Why Support Is the First Check
Before fitting a claim model, check what values the response can actually take. A claim count is a nonnegative integer; a single positive claim severity is a positive amount; and an aggregate loss may be zero or a positive amount. These are different kinds of outcomes, so a distribution that assigns probability to the wrong values is not an appropriate model for the response as recorded.
Support is a necessary screening tool, not a complete selection rule. For example, Gamma and inverse Gaussian distributions both describe strictly positive amounts, so observing positive severities alone does not distinguish between them. Distribution choice also depends on the data-generating process and how well candidate models describe the observed data.
Support of the Main Candidate Families
For count responses, the Poisson and negative binomial distributions have support \(0,1,2,\ldots\). A Poisson model with parameter \(\lambda>0\) has mean and variance \(\lambda\). One common negative binomial parameterization uses mean \(\mu>0\) and a nonnegative overdispersion parameter \(\alpha\), with variance \(\mu+\alpha\mu^2\). Both families can assign positive probability to zero, but the negative binomial allows variance greater than the mean when \(\alpha>0\).
For strictly positive continuous responses, the Gamma and inverse Gaussian distributions are common candidates. Under a shape-scale Gamma parameterization, shape \(a>0\) and scale \(b>0\) give mean \(ab\) and variance \(ab^2\). Under an inverse Gaussian parameterization with mean \(\mu>0\) and shape parameter \(\lambda>0\), the variance is \(\mu^3/\lambda\). Both distributions have support \(y>0\); neither gives an exact zero positive probability.
A Tweedie model is part of a family of exponential-dispersion models whose variance function is proportional to \(\mu^p\). In the compound Poisson-Gamma range \(1<p<2\), its response can be zero or positive: it represents a Poisson number of positive Gamma-distributed components summed to form an aggregate. At the boundaries commonly used in ratemaking, \(p=1\) corresponds to a Poisson variance function and \(p=2\) to a Gamma variance function; the inverse Gaussian has \(p=3\). The support and interpretation still depend on the response being modeled.
| Response being modeled | Possible values | Candidate families by support | Important limitation |
|---|---|---|---|
| Claim count | 0, 1, 2, ... | Poisson; negative binomial | A continuous severity model does not describe integer counts. |
| Single positive claim severity | Positive amounts only | Gamma; inverse Gaussian | Neither ordinary family allows an exact zero. |
| Aggregate loss with possible no-loss years | Zero or positive amounts | Compound Poisson-Gamma Tweedie may be suitable | Zeros alone do not establish the compound model’s assumptions. |
The response definition matters as much as the product. “Claim severity” might mean the amount conditional on a claim occurring, while “annual loss per policy” might include years with no claims. Those responses have different supports, even if both are recorded in dollars.
Worked Example: Screening Count Models
Worked Example: Screening Count Models
A hypothetical book has eight policy-year claim counts: 0, 0, 0, 1, 1, 2, 4, and 4. Calculate the sample mean and sample variance, then identify what support can and cannot tell you about Poisson versus negative binomial models.
Method. Treat each observation as a nonnegative integer count. Both candidate families support zero and all positive integer counts, so first use the moments only as a descriptive check on whether the Poisson mean-variance relationship appears plausible.
Step 1: Calculate the sample mean. The total count is \(0+0+0+1+1+2+4+4=12\). With \(n=8\), the sample mean is \(12/8=1.5\) claims per policy-year.
Step 2: Calculate the sample variance. The sum of squared counts is \(0+0+0+1+1+4+16+16=38\). Using the sample-variance formula, \(s^2=[38-8(1.5)^2]/(8-1)=(38-18)/7=20/7\approx2.8571\).
Step 3: Compare with the Poisson benchmark. A Poisson model fitted by the sample mean has \(\lambda=1.5\) and model variance \(1.5\). The observed sample variance, approximately \(2.8571\), exceeds that benchmark. A negative binomial is a reasonable candidate to investigate for overdispersion, but this small sample does not prove that it fits better.
Step 4: Check support. Every observation is a nonnegative integer, including the zeros. Both Poisson and negative binomial are compatible with that support. If, only for illustration, the negative binomial variance formula \(\mu+\alpha\mu^2\) is matched to these sample moments, then \(\alpha=(20/7-1.5)/(1.5)^2=(19/14)/(9/4)=38/63\approx0.6032\). This is a moment-based descriptive value, not proof that the negative binomial is the right model.
Check and common mistake. The calculated mean is 1.5 and the sample variance exceeds it, as stated. A common mistake is to select the negative binomial just because the sample variance is larger. The correction is to treat that comparison as evidence to assess, not as a support-based decision or a guarantee of fit.
Positive Amounts and the Limits of Support
When each response is a single observed positive severity, Gamma and inverse Gaussian models are both compatible with the support. The support check can rule out count distributions, but it cannot determine which positive continuous family is preferable. Other evidence, including the observed shape and the model’s performance, is needed.
Worked Example: Choosing Among Positive-Severity Candidates
A hypothetical insurer models four individual claim severities, recorded in dollars: 400, 600, 800, and 1,200. The candidate Gamma model has shape \(a=4\) and scale \(b=187.50\). The candidate inverse Gaussian model has mean \(\mu=750\) and shape \(\lambda=3{,}000\). Use the observations and parameterizations to screen the candidates by support and compare their means and variances.
Method. Check the observed values against the support first, then use the stated parameters to compare the candidate moments. Do not use a moment match as proof of distributional fit.
Step 1: Check the observed values. All four severities are strictly positive amounts. Poisson and negative binomial models are unsuitable because they describe integer counts, not dollar severities. Both the Gamma and inverse Gaussian candidates allow every observed value.
Step 2: Calculate the Gamma moments. Its mean is \(ab=4(187.50)=750\) dollars. Its variance is \(ab^2=4(187.50)^2=4(35{,}156.25)=140{,}625\) dollars squared.
Step 3: Calculate the inverse Gaussian moments. Its mean is the stated \(\mu=750\) dollars. Its variance is \(\mu^3/\lambda=750^3/3{,}000=421{,}875{,}000/3{,}000=140{,}625\) dollars squared.
Step 4: Interpret the result. The two candidates have the same mean and variance and both have compatible support. Support and these two moments do not distinguish them. Further assessment of their distributional shapes and fit is needed.
Check and common mistake. Both calculations give a mean of 750 and variance of 140,625, and all observed severities are positive. A common mistake is to claim Gamma is preferred because the observations are positive. The correction is that inverse Gaussian also has strictly positive support; these facts alone do not choose between them.
Why a Compound Tweedie Model Can Include Zero
For an annual aggregate, a zero can arise because there were no claims, while a positive outcome is the sum of one or more positive claim amounts. In a compound Poisson-Gamma model, let \(N\) be a Poisson claim count and let each claim amount be positive and Gamma-distributed. The aggregate \(S\) is the sum of the \(N\) claim amounts, with \(S=0\) when \(N=0\). This construction explains the zero mass directly.
Worked Example: Calculating the Zero Mass of an Aggregate
In a hypothetical annual-loss model, the number of claims \(N\) has a Poisson distribution with parameter \(\lambda=0.5\). Conditional on a claim occurring, each claim amount \(Z\) has a Gamma distribution with shape \(a=2\) and scale \(b=1{,}000\) dollars. Assume claim amounts are independent of one another and of \(N\), and define annual aggregate loss \(S\) as their sum. Calculate \(P(S=0)\), the probability of a positive aggregate, and the mean and variance of \(S\).
Method. Use the compound Poisson construction. A zero aggregate occurs exactly when the claim count is zero; aggregate moments follow from the Poisson count and the severity’s first two moments.
Step 1: Calculate the zero probability. Since \(N\sim\operatorname{Poisson}(0.5)\), \(P(S=0)=P(N=0)=e^{-0.5}\approx0.6065\).
Step 2: Calculate the positive probability. The aggregate is positive whenever at least one claim occurs, so \(P(S>0)=1-P(S=0)=1-0.6065=0.3935\), to four decimal places.
Step 3: Calculate the severity moments. For the stated Gamma shape-scale parameters, \(E[Z]=ab=2(1{,}000)=2{,}000\) dollars and \(\operatorname{Var}(Z)=ab^2=2(1{,}000)^2=2{,}000{,}000\) dollars squared. Therefore \(E[Z^2]=\operatorname{Var}(Z)+(E[Z])^2=2{,}000{,}000+4{,}000{,}000=6{,}000{,}000\) dollars squared.
Step 4: Calculate the aggregate moments. Conditional on \(N\), the aggregate is a sum of \(N\) independent severities. Hence \(E[S]=E[N]E[Z]=0.5(2{,}000)=1{,}000\) dollars. For a compound Poisson sum, \(\operatorname{Var}(S)=\lambda E[Z^2]\), so \(\operatorname{Var}(S)=0.5(6{,}000{,}000)=3{,}000{,}000\) dollars squared.
Check and common mistake. The probabilities of zero and positive aggregate sum to \(0.6065+0.3935=1.0000\). A common mistake is to model \(S\) with an ordinary Gamma distribution because positive claim amounts are Gamma-distributed. The correction is that the response here is an annual sum that can be zero; the compound model includes that zero mass, unlike an ordinary Gamma model.
Diagnose the Source of Zeros
A zero in a data column does not, by itself, establish that the response follows a compound Poisson-Gamma process. First determine what the recorded zero means. It may be a genuine no-claim outcome, a payment below a deductible, a rounded small value, or a code for missing information. Those mechanisms imply different distributions for the observed response.
For example, if a single reported claim payment is zero because the claim falls below a deductible, the recorded payment is not an ordinary strictly positive severity. It has a point mass at zero induced by the contract, with positive payments above zero. That does not automatically make it a Tweedie response: the compound Poisson-Gamma justification specifically concerns a Poisson number of positive components with Gamma severities. Model the response and its recording mechanism, rather than choosing a family solely because zeros appear.
Worked Example: A Zero Created by a Deductible
A hypothetical covered loss \(X\) for a reported claim is uniformly distributed between 0 and 2,000 dollars. The policy has a 500-dollar deductible, and the recorded insurer payment is \(Y=\max(X-500,0)\). Calculate the probability of a zero payment and the expected payment, then assess whether an ordinary Gamma or inverse Gaussian model describes \(Y\) as recorded.
Method. Translate the payment rule into regions of the underlying loss. The payment is zero for losses at or below the deductible and positive above it.
Step 1: Calculate the zero probability. Under the uniform distribution, the interval from 0 to 500 has length 500 out of a total length of 2,000. Thus \(P(Y=0)=P(X\leq500)=500/2{,}000=0.25\).
Step 2: Calculate the expected payment. The payment is \(x-500\) for \(500<x\leq2{,}000\), and zero otherwise. Therefore \(E[Y]=(1/2{,}000)\int_{500}^{2{,}000}(x-500)\,dx\). The integral is the area of a triangle with base 1,500 and height 1,500, or \((1/2)(1{,}500)(1{,}500)=1{,}125{,}000\). Dividing by 2,000 gives \(E[Y]=562.50\) dollars.
Step 3: Assess support. The recorded response has probability 0.25 at zero and otherwise takes positive values up to 1,500 dollars. Neither ordinary Gamma nor ordinary inverse Gaussian can represent its zero probability.
Check and common mistake. The zero payment probability is 0.25, and the expected payment is 562.50 dollars. A common mistake is to treat every zero-and-positive response as a standard Tweedie aggregate. The correction is to recognize that these zeros arise from a deductible applied to a single reported loss; the compound Poisson-Gamma assumption has not been established.
A Practical Screening Sequence
Use support to eliminate incompatible candidates before comparing distributional fit. State the response unit and recording rule first; then classify the response as a count, a strictly positive amount, or an amount that may be zero. For zero-inclusive amounts, investigate the mechanism behind the zero before settling on a family.
Specify whether each observation is a count, an individual severity, a payment, or an aggregate over a defined period.
Integer counts point toward Poisson or negative binomial support; strictly positive continuous amounts are compatible with Gamma or inverse Gaussian support.
Determine whether they represent no claims, a deductible or threshold, rounding, or missing data. Do not treat these mechanisms as interchangeable.
A compound Poisson-Gamma Tweedie model can be a candidate for zero-and-positive aggregate losses when its process assumptions are appropriate. Support does not by itself establish that model or distinguish families with the same support.
For exam problems aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus, keep the response definition in view. A correct model choice must match both the values observed and the process that produces them. Do not confuse the support of a component, such as a positive claim severity, with the support of a sum that includes years with no claims.
Check Your Understanding
For each scenario, identify the response support and state what the support check does and does not justify.
- A hypothetical insurer records annual claim counts of 0, 1, 0, 2, 3, and 0. Name the compatible count families and state whether the zeros rule out either one. (Answer: Nonnegative-integer support is compatible with both Poisson and negative binomial; zeros rule out neither.)
- A severity model is fitted to individual claim amounts, each strictly greater than zero. Can support alone distinguish Gamma from inverse Gaussian? (Answer: No; both have strictly positive support, so support alone cannot distinguish them.)
- A hypothetical annual aggregate loss is zero when no claims occur and positive otherwise; claim count is Poisson and individual severities are positive Gamma variables. Which family range is a natural candidate, and why can the aggregate be zero? (Answer: Compound Poisson-Gamma Tweedie with \(1<p<2\); the aggregate is zero exactly when the Poisson claim count is zero.)
- A database records a zero for every reported claim payment when the underlying loss is no greater than the deductible. Is the observed payment an ordinary positive Gamma response? State the key support issue. (Answer: No; the recorded payment has a point mass at zero, which an ordinary Gamma model cannot represent.)
- A policy-level annual response contains zeros and positive amounts, but the analyst does not know whether the zeros are true no-claim years or missing-value codes. Is a Tweedie choice justified from support alone? (Answer: No; the zero mechanism must be identified before the compound Tweedie assumptions can be assessed.)