Matching Insurance Outcomes To Distributions

Why the Outcome Type Comes First

A distribution is a model for a particular random outcome, not a generic label for “insurance data.” The number of claims in a year, the amount paid on one positive claim, and the total paid for a policy-year have different possible values and different ways of producing zero. A model that is plausible for one of these outcomes may be inappropriate for another.

A useful first pass is to check three features: support (the values the outcome can take), the relationship between its mean and variance, and whether zero is possible or has a distinct point probability. These features narrow the candidate families. They do not, by themselves, establish that a model fits the full data distribution.

Definition: The support of a random variable is the smallest closed set having probability 1; equivalently, it is the set of points for which every open neighborhood has positive probability. For a discrete random variable, the support is the set of values with positive probability. A point can be in the support of a continuous distribution even though the probability of that exact point is zero.

Match the Outcome’s Support

A claim count $N$ is a nonnegative integer. Under a Poisson model with parameter $\lambda>0$, $P(N=n)=e^{-\lambda}\lambda^n/n!$ for $n=0,1,2,\ldots$. A negative binomial count model also has nonnegative integer support. Both models allow zero claims, so observing zeros alone does not distinguish them.

A positive claim severity $X$, when modeled conditional on a claim having occurred, is usually strictly positive and continuous. Gamma and inverse Gaussian distributions are common candidates: each has support on $(0,\infty)$ and no point mass at zero. If a data set of “severities” includes zero, first check what was recorded. A zero may represent a no-claim policy, a rejected or fully deductible claim, or a different outcome definition; it is not automatically a zero draw from a positive-severity Gamma or inverse Gaussian model.

An aggregate loss $S$, the sum of payments during a period, is nonnegative and can be exactly zero when there are no claims. Under a compound Poisson model, that event creates a point mass at zero, while positive aggregate amounts are modeled by the distribution of one or more summed severities. A compound Poisson-Gamma model belongs to the Tweedie family for the relevant range of its power parameter.

Outcome	Typical support	Candidate families and zero feature
Claim count	0, 1, 2, …	Poisson or negative binomial; both can assign positive probability to zero
Positive claim severity	$(0,\infty)$	Gamma or inverse Gaussian; continuous, with no atom at zero
Aggregate loss	$[0,\infty)$	Compound Poisson-Gamma/Tweedie; zero may have positive probability

Use Mean-Variance Patterns as Evidence

The Poisson model has $E[N]=\lambda$ and $\operatorname{Var}(N)=\lambda$. It therefore imposes equality of mean and variance. A standard negative binomial parameterization uses mean $\mu$ and dispersion $\alpha>0$, with variance $\mu+\alpha\mu^2$. This permits variance greater than the mean. When $\alpha$ tends to zero, the variance approaches the Poisson variance.

For a Gamma severity with mean $\mu>0$ and dispersion $\phi>0$, $\operatorname{Var}(X)=\phi\mu^2$. Thus a common Gamma dispersion implies variance proportional to the square of the mean and a constant coefficient of variation. For an inverse Gaussian severity, the corresponding variance function is $\operatorname{Var}(X)=\phi\mu^3$; a common dispersion instead implies variance proportional to the cube of the mean. These are diagnostic patterns across risks or groups, not a rule that one sample variance identifies a family conclusively.

For a Tweedie aggregate-loss model with $1<p<2$, the mean-variance relationship is $\operatorname{Var}(S)=\phi\mu^p$, with $\mu=E[S]$ and dispersion $\phi>0$. In the compound Poisson-Gamma interpretation, this range of $p$ accommodates a zero mass together with continuous positive aggregate losses.

Key takeaway: First identify whether the observation is a count, a positive severity, or an aggregate. Then use the support, mean-variance pattern, and zero behavior together. A familiar variance formula cannot repair a mismatch in what the data represent.

Worked Distribution Matches

Worked Example: Choosing a Model for Claim Counts

A hypothetical portfolio has 200 policy-years. Its observed claim count per policy-year has mean 1.8, sample variance 3.42, and 55 policy-years with zero claims. Compare a Poisson model with a negative binomial model using the method of moments and the predicted zero proportion. Treat the supplied summaries as exact for this diagnostic comparison.

Step 1: Check the outcome. Each observation is a number of claims, so its support is the nonnegative integers. Both candidate models have the right support and both allow zero.

Step 2: Compare the moments. A Poisson model with mean $\lambda=1.8$ requires variance 1.8, not 3.42. For the negative binomial parameterization $\operatorname{Var}(N)=\mu+\alpha\mu^2$, the method-of-moments estimate is

\hat{\alpha} =\frac{3.42-1.8}{1.8^2} =\frac{1.62}{3.24} =0.5.

The estimated variance is $1.8+0.5(1.8^2)=1.8+1.62=3.42$, matching the supplied summary.

Step 3: Compare zero probabilities. For this parameterization, the negative binomial zero probability is $(1+\alpha\mu)^{-1/\alpha}$. It is

P(N=0)=(1+0.5(1.8))^{-2}=1.9^{-2} =\left(\frac{10}{19}\right)^2 =\frac{100}{361} \approx 0.2770.

This predicts $200(0.2770)\approx55.40$ zero-claim policy-years. The observed proportion is $55/200=0.2750$. In contrast, the Poisson model predicts $P(N=0)=e^{-1.8}\approx0.1653$, or about $200(0.1653)=33.06$ zero-claim policy-years.

Conclusion and check. The negative binomial is the more plausible of these two candidates for the supplied summaries: it matches the mean and variance through the fitted dispersion and predicts a zero proportion close to the observed 0.2750. This is a moment-and-zero diagnostic, not a complete goodness-of-fit test.

Common mistake avoided: Choosing Poisson just because the data are counts overlooks its restriction $\operatorname{Var}(N)=E[N]$. Here 3.42 is not 1.8.

Worked Example: Distinguishing Gamma and Inverse Gaussian Severities

For two hypothetical groups of positive, uncensored claim severities, the supplied summaries are: Group A has mean \$10,000 and variance $(\$25,000)^2=250,000,000$ dollars squared; Group B has mean \$20,000 and variance $(\$10,000)^2=100,000,000$ dollars squared. Determine which of the Gamma or inverse Gaussian variance patterns is consistent with a common dispersion across groups.

Step 1: Check support. Both groups contain positive continuous severities, so both Gamma and inverse Gaussian models have appropriate support. Neither assigns a point mass to a zero severity.

Step 2: Test the Gamma pattern. For Group A,

\frac{\operatorname{Var}(X)}{\mu^2} =\frac{250{,}000{,}000}{10{,}000^2} =\frac{250{,}000{,}000}{100{,}000{,}000} =2.5.

For Group B, the same ratio is $100{,}000{,}000/20{,}000^2=100{,}000{,}000/400{,}000{,}000=0.25$. These ratios are not equal, so the two groups do not have a common Gamma dispersion.

Step 3: Test the inverse Gaussian pattern. For Group A, $\operatorname{Var}(X)/\mu^3=250{,}000{,}000/10^{12}=0.00025$. For Group B, the ratio is $100{,}000{,}000/(20{,}000)^3=100{,}000{,}000/8{,}000{,}000{,}000{,}000=0.0000125$. The ratios also differ, so a common inverse Gaussian dispersion is not consistent with these summaries.

Conclusion and check. Neither candidate has a common dispersion across these two groups under the stated summaries. The correct match is therefore neither under the common-dispersion assumption; the data may require group-specific dispersion, another model, or further investigation. The arithmetic checks the scaling directly: doubling the mean would multiply variance by 4 under common Gamma dispersion, or by 8 under common inverse Gaussian dispersion. The observed variance instead falls from 250 million to 100 million.

Common mistake avoided: Comparing only raw variances can be misleading because variance depends on the scale of the mean. Compare $\operatorname{Var}(X)/\mu^2$ and $\operatorname{Var}(X)/\mu^3$ against their respective common-dispersion assumptions.

Worked Example: Matching an Aggregate Loss to a Compound Model

A hypothetical policy-year has a Poisson claim count with rate $\lambda=0.4$. Conditional on a claim, its positive payment is Gamma with shape 1 and scale \$5,000, independently across claims and independently of the count. Match the aggregate loss $S$ to a suitable family, calculate its zero probability, mean, and variance, and check the results from the compound components.

Step 1: Identify support and zero behavior. The aggregate is zero when the Poisson count is zero; otherwise it is a sum of positive payments. Hence its support is $[0,\infty)$, with a point mass at zero. A positive-severity Gamma model alone would not represent this aggregate outcome.

Step 2: Calculate the zero probability. The count has a Poisson distribution, so

P(S=0)=P(N=0)=e^{-0.4}\approx0.6703.

Step 3: Calculate the severity moments. A Gamma distribution with shape 1 and scale \$5,000 is exponential with mean $E[X]=\$5,000$ and variance $\operatorname{Var}(X)=\$5,000^2=\$25,000,000$. Thus $E[X^2]=\operatorname{Var}(X)+E[X]^2=\$25,000,000+\$25,000,000=\$50,000,000$.

Step 4: Calculate aggregate moments. For a compound Poisson sum, $E[S]=\lambda E[X]$ and $\operatorname{Var}(S)=\lambda E[X^2]$. Therefore,

E[S]=0.4(\$5{,}000)=\$2{,}000, \qquad \operatorname{Var}(S)=0.4(\$50{,}000{,}000) =\$20{,}000{,}000.

The Gamma severity shape is $a=1$, so the compound Poisson-Gamma aggregate is a Tweedie distribution with power $p=(a+2)/(a+1)=3/2$, within the compound range $1<p<2$. Its variance function is $\phi\mu^{3/2}$. Using $\mu=\$2,000$, the implied dispersion is $\phi=\$20,000,000/(\$2,000)^{3/2}\approx223.6068$ in the corresponding units.

Conclusion and check. The suitable match is a compound Poisson-Gamma, or Tweedie, aggregate model with $p=1.5$; it has zero probability 0.6703, mean \$2,000, and variance \$20,000,000. The moment check can also be made from $E[S]=\lambda E[X]$ and $\operatorname{Var}(S)=\lambda(\operatorname{Var}(X)+E[X]^2)$, which give the same values.

Common mistake avoided: Applying the Gamma severity variance formula directly to aggregate losses would ignore both the random claim count and the zero-claim probability.

Why the Compound Poisson Formulas Hold

The aggregate calculation follows from conditioning on the number of claims. Assume $N\sim\operatorname{Poisson}(\lambda)$, and let $X_1,X_2,\ldots$ be independent and identically distributed positive severities, independent of $N$. Define $S=\sum_{j=1}^{N}X_j$, with an empty sum equal to zero. Given $N=n$, the conditional mean is $nE[X]$ and the conditional variance is $n\operatorname{Var}(X)$.

The law of total expectation gives $E[S]=E[N]E[X]=\lambda E[X]$. The law of total variance gives

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N]\operatorname{Var}(X) +\operatorname{Var}(N)(E[X])^2\\ &=\lambda\operatorname{Var}(X)+\lambda(E[X])^2\\ &=\lambda E[X^2]. \end{aligned}

The last equality uses $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. Also, because every severity is positive, $S=0$ exactly when $N=0$, so $P(S=0)=e^{-\lambda}$. If payments can be zero, negative, or subject to a more complicated recording rule, that equivalence must be reconsidered before using the zero-probability formula.

Practical Checks and Exam Traps

Use the following sequence before selecting a family. Each step rules out mismatches that a moment comparison alone can miss.

Name the random outcome.
Decide whether each observation is a count, a positive severity, or an aggregate loss. State the unit and time period.

Check support and zeros.
Counts are integer-valued; positive severities are positive and continuous; aggregates can include an atom at zero. Verify what a recorded zero means.

Compare moment restrictions.
Check Poisson mean-variance equality, negative binomial overdispersion, or the appropriate severity variance scaling across comparable risks.

Check the whole description.
A family that matches support and moments is a candidate, not a proven fit. Consider whether its zero behavior and distributional shape are also reasonable.

Do not equate “has zeros” with “is negative binomial.” Poisson counts also allow zero; the mean-variance pattern and zero frequency provide additional evidence.
Do not apply a severity family to all policy-years without accounting for no claims. A positive severity is conditional on a claim; an aggregate includes the possibility of no claim.
Do not infer a family from one variance comparison alone. Sampling variation, different exposure levels, or changing dispersion can distort empirical moment patterns.
Do not call every zero an atom from the same mechanism. A zero count, a zero aggregate, and a zero recorded payment have different interpretations.

The matching discipline is useful in ratemaking because the selected response distribution determines what model assumptions are being made about frequency, severity, or total cost. This tutorial is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus; in applied work, candidate models also require validation beyond the three diagnostic clues used here.

Check Your Understanding

For each original scenario, identify the appropriate distributional clue and calculate the requested quantity. Answers are included for checking.

A hypothetical claim-count model is Poisson with mean 0.7 claims per exposure period. What is its variance and probability of zero claims? (Answer: variance $0.7$; $P(N=0)=e^{-0.7}\approx0.4966$.)
A negative binomial count has mean 2 and variance 6 under $\operatorname{Var}(N)=\mu+\alpha\mu^2$. Find $\alpha$ and its probability of zero claims. (Answer: $\alpha=(6-2)/2^2=1$; $P(N=0)=(1+1(2))^{-1}=0.3333$.)
Two positive-severity groups have means \$4,000 and \$8,000. A common Gamma dispersion is $\phi=0.25$. What are the variances, and does doubling the mean multiply variance by 4 or 8? (Answer: $\$4,000,000$ and $\$16,000,000$; the variance multiplies by 4, consistent with the Gamma square-mean pattern.)
A compound Poisson model has rate $\lambda=0.2$ per policy-year and exponential severity with mean \$10,000. Find the aggregate zero probability, mean, and variance. (Answer: $P(S=0)=e^{-0.2}\approx0.8187$; $E[S]=\$2,000$; $E[X^2]=2(\$10,000)^2=\$200,000,000$, so $\operatorname{Var}(S)=\$40,000,000$.)
A set of policy-year aggregate losses contains many zeros and otherwise positive continuous totals. Explain why a positive-severity Gamma model alone does not describe this outcome, and name a suitable compound model if counts are Poisson and severities are Gamma. (Answer: the severity model has no point mass at zero and omits the no-claim outcome; a compound Poisson-Gamma/Tweedie aggregate model is suitable.)

What You'll Learn