Normal and log-normal approximation of aggregate losses

Welcome to Aggregate Loss Approximations!

Hello there! If you’ve been studying for Exam FAM, you already know that calculating the exact distribution of total claims (what we call Aggregate Loss, or \(S\)) can be a real headache. In the real world, an insurance company might have thousands of claims. Trying to calculate every possible outcome exactly is nearly impossible.

That’s where approximations come in! In this chapter, we learn how to use two famous "shortcuts"—the Normal and Log-normal distributions—to estimate the behavior of aggregate losses. These tools help actuaries determine how much capital a company needs to stay solvent without needing a supercomputer.

1. The Foundation: Mean and Variance of \(S\)

Before we can approximate anything, we need to know the "center" and the "spread" of our aggregate loss \(S\). Remember that \(S\) is the sum of a random number of claims (\(N\)) with random sizes (\(X\)).

The "Must-Know" Formulas:
1. Expected Value (Mean): \(E[S] = E[N] \times E[X]\)
2. Variance: \(Var(S) = E[N] \times Var(X) + Var(N) \times (E[X])^2\)

Analogy: Think of \(E[S]\) as the average height of a pile of bricks. It depends on how many bricks you have on average (\(E[N]\)) and how thick each brick is on average (\(E[X]\)). The variance formula is a bit more complex because it accounts for the uncertainty in both the number of bricks and the size of each brick!

Quick Review: Common Frequency Distributions

Don't forget these for your \(Var(N)\) calculations:
• Poisson: \(E[N] = Var(N) = \lambda\)
• Negative Binomial: \(Var(N) > E[N]\)
• Binomial: \(Var(N) < E[N]\)

2. The Normal Approximation

The Normal Approximation relies on the Central Limit Theorem. It suggests that if you have enough claims, the distribution of the total loss starts to look like a bell curve, regardless of what the individual claims look like.

How to use it:

We assume that \(S\) follows a Normal distribution: \(S \sim N(E[S], Var(S))\).
To find the probability that aggregate losses are less than a certain value \(k\), we use the Z-score formula:
\(Z = \frac{k - E[S]}{\sqrt{Var(S)}}\)

Step-by-Step Process:
1. Calculate \(E[S]\) using the frequency and severity means.
2. Calculate \(Var(S)\) using the formula from Section 1.
3. Plug these into the Z-score formula.
4. Look up the probability in the standard normal distribution table (\(\Phi\) table).

Common Mistake to Avoid:
Students often forget to take the square root of the variance. The Z-score formula requires the standard deviation, \(\sigma_S = \sqrt{Var(S)}\), not the variance itself!

Did you know? The Normal approximation is generally better when the expected number of claims (\(E[N]\)) is very large. However, it has one big flaw: a Normal distribution can technically be negative, but insurance losses are never less than zero!

3. The Log-normal Approximation

Since insurance losses are usually skewed (lots of small claims, a few huge ones) and always positive, the Log-normal Approximation is often a better fit than the Normal one.

The Concept:

We assume that the natural log of our loss, \(\ln(S)\), follows a Normal distribution. This means \(S\) itself is Log-normal. We need to find two parameters: \(\mu\) and \(\sigma^2\). Note: These are not the mean and variance of \(S\); they are the mean and variance of \(\ln(S)\)!

How to solve for the parameters:

To match the Log-normal distribution to our aggregate loss, we use the Method of Moments. You will be given (or will calculate) \(E[S]\) and \(E[S^2]\).
Recall that \(E[S^2] = Var(S) + (E[S])^2\).

Use these two "magic" formulas to find \(\sigma^2\) and \(\mu\):
1. \(\sigma^2 = \ln\left(\frac{E[S^2]}{(E[S])^2}\right) = \ln\left(1 + \frac{Var(S)}{(E[S])^2}\right)\)
2. \(\mu = \ln(E[S]) - 0.5\sigma^2\)

Step-by-Step for Probability:
If you need to find \(Pr(S \le k)\):
1. Find \(\mu\) and \(\sigma\) using the formulas above.
2. Transform the problem into a log-space: \(Pr(\ln(S) \le \ln(k))\).
3. Calculate the Z-score: \(Z = \frac{\ln(k) - \mu}{\sigma}\).
4. Use the Normal table.

Key Takeaway: The Log-normal distribution is "right-skewed," which matches the behavior of most insurance claim data better than the symmetric Normal distribution.

4. Normal vs. Log-normal: Which one to use?

In an exam setting, the question will usually tell you which one to use. However, here are some tips for understanding the difference:

• Use Normal when: The number of claims is very large and the distribution is relatively symmetric.
• Use Log-normal when: The distribution is skewed or when you want to ensure the model doesn't allow for negative losses.

Memory Aid:
Normal is Nice and simple (uses \(E[S]\) and \(Var(S)\) directly).
Log-normal Loves Logs (requires \(\ln\) and matching moments).

5. Summary and Quick Review

1. Aggregate Mean: \(E[S] = E[N]E[X]\).
2. Aggregate Variance: \(Var(S) = E[N]Var(X) + Var(N)E[X]^2\).
3. Normal Z-score: \(Z = (k - E[S]) / \sqrt{Var(S)}\).
4. Log-normal Parameters: Match \(E[S]\) and \(E[S^2]\) to find \(\mu\) and \(\sigma^2\).
5. Log-normal Z-score: \(Z = (\ln(k) - \mu) / \sigma\).

Don't worry if this seems tricky at first! The most common errors are simple arithmetic slips in the variance formula or mixing up the Log-normal parameters. Practice a few problems where you have to calculate \(Var(S)\) first, and the rest will start to feel like second nature.

Pro-tip: In Log-normal problems, always check if the question gives you the Coefficient of Variation (\(CV\)). \(CV = \sqrt{Var(S)} / E[S]\). If they do, the formula for \(\sigma^2\) becomes even easier: \(\sigma^2 = \ln(1 + CV^2)\)!