Excess-of-loss reinsurance: the retained claim, and choosing the retention

CM2 · Financial Engineering & Loss Reserving · 12 min read

What excess of loss does that proportional reinsurance does not

Under proportional reinsurance the insurer keeps a fixed share α of every claim. The retained claim is αX, which has the same shape as X, merely scaled. Its coefficient of variation does not change at all.

Under excess of loss the insurer keeps everything up to a retention M and nothing above it:

In words: the insurer pays the first M of every claim and the reinsurer pays the rest.

That is a different operation entirely. Y is bounded — it cannot exceed M whatever X does — so the tail of the claim distribution has been removed rather than shrunk. Since ruin is caused by the tail, excess of loss is the arrangement that actually addresses the risk, and it is why it is the common form in practice for large or volatile classes.

It is also the reason this is a harder calculation. Proportional reinsurance leaves an exponential claim exponential; excess of loss leaves it as a truncated distribution with an atom of probability sitting at M, and that atom has to be handled explicitly.

The retained claim, for exponential claims

Take X exponential with mean μ. The two pieces are:

In words: the insurer's share of the average claim, and the reinsurer's share. They add to μ, which is the first check to run.

The second formula is worth a moment, because it is cleaner than it has any right to be. The expected amount above a retention is the whole mean multiplied by the probability of exceeding the retention. That is the memoryless property of the exponential distribution: given that a claim has already passed M, the excess over M has the same distribution as a fresh claim. No other claim distribution is that convenient, which is both why examples use the exponential and why real work does not.

The moment generating function of the retained claim needs the integral and the atom together. For r < 1/μ, writing a = 1/μ − r:

The first term is the claims that settle below the retention; the second is the lump of probability at exactly M, which is where every claim above the retention ends up from the insurer's point of view. Forgetting that second term is the single commonest error in the topic, and it understates the retained risk.

The premium the insurer keeps

The reinsurer charges on the expected claims it takes, at its own loading θ_R:

Everything then follows from one function of M, and it is worth writing the expected profit out because it is more informative than it looks:

In words: the loading the insurer earns on the whole portfolio, less the loading the reinsurer earns on the part ceded.

Two things fall out. The profit rises with M, always, because ceding less means paying away less of the reinsurer's margin. And it is negative below a retention of

where the insurer has handed the reinsurer more margin than it has kept. That is a number worth computing before anything else, because every sensible answer lies above it, and it depends only on the ratio of the two loadings — not on the claim rate, and not on the surplus.

The retention that maximises safety

The adjustment coefficient R is still the positive root of the Lundberg equation, now with the retained claim's moment generating function and the net premium in it:

There is no closed form for R here, and there is not meant to be. The method is to solve it numerically for each candidate retention and compare. Two opposed forces are at work and that is why an interior optimum exists:

• Lowering M reduces the retained claims, and especially their tail, which pushes R up.
• Lowering M also reduces the net premium, because more is paid away at the reinsurer's higher loading, which pushes R down.

At a very low retention the second effect wins outright — the net premium falls below the retained expected claims and no positive R exists at all, which is the arithmetic saying ruin is certain. At a very high retention the first effect vanishes and R returns to its no-reinsurance value. Somewhere between, R is maximised.

Why the R-maximising retention is usually the wrong answer

This is the part a question is really testing, and it is a judgement rather than a calculation.

Maximising R maximises safety and takes no account of profit whatsoever. On the worked example below the R-optimal retention is barely above the average claim size: it cedes 41% of the gross premium and keeps 41% of the profit. No insurer would choose it.

What makes the decision tractable is the shape of the two curves near the optimum, and it is worth seeing. R is extremely flat around its maximum while profit is steep. On the example, moving the retention a fifth above the optimum costs 2% of the adjustment coefficient and gains 27% more profit. So the practical recommendation is almost never the maximum; it is a retention comfortably above it, chosen so that the ruin probability stays inside whatever the insurer or its regulator will tolerate.

The right way to present the answer, then, is a table of retentions against both R and profit, with the constraint stated: the lowest retention that keeps the ruin bound under the required figure, and no lower.

Other things the retention has to satisfy

A real decision has constraints the adjustment coefficient knows nothing about, and a good answer names a few:

• The largest claim the insurer could survive, which is a limit on M quite separate from any probability.
• Regulatory capital, which may charge less capital for a reinsured portfolio, so the reinsurance partly pays for itself.
• Reinsurer credit risk. The arithmetic above assumes the reinsurer pays. A cheap reinsurer that may not pay is not cheap, and the whole benefit of a low retention rests on the counterparty.
• Reinstatements and limits. Real excess-of-loss cover is usually capped, so the reinsurer pays the layer between M and some upper limit and the insurer is exposed again above it. The arithmetic here assumes unlimited cover above M.
• Stability of results rather than survival, which is what a shareholder cares about and what the variance of the retained claims measures.

Where marks are lost

Forgetting the atom at M in the moment generating function. The term erMP(X > M) is not optional. Without it the integral describes a claim that disappears when it exceeds the retention.

Scaling the gross premium by the retention. The net premium is the gross premium minus the reinsurance premium, which depends on the reinsurer's loading. It is not the gross premium multiplied by any ratio, and the whole economics of the arrangement lives in that difference.

Using the gross claim's moment generating function with the net premium. Both change. Mixing one of each gives a number that is neither.

Reporting the R-maximising retention as "the optimal retention" with no further comment. It is optimal for one objective. Say which, and say what it costs.

Dropping the restriction r < 1/μ. For exponential claims the moment generating function does not exist beyond that, so a root found outside it is not a root. Check μR < 1.

Questions students ask

Why does excess of loss beat proportional reinsurance for solvency?

Because ruin is a tail event and excess of loss removes the tail while proportional reinsurance only scales it. For the same expected amount ceded, excess of loss takes away the claims that could actually exhaust the fund and leaves the many small ones the premium easily covers. The price of that is that the reinsurer is taking the nastiest part of the distribution and will charge accordingly, which is why θ_R is usually well above θ.

What if the claims are not exponential?

The method is unchanged and the integrals get worse. For a lognormal or Pareto claim there is no closed form for either the retained mean or its moment generating function, so both are computed numerically — and for a Pareto claim heavy enough, the moment generating function of the gross claim does not exist at all, so the adjustment coefficient is undefined without reinsurance and becomes defined once a retention caps the claim. That is a striking point: for some portfolios excess of loss is not an improvement to the model, it is what makes the model exist.

Is there a retention that maximises expected profit?

Yes, and it is no reinsurance at all, as long as the reinsurer's loading exceeds the insurer's. The profit formula is increasing in M without limit. That is precisely why the question has to be posed as a trade-off rather than an optimisation: one objective is maximised at one end and the other in the middle.

How is the retention chosen in practice?

By constraint rather than by optimisation. The insurer decides how much ruin probability, or how much capital, or how large a single loss it is prepared to carry, and then buys the cheapest cover that satisfies it. The adjustment coefficient's role is to turn "how much risk" into a number that can be compared across retentions, not to pick one.

A worked example

An insurer writes a class in which claims arrive as a Poisson process at 10 a year and claim amounts are exponential with a mean of ₹20,000. Its premium loading is 30%. A reinsurer will write excess of loss above any retention M at a loading of 50%. The insurer's surplus is ₹2,00,000.

Every figure is invented for this example. It is not a real portfolio and not a real reinsurance quote.

Step 1: where the insurer starts

With no reinsurance, the adjustment coefficient is the exponential-claims formula from the ruin theory page:

A one-in-ten chance of ruin. That is the figure the reinsurance is being bought to reduce.

Step 2: the pieces of the retained claim

Each row sums to ₹20,000, the whole mean, which is the free check. Work one by hand: at M = 40,000, e−2 = 0.1353353, so the reinsurer's share is 20,000 × 0.1353353 = ₹2,706.71 and the insurer keeps the rest. Note that at a retention of twice the average claim the insurer is still keeping 86% of the average claim — the retention bites much less on the mean than it does on the tail, which is the whole point of it.

These came from the closed forms. Integrating x(1/μ)e−x/μ numerically from 0 to M over 400,000 intervals and adding M e−M/μ for the atom gives 12,642.4112, 15,537.3968 and 17,293.2943 — the same to four decimal places.

Step 3: the moment generating function, checked numerically

The formula is MY(r) = (1/μa)(1 − e−aM) + e−aM with a = 1/μ − r.

Nine decimal places. That matters more here than on most pages, because the Lundberg root is found by searching on this function and an error in it would be invisible in the final answer.

Step 4: the break-even retention

Build the profit from its three pieces, so the formula can be seen rather than quoted:

which is λμ(θ − θRe−M/μ) with the numbers in, and is zero when e−M/20,000 = 0.30/0.50 = 0.6:

Below that retention the insurer pays the reinsurer more margin than it keeps and makes an expected loss. Every sensible retention is above ₹10,216.

Step 5: the trade-off, retention by retention

For each M, solve 10 + c(M)R = 10 MY(R) for the positive root.

A one-rupee grid search puts the maximum at M* = ₹20,832, where R = 0.0000194640 — 1.6869 times the no-reinsurance figure. Check it in the Lundberg equation, which the search did not use:

Reading the table, which is the actual answer

At the optimum the insurer pays the reinsurer ₹1,05,867 — 41% of its gross premium — and keeps ₹24,711 of expected profit against ₹60,000 unreinsured. It has given away 59% of its profit to take the ruin bound from 9.95% to 2.04%.

Now look at the shape of the R column, because this is what a good answer says. Between ₹20,000 and ₹22,000 the adjustment coefficient barely moves — 0.00001944, 0.00001946, 0.00001942 — while profit climbs from ₹23,212 to ₹26,713. R is flat at its maximum and profit is steep. Go further, to a retention of ₹30,000: R falls 7% below its maximum, the ruin bound rises from 2.04% to 2.68%, and profit rises 52% to ₹37,687.

So the recommendation is not ₹20,832. It is: decide what ruin probability is acceptable and buy the cheapest cover that achieves it. If the tolerance is 3%, a retention of ₹32,778 does it exactly and leaves ₹40,581 of expected profit — 68% of the unreinsured figure, against the 41% the R-maximising retention keeps. If it is 2%, nothing achieves it — the best available is 2.04% — and the answer is that this portfolio needs more capital rather than more reinsurance. Saying that last sentence is worth more than finding the maximum.

And a check on the two limits

At a retention of ₹2,00,000 — ten times the average claim — the net premium is ₹2,59,986 against a gross ₹2,60,000 and R is 0.0000115541 against the unreinsured 0.0000115385. The reinsurance has become almost nothing, as it should: the chance of a claim above ten times the mean is e−10 = 0.0000454. At the other end, below ₹10,216 there is no positive root at all, because the net premium no longer covers the retained claims. The model behaves sensibly at both extremes, which is the last thing to check before trusting anything in the middle.

Where this fits in CM2

This page explains one idea from CM2 Financial Engineering & Loss Reserving. The full CM2 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.

Other CM2 topics explained

• Utility theory and risk aversion: non-satiation, A(w) and R(w)
• Measures of investment risk: variance, semi-variance, VaR and Tail VaR
• CAPM and the security market line: beta and the market price of risk
• Ito's lemma explained term by term, and what it does to a share price
• The chain ladder method: estimating outstanding claims from a run-off triangle
• The Bornhuetter-Ferguson method: the chain ladder with a second opinion
• Binomial option pricing: where the risk-neutral probability comes from
• The Black-Scholes formula and put-call parity, read term by term
• Ruin theory: the adjustment coefficient and Lundberg bound explained
• Mean-variance portfolio theory: the efficient frontier and why diversification works
• The term structure: spot rates, forward rates and par yields
• Short-rate models: Vasicek, Cox-Ingersoll-Ross and Hull-White compared
• The Merton model: company equity as an option, and where a credit spread comes from
• The Greeks: delta, gamma, vega, theta and rho, and how a hedge decays
• The inflation-adjusted chain ladder: taking past inflation out and putting a future assumption in
• The average cost per claim method: developing numbers and severity separately
• Valuing a maturity guarantee by simulation, and checking it against the formula
• The Efficient Markets Hypothesis: the three forms, and how each one is tested
• Single-index and multifactor models: getting a covariance matrix out of a few betas
• State price deflators: pricing with real-world probabilities
• What makes a risk insurable, and what to do when it almost is
• Two-state and intensity credit models: backing a default rate out of a bond price
• The martingale representation theorem, and why hedging is possible at all

Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.