0.5}{0.5}=1>0$. The insurer is profitable on average, so ruin is not certain.

2. Calculate $\psi (0)$:
   Using the general formula $\psi (0)=\mu /c$:
   $\psi (0)=0.5/1=0.5$

3. Calculate $\psi (1)$:
   We use the recursion for $u=0$:
   $\psi (0)=f_0\psi (1)+(1-F_0)=f_0\psi (1)+(f_1+f_2+\dots )$
   $0.5=0.6\cdot \psi (1)+(0.3+0.1)=0.6\psi (1)+0.4$
   $0.6\psi (1)=0.1⟹\psi (1)=\frac{0.1}{0.6}=\frac{1}{6}\approx 0.1667$
   The probability of ruin drops from 50% to about 16.7% with just one unit of initial capital.

4. Calculate $\psi (2)$:
   Now use the recursion for $u=1$:
   $\psi (1)=f_0\psi (2)+f_1\psi (1)+(1-F_1)=f_0\psi (2)+f_1\psi (1)+(f_2+f_3+\dots )$
   $\frac{1}{6}=0.6\cdot \psi (2)+0.3\cdot \frac{1}{6}+0.1$
   $\frac{1}{6}=0.6\psi (2)+0.05+0.1=0.6\psi (2)+0.15$
   $0.6\psi (2)=\frac{1}{6}-0.15=\frac{1}{6}-\frac{15}{100}=\frac{1}{6}-\frac{3}{20}=\frac{10-9}{60}=\frac{1}{60}$
   $\psi (2)=\frac{1/60}{0.6}=\frac{1}{36}\approx 0.0278$
   With two units of initial capital, the ruin probability is now less than 3%.

We could continue this process indefinitely to find $\psi (u)$ for any integer surplus $u$.

Finite vs. Infinite Time Horizon

The recursive method we've discussed calculates the probability of ultimate ruin, meaning ruin at any point in an infinite future. A related problem is calculating the probability of ruin within a finite time horizon, $\psi (u,t)$, which is the probability of ruin occurring by time $t$.

The following paper presents algorithms for both infinite and finite time ruin. While the derivations involving generating functions are beyond the scope of this lesson, it's useful to see the structure of the recursive formulas.

The key difference in the finite time recursion is the additional index for time, $t$. The calculation becomes a two-dimensional recursion over surplus $u$ and time $t$, which is computationally more intensive but conceptually similar.

For a practical demonstration, the following video solves a discrete ruin problem from a past exam. It's an excellent way to consolidate your understanding of the step-by-step calculation.

Conclusion

Today, we have added a powerful and precise tool to your ruin theory arsenal. Unlike the bounds and approximations from our last lesson, the recursive method allows for the exact calculation of ruin probabilities, provided the claim amounts are discrete. This highlights a common trade-off in modeling: a simpler model structure (discrete claims) can permit more exact computational methods.

Key Takeaways:

• For a discrete-time risk model with integer claims, the ultimate ruin probability $\psi (u)$ can be calculated exactly.
• The method is based on the recursive formula: $\psi (u)={\sum }_{k=0}^u\psi (u+1-k)f_k+(1-F_u)$.
• To start the recursion, you need a base case, $\psi (0)$. In a model with premium $c=1$ per period and positive safety loading, this is given by $\psi (0)=\mu$, where $\mu$ is the expected claim amount per period.
• By repeatedly applying the formula, we can compute $\psi (1),\psi (2),\psi (3),\dots$ in sequence.
• A similar recursive structure can be used to calculate finite-time ruin probabilities, $\psi (u,t)$.

Preview of the Next Lesson

We have now examined how to calculate ruin probabilities. But what happens if an insurer wants to reduce its risk of ruin? One of the most common strategies is reinsurance. In our next and final lesson of this module, you will learn how different reinsurance arrangements affect the insurer's risk profile. We will see how reinsurance changes the adjustment coefficient and, consequently, the probability of ruin.