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Lundberg adjustment coefficient

The positive exponential root behind Lundberg’s ruin bound. See why zero is excluded, how the root creates a martingale, and when heavy tails prevent it from existing.

The Lundberg adjustment coefficient is the strictly positive exponential root that gives an upper bound on ultimate ruin probability in the classical Cramér–Lundberg ruin model. When it exists, it converts initial capital uu into the bound e−Rue^{-Ru}: a larger RR, in the same monetary units, makes that bound fall faster with capital.

For surplus

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

assume a homogeneous Poisson claim count with rate λ>0\lambda>0, positive iid claim amounts XiX_i independent of that count process, mean claim μ<∞\mu<\infty, and constant premium rate c>λμc>\lambda\mu. Define the claim moment-generating function MX(r)=E[erX]M_X(r)=\mathbb{E}[e^{rX}]. The coefficient solves

λ(MX(R)−1)=cR,R>0,\lambda\bigl(M_X(R)-1\bigr)=cR,\qquad R>0,
with MX(R)<∞M_X(R)<\infty. This is the defining equation in Hald and Schmidli, §2. Premium income and claims must use the same time and money units; RR has units of inverse money, so RuRu is dimensionless.

Zero solves the equation too

Set q(r)=λ(MX(r)−1)−crq(r)=\lambda(M_X(r)-1)-cr. Then

q(0)=0,q′(0)=λμ−c<0,q′′(r)=λE[X2erX]>0q(0)=0,\qquad q'(0)=\lambda\mu-c<0, \qquad q''(r)=\lambda\mathbb{E}[X^2e^{rX}]>0
on the interior of the finite MGF domain. The curve initially moves below zero and bends upward. If it crosses zero again, that crossing is the unique positive root. A numerical solver returning 00 has found the trivial root, which supplies only the bound ψ(u)≤1\psi(u)\le1. Imperial’s insurance-mathematics notes, §5 develop this root geometry.

For a calculation, first identify the finite MGF domain; then bracket a sign change with both endpoints strictly positive. Do not continue the algebraic expression for an MGF outside the region where its expectation is finite.

Consider fictional exponential claims with mean 1 monetary unit, arrival rate 1 per year and premiums of 1.2 units per year. Here
MX(r)=11−r,r<1,q(r)=r(1.2r−0.2)1−r.M_X(r)=\frac{1}{1-r},\quad r<1, \qquad q(r)=\frac{r(1.2r-0.2)}{1-r}.

The roots are 00 and R=1/6≈0.1667R=1/6\approx0.1667 per monetary unit. Both follow from the numerator, and the positive root lies inside the MGF domain. The expected surplus drift is positive at 0.2 units per year.

Fictional exponential claims with mean 1, arrival rate 1 per year and premium rate 1.2. The convex function q has roots 0 and 1/6 and is negative between them. Only the positive root gives a decreasing ruin bound; at capital 12 the bound is 13.53%.
Fictional exponential claims with mean 1, arrival rate 1 per year and premium rate 1.2. The convex function q has roots 0 and 1/6 and is negative between them. Only the positive root gives a decreasing ruin bound; at capital 12 the bound is 13.53%.
Original analytic curve for the fictional model. The vertical scale magnifies the shallow dip below zero; the displayed range ends at r = 0.24, well inside the MGF domain r < 1.
At capital u=12u=12, the bound is e−2=0.1353352832e^{-2}=0.1353352832, or 13.53%. The exact ultimate ruin probability for exponential claims includes an extra factor:
ψ(12)=λμce−R⋅12=56e−2=0.1127794027≈11.28%.\psi(12)=\frac{\lambda\mu}{c}e^{-R\cdot12} =\frac56e^{-2}=0.1127794027\approx11.28\%.
The Charles University summary, page 1 gives this exact exponential-claim formula. The bound therefore leaves a visible margin even in a model where the exponent is exact. It is an infinite-horizon probability bound, not a forecast of annual default.

Why the root produces a bound

Write cumulative net loss as Y(t)=∑i=1N(t)Xi−ctY(t)=\sum_{i=1}^{N(t)}X_i-ct. The compound-Poisson transform gives

E[erY(t)]=etq(r).\mathbb{E}[e^{rY(t)}]=e^{tq(r)}.
At r=Rr=R, the time exponent vanishes. Independent increments then make Z(t)=eRY(t)Z(t)=e^{RY(t)} a nonnegative martingale, starting at 1. This construction and the following stopping argument are developed in Lemvig’s Copenhagen lecture notes, §§6–7.

Let τ=inf⁡{t≥0:U(t)<0}\tau=\inf\{t\ge0:U(t)<0\}, with inf⁡∅=∞\inf\varnothing=\infty. At ruin, Y(τ)>uY(\tau)>u, so Z(τ)>eRuZ(\tau)>e^{Ru}. Apply optional stopping only at the bounded time τ∧t\tau\wedge t:

1=E[Z(τ∧t)]≥E[Z(τ)1{τ≤t}]≥eRuP(τ≤t).1=\mathbb{E}[Z(\tau\wedge t)] \ge\mathbb{E}[Z(\tau)\mathbf{1}_{\{\tau\le t\}}] \ge e^{Ru}\mathbb{P}(\tau\le t).
Letting t→∞t\to\infty gives Lundberg’s inequality:
ψ(u)=P(τ<∞)≤e−Ru,u≥0.\psi(u)=\mathbb{P}(\tau<\infty)\le e^{-Ru},\qquad u\ge0.

Stopping at a finite horizon before taking the limit avoids assuming an unjustified expectation identity at the possibly infinite ruin time. The inequality also discards the size of the overshoot below zero, one reason it need not be an equality.

What the coefficient leaves out

Positive average profit is insufficient for existence. It makes q′(0)<0q'(0)<0, but the curve must return to zero within its finite domain. One sufficient condition is that the MGF diverges at a finite positive right endpoint, as in the exponential example. Hald and Schmidli, §2 state an existence criterion separately from the net profit condition.
For Pareto or nondegenerate lognormal claims, MX(r)=∞M_X(r)=\infty for every r>0r>0. A positive coefficient cannot be obtained from this equation, even when the mean is finite and premiums exceed expected claims. Heavy-tail ruin analysis uses different arguments; see the Charles University summary, page 3 and subexponential distribution. Truncating the claim tail to obtain a root changes the model being bounded.
A decay coefficient is not a complete probability model. Under additional Cramér–Lundberg asymptotic conditions, including E[XeRX]<∞\mathbb{E}[Xe^{RX}]<\infty, one obtains ψ(u)∼Ce−Ru\psi(u)\sim C e^{-Ru} as u→∞u\to\infty. The prefactor CC also depends on the model. Thus comparing RR values compares exponential decay rates; it does not by itself rank exact ruin probabilities at a chosen finite capital. Imperial’s notes, §7 distinguish this asymptotic result from the bound.
The membership lesson Superposition, Thinning, and Compound Poisson Processes develops the aggregate-claim building block; a public preview is available. Its mean and variance calculations describe typical loss behaviour. The adjustment coefficient additionally needs the full positive exponential transform, which can fail even when those ordinary moments exist.

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