Common Unit Errors in Surplus Models

Why Units Matter in a Surplus Model

A surplus calculation can be algebraically correct and still be wrong if its inputs use incompatible units. A claim intensity stated per month cannot be combined directly with a premium rate stated per year. Likewise, a horizon of 18 months is not 18 years when calculating an expected claim count. These mistakes can change an adjustment coefficient or make a calculated ruin bound meaningless.

The classical surplus model makes the time basis explicit. Let $U(t)$ be the insurer's surplus at time $t$, with initial surplus $u$. Claims arrive according to a Poisson process $N(t)$ with intensity $\lambda$, and claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, independent of $N(t)$. Premium income accrues at a constant rate $c$. The surplus is

U(t)=u+ct-\sum_{j=1}^{N(t)}X_j.

The time unit used for $\lambda$, $c$, and $t$ must be the same. For example, if $t$ is measured in years, then $\lambda$ is claims per year and $c$ is dollars per year. Claim amounts and surplus must also use the same monetary unit.

Definition: In the classical surplus model, $\lambda$ is the expected number of claims per chosen time unit, $c$ is premium income per that same time unit, and $X$ is a claim amount in the chosen monetary unit. The expected claim cost per time unit is $\lambda E[X]$. The usual positive-safety-loading condition is $c>\lambda E[X]$.

The adjustment coefficient $R$, when it exists, is the positive solution of the Lundberg equation

\lambda\bigl(E[e^{RX}]-1\bigr)=cR.

If $X$ is measured in dollars, $R$ is measured per dollar, so $RX$ is dimensionless. The left and right sides of the equation are both rates per unit time. This is a useful first check: the time units must agree on both sides, and the exponent must be dimensionless.

Changing the Time Unit Consistently

Suppose the original time unit is a year and the new time unit is a month. The conversion is not just a change to the horizon: both the claim intensity and premium rate must be expressed per month. A year contains 12 months, so

\lambda_{\text{month}}=\frac{\lambda_{\text{year}}}{12}, \qquad c_{\text{month}}=\frac{c_{\text{year}}}{12}, \qquad T_{\text{months}}=12T_{\text{years}}.

Claim amounts are unchanged by this time conversion. Since $\lambda$ and $c$ are divided by the same factor, their ratio $c/\lambda$ is unchanged. The adjustment coefficient is therefore unchanged too, provided the monetary unit and claim-size distribution have not changed.

Key result: A consistent change of time unit changes $\lambda$, $c$, and the numerical value used for the horizon, but does not change $R$. Changing the monetary unit is different: if claim amounts are expressed in a different currency unit, the numerical value of $R$ must change inversely.

For example, changing claims from dollars to thousands of dollars divides every numerical claim amount by 1,000. The numerical adjustment coefficient must then be 1,000 times larger, because the product $RX$ must stay the same. By contrast, changing years to months leaves $X$ unchanged and leaves $R$ unchanged.

Worked Example: Finding an Adjustment Coefficient From Monthly Inputs

A hypothetical insurer has Poisson claim arrivals at 1 claim per month. Claim amounts are exponentially distributed with mean $500, and premium income is $650 per month. Find the positive adjustment coefficient, stating its monetary unit. Use the Lundberg equation and verify the result.

Method and model. Here $\lambda=1$ claim per month, $c=650$ dollars per month, and $X$ is exponential with mean 500 dollars. For $0\leq R<1/500$, its moment-generating function is $E[e^{RX}]=1/(1-500R)$. The safety loading is positive because $c=650>1(500)=\lambda E[X]$.

Substitute into the Lundberg equation:

1\left(\frac{1}{1-500R}-1\right)=650R.

The zero root is not the positive adjustment coefficient. For a positive root, divide both sides by $R$, and simplify:

\frac{500}{1-500R}=650, \qquad 1-500R=\frac{500}{650}=\frac{10}{13}, \qquad 500R=\frac{3}{13}, \qquad R=\frac{3}{6500}.

Thus $R\approx0.0004615$ per dollar. The solution is in the required domain, since $3/6500<1/500$.

Check. At this value, $500R=3/13$, so $1/(1-500R)=13/10$. The left side of the Lundberg equation is $1(13/10-1)=3/10$ per month. The right side is $650(3/6500)=3/10$ per month. The two sides agree.

Common mistake avoided. A tempting but incorrect value is $1/3250$ per dollar. Substitution gives $500R=2/13$, so the left side is $1(2/11)=2/11$ per month, whereas the right side is $650/3250=1/5$ per month. Since $2/11\ne1/5$, that value does not solve the equation.

Deriving the Exponential-Claim Formula

The previous calculation follows a reusable pattern. Suppose claim amounts are exponential with mean $\mu$, the claim intensity is $\lambda$, and premium income is $c$, all on a common time basis. The moment-generating function is $E[e^{RX}]=1/(1-\mu R)$ for $0\leq R<1/\mu$. Substitution into the Lundberg equation gives

\lambda\left(\frac{1}{1-\mu R}-1\right)=cR.

For a positive root, divide by $R$. Since $1/(1-\mu R)-1=\mu R/(1-\mu R)$, this yields

\frac{\lambda\mu}{1-\mu R}=c, \qquad R=\frac{c-\lambda\mu}{c\mu}.

This positive solution exists when $c>\lambda\mu$. It also satisfies the moment-generating function restriction: $R<1/\mu$ because $c-\lambda\mu<c$. The formula makes the unit requirement visible. The ratio $c/\lambda$ has units of money per claim, like $\mu$; the numerator $c-\lambda\mu$ is money per time, and the denominator $c\mu$ is money squared per time. Thus $R$ has units of inverse money.

The formula is specific to exponential claim amounts and to the stated classical model. It should not be applied to a different severity distribution merely because the inputs have been converted correctly. For other distributions, use the appropriate moment-generating function in the Lundberg equation.

Worked Example: Converting Annual Premium Income Before Solving

A hypothetical insurer has an average of 2 claims per month. Claim amounts are exponentially distributed with mean $400. Its premium income is $13,200 per year. Find the adjustment coefficient per dollar and check it using monthly units.

Method and model. The time units initially differ: $\lambda=2$ claims per month, while $c=13{,}200$ dollars per year. Convert premium income to a monthly rate before using the model:

c=\frac{13{,}200}{12}=1{,}100\text{ dollars per month}.

The expected claim cost is $2(400)=800$ dollars per month, so the safety loading is positive. The exponential formula, with $\lambda=2$, $c=1{,}100$, and $\mu=400$, gives

R=\frac{1{,}100-2(400)}{1{,}100(400)} =\frac{300}{440{,}000} =\frac{3}{4{,}400} \approx0.0006818\text{ per dollar}.

Check. We have $400R=3/11$, so $E[e^{RX}]=1/(1-3/11)=11/8$. The left side of the Lundberg equation is $2(11/8-1)=3/4$ per month. The right side is $1{,}100(3/4{,}400)=3/4$ per month.

Common mistake avoided. Treating $13,200 per year as $13,200 per month would give $c/\lambda=6{,}600$ dollars per claim instead of the correct $550$. It would produce $R=(6{,}600-400)/(6{,}600(400))$, which is not the solution to the stated model. Converting the time basis first prevents this error.

Using a Horizon With Compatible Units

A time horizon $T$ does not enter the ultimate-ruin Lundberg equation directly. It does enter finite-time quantities such as the expected number of claims by time $T$, which is $\lambda T$, and premium income over the interval, which is $cT$. These products are meaningful only when $T$ uses the same time unit as the rate.

A related dimensional check is that $\lambda T$ is an expected claim count, while $cT$ is a monetary amount. If a calculation yields an expected claim count with time units remaining, or premium income with units of dollars per year, the conversion has not been completed.

Worked Example: Converting an 18-Month Horizon

Use the hypothetical insurer from the previous example: $\lambda=24$ claims per year, $c=13{,}200$ dollars per year, and exponential claim amounts with mean $400. For a horizon of 18 months, calculate the expected claim count and premium income. Then state the adjustment coefficient and explain why the horizon does not change it.

Method. Since the rates are annual, convert the horizon to years: $T=18/12=1.5$ years. The expected number of claims is

E[N(T)]=\lambda T=24(1.5)=36\text{ claims}.

Premium income during the same interval is

cT=13{,}200(1.5)=19{,}800\text{ dollars}.

The expected aggregate claim amount over this interval is $36(400)=14{,}400$ dollars, so expected premium income less expected claims is $19{,}800-14{,}400=5{,}400$ dollars. This difference is an expected change in surplus, not a ruin probability.

The adjustment coefficient is unchanged from the previous example: $R=3/4{,}400$ per dollar. It is determined by the claim and premium rates and the severity distribution, not by the selected observation horizon. For instance, with initial surplus $u=4{,}400$, the ultimate-ruin Lundberg bound is $e^{-Ru}=e^{-3}\approx0.0498$. This is an ultimate bound; the 18-month horizon does not turn it into a finite-horizon ruin probability.

Common mistake avoided. Using $T=18$ with annual rates gives $\lambda T=24(18)=432$ claims, as though the interval were 18 years. The correct annual-unit horizon is 1.5 years, giving 36 claims. Equivalently, use monthly rates $\lambda=2$ and $c=1{,}100$, with $T=18$ months: the results are again 36 claims and $19,800 of premium income.

A Practical Unit-Audit Procedure

Before solving a surplus-model problem, record the units beside every input. This short audit catches errors before they propagate into a root, a ruin bound, or a finite-time calculation.

Choose one time unit.
Write $\lambda$, $c$, and $T$ in that unit. Convert, for example, annual rates to monthly rates by dividing by 12, and convert years to months by multiplying by 12.

Choose one monetary unit.
Express $u$, $cT$, and claim amounts in compatible money units. If claim sizes are in thousands of dollars, do not insert a surplus measured in dollars without converting it.

Check the model conditions.
Confirm that the assumed claim-arrival process and severity distribution match the equation being used, and check the positive-loading condition $c>\lambda E[X]$ before seeking a positive adjustment coefficient.

Check dimensions and substitute back.
Confirm that $RX$ is dimensionless, that both sides of the Lundberg equation have units of inverse time, and that the proposed positive root satisfies the equation and any moment-generating function domain restriction.

The most reliable final check is substitution into the original equation using the same units as the problem. A rounded value can cause a small numerical discrepancy, but the unrounded value should satisfy the equation. Also verify that a stated bound is being used for the event it actually bounds: an ultimate-ruin bound is not automatically a finite-horizon result.

Key takeaway: In a surplus model, use one time basis for claim intensity, premium rate, and horizon, and one monetary basis for surplus and claims. A consistent time conversion preserves the adjustment coefficient; an inconsistent one can produce a plausible-looking but invalid answer.

Check Your Understanding

For each problem, write down the common time unit before calculating. Use the exponential-claim adjustment formula only when the claim distribution and model assumptions stated in the problem support it.

A hypothetical insurer has 3 claims per week, premium income of $1,800 per week, and exponential claim amounts with mean $400. Find the adjustment coefficient per dollar. (Answer: $R=1/1{,}200\approx0.0008333$ per dollar.)
A hypothetical insurer has 36 claims per year and premium income of $18,000 per year. Claim amounts are exponential with mean $400. Find the positive adjustment coefficient. (Answer: $R=1/2{,}000=0.0005000$ per dollar.)
A hypothetical insurer has 5 claims per day, premium income of $2,500 per 30-day month, and mean claim amount $300. Does a positive adjustment coefficient exist under the classical model? (Answer: No; $c=2{,}500/30\approx83.33$ dollars per day, while $\lambda E[X]=5(300)=1{,}500$ dollars per day, so the safety loading is negative.)
A hypothetical insurer has 1.5 claims per month and annual premium income of $10,800. Claim amounts are exponential with mean $500. Find the adjustment coefficient per dollar. (Answer: $c=900$ dollars per month and $R=(900-1.5(500))/(900(500))=1/3{,}000\approx0.0003333$ per dollar.)
A hypothetical insurer has 8 claims per quarter, premium income of $10,000 per quarter, and exponential claim amounts with mean $1,000. For a nine-month horizon, find the expected claim count and the positive adjustment coefficient. (Answer: $T=3$ quarters, so $E[N(T)]=8(3)=24$; $R=(10{,}000-8(1{,}000))/(10{,}000(1{,}000))=1/5{,}000=0.0002000$ per dollar.)

What You'll Learn