Compound Poisson Claims in the Surplus Model

Introduction

In a surplus model, the insurer’s surplus changes as premium income arrives and claims are paid. To analyze that movement over a fixed period, first identify the total claim amount during the period. When claim arrivals follow a Poisson process and claim amounts are independent and identically distributed, this total has a compound Poisson distribution. Its mean and variance combine two sources of uncertainty: how many claims occur and how large they are.

This tutorial develops those aggregate moments from the model assumptions and applies them to hypothetical insurance portfolios. The results are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: In the classical surplus model, let \(u\) be initial surplus, \(c\) a constant premium rate per unit time, and \(S(t)\) the aggregate amount of claims by time \(t\). The surplus at time \(t\) is \(U(t)=u+ct-S(t)\). For the compound Poisson model, \(N(t)\) is a homogeneous Poisson claim-count process with rate \(\lambda\) per unit time, and \(S(t)=\sum_{j=1}^{N(t)}X_j\), where the claim amounts \(X_1,X_2,\ldots\) are independent and identically distributed, nonnegative, and independent of \(N(t)\).

Model Assumptions and Aggregate Moments

The rate \(\lambda\) is a claim frequency per unit of time, while \(t\) is the length of the interval measured in the same time unit. Therefore \(N(t)\) has a Poisson distribution with parameter \(\lambda t\), so \(E[N(t)]=\operatorname{Var}(N(t))=\lambda t\). Let a single claim amount \(X\) have finite mean \(\mu=E[X]\) and finite variance \(\sigma^2=\operatorname{Var}(X)\). The claim amounts may have any severity distribution satisfying these conditions; a Poisson count alone does not determine the aggregate distribution.

The aggregate amount is zero if no claims occur. If \(n\) claims occur, it is the sum of \(n\) independent severities. The formulas below rely on independence between the claim count and severities, as well as independence among severities. If severity depends on the number of claims, these formulas need not apply.

Theorem: Under the compound Poisson assumptions, for an interval of length \(t\), \(E[S(t)]=\lambda t E[X]\) and \(\operatorname{Var}(S(t))=\lambda t E[X^2]=\lambda t\{\operatorname{Var}(X)+(E[X])^2\}\).

The second moment \(E[X^2]\), not just the severity variance, appears in the aggregate variance. It includes the contribution from random claim counts. The following derivation shows why.

Condition on the claim count.
Given \(N(t)=n\), the aggregate is a sum of \(n\) independent claim amounts. Thus \(E[S(t)\mid N(t)=n]=n\mu\) and \(\operatorname{Var}(S(t)\mid N(t)=n)=n\sigma^2\).

Find the mean.
By iterated expectation, \(E[S(t)]=E[E[S(t)\mid N(t)]]=E[N(t)\mu]=\lambda t\mu\).

Find the variance.
The law of total variance gives \(\operatorname{Var}(S(t))=E[\operatorname{Var}(S(t)\mid N(t))]+\operatorname{Var}(E[S(t)\mid N(t)])\). The first term is \(\lambda t\sigma^2\); the second is \(\mu^2\lambda t\). Their sum is \(\lambda t(\sigma^2+\mu^2)=\lambda tE[X^2]\).

This decomposition is useful in its own right: \(E[N(t)]\sigma^2\) is variation in severities when the count is fixed, and \(\operatorname{Var}(N(t))\mu^2\) is variation arising because the count itself fluctuates. The identity \(E[X^2]=\sigma^2+\mu^2\) then gives the compact formula.

\begin{aligned} E[S(t)] &= \lambda t\mu,\\ \operatorname{Var}(S(t)) &= \lambda t\sigma^2+\lambda t\mu^2 =\lambda tE[X^2]. \end{aligned}

Worked Calculations

For each calculation, first put the frequency rate and interval in matching time units. Next calculate the severity moments, then apply the aggregate formulas. Money amounts below are in dollars unless otherwise stated; variances are in dollars squared.

Worked Example: Exponential Claims Over Six Months

Problem. A hypothetical portfolio has a Poisson claim frequency of \(\lambda=4\) claims per year. Claim amounts are independent exponential random variables with mean \(2{,}500\). Find the mean and variance of aggregate claims over six months, and the aggregate standard deviation.

Method and assumptions. Use the compound Poisson moment formulas. The exponential severity has rate parameter \(1/2{,}500\) per dollar, so \(E[X]=2{,}500\) and \(\operatorname{Var}(X)=2{,}500^2=6{,}250{,}000\). Count and severities are independent.

Step 1: Expected count. Six months is \(t=0.5\) years, so \(\lambda t=4(0.5)=2\). Thus \(N(t)\sim\operatorname{Poisson}(2)\).

Step 2: Aggregate mean. \(E[S(t)]=2(2{,}500)=5{,}000\).

Step 3: Aggregate variance. For an exponential variable, \(E[X^2]=\operatorname{Var}(X)+(E[X])^2=6{,}250{,}000+6{,}250{,}000=12{,}500{,}000\). Therefore \(\operatorname{Var}(S(t))=2(12{,}500{,}000)=25{,}000{,}000\).

Step 4: Standard deviation and check. The standard deviation is \(\sqrt{25{,}000{,}000}=5{,}000\). As a check using conditional variance, the two components are \(2(6{,}250{,}000)=12{,}500{,}000\) and \(2(2{,}500^2)=12{,}500{,}000\), which sum to \(25{,}000{,}000\). The result avoids the mistake of using a full year of expected claims for a six-month interval.

Worked Example: Two-Point Severity and Count Variability

Problem. A hypothetical insurer has a Poisson claim frequency of 12 claims per year. Each claim amount is \(1{,}000\) with probability \(0.7\) and \(4{,}000\) with probability \(0.3\). Find the aggregate mean and variance over one quarter.

Method and assumptions. Calculate the first and second raw moments of the discrete severity distribution, then multiply the second raw moment by the expected claim count. Claims are independent and identically distributed, and are independent of the Poisson count.

Step 1: Expected count. One quarter is \(t=0.25\) years, so \(\lambda t=12(0.25)=3\). Hence \(N(t)\sim\operatorname{Poisson}(3)\).

Step 2: Severity mean. \(E[X]=0.7(1{,}000)+0.3(4{,}000)=700+1{,}200=1{,}900\).

Step 3: Severity second moment. \(E[X^2]=0.7(1{,}000^2)+0.3(4{,}000^2)=700{,}000+4{,}800{,}000=5{,}500{,}000\).

Step 4: Aggregate moments. \(E[S(t)]=3(1{,}900)=5{,}700\), and \(\operatorname{Var}(S(t))=3(5{,}500{,}000)=16{,}500{,}000\).

Step 5: Check by decomposition. The severity variance is \(5{,}500{,}000-1{,}900^2=5{,}500{,}000-3{,}610{,}000=1{,}890{,}000\). The conditional-severity component is \(3(1{,}890{,}000)=5{,}670{,}000\), and the count component is \(3(1{,}900^2)=10{,}830{,}000\). Their sum is \(16{,}500{,}000\).

Common mistake avoided. Multiplying the severity variance by three alone would give \(5{,}670{,}000\), not the aggregate variance. That calculation holds the count variability out; the missing \(10{,}830{,}000\) must also be included.

Worked Example: Gamma Claims and a Nine-Month Interval

Problem. A hypothetical portfolio has a Poisson frequency of 6 claims per year. Individual claim amounts follow a Gamma distribution with shape \(2\) and scale \(500\) dollars. Find the aggregate mean and variance over nine months.

Method and assumptions. For a Gamma distribution with shape \(k\) and scale \(\theta\), \(E[X]=k\theta\) and \(\operatorname{Var}(X)=k\theta^2\). Here \(k=2\) and \(\theta=500\), so the severity mean is \(2(500)=1{,}000\) and severity variance is \(2(500^2)=500{,}000\). Claim amounts are independent of each other and of the claim count.

Step 1: Expected count. Nine months is \(t=0.75\) years, so \(\lambda t=6(0.75)=4.5\).

Step 2: Severity second moment. \(E[X^2]=500{,}000+1{,}000^2=1{,}500{,}000\).

Step 3: Aggregate moments. \(E[S(t)]=4.5(1{,}000)=4{,}500\), and \(\operatorname{Var}(S(t))=4.5(1{,}500{,}000)=6{,}750{,}000\).

Step 4: Check by decomposition. The two variance components are \(4.5(500{,}000)=2{,}250{,}000\) and \(4.5(1{,}000^2)=4{,}500{,}000\), giving \(6{,}750{,}000\). This check confirms that the nine-month count parameter, rather than the annual rate of 6, was used.

Interpreting the Results and Avoiding Traps

The aggregate mean scales with both the expected claim count and the mean severity. The aggregate variance scales with the expected count and the severity second moment. Consequently, two portfolios can have the same aggregate mean but different aggregate variances if their claim severities differ in spread or average size.

A frequent exam error is to substitute \(\operatorname{Var}(X)\) where the formula requires \(E[X^2]\). The distinction can be displayed directly:

\operatorname{Var}(S(t)) =\lambda t\operatorname{Var}(X) +\lambda t(E[X])^2,

so using only \(\lambda t\operatorname{Var}(X)\) omits the variability in the number of claims. Another common error is a time-unit mismatch: if \(\lambda\) is per year and the interval is in months, convert the interval to years before calculating \(\lambda t\). Finally, the mean and variance formulas require a finite severity second moment for the variance to be finite.

If severity amounts are strictly positive, \(S(t)=0\) exactly when \(N(t)=0\), giving \(P(S(t)=0)=e^{-\lambda t}\). If zero-valued severities are possible, the aggregate may also be zero when claims occur, so this equality does not generally hold. Keeping this distinction in mind helps when describing the aggregate distribution, rather than just its moments.

Key takeaway: For a compound Poisson total over an interval of length \(t\), the count parameter is \(\lambda t\), the aggregate mean is \(\lambda tE[X]\), and the aggregate variance is \(\lambda tE[X^2]\). Check time units and include the count-variability component.

Check Your Understanding

For each hypothetical portfolio, use independent Poisson claim counts and independent, identically distributed severities. Calculate the requested aggregate quantities before checking the answer.

A portfolio has a Poisson rate of 5 claims per year and exponential severities with mean \(300\) dollars. Find the aggregate mean and variance over two months. (Answer: mean \(250\) dollars; variance \(150{,}000\) dollars squared.)
Claims arrive at rate 2 per month. Each claim is \(2{,}000\) dollars with probability \(0.8\), or \(8{,}000\) dollars with probability \(0.2\). Find the aggregate mean and variance over three months. (Answer: mean \(19{,}200\) dollars; variance \(96{,}000{,}000\) dollars squared.)
A portfolio has a Poisson rate of \(0.5\) claims per day and Gamma severities with shape \(3\) and scale \(200\) dollars. Find the aggregate mean and variance over eight days. (Answer: mean \(2{,}400\) dollars; variance \(1{,}920{,}000\) dollars squared.)
A hypothetical insurer observes a Poisson rate of \(0.4\) claims per week. Claim severity is uniform on the interval from \(100\) to \(500\) dollars. Find the aggregate mean and variance over ten weeks. (Answer: mean \(1{,}200\) dollars; variance \(413{,}333.33\) dollars squared.)
A portfolio has a Poisson frequency of 9 claims per year. Individual severities have mean \(800\) dollars and variance \(160{,}000\) dollars squared. Over four months, find the aggregate mean, aggregate variance, and probability of no claims. (Answer: mean \(2{,}400\) dollars; variance \(2{,}400{,}000\) dollars squared; probability of no claims \(e^{-3}=0.0498\) to four decimals, assuming strictly positive severities.)

What You'll Learn