Approximating Infinite-Horizon Ruin Probabilities

Why an Infinite-Horizon Ruin Bound Is Useful

An insurer can collect premiums continuously while paying claims at random times. Even when its premium income exceeds expected claim costs, an adverse sequence of claims can exhaust its surplus. The infinite-horizon ruin probability measures this risk without imposing a fixed end date. In many models it is difficult to calculate exactly, but an adjustment coefficient gives a useful exponential upper bound.

The bound depends on the claim-severity distribution and the premium rate, not just on expected annual losses. It can therefore help compare hypothetical portfolios or surplus levels under a consistent model. It must still be interpreted correctly: the Lundberg bound is an upper bound, not automatically the exact ruin probability.

Model and Adjustment Coefficient

Definition: In the classical compound Poisson surplus model, initial surplus is $u\geq 0$, claims arrive according to a Poisson process with rate $\lambda>0$, and claim amounts $X_1,X_2,\ldots$ are independent and identically distributed nonnegative random variables, independent of the arrival process. Premiums are received continuously at rate $c$. The surplus at time $t$ is $$ U(t)=u+ct-\sum_{j=1}^{N(t)}X_j, $$ where $N(t)$ is the number of claims by time $t$. Ruin occurs if $U(t)<0$ at some time. The infinite-horizon ruin probability is $\psi(u)=P(\text{ruin at some }t\geq 0)$.

Write $\mu=E[X]$. The expected claim cost per unit time is $\lambda\mu$. A positive safety loading means $c>\lambda\mu$; when premiums are set using a loading $\theta>0$, the usual choice is $c=(1+\theta)\lambda\mu$. Define the moment-generating function $M_X(r)=E[e^{rX}]$, where it is finite.

Definition: The adjustment coefficient $R$ is a positive solution of $$ \lambda\bigl(M_X(R)-1\bigr)=cR. $$ For a nondegenerate claim distribution, positive safety loading, and a moment-generating function finite in a neighbourhood of zero, the classical model has a positive solution under the usual regularity conditions. Use the positive root, not the trivial root $r=0$.

The root balances the exponential growth of claim risk, represented by $M_X(R)$, against the premium income. If the severity distribution is bounded above by $b$, then $M_X(r)$ is finite for every finite $r$. A stated maximum claim amount is useful for checking that condition, but it is not enough by itself to determine $R$: the claim-size distribution, premium rate, and arrival rate are also needed.

The Lundberg Inequality

Theorem: If the classical compound Poisson model has $c>\lambda E[X]$ and a positive adjustment coefficient $R$, then $$ \psi(u)\leq e^{-Ru},\qquad u\geq 0. $$

This is the Lundberg inequality. It provides a conservative bound on the infinite-horizon ruin probability. The result does not say that $e^{-Ru}$ is the exact probability, and it does not require claims to have a particular named distribution. It does require the model assumptions and a valid positive root.

Why the Bound Holds

Let $S(t)=\sum_{j=1}^{N(t)}X_j$. The compound Poisson moment-generating function gives $$ E\left[e^{R(S(t)-ct)}\right] =\exp\left(t\left\{\lambda\bigl(M_X(R)-1\bigr)-cR\right\}\right)=1, $$ where the last equality follows from the adjustment-coefficient equation. Thus $Z(t)=e^{-RU(t)}=e^{-Ru}e^{R(S(t)-ct)}$ is a nonnegative martingale with initial value $Z(0)=e^{-Ru}$.

Let $\tau$ be the first ruin time. At ruin, $U(\tau)<0$, so $Z(\tau)=e^{-RU(\tau)}>1$. For any fixed $T$, optional stopping at the bounded time $\tau\wedge T$ gives $E[Z(\tau\wedge T)]=e^{-Ru}$. On the event $\{\tau\leq T\}$, the stopped value is greater than one, and everywhere it is nonnegative. Therefore, $$ P(\tau\leq T)\leq E[Z(\tau\wedge T)]=e^{-Ru}. $$ As $T$ increases, $P(\tau\leq T)$ increases to $\psi(u)$. Taking the limit proves $\psi(u)\leq e^{-Ru}$. The argument also shows why an adjustment coefficient yields an upper bound rather than an equality.

Worked Calculations

Worked Example: Exponential Claims and the Lundberg Bound

A hypothetical insurer has Poisson claim frequency $\lambda=2$ per year. Individual claim amounts are exponential with mean $\beta=\$1{,}000$, and the annual premium rate is $c=(1.25)\lambda\beta$. Initial surplus is $u=\$4{,}000$. Find the adjustment coefficient and the Lundberg bound. Then compare it with the exact ruin probability for this model.

Method and assumptions. The claims are independent exponential amounts with rate $1/\beta$, independent of the Poisson arrivals. The loading is $\theta=0.25$, so $c=1.25(2)(\$1{,}000)=\$2{,}500$ per year. For $0\leq r<1/\beta$, $M_X(r)=1/(1-\beta r)$.

Substitute the moment-generating function in the root equation: $$ 2\left(\frac{1}{1-1000R}-1\right)=2500R. $$ For the positive root, divide by $R$ and simplify: $$ \frac{2000}{1-1000R}=2500,\qquad 1-1000R=0.8,\qquad R=0.0002\text{ per dollar}. $$ This root is less than $1/\beta=0.001$, as required for the moment-generating function to be finite.

The exponent is $Ru=0.0002(4000)=0.8$. Therefore the Lundberg bound is $$ \psi(4000)\leq e^{-0.8}=0.4493. $$ For exponential severities in this model, the exact infinite-horizon ruin probability is $\psi(u)=\frac{1}{1+\theta}e^{-Ru}$. Hence $\psi(4000)=\frac{1}{1.25}(0.4493)=0.3595$, to four decimal places. The exact result is below the bound, as it must be. Common mistake avoided: treating the bound $0.4493$ as the exact probability.

Worked Example: A Bounded Two-Point Claim Severity

A hypothetical portfolio has claim frequency $\lambda=1$ per year. Each claim is equally likely to be $\$1{,}000$ or $\$2{,}000$; thus the maximum severity is $b=\$2{,}000$. The premium rate is $c=2/\ln 2$ thousand dollars per year, and initial surplus is $u=\$2{,}000$. Calculate the adjustment coefficient and the Lundberg bound.

Method and assumptions. Measure claim amounts and surplus in thousands of dollars for the calculation. Then $X$ is 1 or 2, each with probability $1/2$, $E[X]=1.5$, and $M_X(r)=\tfrac12e^r+\tfrac12e^{2r}$. The premium rate is $c=2/\ln2=2.8854$ thousand dollars per year. It exceeds expected claims of $1(1.5)=1.5$ thousand dollars per year.

At $R=\ln2$, the moment-generating function is $$ M_X(\ln2)=\tfrac12(2)+\tfrac12(4)=3. $$ Therefore $\lambda(M_X(R)-1)=1(3-1)=2$, while $cR=(2/\ln2)(\ln2)=2$. The root equation holds, so $R=\ln2$ per thousand dollars.

Since $u=2$ thousand dollars, $Ru=2\ln2=\ln4$. The Lundberg bound is $$ \psi(2000)\leq e^{-2\ln2}=\frac14=0.2500. $$ The severity cap is respected by both possible claim amounts, and the exponent is dimensionless because $R$ is measured per thousand dollars while $u$ is in thousands. Common mistake avoided: inserting $u=2000$ while using an adjustment coefficient expressed per thousand dollars, which would incorrectly multiply the exponent by 1,000.

Asymptotic Approximation and Interpretation

For some models, a sharper estimate for large initial surplus is available. Under the classical Cramér-Lundberg conditions, including a nonlattice claim-size distribution and a finite moment-generating function beyond $R$, the ruin probability has the asymptotic form $$ \psi(u)\sim C e^{-Ru}\quad\text{as }u\to\infty,\qquad C=\frac{c-\lambda E[X]}{\lambda E[Xe^{RX}]-c}. $$ Here $\sim$ means that the ratio of the two sides tends to one as $u$ grows. The coefficient $C$ is not generally one, so this approximation is not the same as the Lundberg bound. It is intended for sufficiently large surplus and does not replace the bound as a guarantee for every $u$. For lattice claim amounts, the standard asymptotic expression can require a lattice adjustment.

For exponential claims, the expression gives $C=1/(1+\theta)$, and the resulting ruin formula is exact. In other severity models, the asymptotic expression is an approximation whose accuracy depends on the model and the surplus level. A finite maximum severity ensures all exponential moments exist, but it does not make the asymptotic estimate exact.

Worked Example: Assessing Surplus and Loading Changes

A hypothetical insurer has $\lambda=4$ claims per year, exponential claim amounts with mean $\beta=\$500$, and loading $\theta=0.20$. Find the Lundberg bound and exact ruin probability at $u=\$3{,}000$. Then recalculate them at the same surplus with $\theta=0.50$, and at twice the original surplus with the original loading.

Method and assumptions. For exponential claims, $R=\theta/((1+\theta)\beta)$, while the exact ruin probability is $\psi(u)=\frac{1}{1+\theta}e^{-Ru}$. The premium rate is always $c=(1+\theta)\lambda\beta$, so changing $\theta$ changes the premium rate as well as $R$.

At $\theta=0.20$, $c=1.20(4)(500)=\$2{,}400$ per year, and $R=0.20/(1.20(500))=0.0003333$ per dollar. At $u=\$3{,}000$, $Ru=1$. The bound is $e^{-1}=0.3679$; the exact probability is $(1/1.20)e^{-1}=0.3066$.

At $\theta=0.50$, $c=1.50(4)(500)=\$3{,}000$ per year, and $R=0.50/(1.50(500))=0.0006667$ per dollar. With $u=\$3{,}000$, $Ru=2$. The bound is $e^{-2}=0.1353$; the exact probability is $(1/1.50)e^{-2}=0.0902$.

Return to $\theta=0.20$ and double surplus to $u=\$6{,}000$. The adjustment coefficient and premium rate remain unchanged, and $Ru=0.0003333(6000)=2$. The bound is $e^{-2}=0.1353$; the exact probability is $(1/1.20)e^{-2}=0.1128$. All probabilities are rounded to four decimal places. The calculations show that both more surplus and greater loading reduce the ruin estimates in this model. Common mistake avoided: changing the loading but leaving the premium rate unchanged; the loading condition requires recalculating $c$.

Practical Checks and Exam Traps

The adjustment coefficient is sensitive to the entire claim distribution, particularly its upper tail. Two portfolios can have the same expected claim cost but different values of $M_X(R)$, and therefore different positive roots. Replacing a severity distribution by its mean generally does not preserve the adjustment coefficient.

Check the premium condition. Confirm $c>\lambda E[X]$ before using the positive-safety-loading result. If the premium rate is inadequate, the standard bound does not apply as stated.
Use the positive root. The equation also has the trivial solution $r=0$; it is not the adjustment coefficient.
Keep units consistent. If claim amounts are measured in thousands of dollars, use surplus in thousands and report $R$ per thousand dollars.
Do not confuse bound and estimate. $e^{-Ru}$ is a guaranteed upper bound under the model assumptions; an exact formula or asymptotic approximation requires additional distributional conditions.
Do not infer a distribution from a maximum claim. Knowing $X\leq b$ establishes bounded severity but does not determine $M_X(R)$ or the root without further information about claim amounts.

Key takeaway: Specify the compound Poisson model, verify positive safety loading, solve $\lambda(M_X(R)-1)=cR$ for the positive root, and then calculate $e^{-Ru}$. Treat that value as an upper bound unless additional assumptions justify a separate exact or asymptotic result.

Check Your Understanding

For each hypothetical scenario, keep the stated units consistent and distinguish the Lundberg bound from any exact probability.

Claims arrive at rate $2$ per year and have exponential amounts with mean $\$2{,}000$. The premium rate uses loading $\theta=0.25$, and initial surplus is $\$5{,}000$. Find $R$ and the Lundberg bound. (Answer: $R=0.0001$ per dollar; bound $e^{-0.5}=0.6065$.)
For exponential claims with mean $\$1{,}000$ and loading $\theta=0.50$, calculate the Lundberg bound and exact ruin probability at $u=\$2{,}000$. (Answer: bound $e^{-2/3}=0.5134$; exact probability $(2/3)e^{-2/3}=0.3423$.)
A portfolio has $\lambda=3$ per year, exponential claim amounts with mean $\$400$, loading $\theta=0.50$, and surplus $u=\$1{,}200$. Find the premium rate, adjustment coefficient, and bound. (Answer: $c=\$1{,}800$ per year; $R=0.0008333$ per dollar; bound $e^{-1}=0.3679$.)
Claims are equally likely to be $\$1{,}000$ or $\$3{,}000$, with $\lambda=1$ per year. The premium rate is $c=4/\ln2$ thousand dollars per year, and surplus is $\$1{,}000$. Verify $R=\ln2$ per thousand dollars and find the bound. (Answer: $M_X(\ln2)=\tfrac12(2)+\tfrac12(8)=5$, so both sides of the root equation equal $4$; bound $e^{-\ln2}=0.5000$.)
For exponential claims with mean $\$500$ and loading $\theta=0.20$, compare the exact ruin probability at surplus $\$3{,}000$ and $\$6{,}000$. (Answer: $R=0.0003333$ per dollar; the exact probabilities are $(1/1.20)e^{-1}=0.3066$ and $(1/1.20)e^{-2}=0.1128$, respectively.)

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