Surplus and Ruin Theory

Analyse when a classical insurer surplus process crosses zero and preserve the model's practical boundaries.

Surplus and Ruin Theory

Aggregate risk describes one period. Ruin theory studies an insurer's path through time: a firm can fail before an apparently favourable long-run average is realised.

1. Classical Cramér–Lundberg model

Let

U(t)=u+ct-\sum_{i=1}^{N(t)}X_i,

where:

Symbol	Meaning
$u\ge0$	initial surplus/capital
$c>0$	premium income rate, net of whatever costs the model includes
$N(t)$	Poisson claim count with intensity $\lambda$
$X_i$	iid positive claim severities, independent of $N(t)$

Between claims, surplus rises linearly; at claim times, it jumps downward.

The time of ruin is

\tau=\inf\{t\ge0:U(t)<0\}.

Two probabilities must not be confused:

\psi(u,T)=P(\tau\le T\mid U(0)=u),

\psi(u)=P(\tau<\infty\mid U(0)=u).

Finite-horizon ruin is usually smaller than ultimate ruin and often closer to an operational planning question.

2. Net profit condition

Expected claim outflow per unit time is $\lambda E[X]$ . For surplus to have positive drift,

c>\lambda E[X].

Write

c=(1+\theta)\lambda E[X],

where $\theta>0$ is premium loading in this simplified model.

Even when expected premium exceeds expected claims, one early large loss can cause ruin. The condition supports a non-trivial ultimate result; it does not imply a small finite-horizon probability.

3. Adjustment coefficient and Lundberg bound

If the severity MGF exists for positive arguments, the adjustment coefficient $R>0$ solves

\lambda\{M_X(R)-1\}=cR.

Under the classical assumptions,

\psi(u)\le e^{-Ru}.

$R$ summarises premium loading and severity-tail behaviour. A larger $R$ gives faster exponential decay of the bound with initial capital.

This tool can fail for heavy-tailed severities whose MGF is infinite for all $t>0$ . Failure of the equation is information about the model, not a numerical inconvenience to ignore.

4. Exact Exponential example

Suppose:

N(t)\sim\operatorname{Poisson}(40t), \qquad E[X]=£5{,}000,

and premium loading is 20%, so

c=1.2(40)(5{,}000)=£240{,}000\text{ per year}.

For Exponential severity with rate $\beta=1/5{,}000$ , the adjustment coefficient is

R=\beta-\frac{\lambda}{c} =0.0002-\frac{40}{240{,}000} =0.00003333.

The exact ultimate ruin probability is

\psi(u)=\rho e^{-Ru}, \qquad \rho=\frac{\lambda E[X]}{c}=\frac{1}{1.2}.

At initial capital $u=£100{,}000$ :

\psi(100{,}000) =\frac{1}{1.2}e^{-3.3333} \approx2.97\%.

The Lundberg bound is $e^{-3.3333}\approx3.57\%$ . The exact result depends on Exponential severity; the bound depends on the broader classical assumptions and existence of $R$ .

5. How assumptions change the answer

Classical assumption	Practical complication	Direction is not automatic
constant premium rate	renewals, rate changes, expenses	higher gross premium may accompany higher exposure
Poisson independent arrivals	catastrophe clustering and seasonality	clustering usually worsens short-horizon paths
iid severity	inflation, limits, mix, trend	changing tail can dominate average trend
no investment return	stochastic assets and liquidity	return can help or introduce market dependence
immediate known payment	reporting and settlement delays	accounting insolvency and cash ruin differ
no reinsurance default	delayed/disputed recoveries	nominal cover can overstate available liquidity

6. Reinsurance in the surplus process

For retained claim $g(X_i)$ and net premium rate $c_{net}$ ,

U^{net}(t)=u+c_{net}t-\sum_{i=1}^{N(t)}g(X_i).

Reinsurance reduces claim severity but also costs premium. Comparing gross $g(X)=X$ with net $g(X)$ while leaving $c$ unchanged overstates the economic benefit.

For occurrence cover, the process must group claims into events before applying $g$ .

7. What classical ruin is—and is not

Ruin theory is valuable for:

understanding path dependence and timing;
studying the interaction of premium loading, capital, and claim tails;
comparing finite and ultimate horizons;
designing stress tests and simulation checks.

It is not by itself IFRS 17 measurement, a Solvency UK SCR calculation, liquidity regulation, or proof of firm viability. Real decisions add reserving, expenses, taxes, assets, management actions, legal entities, and governance.

Proceed to the Finite-Time Simulation Lab to estimate path probabilities directly.

Practice

$\lambda=10$ , $E[X]=£4{,}000$ , and $c=£44{,}000$ per year. Find loading $\theta$ .
Why is $P(U(10)<0)$ not generally equal to $P(\tau\le10)$ ?
If $c\le\lambda E[X]$ , what does the long-run drift imply in the classical model?

Answers

Expected claims are £40,000; $c=(1+\theta)40{,}000$ , so $\theta=10\%$ .
Surplus may cross below zero before year 10 and later recover; ruin records the first crossing, not only the endpoint.
Drift is non-positive; under the usual conditions ultimate ruin occurs with probability one.