How Claim Frequency Affects Ruin

Why Claim Frequency Matters

In a surplus model, claims reduce the insurer’s funds while premium income increases them. If claim severity and the premium rate stay fixed, a higher claim frequency means more opportunities for claims to erode surplus. This suggests that ruin becomes more likely, but the claim can be made precise: a coupling argument shows that increasing the Poisson intensity cannot reduce either finite-horizon or ultimate ruin probability.

Frequency also affects more than the expected total amount of claims. It changes the variability of aggregate claims and the chance of an unusually large claim arriving before a specified time. These are useful diagnostics, but neither the expected surplus nor the variance alone determines the probability of ruin.

Definition: Let $u\geq 0$ be initial surplus, $c\geq 0$ the premium income per unit time, and $N(t)$ a Poisson process with intensity $\lambda$ claims per unit time. Let the claim amounts $X_1,X_2,\ldots$ be independent and identically distributed nonnegative random variables, independent of $N(t)$. The surplus at time $t$ is $$ U(t)=u+ct-\sum_{j=1}^{N(t)}X_j. $$ The time of ruin is $\tau=\inf\{t\geq 0:U(t)<0\}$. The finite-horizon ruin probability is $\psi(u,T;\lambda,c)=P(\tau\leq T)$, and the ultimate ruin probability is $\psi(u;\lambda,c)=P(\tau<\infty)$.

A comparison of claim frequencies means changing $\lambda$ while keeping $u$, $c$, and the entire severity distribution fixed. Time units must agree: if $\lambda$ is measured per year, then $t$ and $T$ must be in years. The premium rate $c$ is not automatically increased when $\lambda$ increases.

The Pathwise Frequency Comparison

Suppose $\lambda_2>\lambda_1$. Construct a model with intensity $\lambda_2$ by adding to the $\lambda_1$ claim process an independent Poisson process of intensity $\lambda_2-\lambda_1$. Give every claim in both processes an independent severity drawn from the same distribution. The combined process is Poisson with intensity $\lambda_2$, and its aggregate claims at each time are at least as large as those in the $\lambda_1$ process.

Use a common baseline.
Let both models have the same initial surplus, premium rate, and baseline claims with intensity $\lambda_1$.

Add claims to the higher-frequency model.
Give it an independent Poisson process of additional claims with intensity $\lambda_2-\lambda_1$, and use the same severity distribution.

Compare surplus paths.
At every time $t$, the added claims can only reduce surplus, so $U_2(t)\leq U_1(t)$.

Compare ruin events.
If the lower-frequency model has negative surplus by time $T$, the higher-frequency model does too. Thus its ruin event contains the lower-frequency model’s ruin event.

\lambda_2\geq\lambda_1 \quad\Longrightarrow\quad \psi(u,T;\lambda_2,c)\geq\psi(u,T;\lambda_1,c) \quad\text{and}\quad \psi(u;\lambda_2,c)\geq\psi(u;\lambda_1,c).

The inequalities follow from inclusion of events under this construction: whenever the lower-frequency path is ruined, the higher-frequency path is also ruined. The result establishes a nondecreasing relationship; it does not say the probabilities must be strictly different for every parameter choice. For example, over a very short horizon, the extra claims may have little effect on a particular ruin event.

Moments: Useful, but Not Ruin Probabilities

For a fixed time $T$, write $S(T)=\sum_{j=1}^{N(T)}X_j$, and suppose the severity has finite mean $\mu=E[X]$ and finite second moment $E[X^2]$. The compound Poisson moments are $$ E[S(T)]=\lambda T\mu,\qquad \operatorname{Var}(S(T))=\lambda T E[X^2]. $$ Consequently, $$ E[U(T)]=u+cT-\lambda T\mu,\qquad \operatorname{Var}(U(T))=\lambda T E[X^2]. $$ The variance is not $\lambda T\operatorname{Var}(X)$: variation in the number of claims also contributes. These moments summarize the surplus at one time; ruin depends on whether the surplus falls below zero at any time, so they do not by themselves give a ruin probability.

Justification: Conditional on $N(T)=n$, the aggregate claims have mean $n\mu$ and variance $n\operatorname{Var}(X)$. Applying the conditional-variance identity gives $$ \operatorname{Var}(S(T)) =E[N(T)]\operatorname{Var}(X)+\operatorname{Var}(N(T))\mu^2 =\lambda T\bigl(\operatorname{Var}(X)+\mu^2\bigr) =\lambda T E[X^2], $$ because a Poisson random variable with mean $\lambda T$ has variance $\lambda T$.

Worked Example: Comparing Surplus Moments at Two Frequencies

A hypothetical insurer has initial surplus $u=\$50{,}000$ and receives premium income at $c=\$12{,}000$ per year. Claims arrive according to a Poisson process. Claim severity is exponential with mean $\$8{,}000$, independently of frequency. Compare expected surplus and variance of surplus after $T=3$ years when $\lambda=0.4$ and $\lambda=0.8$ claims per year.

Method. Use the compound Poisson moment formulas. Since the exponential severity has mean $\mu=\$8{,}000$ and variance $\mu^2$, its second moment is $E[X^2]=2\mu^2=2(\$8{,}000)^2=128{,}000{,}000$ dollars squared.

For $\lambda=0.4$, the expected number of claims over three years is $0.4(3)=1.2$. Expected aggregate claims are $1.2(\$8{,}000)=\$9{,}600$. Premium income over the period is $3(\$12{,}000)=\$36{,}000$, so $$ E[U(3)]=\$50{,}000+\$36{,}000-\$9{,}600=\$76{,}400. $$ The surplus variance is $1.2(128{,}000{,}000)=153{,}600{,}000$ dollars squared.

For $\lambda=0.8$, the expected number of claims is $0.8(3)=2.4$, and expected aggregate claims are $2.4(\$8{,}000)=\$19{,}200$. Therefore $$ E[U(3)]=\$50{,}000+\$36{,}000-\$19{,}200=\$66{,}800, $$ and the surplus variance is $2.4(128{,}000{,}000)=307{,}200{,}000$ dollars squared.

The higher frequency lowers expected surplus and doubles its variance; the premium rate and severity distribution have not changed. These moment calculations are not ruin probabilities. Common mistake avoided: treating the expected surplus as if it were a measure of the probability that surplus stayed nonnegative throughout the three years.

A Lower Bound from Oversized Claims

A useful finite-horizon calculation isolates claims that are large enough to cause ruin even in the most favorable circumstance allowed by the model. Suppose $c\geq0$, and consider any claim larger than $u+cT$. Before that claim arrives at time $t\leq T$, surplus is at most $u+ct$, because earlier claims are nonnegative. Since $u+ct\leq u+cT$, a claim larger than $u+cT$ makes surplus negative at arrival.

The claims exceeding a threshold $M$ form a thinned Poisson process with intensity $\lambda P(X>M)=\lambda S_X(M)$, where $S_X(M)=P(X>M)$. Thus the probability of at least one such claim by time $T$ gives a lower bound:

\psi(u,T;\lambda,c)\geq 1-\exp\!\left[-\lambda T S_X(u+cT)\right].

This is a lower bound, not usually an exact ruin probability: ruin can also result from several claims whose individual amounts are each below the threshold. For a continuous severity distribution, writing $P(X>u+cT)$ or $S_X(u+cT)$ gives the same tail probability. For distributions with atoms, retain the strict inequality because a claim equal to the threshold need not cause negative surplus.

Worked Example: An Oversized-Claim Bound at Two Frequencies

A hypothetical insurer has $u=\$20{,}000$, premium rate $c=\$10{,}000$ per year, and a two-year horizon. Claim sizes are exponential with mean $\$10{,}000$. Compare the oversized-claim lower bounds for $\lambda=0.5$ and $\lambda=1$ claims per year.

Method. The threshold is the largest surplus the insurer could have just before a claim by the end of the horizon: $$ u+cT=\$20{,}000+(\$10{,}000)(2)=\$40{,}000. $$ For an exponential severity with mean $\$10{,}000$, the survival function is $S_X(x)=e^{-x/10{,}000}$ for $x\geq0$. Hence $$ S_X(\$40{,}000)=e^{-4}=0.0183156. $$

At $\lambda=0.5$, the expected number of claims exceeding the threshold over two years is $$ (0.5)(2)(0.0183156)=0.0183156. $$ The lower bound is $1-e^{-0.0183156}=0.0181489$, or 0.0181 to four decimal places.

At $\lambda=1$, that expected number is $$ (1)(2)(0.0183156)=0.0366312. $$ The lower bound is $1-e^{-0.0366312}=0.0359685$, or 0.0360 to four decimal places. The probability that an individual claim exceeds the threshold remains $0.0183156$; it is the expected number of such claims over the horizon that changes with $\lambda T$.

The higher frequency produces the larger lower bound, consistent with the pathwise comparison. Common mistake avoided: using $0.0366312$ as the severity-tail probability. It is the Poisson mean for oversized claims at $\lambda=1$, after multiplying the tail probability by $\lambda T$.

An Exact Finite-Horizon Comparison in a Special Case

Some models permit an exact ruin calculation without solving a general ruin equation. If every claim has the same fixed size, the initial surplus and premium rate can be chosen so that ruin occurs exactly at a particular claim count within the horizon. Then a Poisson count probability gives the answer directly.

Worked Example: Ruin Exactly at the Second Claim

A hypothetical fund starts with $u=\$100$, receives premium income at $c=\$5$ per year, and pays exactly $\$60$ for each claim. Claims follow a Poisson process. Find the probability of ruin by $T=1$ year for $\lambda=0.5$ and $\lambda=1.5$ claims per year.

Method. Before any claim, surplus is nonnegative. After the first claim, surplus is at least $\$100-\$60=\$40$, so one claim cannot cause ruin. By the end of the year, the maximum surplus before a claim, ignoring any previous claims, is $\$100+\$5=\$105$. After two claims, even at that latest possible time, surplus is at most $\$105-2(\$60)=-\$15$. Thus ruin by year 1 occurs exactly when at least two claims arrive.

For a Poisson count with mean $\lambda T$, $P(N(T)\geq2)=1-P(N(T)=0)-P(N(T)=1)=1-e^{-\lambda T}(1+\lambda T)$.

At $\lambda=0.5$, the mean count is $0.5$, giving $$ P(N(1)\geq2)=1-e^{-0.5}(1+0.5) =1-(0.6065307)(1.5)=0.0902 $$ to four decimal places.

At $\lambda=1.5$, the mean count is $1.5$, giving $$ P(N(1)\geq2)=1-e^{-1.5}(1+1.5) =1-(0.2231302)(2.5)=0.4422 $$ to four decimal places. The higher-frequency model has the higher ruin probability. Common mistake avoided: counting only paths with exactly two claims; paths with three or more claims are also ruined and must be included.

What Must Stay Fixed in the Comparison?

The monotonicity result depends on comparing models with the same premium rate and severity distribution. If the premium rate is changed along with claim frequency, the pathwise argument no longer compares the same premium income. For example, increasing premium income in the higher-frequency model could offset some or all of the effect of the additional claims. A claim-frequency comparison must therefore state clearly which quantities are held fixed.

A related trap is to use the expected number of claims, $\lambda T$, as though it were the ruin probability. It is a count mean and may exceed one; ruin is an event about the surplus path. Use a direct probability calculation only when the model supports it, or identify a bound as a bound. These frequency comparisons are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Key takeaway: With initial surplus, premium rate, and claim severity distribution fixed, adding an independent stream of Poisson claims can only lower the surplus path. Therefore, higher claim frequency cannot reduce finite-horizon or ultimate ruin probability. Moment calculations and oversized-claim bounds help quantify the effect, but they are not interchangeable with an exact ruin probability.

Check Your Understanding

For each problem, keep track of the time unit and distinguish a ruin probability from a moment or lower bound.

A hypothetical insurer has $u=\$30{,}000$, $c=\$8{,}000$ per year, $\lambda=0.25$ claims per year, mean claim size $\$4{,}000$, and horizon $T=2$ years. Find the expected surplus at time $T$. (Answer: $\$44{,}000$.)
Initial surplus is $\$10{,}000$, premium income is $\$5{,}000$ per year, $T=2$ years, and severity is exponential with mean $\$5{,}000$. At $\lambda=0.6$ claims per year, calculate the oversized-claim lower bound, to four decimals. (Answer: $1-e^{-0.0219788}\approx0.0217$.)
A fund has $u=\$80$, $c=\$10$ per year, and deterministic claims of $\$50$. Claims are Poisson with $\lambda=1$ per year. Find the exact probability of ruin by $T=1$. (Answer: $1-2e^{-1}\approx0.2642$.)
Two models have the same initial surplus, premium rate, and severity distribution, but their intensities are $0.4$ and $0.9$ claims per year. Without computing a numerical ruin probability, state what can be concluded about their finite-horizon ruin probabilities and justify the conclusion. (Answer: The probability for intensity $0.9$ is at least as large, by coupling the additional claims as an independent Poisson process of intensity $0.5$ per year.)
A hypothetical insurer has $u=\$15{,}000$, $c=\$20{,}000$ per year, a three-month horizon, and exponential claim sizes with mean $\$5{,}000$. Claims arrive at $\lambda=12$ per year. Calculate the oversized-claim lower bound to four decimals. (Answer: $1-e^{-0.0549469}\approx0.0535$.)

What You'll Learn