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

Applications of Reinsurance and Ruin in Risk Modelling

Updated 11 October 2026 · Fact-checked

Reinsurance changes each claim X into a retained part and a reinsured part. You then rebuild the aggregate retained claims S = ΣY as a compound sum using the new claim size Y. Proportional gives Y = αX. Excess of loss gives Y = min(X, M). Ruin assessment then uses the retained claims and the reduced premium.

Understand Applications: Reinsurance and Ruin in Risk Modelling

A compound model says total claims S = X₁ + X₂ + … + X_N, where N is the number of claims and the Xᵢ are independent, identically distributed claim sizes, independent of N. Reinsurance does not change N for the insurer's own count of claims. It changes the size of each claim the insurer pays.

Under proportional reinsurance, the insurer keeps a fixed share α of every claim. The retained claim is Y = αX and the reinsurer pays (1 − α)X. Every claim is shared, big or small.

Under individual excess of loss reinsurance with retention M, the insurer pays the first M of each claim. The retained claim is Y = min(X, M). The reinsurer pays max(X − M, 0). Small claims stay fully with the insurer. Only large claims are shared.

The key idea: the retained aggregate S_R = Y₁ + … + Y_N is again a compound distribution. The number of claims N is the same, but the claim size is now Y. So you use the usual compound moment formulas with E[Y] and E[Y²]. For compound Poisson with rate λ, E[S_R] = λE[Y] and Var(S_R) = λE[Y²].

One subtle point: under excess of loss, the reinsurer sees only claims that exceed M. If N is Poisson(λ), the number of claims exceeding M is Poisson(λP(X > M)). This is the thinning property. The reinsurer's aggregate claims are compound Poisson with this rate and claim size (X − M) given X > M.

Ruin assessment asks whether the insurer's surplus stays positive. With reinsurance, the insurer pays a reinsurance premium out of income, so the net premium income falls. The insurer needs the retained premium loading to stay positive, or ruin is certain in the long run. A common check is the adjustment coefficient R, the positive solution of the equation below.

Key rules to remember

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.

How to solve Applications: Reinsurance and Ruin in Risk Modelling questions

Use this order for any question that mixes compound distributions, reinsurance and ruin.

  1. 1Write the original model: the distribution of N and of X, and the time period.
  2. 2Identify the reinsurance type and the retention: proportional with α, or excess of loss with M.
  3. 3Define the retained claim Y (and the reinsurer's claim Z) in terms of X.
  4. 4Find E[Y] and E[Y²] by integration or from known results. Check the limits: M = 0 and M = ∞.
  5. 5Apply the compound formulas using N and Y. For compound Poisson use λE[Y] and λE[Y²].
  6. 6If the reinsurer is asked about, thin the claim count: λ·P(X > M), and use Z given X > M.
  7. 7For premiums, use the stated premium principle (for example expected value with loading θ) separately for insurer and reinsurer.
  8. 8For ruin, compute retained premium income, check the net profit condition, then solve for R or compute the stated probability. State the assumptions.

Quickest way: Retained claim first, compound formula second

When to use it: Use when time is short and the question asks for mean, variance or a probability of retained aggregate claims.

  1. Write Y in one line: αX or min(X, M).
  2. Compute only E[Y] and E[Y²]. Use E[min(X,M)] = E[X] − E[Z] if E[Z] is easier.
  3. Plug into λE[Y] and λE[Y²] for compound Poisson.
  4. Sanity check: retained mean must be below the original mean, and the retained variance too.
  5. For ruin, compare c with λE[Y] before doing any heavy algebra.

Common mistakes in Applications: Reinsurance and Ruin in Risk Modelling

  • Using Var(Y) instead of E[Y²] in the compound Poisson variance.

    The general formula uses Var(X), so students carry it over without checking the Poisson case.

    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.

    The two types are mixed up because both reduce the insurer's claims.

    Fix: Proportional: Y = αX, all claims. Excess of loss: Y = min(X, M), only the part above M is passed on.

  • Forgetting that the reinsurer's claim count is smaller than N.

    Students keep λ unchanged when moving to the reinsurer's side.

    Fix: Use λ·P(X > M) as the reinsurer's claim rate, with claim size X − M given X > M.

  • Using full premium income in the ruin calculation after reinsurance.

    The reinsurance premium is easy to forget when the question focuses on claims.

    Fix: Subtract the reinsurance premium from income first. Then test c > λE[Y].

  • Computing E[Z] as E[X] − M.

    It looks like 'the amount above M', but it ignores claims below M, which give Z = 0.

    Fix: Use E[Z] = ∫_M^∞ (1 − F(x)) dx, or E[X] − E[min(X, M)].

  • Taking the negative or zero root when solving for the adjustment coefficient.

    The equation always has R = 0 as a root.

    Fix: State that you need the strictly positive root, and check it lies within the range where the MGF exists.

Worked examples

Example 1

Annual claim numbers for an insurer are Poisson with mean 50. Claim sizes X are exponential with mean ₹40,000. The insurer has an individual excess of loss treaty with retention ₹1,00,000. Find (a) the expected annual aggregate claims retained by the insurer, (b) the expected number of claims reaching the reinsurer, and (c) the expected annual aggregate claims paid by the reinsurer.

Show the solution
  1. Let μ = 1/40,000 per rupee. Retained claim Y = min(X, 1,00,000).
  2. E[Y] = (1 − e^(−μM)) ÷ μ with μM = 1,00,000 ÷ 40,000 = 2.5.
  3. e^(−2.5) = 0.082085. So E[Y] = 40,000 × (1 − 0.082085) = 40,000 × 0.917915 = ₹36,717 (rounded).
  4. (a) E[S_R] = 50 × 36,717 = ₹18,35,830 (rounded).
  5. (b) P(X > M) = e^(−2.5) = 0.082085. Expected number reaching the reinsurer = 50 × 0.082085 = 4.104.
  6. (c) E[Z] = e^(−2.5) × 40,000 = 0.082085 × 40,000 = ₹3,283 (rounded). Reinsurer's expected aggregate = 50 × 3,283 = ₹1,64,170 (rounded).
  7. Check: 18,35,830 + 1,64,170 = 20,00,000 = 50 × 40,000, the original expected aggregate.

Answer: (a) about ₹18,35,830; (b) about 4.10 claims; (c) about ₹1,64,170.

Example 2

Aggregate claims S are compound Poisson with λ = 100 per year. Claim sizes X are uniform on (0, 10) in ₹ lakh. The insurer buys proportional reinsurance and retains 60% of each claim. Find the mean and variance of the retained aggregate claims, and the percentage reduction in the standard deviation compared with no reinsurance.

Show the solution
  1. E[X] = 5 and E[X²] = 10² ÷ 3 = 33.3333 (₹ lakh squared).
  2. Retained claim Y = 0.6X. So E[Y] = 3 and E[Y²] = 0.36 × 33.3333 = 12.
  3. Mean of retained aggregate = λE[Y] = 100 × 3 = 300 (₹ lakh).
  4. Variance of retained aggregate = λE[Y²] = 100 × 12 = 1,200.
  5. Without reinsurance: variance = 100 × 33.3333 = 3,333.33, standard deviation = 57.735.
  6. With reinsurance: standard deviation = √1,200 = 34.641.
  7. Ratio = 34.641 ÷ 57.735 = 0.6, so the reduction is 40%.
  8. This agrees with the rule that standard deviation scales by α for proportional cover.

Answer: Mean ₹300 lakh, variance 1,200 (₹ lakh squared), standard deviation falls by 40%.

Exam tips

  • Always define Y and Z in symbols before any calculation. Marks are given for the set-up.
  • Check your answer with E[Y] + E[Z] = E[X]. It takes ten seconds and catches most errors.
  • Write 'compound Poisson, so Var(S) = λE[Y²]' explicitly. Examiners look for the right formula.
  • In ruin parts, state the net profit condition and the premium after reinsurance cost. Then solve for R.
  • In Paper B (R), show the code logic: simulate N, simulate X, apply min(X, M), sum, and repeat. Report the mean and variance next to the theoretical values.

Practice questions from Compound distributions and their applications in risk modelling

Applications: Reinsurance and Ruin in Risk Modelling in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Applications: Reinsurance and Ruin in Risk Modelling: frequently asked questions

What is the difference between proportional and excess of loss reinsurance in CS2?

Proportional reinsurance shares every claim in a fixed ratio, so Y = αX. Excess of loss reinsurance caps each claim at the retention, so Y = min(X, M). Proportional cover scales the variance by α². Excess of loss cuts the large claims and so reduces the tail more directly.

How do I find aggregate claims after excess of loss reinsurance?

Replace X by Y = min(X, M) in the compound model. Find E[Y] and E[Y²], then use the compound formulas with the same claim number distribution. For compound Poisson, E[S] = λE[Y] and Var(S) = λE[Y²].

Why does the reinsurer's claim count differ from the insurer's?

Only claims larger than M produce a payment under excess of loss. If N is Poisson(λ), the number exceeding M is Poisson(λP(X > M)). That is thinning of the Poisson process.

How does reinsurance affect ruin probability?

It lowers the variability of claims, which can reduce ruin probability. It also costs premium, which lowers income. You must check both effects, using the retained premium and the retained claim distribution, for example through the adjustment coefficient.