Compound Distributions: Frequency and Severity Models

The Compound (Collective Risk) Model

In the compound model, the aggregate loss is S = X1 + X2 + ... + XN, where N is the random number of claims (the frequency component) and X1, X2, ..., XN are independent and identically distributed claim amounts (the severity component). The key assumptions are: (1) the Xi are iid, (2) the Xi are independent of N, and (3) if N = 0, then S = 0.

This model separates "how many claims occur" from "how large each claim is," which is the natural way actuaries think about insurance losses. The frequency distribution (N) captures the randomness in the number of events, while the severity distribution (Xi) captures the randomness in the size of each event.

The compound model is sometimes called the collective risk model or the random sum model. It is the workhorse of short-term insurance modeling and appears extensively on Exam P and later actuarial exams.

Mean and Variance of Aggregate Losses

The expected aggregate loss is derived using the tower property: E[S] = E[E[S|N]] = E[N * E[X]] = E[N] * E[X]. Conditioning on N, the sum of N iid variables has mean N * E[X], and taking the expectation over N gives the result.

The variance is derived using the total variance formula: Var(S) = E[Var(S|N)] + Var(E[S|N]). Since Var(S|N) = N * Var(X) and E[S|N] = N * E[X]:

• E[Var(S|N)] = E[N] * Var(X)
• Var(E[S|N]) = (E[X])^2 * Var(N)

Therefore Var(S) = E[N] * Var(X) + Var(N) * (E[X])^2. This formula is one of the most important on Exam P. Notice that both the variability in frequency (Var(N)) and the variability in severity (Var(X)) contribute to the aggregate variance.

The Poisson Frequency Shortcut

When N ~ Poisson(lambda), an important simplification occurs. Since E[N] = Var(N) = lambda, the variance formula becomes Var(S) = lambda * Var(X) + lambda * (E[X])^2 = lambda * (Var(X) + (E[X])^2) = lambda * E[X^2]. This elegant result means you only need the second moment of the severity distribution, not the mean and variance separately.

Example: Claims arrive at a Poisson rate of lambda = 5 per month. Each claim is Exponential with mean 1000. Find E[S] and Var(S). E[S] = 5 * 1000 = 5000. For the exponential, E[X^2] = 2/lambda^2 = 2 * 1000^2 = 2,000,000 (using the variance 1,000,000 plus mean squared 1,000,000). Var(S) = 5 * 2,000,000 = 10,000,000. Standard deviation of S = sqrt(10,000,000) = 3162.

The compound Poisson model has additional nice properties. Its MGF is M_S(t) = exp(lambda * (M_X(t) - 1)), where M_X(t) is the MGF of the severity distribution. This formula comes from M_S(t) = M_N(ln(M_X(t))) where M_N(s) = exp(lambda(e^s - 1)) is the Poisson MGF.

Non-Poisson Frequency Models

If N ~ Binomial(m, q), then E[N] = mq and Var(N) = mq(1-q). The aggregate loss formulas are E[S] = mq * E[X] and Var(S) = mq * Var(X) + mq(1-q) * (E[X])^2 = mq * [Var(X) + (1-q) * (E[X])^2]. The binomial frequency model arises when there is a fixed number m of exposures, each with independent claim probability q.

If N ~ Negative Binomial(r, p) with E[N] = r(1-p)/p and Var(N) = r(1-p)/p^2, the general compound formulas apply directly. The negative binomial is often used as the marginal frequency distribution in Poisson-Gamma mixture models, as discussed in the mixture distributions article.

For any frequency distribution, the MGF of S can be computed as M_S(t) = G_N(M_X(t)), where G_N(z) = E[z^N] is the probability generating function of N. This relationship is occasionally tested on Exam P.

Normal Approximation for Aggregate Losses

For large expected claim counts, the CLT implies S is approximately Normal(E[S], Var(S)). This allows you to compute approximate probabilities like P(S > c). The steps are: (1) compute E[S] and Var(S) using the compound model formulas, (2) standardize: Z = (c - E[S]) / sqrt(Var(S)), (3) use the normal table.

Example (continuing): Find P(S > 8000). Z = (8000 - 5000)/3162 = 0.949. P(S > 8000) = 1 - Phi(0.949) = 1 - 0.8286 = 0.171. The normal approximation is reasonable here because E[N] = 5 and the severity distribution is not too skewed.

When the frequency is Poisson and the expected number of claims is small (say lambda < 5), the normal approximation may be poor. In such cases, exact calculations or simulation may be needed. On Exam P, the problem will typically specify "use the normal approximation" when it intends for you to use it.

Stop-Loss Premiums

The stop-loss premium for a retention d is E[(S - d)+] = E[S] - E[min(S, d)]. It represents the expected amount the reinsurer pays when the aggregate loss exceeds the retention d. Computing exact stop-loss premiums requires the distribution of S, which is generally complicated. On Exam P, you may be asked to approximate the stop-loss premium using the normal approximation.

Using the normal approximation: E[(S - d)+] is approximately (E[S] - d) * Phi((E[S] - d)/sigma_S) + sigma_S * phi((d - E[S])/sigma_S), where phi is the standard normal PDF and Phi is the CDF. However, on Exam P, a simpler approach is often used: if the problem asks for E[(S - d)+] and the normal approximation applies, compute P(S > d) and use integration of the normal tail.

Common Pitfalls

The most frequent error is confusing the compound model formula with the individual risk model formula. In the compound model, Var(S) = E[N] * Var(X) + Var(N) * (E[X])^2. Some students mistakenly use Var(S) = N * Var(X) (treating N as a constant) or Var(S) = Var(N) * Var(X) (ignoring the cross terms).

Another common mistake is forgetting the Poisson shortcut. When N is Poisson, Var(S) = lambda * E[X^2], not lambda * Var(X). These are different because E[X^2] = Var(X) + (E[X])^2. Using Var(X) instead of E[X^2] underestimates the aggregate variance.

Students also sometimes confuse the aggregate loss S with a single claim X. The problem asks for E[S] and Var(S), which depend on both the frequency and severity, not just the severity alone. Always identify both distributions before computing.

Study Tips

For Exam P, memorize the compound model mean and variance formulas, especially the Poisson shortcut Var(S) = lambda * E[X^2]. Practice identifying the frequency and severity distributions from problem descriptions. Work through problems that combine compound model calculations with the normal approximation. Know the MGF of the compound Poisson model. These problems typically take four to five minutes and appear in two to three questions per exam. Build speed by practicing the systematic approach: identify N and X distributions, compute their moments, apply the compound model formulas, and approximate if needed.