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

Moments of Compound Distributions: Mean and Variance of S

Updated 11 October 2026 · Fact-checked

A compound random variable is S = X₁ + … + X_N, where N is a random number of claims and the X's are iid claim sizes independent of N. Use conditioning on N: E[S] = E[N]E[X] and Var[S] = E[N]Var[X] + Var[N](E[X])². Apply these directly to find the moments.

Understand Moments of Compound Distributions

In risk modelling, total claims in a period are a sum of a random number of random claim sizes. Write S = X₁ + X₂ + … + X_N. Here N is the claim number and the X_i are the claim amounts. If N = 0, then S = 0.

The standard model has two assumptions. The X_i are independent and identically distributed. And N is independent of the X_i. State both in your answer. The results below fail without them.

The idea is to fix N first. Given N = n, S is a sum of n fixed iid terms, so its mean is nE[X] and its variance is nVar[X]. Then you average over N. The tools are the tower law, E[S] = E[E[S | N]], and the law of total variance, Var[S] = E[Var(S | N)] + Var(E[S | N]).

Applying them: E[S | N] = N·E[X], so E[S] = E[N]E[X]. Also Var(S | N) = N·Var[X] and Var(E[S | N]) = Var(N·E[X]) = (E[X])²Var[N]. Adding gives the variance formula. The first part is variation from claim sizes. The second part is variation from claim numbers.

For higher moments, the most direct route is the moment generating function: M_S(t) = M_N(ln M_X(t)) = E[(M_X(t))^N]. Differentiating at t = 0 gives the moments. For the third central moment, a known result for the compound Poisson is that the third central moment equals λE[X³].

Key rules to remember

Model
S = X₁ + … + X_N, with X_i iid, independent of N
S = 0 when N = 0. State these assumptions.
Mean of S
E[S] = E[N] × E[X]
Follows from the tower law E[S] = E[E[S | N]].
Variance of S
Var[S] = E[N] × Var[X] + Var[N] × (E[X])²
From the law of total variance. Needs independence of N and the X's.
Law of total variance
Var[S] = E[Var(S | N)] + Var(E[S | N])
Works for any pair of random variables with finite variance.
Compound Poisson
N ~ Poisson(λ): E[S] = λE[X], Var[S] = λE[X²]
Since E[N] = Var[N] = λ, the variance simplifies using Var[X] + (E[X])² = E[X²].
Compound MGF
M_S(t) = M_N(ln M_X(t))
Also equals G_N(M_X(t)) where G_N is the pgf of N. Valid where M_X(t) exists.
Compound Poisson MGF and third central moment
M_S(t) = exp(λ(M_X(t) − 1)); E[(S − E[S])³] = λE[X³]
The third central moment result is for compound Poisson only.

How to solve Moments of Compound Distributions questions

Use this method for any question asking for moments of a compound variable S.

  1. 1Write the model: define N, X and S. State that the X's are iid and independent of N.
  2. 2Find E[N] and Var[N] from the distribution of N. For Poisson both equal λ.
  3. 3Find E[X] and Var[X]. If you are given E[X²], use Var[X] = E[X²] − (E[X])².
  4. 4Calculate E[S] = E[N]E[X].
  5. 5Calculate Var[S] = E[N]Var[X] + Var[N](E[X])². For compound Poisson use λE[X²] as a check.
  6. 6For higher moments or a full distribution, use M_S(t) = M_N(ln M_X(t)) and differentiate at t = 0.
  7. 7Check units and sense: variance should be positive, and a larger λ should give a larger mean and variance.

Quickest way: Plug into the two formulas

When to use it: Use when N and X have known means and variances and you need only E[S] and Var[S], especially in MCQs.

  1. Write down E[N], Var[N], E[X], Var[X] in a row.
  2. Compute E[S] = E[N]E[X].
  3. Compute Var[S] = E[N]Var[X] + Var[N](E[X])².
  4. If N is Poisson, cross-check with Var[S] = λE[X²].
  5. If the question gives a coefficient of variation or a mixture, convert it to variance first.

Common mistakes in Moments of Compound Distributions

  • Using Var[S] = E[N]Var[X] only.

    Students forget that the number of claims is also random.

    Fix: Always add the second term Var[N](E[X])². It is zero only when N is fixed.

  • Writing Var[S] = λVar[X] for compound Poisson.

    Students copy the general form and drop the second term, or confuse Var[X] with E[X²].

    Fix: For compound Poisson the variance is λE[X²], the second moment, not the variance.

  • Using the standard deviation of X in place of the variance.

    The question gives the standard deviation and the student plugs it in directly.

    Fix: Square the standard deviation first to get Var[X].

  • Forgetting to state independence assumptions.

    Students treat the formulas as always true.

    Fix: Say that the X's are iid and independent of N before using the formulas. Marks are given for this.

  • Confusing E[N²] with (E[N])².

    Students use E[S²] = E[N]E[X²] without the cross terms.

    Fix: Use E[S²] = Var[S] + (E[S])² from the correct variance formula, or use the MGF.

Worked examples

Example 1

Annual claim numbers N follow a Poisson distribution with mean 50. Claim sizes X are independent of N with mean ₹20,000 and standard deviation ₹30,000. Find the mean and standard deviation of annual aggregate claims S.

Show the solution
  1. Model: S = X₁ + … + X_N, compound Poisson with λ = 50.
  2. E[X] = 20,000. Var[X] = 30,000² = 900,000,000.
  3. E[X²] = 900,000,000 + 20,000² = 900,000,000 + 400,000,000 = 1,300,000,000.
  4. E[S] = 50 × 20,000 = 1,000,000.
  5. Var[S] = λE[X²] = 50 × 1,300,000,000 = 65,000,000,000.
  6. Check with the general formula: E[N]Var[X] + Var[N](E[X])² = 50 × 900,000,000 + 50 × 400,000,000 = 45,000,000,000 + 20,000,000,000 = 65,000,000,000. Same.
  7. Standard deviation = √65,000,000,000 ≈ 254,951.

Answer: E[S] = ₹10,00,000 and the standard deviation is about ₹2,54,951.

Example 2

N has mean 4 and variance 8. Claim sizes have mean 500 and variance 40,000, independent of N. Derive Var[S] using the law of total variance and evaluate it.

Show the solution
  1. Condition on N. E[S | N] = 500N and Var(S | N) = 40,000N, because the X's are iid and independent of N.
  2. E[Var(S | N)] = 40,000 × E[N] = 40,000 × 4 = 160,000.
  3. Var(E[S | N]) = 500² × Var[N] = 250,000 × 8 = 2,000,000.
  4. Var[S] = 160,000 + 2,000,000 = 2,160,000.
  5. E[S] = 4 × 500 = 2,000, so the standard deviation is √2,160,000 ≈ 1,469.7.

Answer: Var[S] = 21,60,000 (in square rupee units); E[S] = 2,000.

Exam tips

  • Show the conditioning step in derivation questions: E[S | N] and Var(S | N), then the two laws. Method marks depend on it.
  • Write the independence assumptions before using any formula.
  • For compound Poisson, use λE[X²] as a fast answer and a check on your general formula.
  • If the question asks for a skewness or third moment, use the MGF or cumulants. Keep the algebra tidy and differentiate at t = 0.
  • In Paper B, check the simulated mean and variance of S against these formulas to validate your code.

Practice questions from Compound distributions and their applications in risk modelling

Moments of Compound Distributions: frequently asked questions

How do I find the mean and variance of a compound Poisson distribution?

With N ~ Poisson(λ), E[S] = λE[X] and Var[S] = λE[X²]. Find E[X²] from Var[X] + (E[X])² if you are given the variance.

Why does the variance have two terms?

One term comes from variation in claim sizes given N, which is E[N]Var[X]. The other comes from variation in the number of claims, which is Var[N](E[X])². The law of total variance splits them.

Do I need to derive the formulas in the exam?

Sometimes the question says show or derive, so you should be able to. Know the tower law and the law of total variance, and write each line clearly.

Can I use these formulas if claim sizes and numbers are dependent?

No. The results assume the X's are iid and independent of N. If that fails, you must go back to conditioning and work from first principles.