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Risk Modelling and Survival Analysis · Compound distributions and their applications in risk modelling

Compound Distributions and Aggregate Claims Basics Explained

Updated 11 October 2026 · Fact-checked

A compound distribution describes a random sum S = X1 + X2 + ... + XN, where N is the random number of claims (frequency) and each Xi is a claim amount (severity). If the Xi are independent, identically distributed and independent of N, you find the mean, variance and distribution of S from those of N and X.

Understand Compound Distributions and Aggregate Claims Basics

An insurer does not know how many claims will arrive in a year, nor how large each will be. Total claims S depend on both. The aggregate claims model captures this: S = X1 + X2 + ... + XN, with S = 0 when N = 0.

Frequency is the distribution of N, the number of claims. It is discrete and takes values 0, 1, 2, ... Common choices are Poisson, binomial and negative binomial. Severity is the distribution of a single claim amount X. It is usually continuous, positive and right-skewed, for example exponential, gamma, lognormal or Pareto.

The standard model (often called the collective risk model) makes three assumptions: the Xi are independent of each other, the Xi all have the same distribution as X, and N is independent of every Xi. State these in any answer. If one fails, the simple formulas no longer hold. For example, claims from one flood are not independent, and inflation can change the severity distribution with time.

The word compound means the sum has a random number of terms. S has a compound distribution, such as compound Poisson when N is Poisson. Compare this with the individual risk model, where a fixed number of policies each give a claim of zero or more. There the number of terms is fixed, not random.

The key idea is to condition on N. Given N = n, S is a sum of n independent claims, so its distribution is the n-fold convolution of the severity distribution. Then you average over n using the frequency probabilities. This gives the distribution function: FS(x) = Σ P(N = n) × FX*n(x).

Key rules to remember

Aggregate claims
S = X1 + X2 + ... + XN, with S = 0 if N = 0
N is random. Xi are i.i.d. and independent of N.
Distribution function of S
FS(x) = Σ (n = 0 to ∞) P(N = n) × FX*n(x)
FX*n is the n-fold convolution of FX. FX*0 is a point mass at 0.
Probability of zero aggregate claims
P(S = 0) = P(N = 0) if X > 0 always
If claims are strictly positive, S = 0 only when there are no claims.
Mean of S
E[S] = E[N] × E[X]
Needs the Xi identically distributed and independent of N.
Variance of S
Var(S) = E[N] × Var(X) + Var(N) × (E[X])²
Needs independence of Xi from each other and from N. Derived by conditioning on N.
Moment generating function of S
MS(t) = MN(ln MX(t))
Equivalent to MS(t) = PN(MX(t)), where PN is the pgf of N. Valid where the mgfs exist.

How to solve Compound Distributions and Aggregate Claims Basics questions

Use this order for any question on the aggregate claims model.

  1. 1Write the model: S = X1 + ... + XN, and note that S = 0 when N = 0.
  2. 2Identify the frequency distribution of N and the severity distribution of X. Write down the parameters of each.
  3. 3State the assumptions: Xi i.i.d., and N independent of the Xi. If the question gives different conditions, say so.
  4. 4Compute the basic quantities you need: E[N], Var(N), E[X], Var(X). Use the standard results for the named distributions.
  5. 5Apply the formulas for E[S] and Var(S), or condition on N for probabilities, or use the mgf if the question asks for the distribution.
  6. 6For P(S = 0) or small values of S, list the cases by N directly.
  7. 7Check the answer: units in rupees, variance not negative, and Var(S) at least E[N] × Var(X).
  8. 8Write the final result with a short interpretation if asked.

Quickest way: Mean and variance in four lines

When to use it: When the question gives you moments of N and X and asks for E[S], Var(S) or a normal approximation, and does not need the full distribution.

  1. Write E[N], Var(N), E[X], Var(X) in a row.
  2. Multiply to get E[S] = E[N] × E[X].
  3. Compute Var(S) = E[N] × Var(X) + Var(N) × (E[X])². Keep the two terms separate to avoid slips.
  4. If N is Poisson, then Var(N) = E[N], so Var(S) = E[N] × E[X²]. This is a useful shortcut that needs only the second moment.

Common mistakes in Compound Distributions and Aggregate Claims Basics

  • Treating N as fixed and using Var(S) = n × Var(X).

    It is the formula for a sum of a fixed number of terms, and it is familiar from statistics.

    Fix: Always include the second term Var(N) × (E[X])². It captures the variation from the random claim count.

  • Forgetting that S = 0 when N = 0, so P(S = 0) is missed.

    Students focus on the continuous severity and forget the point mass at zero.

    Fix: Write P(S = 0) = P(N = 0) first when claims are positive. Treat S as having a mixed distribution.

  • Using E[S] = E[N] × E[X] without checking independence.

    The formula looks general and is memorised without its conditions.

    Fix: State the assumptions in the answer. Dependence between N and X can change even the mean.

  • Confusing frequency and severity, for example using the Poisson for claim size.

    Both are described by named distributions, and the roles blur.

    Fix: Ask: is it a count (N, discrete) or an amount (X, positive, usually continuous)? Label each in your working.

  • Writing the distribution of S as the distribution of N times X.

    Students multiply random variables instead of summing N copies.

    Fix: S is a sum of N independent copies of X, not N × X. The n-fold convolution is needed. Only if all claims were the same value X would S = N × X. Then Var(S) = E[N²] × Var(X) + Var(N) × (E[X])², which is larger than the variance of the i.i.d. sum.

  • Mixing up E[X²] and (E[X])² in the Poisson shortcut.

    Both look like 'the mean squared' under time pressure.

    Fix: For compound Poisson use Var(S) = λ × E[X²], where E[X²] = Var(X) + (E[X])². Compute E[X²] explicitly.

Worked examples

Example 1

The number of claims N in a year has mean 20 and variance 30. Each claim X has mean ₹5,000 and standard deviation ₹4,000. Assuming the standard model, find E[S] and Var(S) and the standard deviation of S.

Show the solution
  1. Assumptions: the Xi are i.i.d. and independent of N.
  2. E[N] = 20, Var(N) = 30, E[X] = 5,000, Var(X) = 4,000² = 16,000,000.
  3. E[S] = 20 × 5,000 = ₹1,00,000.
  4. Var(S) = E[N] × Var(X) + Var(N) × (E[X])².
  5. First term: 20 × 16,000,000 = 320,000,000.
  6. Second term: 30 × 25,000,000 = 750,000,000.
  7. Var(S) = 1,070,000,000.
  8. SD(S) = √1,070,000,000 ≈ 32,711.

Answer: E[S] = ₹1,00,000, Var(S) = 1,070,000,000 (₹²), and SD(S) ≈ ₹32,711.

Example 2

Claims arrive as a Poisson number N with mean 2. Each claim is independently ₹10,000 with probability 0.6 and ₹20,000 with probability 0.4. Find E[S], Var(S) and P(S = 0).

Show the solution
  1. N is Poisson with λ = 2, so E[N] = Var(N) = 2.
  2. E[X] = 0.6 × 10,000 + 0.4 × 20,000 = 6,000 + 8,000 = 14,000.
  3. E[X²] = 0.6 × 10⁸ + 0.4 × 4 × 10⁸ = 0.6 × 10⁸ + 1.6 × 10⁸ = 2.2 × 10⁸.
  4. E[S] = 2 × 14,000 = ₹28,000.
  5. For compound Poisson, Var(S) = λ × E[X²] = 2 × 2.2 × 10⁸ = 4.4 × 10⁸.
  6. Check by the general formula: Var(X) = 2.2 × 10⁸ − 1.96 × 10⁸ = 0.24 × 10⁸. Then 2 × 0.24 × 10⁸ + 2 × 1.96 × 10⁸ = 0.48 × 10⁸ + 3.92 × 10⁸ = 4.4 × 10⁸. This matches.
  7. Claims are always positive, so S = 0 only if N = 0.
  8. P(S = 0) = e^(−2) ≈ 0.1353.

Answer: E[S] = ₹28,000, Var(S) = 4.4 × 10⁸ (₹²), and P(S = 0) = e⁻² ≈ 0.1353.

Exam tips

  • Write the assumptions (i.i.d. claims, independence from N) in one line. Examiners award marks for stating them.
  • Show the two-term variance formula before substituting numbers. Part marks depend on method.
  • For compound Poisson questions, use Var(S) = λ × E[X²] and cross-check once with the general formula if time allows.
  • In Paper B (R or Excel), simulate S by drawing N, then summing N claim values, repeated many times. Use the simulated mean and variance to check your formulas.
  • Always give units in rupees and say which quantity is a variance (₹²) and which is a standard deviation (₹).

Practice questions from Compound distributions and their applications in risk modelling

Compound Distributions and Aggregate Claims Basics: frequently asked questions

What is a compound distribution in actuarial science?

It is the distribution of a sum of a random number of random variables. In the aggregate claims model, S = X1 + ... + XN where N is the claim count and Xi are the claim amounts. If N is Poisson, S is compound Poisson.

What is the difference between frequency and severity?

Frequency is the distribution of the number of claims N, which is discrete. Severity is the distribution of the size of one claim X, which is usually continuous and skewed. The aggregate claims S combines both.

Why must N be independent of the claim amounts?

The formulas for E[S] and Var(S) come from conditioning on N and treating the Xi as unaffected by N. If claim size depends on the count, for example in a catastrophe year, those formulas no longer hold.

Is the aggregate claims distribution continuous?

Not entirely. If claims are positive, S has a point mass at 0 equal to P(N = 0), and a continuous part for S > 0 when X is continuous. This is a mixed distribution.