Why Dispersion Is Not the Mean
In a ratemaking model, the mean describes the expected outcome at the modeled risk characteristics. Dispersion describes how widely individual outcomes vary around that mean. These are different features: holding the mean fixed, changing dispersion changes variability, not the expected value. Confusing them can lead to incorrect predictions of risk variability or misleading comparisons between candidate distributions.
The word dispersion does not mean exactly the same parameter in every family. In some distributions it is a separate multiplier of a variance function; in others it is represented by a shape or size parameter. The parameterization also matters: the negative binomial, for example, is written with either a size parameter or its reciprocal, an overdispersion parameter.
Throughout this tutorial, \(Y\) denotes a response and \(\mu=E[Y]\) its mean. No interest or mortality assumptions are involved. The results concern probability distributions for insurance outcomes, not the discounting or survival of cash flows.
A General Mean–Variance Framework
Many generalized linear model response distributions can be expressed as exponential dispersion models. In one common form, their density or probability mass function is proportional to \(\exp\{[y\theta-b(\theta)]/a(\phi)+c(y,\phi)\}\). Here \(\theta\) is a canonical parameter and \(\phi\) is a dispersion parameter. For this form, the mean and variance are \(E[Y]=b'(\theta)\) and \(\operatorname{Var}(Y)=a(\phi)b''(\theta)\). When \(a(\phi)=\phi\), this is commonly written \(\operatorname{Var}(Y)=\phi V(\mu)\), where \(V(\mu)\) is the family’s variance function.
This relationship explains the role of dispersion: at fixed \(\mu\), changing \(\phi\) multiplies the variance function, rather than changing the mean. Not every family has an estimated, freely varying \(\phi\). In particular, the ordinary Poisson model fixes its dispersion at 1; the negative binomial instead includes an additional variance term.
The formulas below use specific, stated parameterizations. In practice, always check the parameterization being used before interpreting a reported “size,” “shape,” or “dispersion” estimate.
Poisson and Negative Binomial Counts
For a Poisson count \(Y\sim\operatorname{Poisson}(\mu)\), the mean and variance are both \(\mu\). The ordinary Poisson distribution has no separate dispersion parameter: once \(\mu\) is specified, its variance is specified as well. A count with a larger mean therefore also has a larger variance under this model.
For the negative binomial, use the NB2 parameterization \(Y\sim\operatorname{NB}(\mu,r)\), where \(\mu>0\) is the mean and \(r>0\) is the size parameter. Equivalently, define \(\alpha=1/r\) as the overdispersion parameter. The mean and variance are
The extra term is nonnegative, so this NB2 variance is at least the Poisson variance at the same mean. Larger \(\alpha\), or equivalently smaller \(r\), means greater variance. Other negative binomial parameterizations exist; do not assume their parameter called “size” or “dispersion” has the same formula.
Worked Example: Comparing Poisson and Negative Binomial Count Variability
A hypothetical policy segment has a modeled annual claim-count mean of 6. Compare a Poisson model with an NB2 model having size \(r=4\). For the negative binomial, use the convention that \(P(Y=y)=\binom{y+r-1}{y}p^r(1-p)^y\), with \(p=r/(r+\mu)\). Find the variance and standard deviation under each model. No interest or mortality assumption applies.
Method and parameters. Under the Poisson assumption, \(Y\sim\operatorname{Poisson}(6)\). Under the NB2 assumption, \(\mu=6\), \(r=4\), and \(\alpha=1/r=0.25\). The stated probability convention gives \(p=4/(4+6)=0.4\), consistent with mean \(r(1-p)/p=4(0.6)/0.4=6\).
Poisson calculation. The variance equals the mean: \(\operatorname{Var}(Y)=6\). The standard deviation is \(\sqrt{6}=2.4495\), rounded to four decimal places.
Negative binomial calculation. \(\operatorname{Var}(Y)=6+6^2/4=6+36/4=6+9=15\). The standard deviation is \(\sqrt{15}=3.8730\), rounded to four decimal places.
Check and trap avoided. Both models have the same mean, 6, while the NB2 variance, 15, exceeds the Poisson variance, 6. A common error is to use \(r\) as if it were \(\alpha\), calculating \(6+4(36)=150\). The correction is \(\alpha=1/r=0.25\), so the correct variance is 15.
Gamma and Inverse Gaussian Dispersion
For a Gamma response parameterized by its mean \(\mu>0\) and dispersion \(\phi>0\), use \(\operatorname{Var}(Y)=\phi\mu^2\). In the corresponding shape-scale parameterization, the shape is \(1/\phi\) and the scale is \(\phi\mu\). Thus, at fixed mean, larger \(\phi\) means greater variance and a smaller shape parameter. The coefficient of variation is \(\sqrt{\operatorname{Var}(Y)}/\mu=\sqrt{\phi}\).
For an inverse Gaussian response with mean \(\mu>0\) and dispersion \(\phi>0\), the variance is \(\phi\mu^3\). In the conventional inverse Gaussian parameterization with mean \(\mu\) and shape \(\lambda\), \(\phi=1/\lambda\), so the variance is \(\mu^3/\lambda\). Consequently, larger \(\phi\), or smaller \(\lambda\), means greater variability at fixed mean.
A Gamma dispersion and an inverse Gaussian dispersion are not directly comparable just because their numerical values look similar. Their variance functions differ: one multiplies \(\mu^2\), the other \(\mu^3\). Their units also differ when \(Y\) is measured in dollars.
Worked Example: Reading Gamma and Inverse Gaussian Shape
Two hypothetical models describe positive claim severities, each with mean \(\mu=\$100\). The Gamma model has dispersion \(\phi=0.04\); the inverse Gaussian model has shape \(\lambda=2{,}500\). Calculate each model’s variance and standard deviation, and verify the implied shape or dispersion conversion. No interest or mortality assumption applies.
Gamma model. \(Y\) is Gamma with mean \(\mu=100\) and dispersion \(\phi=0.04\). Its shape is \(1/\phi=1/0.04=25\), and its scale is \(\phi\mu=0.04(100)=4\). Therefore, \(\operatorname{Var}(Y)=0.04(100^2)=0.04(10{,}000)=400\) dollars squared. Its standard deviation is \(\sqrt{400}=\$20\).
Inverse Gaussian model. \(Y\) is inverse Gaussian with mean \(\mu=100\) and shape \(\lambda=2{,}500\). The equivalent dispersion is \(\phi=1/\lambda=1/2{,}500=0.0004\). Therefore, \(\operatorname{Var}(Y)=100^3/2{,}500=1{,}000{,}000/2{,}500=400\) dollars squared. Its standard deviation is \(\sqrt{400}=\$20\).
Check and trap avoided. Both models produce the same mean and variance for these parameter choices, but that does not make the distributions identical. A common error is to apply the Gamma variance formula \(\phi\mu^2\) to the inverse Gaussian. The inverse Gaussian formula uses \(\phi\mu^3\), or equivalently \(\mu^3/\lambda\); the specified \(\lambda\) verifies the computed variance.
Tweedie Dispersion and Variance Power
A Tweedie model is commonly parameterized by a mean \(\mu>0\), a dispersion \(\phi>0\), and a variance power \(p\), with \(\operatorname{Var}(Y)=\phi\mu^p\). Here \(\phi\) scales the variance at a fixed mean and power, while \(p\) determines how the variance changes with the mean. They play distinct roles. The power is not an alternative name for dispersion.
For \(1<p<2\), the Tweedie distribution has a compound Poisson-Gamma form and can represent nonnegative outcomes with a point mass at zero and positive continuous values. In this range, changing \(\phi\) while holding \(\mu\) and \(p\) fixed changes the variability and the distributional components; it need not leave the probability of zero unchanged. It nevertheless leaves the specified mean \(\mu\) unchanged. The variance formula alone does not determine every feature of the distribution.
The numerical value of \(\phi\) depends on the units of \(Y\) and the selected power. Therefore, do not compare Tweedie dispersion values across models without checking both the outcome units and \(p\). Compare the implied variance at a relevant mean instead.
Worked Example: Separating Tweedie Dispersion From Power
A hypothetical aggregate-loss response is modeled as Tweedie with mean \(\mu=100\) dollars and power \(p=1.5\). Calculate the variance and standard deviation for dispersion \(\phi=0.02\), then calculate the variance if the dispersion is increased to \(0.08\) without changing the mean or power. No interest or mortality assumption applies.
Initial model. The distribution is Tweedie with \(\mu=100\), \(p=1.5\), and \(\phi=0.02\). Since \(100^{1.5}=100\sqrt{100}=1{,}000\), \(\operatorname{Var}(Y)=0.02(1{,}000)=20\) dollars squared. The standard deviation is \(\sqrt{20}=\$4.4721\), rounded to four decimal places.
Increased dispersion. With \(\mu=100\) and \(p=1.5\) unchanged, but \(\phi=0.08\), the variance is \(0.08(1{,}000)=80\) dollars squared. The standard deviation is \(\sqrt{80}=\$8.9443\), rounded to four decimal places. The mean remains \(\$100\).
Check and trap avoided. The dispersion has been multiplied by \(0.08/0.02=4\), and the variance has also increased by a factor of \(80/20=4\), as the variance formula requires. A common error is to treat \(p=1.5\) as the dispersion or to change the mean when \(\phi\) changes. Here, \(p\) sets the power of the mean, \(\phi\) scales the variance, and \(\mu\) remains the expected value.
How to Interpret a Reported Parameter
When a model output reports a parameter called “dispersion,” “shape,” or “size,” first identify its family and exact parameterization. Then write the mean and variance formulas before interpreting the estimate. The following summary uses the conventions established above.
| Family and parameterization | Mean | Variance | Effect of greater dispersion |
|---|---|---|---|
| Poisson with mean \(\mu\) | \(\mu\) | \(\mu\) | No free dispersion parameter in the ordinary model |
| NB2 with size \(r\), or \(\alpha=1/r\) | \(\mu\) | \(\mu+\mu^2/r=\mu+\alpha\mu^2\) | Greater \(\alpha\), or smaller \(r\), increases variance |
| Gamma with dispersion \(\phi\) | \(\mu\) | \(\phi\mu^2\) | Greater \(\phi\) increases variance; shape \(1/\phi\) decreases |
| Inverse Gaussian with dispersion \(\phi=1/\lambda\) | \(\mu\) | \(\phi\mu^3=\mu^3/\lambda\) | Greater \(\phi\), or smaller \(\lambda\), increases variance |
| Tweedie with dispersion \(\phi\) and power \(p\) | \(\mu\) | \(\phi\mu^p\) | At fixed \(p,\mu\), greater \(\phi\) increases variance |
A Poisson model is sometimes extended using a quasi-likelihood or quasi-Poisson variance relation with an extra scale multiplier. That extension permits a variance different from \(\mu\), but it is not the ordinary Poisson probability distribution. Keep that distinction clear when interpreting model assumptions.
Common Interpretation Traps
- Assuming every family has a free dispersion parameter. Ordinary Poisson fixes \(\operatorname{Var}(Y)=\mu\). Do not attach an estimated scale parameter unless the model specification allows it.
- Confusing size with overdispersion. In NB2, \(r\) and \(\alpha\) are reciprocals. Increasing \(r\) reduces the extra variance; increasing \(\alpha\) increases it.
- Assuming a larger shape means more variability. In the Gamma and inverse Gaussian conventions here, shape is the reciprocal of dispersion. Larger shape means smaller variance at fixed mean.
- Comparing dispersion across different variance functions. A Gamma \(\phi\) and an inverse Gaussian \(\phi\) do not multiply the same power of \(\mu\). Compare the implied variance at the relevant mean, not the raw parameter values.
- Calling the Tweedie power a scale parameter. The power \(p\) determines the mean exponent in the variance function; \(\phi\) is the multiplier.
For ratemaking, these distinctions matter because the mean often drives the indicated expected cost, while variability affects uncertainty and the distribution of outcomes. The variance formula is a statement about the assumed response distribution conditional on its mean; it does not by itself determine a rating factor or premium. This tutorial is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
Check Your Understanding
For each problem, identify which parameter controls the mean and calculate the requested variability under the stated parameterization.
- A hypothetical claim-count model has mean \(\mu=8\) and uses NB2 size \(r=2\). Find \(\alpha\), the variance, and the standard deviation. (Answer: \(\alpha=0.5\), variance \(40\), standard deviation \(\sqrt{40}\approx6.3249\); the mean is 8.)
- A hypothetical Gamma severity model has mean \(\$250\) and dispersion \(\phi=0.16\). Find its shape, scale, variance, and coefficient of variation. (Answer: shape \(6.25\), scale \(\$40\), variance \(\$10{,}000\), coefficient of variation \(0.4\).)
- A hypothetical inverse Gaussian severity model has mean \(\$40\) and shape \(\lambda=800\). Find its dispersion and variance. (Answer: \(\phi=1/800=0.00125\), variance \(40^3/800=\$80\) squared.)
- A hypothetical Tweedie model has mean \(\$64\), power \(p=1.5\), and dispersion \(\phi=0.05\). Find its variance and standard deviation; then find the variance if \(\phi\) doubles while the other parameters remain fixed. (Answer: variance \(25.6\) dollars squared and standard deviation \(\sqrt{25.6}\approx\$5.0596\); doubled-dispersion variance \(51.2\) dollars squared.)
- At mean \(\mu=12\), compare an ordinary Poisson count with an NB2 count of size \(r=6\). Calculate both variances and their ratio. (Answer: Poisson variance \(12\), NB2 variance \(12+144/6=36\), and NB2-to-Poisson variance ratio \(3\); both means are 12.)