The question
An insurer starts the year with some money of its own, collects premiums at a steady rate, and pays claims whenever they happen. Premiums arrive smoothly; claims arrive in lumps at random times. However profitable the business is on average, there is always a sequence of bad luck that empties the fund before the premiums catch up.
Ruin theory asks: what is the chance of that, and what can be done about it? It is the one topic in CM2 that is explicitly about survival rather than value, and it is the mathematical ancestor of every solvency capital calculation in use today.
The surplus process
Write U for the insurer's initial surplus, c for the rate at which premium income arrives, and S(t) for the total claims paid up to time t. Then the surplus at time t is
In words: what you started with, plus what has come in, minus what has gone out.
Ruin means U(t) falling below zero at some point, and the probability of that ever happening is written ψ(U). Note what ruin is not: it is not the insurer ending the year with less than it started. The surplus has only to touch zero once, at any time, however briefly.
The claims half: a compound Poisson process
Two assumptions describe the claims, and they are the standard model:
- claims arrive as a Poisson process with rate λ a year, so the number in a year is Poisson(λ) and the gaps between claims are exponential with mean 1/λ;
- claim amounts X1, X2, … are independent of each other and of the arrivals, with mean μ and moment generating function MX.
Two results follow immediately and are worth knowing cold, because a question will often ask for them before anything harder:
Var[S(t)] = λt · E[X2]
The variance formula catches people out. It is λt E[X2], the second moment, not λt Var[X] — because both the number of claims and their sizes are random, and the two sources of variation add. Expanding, Var[S] = λt(Var[X] + E[X]2), which shows the extra term explicitly.
The premium: a loading on the expected claims
In words: charge the expected claims and then add a margin of θ on top.
θ is the premium loading factor, and the condition θ > 0 is what makes the problem interesting. If θ were zero, premium income would exactly match expected claims, the surplus would be a process with no drift, and ruin would be certain — because a random walk with no drift reaches every level eventually, however high the starting point. That is the single most important qualitative fact in the topic: positive loading is not a profit target, it is the condition for survival.
The adjustment coefficient
The probability of ruin has no simple closed form for a general claim distribution. What there is instead is a bound, and it is governed by one number: the adjustment coefficient R, the unique positive root of
In words: find the positive value of R at which the straight line on the left meets the moment generating function on the right.
Both sides equal λ at R = 0, so zero is always a root and is not the one wanted. The left side rises as a straight line; the right side is convex and rises faster and faster. They therefore cross exactly once more at a positive R, and that crossing is the adjustment coefficient. Sketching those two curves is the quickest way to remember both the equation and why the root is unique.
For exponential claim amounts with mean μ, the moment generating function is MX(r) = 1/(1 − μr) for r < 1/μ, and the equation can be solved by hand. Substituting and cancelling one factor of R gives
That is worth deriving once, because it shows what R depends on and what it does not. It rises with the loading θ and falls with the average claim size μ. It does not involve λ at all — writing twice as much business of the same kind changes the premium and the claims in the same proportion and leaves R untouched.
Lundberg inequality
In words: the probability of ever being ruined is at most an exponential in the initial surplus, with the adjustment coefficient as the rate.
This is the result the whole topic exists to reach, and three readings of it are examinable.
Capital works, and it works exponentially. Doubling U squares the bound. On the example below, five lakh of surplus caps the ruin probability at 1.8% and ten lakh caps it at 0.034% — a factor of fifty-four for twice the money.
R is a measure of safety, so bigger is better. Anything that raises R lowers the bound: a larger loading, smaller claims, or a claim distribution with a lighter tail. That is what makes R worth calculating even when the ruin probability itself is not wanted.
It is only a bound. For exponential claims the exact answer happens to be known, and it is ψ(U) = e−RU/(1 + θ) — the bound divided by 1 + θ. So the bound overstates the risk by exactly the loading factor, which is a useful calibration: at a 25% loading it is 25% too pessimistic, at every value of U.
What reinsurance does
Under proportional reinsurance the insurer keeps a fraction α of every claim and passes on the rest, paying the reinsurer a premium based on its own, usually higher, loading θR:
net premium = c − reinsurance premium
Both sides of the surplus process shrink: the claims the insurer meets are smaller, and so is the premium left to meet them. Which effect wins is not obvious and is the point of the question. Work it out by finding the effective loading on the retained business — net premium divided by retained expected claims, minus one — and putting that into the same formula for R.
The trade-off is real and runs in both directions at once. Reinsurance almost always raises R and so lowers the ruin bound, because it cuts the size of the claims that could do the damage. It almost always lowers expected profit, because the reinsurer charges more per unit of risk than the insurer does. An exam answer that reports only one of those two has answered half the question.
Where marks are lost
Using Var[X] where E[X2] belongs. The variance of aggregate claims is λt E[X2]. For an exponential claim amount with mean 25,000, E[X2] = 2μ2, which is twice the square of the mean — so using the variance instead halves the answer.
Taking R = 0. Zero always solves the Lundberg equation and is never the answer. If your algebra gives R = 0, you have cancelled a factor of R too few times.
Forgetting the restriction on the moment generating function. For exponential claims MX(r) exists only for r < 1/μ, and R must lie below that. Checking μR < 1 takes five seconds and catches a root found on the wrong branch.
Treating the Lundberg bound as the ruin probability. It is an upper bound. Say so, and say which direction the error goes in.
Reporting a new R after reinsurance without recomputing the premium. The net premium is the gross premium minus the reinsurance premium, not the gross premium scaled by α. Those two differ precisely because the reinsurer's loading differs from the insurer's, which is the whole economics of the arrangement.
Questions students ask
Why is ruin certain when the loading is zero?
Because a process with zero drift and positive variance has no tendency to return anywhere. Over a long enough horizon it wanders arbitrarily far in both directions, so it eventually goes below any fixed level, including zero, whatever the starting surplus. The loading provides the upward drift that makes survival possible, and the adjustment coefficient measures how much drift there is relative to the randomness.
Is an insurer with a 2% ruin probability in trouble?
The model's "ruin" is not insolvency in the regulatory sense. It assumes the business runs for ever on unchanged terms, that no capital is ever injected, that premiums are never repriced and that nothing is earned on the fund. Real insurers do all four. The number is best read as a comparison between options — this retention against that one, this loading against that one — rather than as a forecast.
What happens with excess-of-loss reinsurance instead?
The same method, with a harder integral. Under a retention limit M the insurer keeps min(X, M), so the moment generating function of the retained claim has to be computed for that truncated variable before the Lundberg equation can be solved, and it usually has to be solved numerically. The structure of the answer is unchanged: find the effective loading on the retained business, find R, compare.
Does the Poisson assumption matter much?
It matters for the mathematics more than for the conclusions. Poisson arrivals make the claim count's mean equal its variance, which is often optimistic for real portfolios where claims cluster, because one storm causes many claims at once. A mixed or negative binomial count has a larger variance and so a worse tail, and the qualitative lessons — that capital helps exponentially and that a positive loading is essential — survive unchanged.
A worked example
An insurer writes a line of business in which claims arrive as a Poisson process at 8 a year, and claim amounts are exponentially distributed with a mean of ₹25,000. The premium carries a loading of 25%, and the insurer holds an initial surplus of ₹5,00,000.
Every figure is invented for this example. It is not a real portfolio and not a real reinsurance quote.
Step 1: the premium and the expected result
c = (1 + 0.25) × 2,00,000 = ₹2,50,000 a year
expected annual profit = 2,50,000 − 2,00,000 = ₹50,000
Step 2: the adjustment coefficient
That is 8 × 10−6 per rupee. The units matter: R is per unit of money, so it must be multiplied by a surplus in rupees to give a dimensionless exponent.
The check: put it back into the Lundberg equation
Both sides must be equal, and neither side was used to produce the answer.
right: λ/(1 − μR) = 8 / (1 − 25,000 × 0.000008) = 8 / (1 − 0.2) = 8 / 0.8 = 10
Equal, exactly. And μR = 0.2, comfortably below 1, so the moment generating function exists at R and the root is on the right branch.
A second, independent route: solve 8 + 2,50,000 r = 8/(1 − 25,000 r) numerically by bisecting on the interval from 0 to 1/25,000. Three hundred bisections return 0.0000080000. The closed form and the numerical root agree to ten decimal places.
Step 3: the Lundberg bound
ψ(5,00,000) ≤ e−4 = 0.0183156, that is 1.83%
And because the claims are exponential the exact figure is available too:
The bound divided by the exact answer is 1.25, which is 1 + θ to four figures — so the bound is 25% pessimistic here, and would be at any surplus.
What capital buys
| Initial surplus | RU | Lundberg bound | Exact |
|---|---|---|---|
| 0 | 0 | 1.0000 | 0.8000 |
| ₹5,00,000 | 4.0 | 0.018316 | 0.014652 |
| ₹10,00,000 | 8.0 | 0.000335 | 0.000268 |
Doubling the surplus from five lakh to ten divides the bound by about fifty-four, because the exponent doubles. That is the exponential in Lundberg inequality doing its work, and it is why a modest amount of capital goes a long way against a portfolio of small claims.
Note the top row. With no initial surplus at all the exact ruin probability is 0.8, which is 1/(1 + θ) — so even a company with nothing behind it survives one time in five on the strength of its loading alone, and the bound of 1.0 is useless there.
Step 4: what the loading is worth
| Loading | R | Bound at ₹5,00,000 |
|---|---|---|
| 10% | 0.00000364 | 16.23% |
| 25% | 0.00000800 | 1.83% |
| 50% | 0.00001333 | 0.13% |
Cutting the loading from 25% to 10% multiplies the ruin bound by about nine. Pricing and solvency are the same question asked twice.
Step 5: proportional reinsurance
The insurer now retains 80% of every claim and reinsures the rest with a reinsurer whose loading is 40%.
net premium income = 2,50,000 − 56,000 = ₹1,94,000
retained expected claims = 8 × 0.80 × 25,000 = ₹1,60,000
net expected profit = 1,94,000 − 1,60,000 = ₹34,000
The retained claim is exponential with mean 0.8 × 25,000 = ₹20,000, and the effective loading on the retained business is
Check it in the Lundberg equation for the retained process, as before:
right: 8 / (1 − 20,000 × 0.0000087629) = 8 / 0.8247423 = 9.7
Equal again, and bisection on the retained equation returns 0.0000087629 as well.
Reading the trade-off
| No reinsurance | Retain 80% | |
|---|---|---|
| Expected annual profit | ₹50,000 | ₹34,000 |
| Effective loading | 25.00% | 21.25% |
| Adjustment coefficient | 0.00000800 | 0.00000876 |
| Lundberg bound at ₹5,00,000 | 1.83% | 1.25% |
| Exact ruin probability | 1.47% | 1.03% |
So the insurer gives up 32% of its expected profit and buys roughly a third off its chance of ruin — 32% off the bound and 30% off the exact figure. R rose by a factor of 1.0954, which does not sound like much until it is put through the exponential: RU went from 4.0000 to 4.3814, and the bound fell by a factor of 1.4644.
Notice why this works at all, because it is the heart of the question. The effective loading went down, from 25% to 21.25%, which on its own would make things worse. R still went up, because the claims were cut by a fifth and R is inversely proportional to the average claim size. The claim-size effect beat the loading effect. With a greedier reinsurer it would not: at a high enough θR the net premium falls far enough to push R down, and the insurer would be paying for an arrangement that makes it less safe as well as less profitable. Finding the retention that maximises R is the standard follow-up, and it is a one-variable optimisation on exactly the arithmetic above.
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
- 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
- Excess-of-loss reinsurance: the retained claim, and choosing the retention
- 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.