Exact Ultimate Ruin with Exponential Claims

Why Exponential Claims Give an Exact Ruin Probability

Lundberg’s inequality gives an upper bound on ultimate ruin, but under a particular claim-size model the probability itself has a closed form. For exponential claim amounts, that form is especially useful: it separates the effect of initial capital from a constant determined by the premium and claim parameters. The constant matters. Using only the exponential decay term gives a bound, not the exact probability.

We use the classical surplus process with initial surplus $u$, constant premium income rate $c$, and a Poisson claim arrival process with rate $\lambda$. Claim amounts are independent and identically distributed, independent of arrivals, and exponential with rate $\beta$. There is no interest credited to surplus and no mortality or other decrement model. Ruin occurs when surplus becomes strictly negative.

Definition: Write $U(t)=u+ct-\sum_{j=1}^{N(t)}X_j$, where $N(t)$ is Poisson with rate $\lambda$, and $X_j$ are independent exponential claim amounts with density $f_X(x)=\beta e^{-\beta x}$ for $x\geq 0$. The ultimate ruin probability is $\psi(u)=P(T<\infty\mid U(0)=u)$, where $T=\inf\{t\geq 0:U(t)<0\}$.

The exponential rate $\beta$ has units inverse to the claim amount, and $E[X]=1/\beta$. The expected claim cost per unit time is $\lambda/\beta$. We assume the net profit condition $c>\lambda/\beta$, and define the safety loading $\theta$ by $c=(1+\theta)\lambda/\beta$, so $\theta>0$. These assumptions are essential to the nontrivial formula below.

Exact Formula and Its Parameters

Theorem: In the classical model above, if claim amounts are exponential with rate $\beta$ and $c>\lambda/\beta$, then for every $u\geq 0$, $$ \psi(u)=\frac{\lambda}{c\beta}\exp\left[-\left(\beta-\frac{\lambda}{c}\right)u\right]. $$ Equivalently, with $c=(1+\theta)\lambda/\beta$, $$ \psi(u)=\frac{1}{1+\theta}\exp\left(-\frac{\beta\theta}{1+\theta}u\right). $$

The multiplicative constant is $\lambda/(c\beta)=\lambda E[X]/c=1/(1+\theta)$. The exponential decay rate is $R=\beta-\lambda/c=\beta\theta/(1+\theta)$. Units provide a useful check: $R$ is inverse dollars when $u$ is measured in dollars, so $Ru$ is dimensionless. The constant is a probability-scale factor, not an additional exponent.

The adjustment coefficient equation is $cR=\lambda(M_X(R)-1)$, where $M_X(r)=\beta/(\beta-r)$ for $r<\beta$. Solving its positive root gives $R=\beta-\lambda/c$. That equation identifies the decay rate, but by itself it does not supply the exact multiplicative constant.

Why the Formula Holds

A short-time conditioning argument gives the integro-differential equation for ultimate ruin. Over a small interval of length $h$, a claim arrives with probability $\lambda h+o(h)$. If no claim arrives, surplus increases to approximately $u+ch$. If a claim arrives, ruin occurs when its amount exceeds the current surplus; otherwise the process continues from the remaining surplus. Taking the first-order terms as $h$ tends to zero gives

c\psi'(u)=\lambda\left[\psi(u)-S_X(u)-\int_0^u\psi(u-x)f_X(x)\,dx\right].

For exponential claims, $S_X(u)=e^{-\beta u}$. Define the convolution $g(u)=\int_0^u\psi(u-x)\beta e^{-\beta x}\,dx$. Rewriting it as $g(u)=\beta e^{-\beta u}\int_0^u\psi(y)e^{\beta y}\,dy$ and differentiating yields $g'(u)=\beta\psi(u)-\beta g(u)$. The integro-differential equation becomes

c\psi'(u)=\lambda\left[\psi(u)-e^{-\beta u}-g(u)\right].

Differentiate this equation and use $g'(u)=\beta\psi(u)-\beta g(u)$. The original equation also gives $g(u)=\psi(u)-e^{-\beta u}-(c/\lambda)\psi'(u)$. Substitution cancels the terms involving $e^{-\beta u}$ and $\psi(u)$, leaving

c\psi''(u)=(\lambda-c\beta)\psi'(u), \qquad \psi''(u)=-R\psi'(u), \qquad R=\beta-\frac{\lambda}{c}.

Since the net profit condition implies $R>0$, the solutions are a constant plus a multiple of $e^{-Ru}$. Ultimate ruin tends to zero as initial surplus tends to infinity, so the constant term must be zero. Write $\psi(u)=Be^{-Ru}$. At $u=0$, the convolution is zero and $S_X(0)=1$, so the integro-differential equation gives $c\psi'(0)=\lambda[\psi(0)-1]$. Substituting $\psi(0)=B$ and $\psi'(0)=-RB$ gives

-cRB=\lambda(B-1), \qquad (\lambda+cR)B=\lambda, \qquad B=\frac{\lambda}{\lambda+cR}=\frac{\lambda}{c\beta}.

This determines both parts of the exact formula. The derivation depends on the exponential density through the simple derivative identity for $g$; it does not establish the same constant for an arbitrary claim-size distribution.

Using the Formula

Keep the time unit consistent: $\lambda$ and $c$ must use the same time unit, while $\beta$ and $u$ must use reciprocal and ordinary units of the same currency. A reliable calculation first finds $E[X]$, checks the net profit condition, and then calculates the constant and $R$ separately.

Worked Example: Exact Ruin Probability at a Given Reserve

A hypothetical insurer receives premium income at $c=\$12{,}500$ per year. Claims arrive as a Poisson process at rate $\lambda=2$ per year, and each claim amount is exponential with mean $\$5{,}000$. Find ultimate ruin probability with initial surplus $u=\$10{,}000$. There is no interest credited and no mortality model.

Method. Calculate the claim rate parameter, check the net profit condition, and substitute the constant and decay rate into the exact formula. The exponential distribution has mean $1/\beta$, so $\beta=1/5{,}000=0.0002$ per dollar. Expected claims per year are $2(5{,}000)=\$10{,}000$, which is below the premium rate of $\$12{,}500$.

The safety loading is $\theta=12{,}500/10{,}000-1=0.25$. The multiplicative constant and decay rate are

\frac{\lambda}{c\beta}=\frac{2}{12{,}500(0.0002)}=0.8, \qquad R=0.0002-\frac{2}{12{,}500}=0.00004\text{ per dollar}.

Thus $Ru=0.00004(10{,}000)=0.4$, and

\psi(10{,}000)=0.8e^{-0.4} =0.8(0.6703200\ldots)=0.5362560\ldots.

To four decimal places, the ultimate ruin probability is 0.5363. As a check, the safety-loading form gives $R=0.0002(0.25/1.25)=0.00004$ and the same constant $1/1.25=0.8$. The common mistake avoided here is omitting the constant and reporting $e^{-0.4}$, which is not the exact probability.

Worked Example: Capital for a Target Ruin Probability

A hypothetical portfolio has Poisson claim arrivals at $\lambda=3$ per year and exponential claim amounts with mean $\$2{,}000$. Its premium rate is $c=\$7{,}500$ per year. Find the smallest reserve in whole cents for which ultimate ruin probability is at most $0.08$. Assume no interest is credited and no mortality model applies.

Method. Invert the exact formula, then round the required capital upward to the next cent so the target is met. Here $\beta=1/2{,}000=0.0005$ per dollar, and expected claims per year are $3(2{,}000)=\$6{,}000$. Since $7{,}500>6{,}000$, the net profit condition holds. The safety loading is $0.25$, giving constant $1/1.25=0.8$ and

R=0.0005-\frac{3}{7{,}500}=0.0001\text{ per dollar}.

Solve the target inequality rather than treating it as an equality for a rounded reserve:

0.8e^{-0.0001u}\leq 0.08 \quad\Longleftrightarrow\quad e^{-0.0001u}\leq 0.1 \quad\Longleftrightarrow\quad u\geq\frac{\ln(10)}{0.0001} =23{,}025.85093\ldots.

The smallest whole-cent reserve meeting the target is therefore $\$23{,}025.86$. At this rounded-up amount, $0.0001u=2.302586$, so $0.8e^{-2.302586}$ is just below $0.08$. Rounding to $\$23{,}025.85$ would round down from the threshold and fail the stated “at most” condition. The common mistake avoided is ordinary nearest-cent rounding when the required answer is a minimum reserve.

Worked Example: Checking the Premium and Claim Units

A hypothetical insurer writes a product with premium income $c=\$12{,}000$ per month. Claims arrive at rate $\lambda=4$ per month, and claim amounts are exponential with mean $\$2{,}500$. Calculate ultimate ruin probability from $u=\$15{,}000$. There is no interest credited and no mortality model.

Method. Work in monthly units throughout, then use the exact formula. The rate parameter is $\beta=1/2{,}500=0.0004$ per dollar. Expected claim cost per month is $4(2{,}500)=\$10{,}000$, so the net profit condition holds and $\theta=12{,}000/10{,}000-1=0.2$.

The constant and decay rate are

\frac{1}{1+\theta}=\frac{1}{1.2}=\frac{5}{6}, \qquad R=\frac{0.0004(0.2)}{1.2} =0.0000666667\text{ per dollar}.

For $u=\$15{,}000$, $Ru=1$. Hence

\psi(15{,}000)=\frac{5}{6}e^{-1} =\frac{5}{6}(0.3678794\ldots) =0.3065662\ldots.

To four decimal places, the answer is 0.3066. As a check, $R=\beta-\lambda/c=0.0004-4/12{,}000=0.0000666667$, agreeing with the loading calculation. The common mistake avoided is mixing a monthly claim rate with an annual premium rate; both rates here are monthly.

Assumptions, Interpretation, and Exam Traps

The formula is exact only for the stated classical model: Poisson arrivals, independent exponential claim amounts with a common rate, constant premium income, and the net profit condition. It assumes the premium and claim rates use compatible time units and claim amounts and reserve use the same currency units. It also assumes no interest or other cash flows in the surplus process. If $c\leq\lambda E[X]$, the positive adjustment rate used in this derivation is unavailable; for the nondegenerate classical model, ultimate ruin has probability one. The displayed exact formula should not be used in that case.

A second important distinction is between exact probability and the Lundberg bound. Under the same positive adjustment coefficient, the bound is $\psi(u)\leq e^{-Ru}$. For exponential claims the exact probability is $(\lambda/(c\beta))e^{-Ru}$, and the multiplier is less than one under the net profit condition. The bound and the exact probability share the decay rate but are not interchangeable.

Do not drop the constant. $e^{-Ru}$ is the bound, not the exact exponential-claim probability.
Do not use the claim mean as the rate. If the mean is $m$, then $\beta=1/m$.
Check the net profit condition first. The formula requires $c>\lambda/\beta$, not merely positive premium income.
Check units before exponentiating. $R$ must be inverse currency units when $u$ is a currency amount.

Key takeaway: For exponential claims in the classical model with positive safety loading, ultimate ruin is exactly a constant times an exponential in initial surplus: $\psi(u)=(1/(1+\theta))e^{-\beta\theta u/(1+\theta)}$. Both the multiplicative constant and the decay rate are required for an exact calculation.

These derivations and calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume a classical surplus model with no interest credited and no mortality model. Claims arrive according to a Poisson process, and claim amounts are independent exponential random variables. Give probabilities to four decimal places and reserves to the nearest cent where requested.

A hypothetical insurer has $\lambda=1$ per year, mean claim amount $\$1{,}000$, premium rate $c=\$1{,}250$ per year, and reserve $u=\$5{,}000$. Find the exact ultimate ruin probability. (Answer: 0.2943)
A hypothetical portfolio has $\lambda=2$ per year, mean claim amount $\$4{,}000$, premium rate $c=\$10{,}000$ per year, and reserve $u=\$20{,}000$. Find the multiplicative constant, decay rate per dollar, and ultimate ruin probability. (Answer: constant $0.8000$, decay rate $0.00005$ per dollar, probability $0.2943$)
A hypothetical insurer has $\lambda=3$ per year, mean claim amount $\$3{,}000$, and premium rate $c=\$12{,}000$ per year. What is the smallest reserve to the nearest cent that gives ultimate ruin probability at most $0.10$? (Answer: $\$24{,}178.84$)
A hypothetical classical model has positive claim frequency and mean claim amount, but its premium rate is $c=\lambda E[X]$. Is the exponential-claim exact formula in this tutorial applicable, and what is ultimate ruin probability under this net profit boundary? (Answer: The formula is not applicable; ultimate ruin probability is 1.)
A hypothetical insurer has $\lambda=5$ per week, mean claim amount $\$600$, premium rate $c=\$3{,}750$ per week, and reserve $u=\$3{,}000$. Find the exact ultimate ruin probability. (Answer: 0.2943)

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