Why Mean and Variance Must Be Kept Separate
A response distribution describes more than the expected insurance outcome. It also describes how much individual outcomes vary around that expectation. In a ratemaking model, the mean is the level being predicted, while the variance describes the response’s variability at that level. A variance function connects the two; a dispersion parameter scales that relationship.
This distinction matters when comparing response distributions and interpreting fitted models. Two portfolios can have the same expected claim amount but different variability. Likewise, the same distribution family can imply different variances when its dispersion parameter changes. The calculations below use hypothetical data and standard exponential-dispersion parameterizations.
The variance and standard deviation are not interchangeable. Variance is measured in squared response units; standard deviation is measured in the response’s original units. When comparing variability, it is sometimes useful to calculate the coefficient of variation, \(CV=\sqrt{\operatorname{Var}(Y)}/\mu\), for a positive mean. It is a unit-free measure of standard deviation relative to the mean.
Where the Variance Function Comes From
For an exponential-dispersion model, write the density or probability function in the form
Here \(\theta\) is the canonical parameter, \(b\) is the cumulant function, and \(a(\phi)\) controls dispersion. The standard moment identities are \(E[Y]=b'(\theta)\) and \(\operatorname{Var}(Y)=a(\phi)b''(\theta)\). In the common scaling \(a(\phi)=\phi\), define \(V(\mu)=b''(\theta)\), expressing the second derivative as a function of the mean. Substitution gives \(\operatorname{Var}(Y)=\phi V(\mu)\).
This representation does not mean that \(\phi\) can be freely changed for every named probability distribution while preserving that exact distribution. For the standard Poisson distribution, \(\phi=1\). Gamma and inverse Gaussian models allow a positive dispersion parameter in their standard exponential-dispersion forms.
Poisson: Variance Equals the Mean
If \(Y\) has a Poisson distribution with parameter \(\mu>0\), then its mean is \(\mu\) and its variance is also \(\mu\). In the exponential-dispersion notation, its variance function is \(V(\mu)=\mu\), and its standard dispersion is \(\phi=1\). Consequently, a standard Poisson response has variance equal to its mean; it does not have a separate free dispersion parameter.
Worked Example: Total Count Across Two Independent Segments
A hypothetical insurer models annual claim counts for two separate policy segments. Assume \(N_A\sim\operatorname{Poisson}(1.4)\) and \(N_B\sim\operatorname{Poisson}(2.1)\), and assume the counts are independent. Calculate the mean, variance, standard deviation, and coefficient of variation of \(N_A+N_B\). Use the standard Poisson model, with dispersion one.
Method and assumptions. For each Poisson count, the variance equals its mean. Independence means variances add when counts are summed. The total \(N=N_A+N_B\) therefore has mean \(E[N]=1.4+2.1=3.5\).
The segment variances are \(\operatorname{Var}(N_A)=1.4\) and \(\operatorname{Var}(N_B)=2.1\). Therefore, \(\operatorname{Var}(N)=1.4+2.1=3.5\).
The standard deviation is \(\sqrt{3.5}=1.8708\) claims, to four decimal places. The coefficient of variation is \(1.8708/3.5=0.5345\), to four decimal places. As a check, the sum of independent Poisson counts is Poisson with parameter equal to the sum of the parameters, 3.5, which also has variance 3.5.
Common mistake avoided. The total’s standard deviation is not found by adding the segment standard deviations. Add the variances under independence, then take the square root.
Gamma: Variance Is Proportional to the Square of the Mean
For a positive Gamma response with mean \(\mu>0\) and dispersion \(\phi>0\), the variance function is \(V(\mu)=\mu^2\), so \(\operatorname{Var}(Y)=\phi\mu^2\). One standard parameterization uses a Gamma shape parameter \(k=1/\phi\) and scale parameter \(\theta=\phi\mu\). Under this parameterization, the mean is \(k\theta=\mu\), and the variance is \(k\theta^2=\phi\mu^2\). The response is positive; this parameterization does not describe a response with a point mass at zero.
Worked Example: Gamma Severity With a Given Dispersion
For a hypothetical positive claim severity \(Y\), suppose the mean is \(\mu=\$1{,}200\), and the Gamma dispersion is \(\phi=0.09\). Calculate the variance, standard deviation, Gamma shape and scale, and coefficient of variation. Assume the stated Gamma parameterization, with shape \(1/\phi\) and scale \(\phi\mu\).
Method. Apply the Gamma variance function \(V(\mu)=\mu^2\). First square the mean: \(\mu^2=1{,}200^2=1{,}440{,}000\) dollars squared. Then multiply by the dispersion: \(\operatorname{Var}(Y)=0.09(1{,}440{,}000)=129{,}600\) dollars squared.
The standard deviation is \(\sqrt{129{,}600}=\$360\). The shape is \(k=1/0.09=100/9\), approximately 11.1111. The scale is \(\theta=0.09(\$1{,}200)=\$108\).
Check the parameterization: \(k\theta=(100/9)(108)=\$1{,}200\), as specified. Also, \(k\theta^2=(100/9)(108^2)=(100/9)(11{,}664)=129{,}600\) dollars squared, matching the variance-function calculation. Finally, \(CV=360/1{,}200=0.30\), or 30%.
Common mistake avoided. The dispersion \(0.09\) is not the variance. The variance is \(0.09\) times the square of the mean, so it is \(129{,}600\) dollars squared.
Inverse Gaussian: Variance Is Proportional to the Cube of the Mean
For a positive inverse Gaussian response with mean \(\mu>0\) and dispersion \(\phi>0\), the variance function is \(V(\mu)=\mu^3\). Thus \(\operatorname{Var}(Y)=\phi\mu^3\). In the common inverse Gaussian parameterization using shape parameter \(\lambda>0\), the variance is \(\mu^3/\lambda\). These forms agree when \(\phi=1/\lambda\). The mean and dispersion must be expressed in compatible units; if \(Y\) is measured in dollars, \(\phi\) has units of inverse dollars and \(\lambda\) has units of dollars.
Worked Example: Inverse Gaussian Severity and Shape Parameter
A hypothetical positive claim severity \(Y\) is modeled as inverse Gaussian with mean \(\mu=\$800\) and dispersion \(\phi=0.0002\) per dollar. Calculate the variance, standard deviation, coefficient of variation, and the equivalent shape parameter \(\lambda\). Assume the inverse Gaussian exponential-dispersion parameterization just described.
Method. Cube the mean: \(\mu^3=800^3=512{,}000{,}000\) dollars cubed. Multiply by the dispersion: \(\operatorname{Var}(Y)=0.0002(512{,}000{,}000)=102{,}400\) dollars squared.
The standard deviation is \(\sqrt{102{,}400}=\$320\). The coefficient of variation is \(320/800=0.40\), or 40%. Since \(\phi=1/\lambda\), the equivalent shape is \(\lambda=1/0.0002=\$5{,}000\).
Check using the shape parameter: \(\mu^3/\lambda=512{,}000{,}000/5{,}000=102{,}400\) dollars squared, which agrees with the variance-function calculation.
Common mistake avoided. Do not use \(\lambda=\mu/\phi\). In this parameterization, dispersion and shape satisfy \(\phi=1/\lambda\); the mean is not part of that conversion.
Comparing the Three Variance Relationships
| Response family | Variance function \(V(\mu)\) | Variance | Dispersion convention |
|---|---|---|---|
| Poisson | \(\mu\) | \(\mu\) | Standard Poisson has \(\phi=1\) |
| Gamma | \(\mu^2\) | \(\phi\mu^2\) | \(\phi>0\) |
| Inverse Gaussian | \(\mu^3\) | \(\phi\mu^3\) | \(\phi>0\), with \(\phi=1/\lambda\) in the shape parameterization |
The powers of the mean have practical consequences. If the mean doubles while dispersion stays fixed, the Poisson variance doubles, the Gamma variance increases by a factor of four, and the inverse Gaussian variance increases by a factor of eight. This comparison concerns the variance implied by each family; it does not claim that every data set should follow one of these relationships.
Variance of a Total and a Frequent Calculation Trap
The variance function describes one response at a specified mean. For a total across responses, an additional assumption matters. If \(Y_1,\ldots,Y_n\) are independent, then \(\operatorname{Var}(\sum_jY_j)=\sum_j\operatorname{Var}(Y_j)\). Without independence, covariance terms are also present: \(\operatorname{Var}(\sum_jY_j)=\sum_j\operatorname{Var}(Y_j)+2\sum_{j<k}\operatorname{Cov}(Y_j,Y_k)\). It is therefore incorrect to add variances when dependence has not been ruled out or otherwise accounted for.
A related trap is confusing the sum of variances with the standard deviation of the sum. For example, suppose two independent quantities have variances 2.4 and 3.6. Their total variance is \(2.4+3.6=6.0\), and the standard deviation of their sum is \(\sqrt{6.0}\approx2.45\). The individual standard deviations are \(\sqrt{2.4}\) and \(\sqrt{3.6}\); adding those standard deviations does not give the standard deviation of the sum. The correction is to add variances first, then take the square root.
Check that the response is being modeled as Poisson, Gamma, or inverse Gaussian, and note any independence or support assumptions.
Use \(\mu\) for Poisson, \(\mu^2\) for Gamma, or \(\mu^3\) for inverse Gaussian.
Multiply by \(\phi\) for Gamma and inverse Gaussian; for the standard Poisson distribution, use \(\phi=1\).
Take the square root for standard deviation, and divide standard deviation by a positive mean for coefficient of variation.
Why This Matters in Ratemaking
A response mean alone does not describe the uncertainty of individual outcomes. The variance function specifies how that uncertainty changes with the mean, while dispersion controls its scale where the family permits a separate dispersion parameter. This distinction helps an actuary interpret a model’s response assumptions and check whether calculated variability has sensible units and magnitude.
A useful exam check is dimensional: if the response is measured in dollars, its variance must be in dollars squared and its standard deviation in dollars. A second check is parameterization: record whether a problem supplies dispersion \(\phi\) or a distribution-specific shape parameter such as Gamma shape \(k\) or inverse Gaussian shape \(\lambda\). Apply the conversion for the stated convention rather than treating parameter names as interchangeable.
Check Your Understanding
For each hypothetical problem, identify the applicable variance relationship before calculating.
- A standard Poisson response has mean 6.25. Find its variance and standard deviation. (Answer: variance \(=6.25\); standard deviation \(=2.5\).)
- A positive Gamma response has mean \(\$500\) and dispersion \(\phi=0.04\). Find its variance and standard deviation. (Answer: variance \(=\$10{,}000\) squared; standard deviation \(=\$100\).)
- An inverse Gaussian response has mean \(\$400\) and dispersion \(0.00125\) per dollar. Find its variance and equivalent shape parameter \(\lambda\). (Answer: variance \(=\$80{,}000\) squared; \(\lambda=\$800\).)
- Two independent hypothetical Gamma responses each have mean 10 and dispersion 0.2. Find the variance and standard deviation of their sum. (Answer: variance \(=40\); standard deviation \(=\sqrt{40}\approx6.3249\).)
- A hypothetical inverse Gaussian response has mean \(\$250\) and shape parameter \(\lambda=\$1{,}000\). Calculate dispersion, variance, and standard deviation. (Answer: \(\phi=0.001\) per dollar; variance \(=\$15{,}625\) squared; standard deviation \(=\$125\).)