Worked Calculations with Exponential-Claim Ruin

From Claim Severity to Ultimate Ruin

In a classical surplus model, several calculations fit together: the claim-size distribution determines the adjustment coefficient, and that coefficient feeds into both an exact ultimate ruin probability and an exponential upper bound. A reliable solution keeps the parameters and units visible throughout. In particular, the coefficient multiplying the exponential in the exact probability is not the same as the safety loading.

Unless a problem states otherwise, the examples below use a hypothetical insurer with claims arriving according to a Poisson process, a constant premium income rate, and no interest or mortality. The results concern ultimate ruin: ruin at some time in an unlimited future, not ruin by a specified finite date.

Model and notation: Let $N(t)$ be a Poisson process with rate $\lambda$ claims per year. Claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, independent of $N(t)$, and have exponential distribution with rate $\beta$ per dollar. Thus $E[X]=1/\beta$, and the moment-generating function is $M_X(r)=\beta/(\beta-r)$ for $r<\beta$. With initial surplus $u$ dollars and premium income $c$ dollars per year, surplus is $U(t)=u+ct-\sum_{j=1}^{N(t)}X_j$. The adjustment coefficient is denoted by $R$.

The net profit condition is $c>\lambda E[X]=\lambda/\beta$. It says that premium income per year exceeds expected claim outgo per year. Under this condition the positive adjustment coefficient is $R=\beta-\lambda/c$, measured in inverse dollars. Define $\alpha=\lambda/(c\beta)$. Then $0<\alpha<1$, and $R=\beta(1-\alpha)$.

Theorem: For the classical model above, when $c>\lambda/\beta$, the adjustment coefficient is $R=\beta-\lambda/c>0$. The exact ultimate ruin probability and the Lundberg bound are

\psi(u)=\alpha e^{-Ru} =\frac{\lambda}{c\beta}e^{-Ru}, \qquad \psi(u)\leq e^{-Ru}, \qquad u\geq 0.

Here $\alpha=\lambda/(c\beta)$, not $1-\alpha$. The bound is an upper bound; it is not generally the exact probability.

Why the Formula Has This Coefficient

The adjustment coefficient solves the positive-root equation $cR=\lambda(M_X(R)-1)$. For exponential claims, substituting $M_X(r)=\beta/(\beta-r)$ gives $cR=\lambda R/(\beta-R)$. The positive root therefore satisfies $R=\beta-\lambda/c$, provided the net profit condition holds.

The coefficient in the exact probability can also be checked from the ruin integro-differential equation. For $u>0$, the ultimate ruin probability satisfies $c\psi'(u)=\lambda\{\psi(u)-\int_0^u\psi(u-x)f_X(x)\,dx-S_X(u)\}$, where $f_X(x)=\beta e^{-\beta x}$ and $S_X(u)=e^{-\beta u}$. This equation accounts for a claim that leaves positive surplus, as well as a claim that causes immediate ruin.

To verify the exponential form, try $\psi(u)=A e^{-Ru}$, where $A$ is a constant. The convolution term is $\int_0^u A e^{-R(u-x)}\beta e^{-\beta x}\,dx =A\beta(e^{-Ru}-e^{-\beta u})/(\beta-R)$. Substitution into the equation gives a term proportional to $e^{-\beta u}$. For it to cancel the tail term $-S_X(u)$, its coefficient must satisfy $A\beta/(\beta-R)=1$. Thus $A=(\beta-R)/\beta$. Since $\beta-R=\lambda/c$, this is $A=\lambda/(c\beta)=\alpha$. The remaining terms satisfy the equation because $R=\beta-\lambda/c$. The resulting decaying solution is the ultimate ruin probability for this model.

The Lundberg bound follows immediately from the exact expression here: since $0<\alpha<1$, $\alpha e^{-Ru}\leq e^{-Ru}$. More generally, the bound is useful even in models where an exact formula is not available, provided a positive adjustment coefficient exists.

Worked Calculations

Worked Example: Calculate the Exact Probability and Bound

A hypothetical insurer has Poisson claim frequency $\lambda=2$ claims per year, exponential claim amounts with rate $\beta=0.001$ per dollar, and premium income $c=\$2{,}500$ per year. Initial surplus is $u=\$3{,}000$. Calculate the adjustment coefficient, exact ultimate ruin probability, and Lundberg bound. Assume no interest or mortality.

Method and model check. The mean claim is $1/\beta=1/0.001=\$1{,}000$, so expected claim outgo is $2(1{,}000)=\$2{,}000$ per year. Since $c=\$2{,}500>\$2{,}000$, the net profit condition holds.

Adjustment coefficient. The frequency-to-premium ratio is $\lambda/c=2/2500=0.0008$ per dollar. Therefore $R=\beta-\lambda/c=0.001-0.0008=0.0002$ per dollar.

Exact probability. The multiplier is $\alpha=\lambda/(c\beta)=2/(2500(0.001))=2/2.5=0.8$. Also, $Ru=0.0002(3000)=0.6$, so $e^{-Ru}=e^{-0.6}=0.5488116$. Hence $\psi(3000)=0.8(0.5488116)=0.4390493$, or 0.4390 to four decimal places.

Lundberg bound. The bound is $e^{-Ru}=e^{-0.6}=0.5488116$, or 0.5488 to four decimal places. It exceeds the exact probability, as it must.

Check. The adjustment coefficient has units of inverse dollars, making $Ru$ dimensionless. The exact probability is below the bound and lies between 0 and 1.

Worked Example: Find Capital for an Exact Ruin Target

A hypothetical insurer has $\lambda=3$ claims per year, exponential severity with rate $\beta=0.002$ per dollar, and premium income $c=\$2{,}000$ per year. Find the smallest whole-dollar initial surplus for which the exact ultimate ruin probability is no greater than 0.10. Also calculate the bound at that capital. Assume no interest or mortality.

Method and model check. The mean claim is $1/0.002=\$500$, so expected annual claims are $3(500)=\$1{,}500$. Premium income exceeds this amount, and the net profit condition holds.

The adjustment coefficient is $R=0.002-3/2000=0.002-0.0015=0.0005$ per dollar. The exact-probability multiplier is $\alpha=3/(2000(0.002))=3/4=0.75$.

To meet the target, solve $0.75e^{-0.0005u}\leq 0.10$. Dividing by 0.75 and taking logarithms gives $u\geq-\ln(0.10/0.75)/0.0005=\ln(7.5)/0.0005$. Since $\ln(7.5)=2.0149030$, the threshold is $u\geq \$4{,}029.8060$. The smallest whole-dollar amount is therefore \$4,030.

At $u=\$4{,}030$, $Ru=0.0005(4030)=2.015$, so the Lundberg bound is $e^{-2.015}=0.133320$, or 0.1333 to four decimal places. The exact probability is $0.75e^{-2.015}=0.099990$, or 0.1000 to four decimal places. The unrounded value is no greater than 0.10. At \$4,029 the capital is below the threshold, so its exact probability is above 0.10.

Check. The exact probability is below the bound. Rounding the threshold down would fail the target, which is why the whole-dollar answer is rounded up.

Worked Example: Keep the Exact Multiplier and the Bound Distinct

A hypothetical insurer has $\lambda=3$ claims per year, exponential claim amounts with rate $\beta=0.002$ per dollar, premium income $c=\$1{,}800$ per year, and initial surplus $u=\$1{,}200$. Calculate the exact ultimate ruin probability and the Lundberg bound. Assume no interest or mortality.

Method and model check. The mean claim is $1/0.002=\$500$, so expected annual claim outgo is $3(500)=\$1{,}500$. Because $1800>1500$, the net profit condition holds.

The adjustment coefficient is $R=0.002-3/1800=0.002-0.0016666667=0.0003333333$ per dollar. The exact multiplier is $\alpha=3/(1800(0.002))=3/3.6=0.8333333$. The exponent is $Ru=0.0003333333(1200)=0.4$.

Since $e^{-0.4}=0.6703200$, the exact probability is $\psi(1200)=0.8333333(0.6703200)=0.5586000$, or 0.5586 to four decimal places. The Lundberg bound is $e^{-0.4}=0.6703200$, or 0.6703 to four decimal places.

Check. The exact probability uses the multiplier $\alpha$; the bound does not. The exact probability is consequently smaller than the bound.

Common Exam Traps and Practical Checks

The most consequential error is confusing the exact-probability multiplier with its complement. If $\alpha=\lambda/(c\beta)$, the exact formula is $\alpha e^{-Ru}$. For the first example, using $1-\alpha=0.2$ instead would give $0.2e^{-0.6}=0.1098$, rather than the correct $0.8e^{-0.6}=0.4390$. The complementary quantity $1-\alpha$ is related to the safety loading under one convention; it is not the multiplier in this exact formula.

Other checks help catch errors before they propagate:

Check net profit first. If $c\leq\lambda/\beta$, the positive adjustment coefficient formula used here does not apply.
Check the units. Both $\beta$ and $\lambda/c$ are per dollar, so $R$ is per dollar and $Ru$ has no units. Do not combine a claim rate per month with premium income per year without converting the time unit.
Do not call the bound exact. The exponential bound omits the multiplier $\alpha$; for exponential claims the exact formula includes it.
Do not confuse time horizons. These formulas give ultimate ruin probability. They do not calculate the probability of ruin by a specified time.
Round at the end. Keep $R$, $\alpha$, and the exponential value unrounded while finding a capital threshold or probability.

These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In a multi-step problem, write down the model, verify the net profit condition, calculate $R$, and then choose the exact formula or the bound according to what is requested.

Check Your Understanding

For each hypothetical model, assume no interest or mortality. Keep time units consistent, and round probabilities to four decimal places.

A model has $\lambda=1.5$ claims per year, exponential claim rate $\beta=0.0015$ per dollar, premium rate $c=\$1{,}200$ per year, and initial surplus $u=\$800$. Check the net profit condition, then find $R$, the exact ultimate ruin probability, and the Lundberg bound. (Answer: Net profit condition holds; $R=0.00025$ per dollar; exact probability $0.6823$; bound $0.8187$.)
For $\lambda=2$ claims per year, $\beta=0.002$ per dollar, $c=\$1{,}250$ per year, and $u=\$1{,}000$, calculate expected annual claim outgo, $R$, and both ruin measures. (Answer: Expected annual claim outgo $\$1{,}000$; $R=0.0004$ per dollar; exact probability $0.5363$; bound $0.6703$.)
A model has $\lambda=2$, $\beta=0.001$, and $c=\$2{,}500$, with rates expressed per year and per dollar as appropriate. Find the smallest whole-dollar initial surplus that makes the exact ultimate ruin probability at most 0.05. (Answer: $\$13{,}863$.)
For $\lambda=4$ claims per year, $\beta=0.003$ per dollar, $c=\$2{,}000$ per year, and $u=\$500$, a candidate uses $1-\lambda/(c\beta)$ as the exact-probability multiplier. Identify the correct multiplier and calculate the exact probability and bound. (Answer: Correct multiplier $2/3$; exact probability $0.4044$; bound $0.6065$.)
A hypothetical insurer has $\lambda=3$ claims per year, $\beta=0.0025$ per dollar, and $c=\$1{,}500$ per year. For $u=\$2{,}000$, calculate expected annual claim outgo, verify the net profit condition, and find the exact probability and bound. (Answer: Expected annual claim outgo $\$1{,}200$; net profit condition holds; $R=0.0005$ per dollar; exact probability $0.2943$; bound $0.3679$.)

What You'll Learn