Economic Modelling · Ruin theory

Effect of Reinsurance on Adjustment Coefficient and Ruin Probability

Updated 11 October 2026 · Fact-checked

Reinsurance changes the insurer's retained claims and its net premium income. You rewrite the adjustment coefficient equation using retained claims and net premium, then solve for R. A larger R gives a smaller Lundberg bound e^(−Ru) on ruin probability. Reinsurance helps only if the net premium still exceeds expected retained claims.

Understand Reinsurance and Ruin Probability

In ruin theory the surplus is U(t) = u + ct − S(t). Here u is the initial surplus, c is the premium rate and S(t) is aggregate claims. Claims arrive as a Poisson process with rate λ. Each claim X has moment generating function M_X(r). Premium is c = (1 + θ)λE[X], where θ is the insurer's premium loading. The adjustment coefficient R is the positive solution of λ + cR = λM_X(R). Lundberg's inequality says ψ(u) ≤ e^(−Ru). A bigger R means a lower upper bound on ruin probability.

Reinsurance changes two things. It reduces the claims the insurer pays. It also reduces premium income, because the insurer pays the reinsurer a premium. The reinsurer charges a loading ξ on its expected payments. So the insurer's net premium rate is the original premium minus (1 + ξ) times the reinsurer's expected claim payments per unit time.

With proportional reinsurance, the insurer keeps a fraction α of every claim. It pays αX, and the reinsurer pays (1 − α)X. The retained claim has the same shape as X but is scaled. With excess of loss reinsurance and retention M, the insurer pays min(X, M) and the reinsurer pays (X − M)+. This cuts off the large claims, which are the ones that drive ruin.

The loading matters. If ξ = θ, the insurer gives up profit at the same rate it gives up risk, and reinsurance usually improves R. If ξ > θ, reinsurance is expensive. Buying too much can leave a net margin that is zero or negative, and then no positive R exists. In that case ruin is certain, and Lundberg's bound is of no use.

To compare arrangements, compute R for each and prefer the larger one. Also check that the net premium exceeds expected retained claims. State this assumption in your answer.

Key rules to remember

Original adjustment coefficient equation: λ + cR = λ M_X(R), with c = (1 + θ)λE[X]; R is the smallest positive root. It is the equation with no reinsurance.
Lundberg's inequality: ψ(u) ≤ e^(−Ru); Valid when a positive R exists. It is an upper bound, not the exact ruin probability.
Proportional reinsurance: net premium rate: c_α = c − (1 + ξ)(1 − α)λE[X]; α is the proportion retained by the insurer, and ξ is the reinsurer's loading.
Proportional reinsurance: equation for R: λ + c_α R = λ M_X(αR); The retained claim αX has mgf M_X(αR). Solve for the positive root.
Proportional reinsurance: condition for a positive R: c_α > αλE[X], that is θ > ξ(1 − α); If ξ > θ, this requires α > 1 − θ/ξ. If ξ ≤ θ, it holds for every α > 0.
Excess of loss: net premium rate: c_M = c − (1 + ξ)λE[(X − M)+]; M is the retention. For exponential claims with mean m, E[(X − M)+] = m e^(−M/m).
Excess of loss: equation for R: λ + c_M R = λ E[e^(R·min(X, M))]; The retained claim is min(X, M). It is bounded above by M.
Exponential claims, mean m, net premium c' and retained mean m': R = (c' − λm')/(c'·m'), when λ = 1 per unit time; This is a shortcut for exponential claims, valid for claim rate λ = 1 and exponential retained claims. Here c' and m' are the retained values.

How to solve Reinsurance and Ruin Probability questions

Use this method for any question that asks how reinsurance changes R or ruin probability. State the Poisson and independence assumptions first.

1Write down the original model: λ, the claim distribution, the loading θ and the premium c = (1 + θ)λE[X].
2Identify the reinsurance type and its parameters: the retention α for proportional, or the retention M for excess of loss. Identify the reinsurer's loading ξ.
3Find the expected amount the reinsurer pays per unit time. For proportional it is (1 − α)λE[X]. For excess of loss it is λE[(X − M)+].
4Calculate the net premium rate: c minus (1 + ξ) times the expected reinsurer payment.
5Check that the net premium exceeds the insurer's expected retained claims per unit time. If it does not, no positive R exists.
6Write the adjustment coefficient equation with the net premium and the retained claim mgf. Solve for the positive root R, algebraically or numerically.
7Give the Lundberg bound e^(−Ru) for the stated u. Compare it with the no-reinsurance bound.
8Comment in words: R rose or fell, and why this depends on ξ compared with θ and on the retention.

Quickest way: Quick route for exponential claims

When to use it: Use it when claims are exponential and the reinsurance is proportional. The retained claim αX is then exponential with mean αm, so the equation reduces to a quadratic.

Compute the net premium c_α and the retained mean αm.
Check that c_α − αλm > 0. If it is not, stop and say no positive R exists.
Write λ + c_α R = λ/(1 − αmR). Cross-multiply and cancel the constant term.
The remaining equation has a zero root and one positive root. Divide through by R and solve the linear equation.
Compute e^(−Ru) for the bound. For exponential claims you can also give the exact ψ(u) = (1/(1 + θ'))e^(−Ru), where θ' is the net loading on retained claims.

Common mistakes in Reinsurance and Ruin Probability

Using the original premium c in the equation after reinsurance.
Students change the claim mgf but forget that the insurer now pays reinsurance premium.
Fix: Always compute the net premium c_α or c_M first. Write it down before forming the equation.
Using M_X(R) instead of M_X(αR) for proportional reinsurance.
The retained claim is αX, but the mgf argument is not scaled.
Fix: The retained claim is αX, so its mgf at R is M_X(αR). Check that the left side uses c_α.
Assuming more reinsurance always raises R.
Students remember that reinsurance reduces risk.
Fix: Compare ξ with θ. If ξ > θ, the net margin shrinks as α falls and R can fall to zero. Check the positivity condition first.
Computing the reinsurer's loading on the insurer's retained claims.
The two sides are confused.
Fix: The loading ξ applies to what the reinsurer expects to pay: (1 − α)λE[X] for proportional, and λE[(X − M)+] for excess of loss.
Treating e^(−Ru) as the exact ruin probability.
Lundberg's inequality is quoted loosely.
Fix: Say ψ(u) ≤ e^(−Ru). Only for special cases such as exponential claims do you have an exact formula.
Using the claim mean instead of the retained claim mean when finding the net loading.
Students forget that the retained claim distribution changed.
Fix: Net loading on retained claims is (c' − λm')/(λm'). Use the retained mean m' throughout.

Worked examples

Example 1

Claims arrive as a Poisson process with rate λ = 1 per year. Claim sizes are exponential with mean ₹1 lakh. The premium loading is θ = 0.5. The insurer buys proportional reinsurance, retaining α = 0.5, with reinsurer loading ξ = 0.6. Find the adjustment coefficient with and without reinsurance. Give the Lundberg bound for initial surplus u = ₹5 lakh in each case.

Show the solution

Work in ₹ lakh. Without reinsurance, c = 1.5 × 1 × 1 = 1.5.
Equation: 1 + 1.5R = 1/(1 − R). Multiply out: (1 + 1.5R)(1 − R) = 1, so 0.5R − 1.5R² = 0. The positive root is R = 1/3 = 0.3333.
Bound without reinsurance: e^(−5/3) = e^(−1.6667) = 0.1889.
With reinsurance, the reinsurer pays an expected (1 − 0.5) × 1 × 1 = 0.5 per year. The reinsurance premium is 1.6 × 0.5 = 0.8. So c_α = 1.5 − 0.8 = 0.7.
Retained claims have mean 0.5. The margin is 0.7 − 0.5 = 0.2 > 0, so a positive R exists.
Equation: 1 + 0.7R = 1/(1 − 0.5R). Multiply out: (1 + 0.7R)(1 − 0.5R) = 1, so 0.2R − 0.35R² = 0. The positive root is R = 0.2/0.35 = 4/7 = 0.5714.
Bound with reinsurance: e^(−5 × 0.5714) = e^(−2.857) = 0.0574.

Answer: R rises from 0.3333 to 0.5714 per ₹ lakh. The Lundberg bound at u = ₹5 lakh falls from 0.1889 to 0.0574. Here ξ is above θ, but the margin is still positive, so reinsurance helps.

Example 2

Using the same data as the previous example, find the smallest retention α for which a positive adjustment coefficient exists under proportional reinsurance. Then find R and the Lundberg bound at u = ₹5 lakh when α = 0.25.

Show the solution

Net premium: c_α = 1.5 − 1.6(1 − α).
Expected retained claims: α × 1 × 1 = α.
A positive R needs c_α > α, so 1.5 − 1.6(1 − α) > α. This gives −0.1 + 1.6α > α, so 0.6α > 0.1 and α > 1/6.
For α = 0.25, c_α = 1.5 − 1.6 × 0.75 = 1.5 − 1.2 = 0.3. The retained mean is 0.25 and the margin is 0.05 > 0.
Equation: 1 + 0.3R = 1/(1 − 0.25R). Multiply out: (1 + 0.3R)(1 − 0.25R) = 1, so 0.05R − 0.075R² = 0. The positive root is R = 0.05/0.075 = 2/3 = 0.6667.
Check with the shortcut: R = (0.3 − 0.25)/(0.3 × 0.25) = 0.05/0.075 = 0.6667.
Bound: e^(−5 × 0.6667) = e^(−3.333) = 0.0357.

Answer: A positive R exists only if α > 1/6. At α = 0.25, R = 0.6667 and the Lundberg bound at u = ₹5 lakh is 0.0357. As α falls towards 1/6 the margin tends to zero, so R tends to zero. Reinsurance helps only up to a point when ξ > θ.

Exam tips

Write the net premium line first. Examiners give marks for c_α or c_M, and many students lose them here.
Always check the positivity condition. A one-line check shows you understand why a positive R may not exist.
For a comparison question, compute R for both types and then explain in words. Say that excess of loss removes the large claims that drive ruin, and that proportional reinsurance scales every claim.
For excess of loss with non-exponential or capped claims, you may be asked only to set up the equation and solve it numerically. Show the form λ + c_M R = λE[e^(R·min(X, M))] clearly.
On a computer-based paper, solve for R with a root finder such as uniroot in R. Start the search just above zero, because R = 0 is always a trivial root.

Practice questions from Ruin theory

Reinsurance and Ruin Probability: frequently asked questions

How does reinsurance reduce the probability of ruin?

It cuts the claims the insurer pays, and it costs a reinsurance premium. If the net premium still exceeds expected retained claims, the adjustment coefficient R usually rises, and the bound e^(−Ru) on ruin probability falls. Retaining a smaller share of the large claims reduces the variability of the surplus.

Is proportional or excess of loss reinsurance better for ruin probability?

For the same expected reinsurance cost, excess of loss usually gives a larger R. It removes the large claims, which are the ones that cause ruin. Proportional reinsurance scales every claim, including the small ones, so it gives up more premium for the same cut in risk. Check the numbers in the question rather than assuming.

What happens to R if the reinsurer's loading ξ is higher than the insurer's loading θ?

The net margin shrinks as the insurer reinsures more. If the retention is too low, the net premium falls below expected retained claims and no positive R exists. Ruin is then certain. Always check the positivity condition.

Does a larger adjustment coefficient always mean a lower ruin probability?

A larger R gives a smaller Lundberg upper bound e^(−Ru). For exponential claims the exact ruin probability also falls as R rises, if the net loading is positive. In general, use R to compare bounds and say that it is an upper bound.