Compound distributions and their applications in risk modelling: formula sheet

Key formulas

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.

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.

Aggregate claims

S = X₁ + X₂ + … + X_N, S = 0 if N = 0

X's are i.i.d. and independent of N.

General mgf of S

M_S(t) = P_N(M_X(t)) = M_N(ln M_X(t))

Valid where M_X(t) exists. P_N(z) = E[z^N].

Mean and variance of S

E[S] = E[N]E[X]; Var(S) = E[N]Var(X) + Var(N)(E[X])²

Holds for any compound distribution under the standard assumptions.

Compound Poisson mgf

M_S(t) = exp{λ(M_X(t) − 1)}

Because P_N(z) = exp{λ(z − 1)} for Poisson(λ).

Compound Poisson moments

E[S] = λE[X]; Var(S) = λE[X²]

Mean and variance of N are both λ. Third central moment is λE[X³].

Compound binomial mgf

M_S(t) = (1 − p + p M_X(t))ⁿ

N ~ Binomial(n, p). Mean np E[X]; variance np Var(X) + np(1 − p)(E[X])².

Compound negative binomial mgf

M_S(t) = (p / (1 − q M_X(t)))^k, with q = 1 − p

N is the number of failures before the kth success, P(N = j) = C(j + k − 1, j) p^k q^j. E[N] = kq/p, Var(N) = kq/p².

Sum of independent compound Poisson

S = S₁ + … + S_m, S_i ~ CP(λ_i, F_i) ⇒ S ~ CP(λ, F), λ = Σλ_i, F(x) = Σ(λ_i/λ)F_i(x)

Requires independence of the S_i. The claim size is a mixture with weights λ_i/λ.

(a, b, 0) class condition

P(N = k) = (a + b ÷ k) × P(N = k − 1), for k = 1, 2, 3, …

Use it to find a and b by dividing consecutive probabilities if the question does not give them.

Parameters a and b

Poisson(λ): a = 0, b = λ. Binomial(n, p), q = 1 − p: a = −p ÷ q, b = (n + 1)p ÷ q. Negative binomial (k, p), q = 1 − p, P(N = x) = C(x + k − 1, x) pᵏ qˣ: a = q, b = (k − 1)q.

Binomial has a negative. Geometric is the negative binomial with k = 1, so a = q, b = 0. Check the parametrisation against the Tables.

Panjer starting value

g₀ = P(S = 0) = G_N(f₀), where G_N is the PGF of N and f₀ = P(X = 0)

If f₀ = 0, then g₀ = P(N = 0). For Poisson, g₀ = exp(λ(f₀ − 1)).

Panjer recursion

g_s = [1 ÷ (1 − a f₀)] × Σ (j = 1 to s) (a + b j ÷ s) f_j g_{s−j}, for s = 1, 2, 3, …

If f₀ = 0 the factor in front is 1. For compound Poisson this becomes g_s = (λ ÷ s) Σ j f_j g_{s−j}.

Convolution formula

P(S = s) = Σ (n ≥ 0) P(N = n) f^{*n}(s), with f^{*n}(s) = Σ_y f(y) f^{*(n−1)}(s − y)

f^{*0}(0) = 1. For continuous claims, F_S(x) = Σ P(N = n) F^{*n}(x).

Mean and variance of S

E[S] = E[N]E[X]; Var[S] = E[N]Var[X] + Var[N](E[X])²

For compound Poisson with parameter λ, Var[S] = λE[X²].

Normal approximation

P(S ≤ s) ≈ Φ((s − E[S]) ÷ √Var[S])

For integer-valued S, use s + 0.5 for P(S ≤ s) and s − 0.5 for P(S ≥ s) (continuity correction).

Compound moments (general)

E[S] = E[N]·E[X]; Var(S) = E[N]·Var(X) + Var(N)·(E[X])²

Use with Y in place of X for retained claims. Needs independent, identically distributed claim sizes independent of N.

Compound Poisson moments

E[S] = λE[X]; Var(S) = λE[X²]

Here λ is the expected number of claims in the period. Note E[X²], not Var(X).

Proportional reinsurance

Y = αX; E[Y] = αE[X]; Var(Y) = α²Var(X)

α is the proportion retained by the insurer. Reinsurer pays (1 − α)X.

Excess of loss retained claim

Y = min(X, M); Z = max(X − M, 0) = X − Y

Y is paid by the insurer, Z by the reinsurer. M is the retention.

Expected retained claim

E[min(X, M)] = ∫₀^M (1 − F(x)) dx

Also E[X] − E[Z]. Use whichever is easier for the given distribution.

Exponential claims, excess of loss

If X ~ Exp(μ) with mean 1/μ: E[min(X,M)] = (1 − e^(−μM)) ÷ μ; E[Z] = e^(−μM) ÷ μ

Memoryless property: given X > M, X − M is again Exp(μ).

Pareto claims, excess of loss

If X ~ Pareto(α, λ) with α > 1: E[Z] = λ^α ÷ ((α − 1)(λ + M)^(α − 1))

Here Z = max(X − M, 0) and the survival function is (λ ÷ (λ + x))^α.

Thinning of Poisson claims

Number of claims exceeding M ~ Poisson(λ·P(X > M))

Valid when claim numbers are Poisson and claim sizes are independent.

Adjustment coefficient

λ·(M_Y(R) − 1) = c·R

c is premium income per unit time and M_Y(R) is the MGF of the claim size. Take the positive root R. Ruin probability ψ(u) ≤ e^(−Ru), where u is initial surplus.

Net profit condition

c > λE[Y]

Retained premium must exceed expected retained claims per unit time, otherwise ruin is certain.

Quick revision

• S = X₁ + … + X_N, with the X's independent and identically distributed and independent of N.
• E[S] = E[N]·E[X].
• Var(S) = E[N]·Var(X) + Var(N)·(E[X])².
• MGF of S: M_S(t) = M_N(ln M_X(t)), where it exists.
• Compound Poisson with rate λ: E[S] = λ·E[X] and Var(S) = λ·E[X²].
• For Poisson N, the mean and variance of N are equal; binomial has variance below the mean; negative binomial has variance above the mean.
• Panjer recursion applies when the claim number distribution is in the (a, b, 0) class, with discrete claim sizes.
• A normal approximation to S needs the exact mean and variance, and may be poor when claims are few or skewed.
• Under excess of loss with retention M, the insurer pays min(X, M) per claim and the reinsurer pays max(X − M, 0).
• Under stop loss with retention d, the reinsurer pays max(S − d, 0).
• Always state your assumptions before you start a calculation.

Common mistakes

• Treating N as fixed and using Var(S) = n × Var(X). 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. Fix: Write P(S = 0) = P(N = 0) first when claims are positive. Treat S as having a mixed distribution.
• Using Var[S] = E[N]Var[X] only. 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. Fix: For compound Poisson the variance is λE[X²], the second moment, not the variance.
• Writing the compound Poisson variance as λVar(X). Fix: Use Var(S) = λE[X²]. Check: λVar(X) + λ(E[X])² = λE[X²].
• Substituting t into the count mgf instead of ln M_X(t). Fix: If you use the pgf, substitute M_X(t) for z. If you use the mgf, substitute ln M_X(t) for t.
• Using the wrong a and b for the binomial or negative binomial. Fix: Derive a and b from P(N = k) ÷ P(N = k − 1) if unsure. Binomial has a negative a. For negative binomial check the Tables for which parameter is p and which is q.
• Starting with g₀ = P(N = 0) when f₀ is not zero. Fix: Always write g₀ = G_N(f₀). Also remember the 1 ÷ (1 − a f₀) factor in the recursion.
• Using Var(Y) instead of E[Y²] in the compound Poisson variance. Fix: For compound Poisson, always write Var(S) = λE[Y²]. Compute E[Y²] directly.
• Treating excess of loss as a share of every claim, or proportional as only for large claims. Fix: Proportional: Y = αX, all claims. Excess of loss: Y = min(X, M), only the part above M is passed on.

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 (₹).
• 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.