Choosing Initial Capital from a Ruin Bound

Introduction

Lundberg’s inequality gives an upper bound on ultimate ruin probability that decreases exponentially with initial capital. A practical question is therefore: how much initial capital is sufficient to make that bound no greater than a specified target? The answer is a direct rearrangement of the bound, but it is only valid when the risk model has a positive adjustment coefficient satisfying the required equation.

The resulting capital is a conservative threshold, not generally an estimate of the exact capital at which the actual ruin probability equals the target. This distinction matters when the bound is used for planning, and when its units or rounding are handled incorrectly.

Model and Capital Threshold

Use the classical compound Poisson surplus model. Claims arrive according to a Poisson process with rate $\lambda>0$ per unit time. Claim amounts $X_1,X_2,\ldots$ are independent and identically distributed nonnegative random variables, independent of the arrival process, with finite mean $E[X]$. Premium income arrives continuously at rate $c$ per unit time. With initial capital $u\geq 0$, 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 $t$. Ultimate ruin is the event that $U(t)<0$ at some time $t\geq 0$, and its probability is denoted by $\psi(u)$. Assume the net profit condition $c>\lambda E[X]$.

Write $M_X(r)=E[e^{rX}]$ for the claim-size moment-generating function where it is finite. A positive adjustment coefficient $R$ is a positive solution in the finite domain of the moment-generating function to

\lambda\bigl(M_X(R)-1\bigr)=cR.

The net profit condition alone does not guarantee that such a positive solution exists. For this tutorial, whenever Lundberg’s bound is applied, assume that a positive solution $R$ exists and that $M_X(R)$ is finite. If no such solution exists, this bound cannot be used with a positive adjustment coefficient.

Theorem: Under the compound Poisson assumptions above, if $R>0$ is a finite-domain solution of $\lambda(M_X(R)-1)=cR$, then Lundberg’s inequality gives $\psi(u)\leq e^{-Ru}$ for $u\geq 0$. For a desired upper-bound target $\alpha$, where $0<\alpha<1$, it is sufficient to choose $$ u\geq \frac{\ln(1/\alpha)}{R}. $$

The threshold follows by requiring the bound itself to be no greater than the target:

e^{-Ru}\leq \alpha \quad\Longleftrightarrow\quad -Ru\leq \ln(\alpha) \quad\Longleftrightarrow\quad u\geq \frac{-\ln(\alpha)}{R} =\frac{\ln(1/\alpha)}{R}.

The second equivalence reverses the inequality when dividing by $-R<0$. It also shows why a small target requires more capital: as $\alpha$ decreases, $\ln(1/\alpha)$ increases.

Worked Applications

Worked Example: Capital for a Five-Percent Bound

A hypothetical insurer has claim arrivals at rate $\lambda=2$ per year. Claim amounts are exponential with mean $\theta=\$1{,}000$, and premium income is $c=\$2{,}500$ per year. Find the smallest capital, to the nearest cent rounded upward, for which Lundberg’s bound is at most $0.05$.

Step 1: Check the model conditions. The expected claim amount per year is $2(\$1{,}000)=\$2{,}000$. Thus $c=\$2{,}500$ per year exceeds expected claims per year, so the net profit condition holds.

Step 2: Find the adjustment coefficient. For an exponential claim amount with mean $\theta$, $M_X(r)=1/(1-\theta r)$ for $0\leq r<1/\theta$. Here the adjustment equation is

2\left(\frac{1}{1-1000r}-1\right)=2500r, \qquad 0<r<0.001.

For a positive solution, divide both sides by $r$, and use $1/(1-1000r)-1=1000r/(1-1000r)$:

\frac{2000}{1-1000r}=2500, \qquad 1-1000r=\frac{2000}{2500}=0.8, \qquad 1000r=0.2, \qquad R=0.0002\text{ per dollar}.

The root is in the required domain $0<R<0.001$, so the moment-generating function is finite there.

Step 3: Convert the target to a capital threshold.

u\geq \frac{\ln(1/0.05)}{0.0002} =\frac{\ln(20)}{0.0002} =\frac{2.995732\ldots}{0.0002} =\$14{,}978.661\ldots.

Rounding upward to the nearest cent gives $u=\$14{,}978.67$. The exponent is $Ru=0.0002(14{,}978.67)=2.995734$, so the bound is $e^{-2.995734}$, approximately $0.0500$. The selected capital therefore meets the target; the unrounded threshold is the equality point for the bound.

Method and trap avoided. The method is to solve for $R$, then rearrange the upper bound. In the adjustment-equation algebra, $1-1000R=2000/2500=0.8$, so $1000R=0.2$; confusing these two quantities gives an incorrect $R$.

Worked Example: Capital Restricted to $100 Increments

A hypothetical portfolio has a positive adjustment coefficient $R=0.0005$ per dollar, obtained from a model for which the moment-generating-function condition holds. Capital must be selected in increments of $\$100$. Find the smallest permitted capital for which the Lundberg upper bound is no greater than $0.01$.

The continuous threshold is

u\geq \frac{\ln(1/0.01)}{0.0005} =\frac{\ln(100)}{0.0005} =\frac{4.605170\ldots}{0.0005} =\$9{,}210.340\ldots.

The smallest multiple of $\$100$ that is at least $\$9{,}210.340\ldots$ is $\$9{,}300$. At this capital,

e^{-Ru}=e^{-0.0005(9300)} =e^{-4.65} \approx 0.0096.

This is below $0.01$. The preceding permitted amount, $\$9{,}200$, gives $e^{-0.0005(9200)}=e^{-4.6}\approx0.0101$, which exceeds $0.01$. Thus $\$9{,}300$ is the smallest permitted amount that meets the bound.

Method and trap avoided. First calculate the continuous threshold, then round up to the permitted capital increment. Rounding to the nearest increment could select $\$9{,}200$, which does not satisfy the requested upper-bound target.

Worked Example: Checking Existing Capital and Finding the Shortfall

A hypothetical insurer has a positive adjustment coefficient $R=0.000125$ per dollar, with the required model conditions satisfied. It currently holds $\$30{,}000$ of initial capital and wants the Lundberg upper bound to be at most $0.01$. Determine whether it meets the target and, if not, how much additional capital is needed. Report the minimum additional amount to the nearest whole dollar, rounded upward.

At current capital, the bound is

e^{-Ru}=e^{-0.000125(30{,}000)} =e^{-3.75} \approx 0.0235.

Since $0.0235>0.01$, the current capital does not meet the target. The total threshold is

u_{\text{required}} =\frac{\ln(100)}{0.000125} =\frac{4.605170\ldots}{0.000125} =\$36{,}841.361\ldots.

The additional capital required is therefore at least $\$36{,}841.361\ldots-\$30{,}000=\$6{,}841.361\ldots$. Rounded upward to the nearest whole dollar, the minimum addition is $\$6{,}842$, giving total capital of $\$36{,}842$. At that total, $Ru=0.000125(36{,}842)=4.60525$, and $e^{-4.60525}$ is approximately $0.0099992$, just below $0.01$.

Method and trap avoided. Compare the current bound with the target before calculating a shortfall. The threshold is a total-capital amount; it is not itself the additional capital to inject.

Interpreting the Result

The threshold answers a specific sufficient-condition question: what capital makes the Lundberg upper bound no greater than the chosen target? It does not assert that actual ultimate ruin probability equals $e^{-Ru}$, or that the actual probability equals the target at the threshold. The bound may be conservative, so a model’s true ruin probability can be smaller.

The adjustment coefficient has inverse-currency units. In the first example, $R$ was measured per dollar, so $Ru$ was dimensionless only when $u$ was entered in dollars. If capital is instead recorded in thousands of dollars, the numerical value of $R$ must be expressed per thousand dollars. For example, $0.0002$ per dollar is $0.2$ per thousand dollars; multiplying $0.0002$ by a capital value stated as $15$ thousand dollars would mix units and give the wrong exponent.

For practical use, record the target, the units of $R$, the units of $u$, the unrounded threshold, and the rounding rule. If capital is constrained to a grid, test the selected amount in $e^{-Ru}$. This check catches both unit errors and rounding in the wrong direction.

Common Mistakes

Reversing the inequality: From $e^{-Ru}\leq\alpha$, the conclusion is $u\geq\ln(1/\alpha)/R$, not $u\leq\ln(1/\alpha)/R$. Dividing by the negative value $-R$ reverses the inequality.
Treating a bound as an exact probability: Lundberg’s inequality provides $\psi(u)\leq e^{-Ru}$. Meeting the target with the bound is sufficient, but the bound need not equal the actual ruin probability.
Assuming the net profit condition guarantees $R$: The condition $c>\lambda E[X]$ is not by itself a guarantee of a positive solution. Verify that a positive solution exists in the finite domain of $M_X$ before applying the bound.
Rounding down: A threshold rounded below $\ln(1/\alpha)/R$ can leave the exponential bound above the target. Round upward when a discrete capital amount is required.
Mixing monetary units: The product $Ru$ must be dimensionless. Match the units used for $R$ and capital.

Key takeaway: Given a valid positive adjustment coefficient $R$, the sufficient initial-capital threshold for a Lundberg upper bound of at most $\alpha$ is $\ln(1/\alpha)/R$. It is a conservative bound-based threshold, and its use depends on a valid model, a finite moment-generating function at $R$, consistent units, and appropriate rounding.

Check Your Understanding

For each problem, use the Lundberg upper bound and keep monetary units consistent.

A model has a valid adjustment coefficient of $0.0004$ per dollar. Find the minimum continuous capital for an upper-bound target of $0.02$. (Answer: $\$9{,}780.06$, rounded to the nearest cent.)
Claim amounts are exponential with mean $\$800$, claim arrivals occur at rate $1.5$ per year, and premium income is $\$1{,}500$ per year. Find $R$, then find the continuous capital threshold for an upper-bound target of $0.05$. (Answer: $R=0.00025$ per dollar; $\$11{,}982.93$, rounded to the nearest cent.)
A valid model has $R=0.0002$ per dollar, and capital must be a whole number of dollars. Find the smallest capital for which the bound is at most $0.005$. (Answer: $\$26{,}492$.)
A portfolio has $R=0.00015$ per dollar and initial capital of $\$20{,}000$. Find its Lundberg bound and determine whether it meets a target of $0.05$. (Answer: $e^{-3}\approx0.0498$; yes, it meets the target.)
A valid model has $R=0.0003$ per dollar, and available capital is $\$15{,}000$. For a target of $0.01$, calculate the bound at the available capital, determine whether it meets the target, and find the additional whole-dollar capital required if it does not. (Answer: The bound is $e^{-4.5}\approx0.0111$, so the target is not met; the total threshold is $\$15{,}350.57$, requiring an additional $\$351$ when rounded upward to whole dollars.)

What You'll Learn