Mean and Variance of Compound Poisson Loss

Why Compound Loss Moments Matter

An aggregate loss total combines two sources of uncertainty: how many claims occur and how large each claim is. Even when individual claim amounts have modest variability, a random claim count contributes additional variation to the total. A useful first summary of an aggregate model is therefore its mean and variance.

This tutorial derives those moments for a compound Poisson model and applies them to discrete severities, exponential severities, and multiple exposures. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$ be a claim count, and let $X_1,X_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where an empty sum is defined to be zero. Write $\mu_X=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$. Assume $E[X^2]$ is finite. Then $S$ has a compound Poisson distribution.

Here $\lambda$ is the expected number of claims over the specified exposure period. The severity distribution describes the amount of one claim, not the aggregate total. The independence assumptions mean that the number of claims does not affect their severity distribution and that different claims have independent severities.

Unless a problem specifies otherwise, examples below treat losses as undiscounted amounts in dollars for the stated period. No interest adjustment or mortality decrement is used; the claim-count and severity assumptions are stated separately in each example. If severities or claim counts are dependent, or if the severity distribution has no finite second moment, the formulas below cannot be applied as stated.

The Mean and Variance

The compound Poisson moments follow from conditioning on the claim count. Given $N=n$, aggregate loss is the sum of $n$ independent severities. Its conditional mean is $n\mu_X$, and its conditional variance is $n\sigma_X^2$. The unconditional variance must also include variability in the conditional mean as $N$ changes.

Theorem: Under the compound Poisson assumptions, $$ E[S]=\lambda E[X]=\lambda\mu_X $$ and $$ \operatorname{Var}(S)=\lambda E[X^2] =\lambda\left(\operatorname{Var}(X)+(E[X])^2\right). $$ Equivalently, $E[S]=\lambda\mu_X$ and $\operatorname{Var}(S)=\lambda(\sigma_X^2+\mu_X^2)$.

The derivation makes clear why the second moment of severity, rather than only its variance, appears in the aggregate variance.

Condition on the number of claims.
For $N=n$, the aggregate is a sum of $n$ independent severities. Thus $E[S\mid N=n]=n\mu_X$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$.

Take the unconditional mean.
By conditional expectation, $E[S]=E[E[S\mid N]]=E[N\mu_X]=\mu_X E[N]=\lambda\mu_X$, since a Poisson variable with parameter $\lambda$ has mean $\lambda$.

Apply the law of total variance.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. The first term is $E[N\sigma_X^2]=\lambda\sigma_X^2$; the second is $\operatorname{Var}(N\mu_X)=\mu_X^2\lambda$, because a Poisson variable has variance $\lambda$.

Combine the two sources of variation.
Adding the terms gives $\operatorname{Var}(S)=\lambda(\sigma_X^2+\mu_X^2)=\lambda E[X^2]$, using $E[X^2]=\operatorname{Var}(X)+(E[X])^2$.

The first component of aggregate variance, $\lambda\sigma_X^2$, reflects variation in individual severities. The second, $\lambda\mu_X^2$, reflects variation in the Poisson claim count. This decomposition is a useful calculation check.

Worked Example: Two-Point Claim Severities

A hypothetical one-year insurance portfolio has a Poisson claim count with parameter $\lambda=2$. Each claim is independently $100$ dollars with probability $0.4$, or $500$ dollars with probability $0.6$, independently of the count. Find the mean and variance of annual aggregate loss. Losses are undiscounted; no interest rate or mortality assumption is used.

Method. Calculate the first and second moments of one claim, then apply the compound Poisson formulas.

The mean severity is $$ E[X]=(0.4)(100)+(0.6)(500)=40+300=340\text{ dollars}. $$

The second raw moment is $$ E[X^2]=(0.4)(100^2)+(0.6)(500^2) =(0.4)(10{,}000)+(0.6)(250{,}000) =4{,}000+150{,}000=154{,}000\text{ dollars}^2. $$

Therefore, $$ E[S]=2(340)=680\text{ dollars}, $$ and $$ \operatorname{Var}(S)=2(154{,}000)=308{,}000\text{ dollars}^2. $$

Check. The severity variance is $154{,}000-340^2=154{,}000-115{,}600=38{,}400$ dollars squared. The decomposition gives $2(38{,}400)+2(340^2)=76{,}800+231{,}200=308{,}000$ dollars squared, matching the result. No rounding is needed.

Common mistake avoided: The expected aggregate loss is $\lambda E[X]$, not a claim count multiplied by an incorrectly averaged severity. The severity probabilities must be applied to the corresponding amounts before multiplying by $\lambda$.

Worked Example: Exponential Claim Severity

For a hypothetical one-year coverage, $N\sim\operatorname{Poisson}(4)$. Given a claim, its severity has an exponential distribution with density $f_X(x)=\frac{1}{250}e^{-x/250}$ for $x\geq 0$, so its mean parameter is $250$ dollars. Determine the aggregate mean and variance. Losses are undiscounted, with no interest adjustment or mortality assumption.

Method. For an exponential severity with mean $\theta$, $E[X]=\theta$ and $E[X^2]=2\theta^2$. The second identity follows from the exponential moments, or by integrating $x^2\theta^{-1}e^{-x/\theta}$ over $x\geq0$.

Here $\theta=250$, so $$ E[X]=250\text{ dollars},\qquad E[X^2]=2(250^2)=2(62{,}500)=125{,}000\text{ dollars}^2. $$

With $\lambda=4$, $$ E[S]=4(250)=1{,}000\text{ dollars} $$ and $$ \operatorname{Var}(S)=4(125{,}000)=500{,}000\text{ dollars}^2. $$

Check. An exponential variable with mean $250$ has variance $250^2=62{,}500$. Thus the variance decomposition gives $4(62{,}500)+4(250^2)=250{,}000+250{,}000=500{,}000$ dollars squared, as required. No rounding is needed.

Common mistake avoided: Using only the severity variance would give $4(62{,}500)=250{,}000$, omitting the variability caused by the random claim count. The correct aggregate variance is twice that value in this example.

Combining Independent Exposures

If independent exposures each have Poisson claim counts, their total count is also Poisson, with parameter equal to the sum of their parameters. When their claims share the same severity distribution and all claims are independent, the combined aggregate can therefore be analyzed using the total Poisson parameter. This is a direct application of the Poisson model and the compound Poisson moment formulas.

Worked Example: A Portfolio of Identical Exposures

A hypothetical portfolio has $8$ independent policies. Each policy produces a Poisson number of annual claims with parameter $1.5$. Each claim is $200$ dollars with probability $0.75$ and $1{,}000$ dollars with probability $0.25$, independently across claims and policies. Find the portfolio's annual aggregate mean, variance, and standard deviation. Losses are undiscounted; there is no interest or mortality adjustment.

Method. Add the independent Poisson parameters, calculate the severity moments, and apply the aggregate formulas.

The portfolio claim count has parameter $$ \lambda_{\mathrm{total}}=8(1.5)=12. $$

For one claim, $$ E[X]=(0.75)(200)+(0.25)(1{,}000)=150+250=400\text{ dollars}, $$ and $$ E[X^2]=(0.75)(200^2)+(0.25)(1{,}000^2) =(0.75)(40{,}000)+(0.25)(1{,}000{,}000) =30{,}000+250{,}000=280{,}000\text{ dollars}^2. $$

Thus the portfolio aggregate mean and variance are $$ E[S]=12(400)=4{,}800\text{ dollars}, \qquad \operatorname{Var}(S)=12(280{,}000)=3{,}360{,}000\text{ dollars}^2. $$

The standard deviation is $$ \sqrt{3{,}360{,}000}=1{,}833.030\ldots\text{ dollars}, $$ or $1{,}833.03$ dollars to the nearest cent.

Check. The single-claim variance is $280{,}000-400^2=120{,}000$. The variance decomposition gives $12(120{,}000)+12(400^2)=1{,}440{,}000+1{,}920{,}000=3{,}360{,}000$ dollars squared. The mean and variance both use the total count parameter $12$, not the per-policy parameter $1.5$.

Common mistake avoided: Applying the per-policy frequency to the entire portfolio understates both moments by a factor of $8$. First combine exposure, then use the resulting total Poisson parameter.

Common Errors and Limits of the Formula

A frequent error is to calculate aggregate variance as $\lambda\operatorname{Var}(X)$. That expression includes severity variation but ignores the randomness in $N$. For the two-point severity example, the incorrect calculation is $$ 2(38{,}400)=76{,}800\text{ dollars}^2. $$ It is incomplete because it omits $\operatorname{Var}(N\mu_X)=2(340^2)=231{,}200$ dollars squared. The correction is $$ \operatorname{Var}(S)=2(38{,}400)+2(340^2)=308{,}000\text{ dollars}^2. $$

This distinction also separates a compound Poisson total from a sum with a fixed number of claims. If the number of claims were fixed at $n$, the sum would have variance $n\operatorname{Var}(X)$. For a Poisson count, the count itself varies, adding the second term $\lambda(E[X])^2$.

Before using the result, verify the model carefully: the count must be Poisson for the displayed form, claim severities must be identically distributed and independent of the count, and a finite second severity moment is needed for a finite aggregate variance. Also keep units consistent: the mean is in dollars, while the variance is in dollars squared. A variance is not itself a typical dollar fluctuation; its square root is the standard deviation in dollars.

Key takeaway: For a compound Poisson aggregate, multiply the severity mean by the expected count to obtain the aggregate mean, and multiply the severity second raw moment by the expected count to obtain aggregate variance. The second raw moment incorporates both claim-size and claim-count variation.

Check Your Understanding

For each problem, identify the Poisson parameter for the stated period and use severity moments to find the requested aggregate quantities.

A hypothetical annual account has $N\sim\operatorname{Poisson}(3)$, with every claim fixed at $80$ dollars. Find the aggregate mean and variance. Losses are undiscounted. (Answer: mean = $240; variance = 19,200 dollars².)
A hypothetical one-year loss count is Poisson with parameter $5$, and individual severities are exponential with mean $100$ dollars. Find the aggregate mean and variance. No interest adjustment is applied. (Answer: mean = $500; variance = 100,000 dollars².)
A hypothetical aggregate has $N\sim\operatorname{Poisson}(2.5)$. Each severity is $50$ dollars with probability $0.2$ and $150$ dollars with probability $0.8$. Find $E[X]$, $E[X^2]$, and the aggregate mean and variance. (Answer: $E[X]=\$130$; $E[X^2]=18,500$ dollars²; mean = $325; variance = 46,250 dollars².)
A hypothetical portfolio has an annual Poisson claim parameter of $8$, with independent claims. Calculate its aggregate moments for a six-month period if severities have mean $300$ dollars and variance $90,000$ dollars squared. Assume the annual rate is constant and losses are undiscounted. (Answer: six-month parameter = 4; mean = $1,200; variance = 720,000 dollars².)
A hypothetical annual aggregate has a Poisson claim parameter of $4$. Individual severities have mean $60$ dollars and variance $400$ dollars squared. Find the aggregate mean and variance, and explain why multiplying the severity variance alone by $4$ is insufficient. (Answer: mean = $240; variance = 16,000 dollars²; multiplying only by $4$ omits the count-variation term $4(60^2)=14,400$ dollars².)

What You'll Learn