Dispersion Parameters And Scale

Why Dispersion Is a Separate Parameter

An exponential-family model has a parameter that determines its mean and, in many cases, another parameter that determines how spread out responses are around that mean. These roles are related but not interchangeable. In a generalized linear model, the natural parameter controls the mean through the cumulant function; the dispersion parameter scales the variance associated with that mean.

The distinction matters when interpreting a model and calculating uncertainty. For example, a Poisson model does not ordinarily estimate an additional dispersion parameter: its variance is tied to its mean. A Gaussian model usually has a dispersion parameter that must be estimated unless it is known in advance. This topic is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: In the exponential-family form $$ f(y\mid\theta,\phi)=\exp\left\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y,\phi)\right\}, $$ $\theta$ is the natural parameter, $\phi$ is the dispersion parameter, $b(\theta)$ is the cumulant function, and $a(\phi)$ is the scale factor in the exponent. A common convention is $a(\phi)=\phi$. Under this convention, the mean is $b'(\theta)$ and the variance is $\phi b''(\theta)$, when the relevant derivatives exist.

Some distributions have a fixed dispersion, commonly $\phi=1$, rather than a dispersion estimated from the data. Others have a free dispersion parameter. In either case, $b''(\theta)$ describes how the variance changes with the mean before the dispersion scale is applied. It is often expressed as a variance function, $V(\mu)$, where $\mu=E[Y]$. Thus the usual GLM variance statement is $\operatorname{Var}(Y)=\phi V(\mu)$.

How Dispersion Scales the Variance

The variance formula follows directly from the exponential-family probability form. Assume a regular exponential family with a moment-generating function in a neighborhood of zero. The part of the log moment-generating function that depends on $t$ is

\log E[e^{tY}] = \frac{b(\theta+a(\phi)t)-b(\theta)}{a(\phi)}.

Differentiating once with respect to $t$ and evaluating at zero gives $E[Y]=b'(\theta)$. Differentiating twice gives the variance:

\begin{aligned} \frac{d}{dt}\log E[e^{tY}] &=b'(\theta+a(\phi)t),\\ \left.\frac{d^2}{dt^2}\log E[e^{tY}]\right|_{t=0} &=a(\phi)b''(\theta) =\operatorname{Var}(Y). \end{aligned}

Because $\mu=b'(\theta)$, the second derivative can also be written as a function of the mean. If $V(\mu)=b''(\theta)$, then $\operatorname{Var}(Y)=a(\phi)V(\mu)$. With the common choice $a(\phi)=\phi$, this becomes $\phi V(\mu)$. The factor $V(\mu)$ determines the mean–variance relationship; dispersion multiplies that relationship rather than replacing it.

Key distinction: The natural parameter $\theta$ indexes the mean through $b'(\theta)$. The dispersion parameter $\phi$ scales the variance. A variance function $V(\mu)$ is not itself a dispersion parameter.

Fixed and Estimated Dispersion

Whether dispersion is fixed depends on the distributional model, not on a general rule that every exponential-family model must estimate $\phi$. In standard Bernoulli, binomial, and Poisson likelihood models, the dispersion is conventionally fixed at one. Their variance is determined by the mean and the distributional form. In Gaussian and gamma models, dispersion is commonly free and estimated, unless the model specifies it as known.

Response model	Typical variance statement	Usual dispersion treatment
Bernoulli with mean $\mu$	$\mu(1-\mu)$	Fixed at $\phi=1$
Binomial count with $n$ trials and success probability $p$	$np(1-p)$	Fixed at $\phi=1$
Poisson count with mean $\mu$	$\mu$	Fixed at $\phi=1$
Gaussian response with mean $\mu$	$\phi$	Usually estimated unless known
Gamma response with mean $\mu$	$\phi\mu^2$	Usually estimated unless known

The table uses the conventional unit-dispersion parameterization for Bernoulli, binomial, and Poisson models. A model that allows overdispersion may introduce an additional scale, but that is a different modeling assumption from the ordinary binomial or Poisson likelihood. In particular, observing variation in data does not by itself mean the standard model has a free dispersion parameter.

Worked Example: Estimating Gaussian Dispersion

Problem. A hypothetical Gaussian regression for annual repair cost has eight observations and three fitted regression coefficients, including the intercept. The residual sum of squares is $72$, in squared dollars. Under the usual Gaussian model with independent errors of common variance $\phi$, calculate the maximum-likelihood estimate of dispersion and the unbiased estimate of error variance.

Method and assumptions. Conditional on the predictors, assume each response is normally distributed with its regression mean and common variance $\phi$. The regression coefficients are estimated from the same eight observations. For this Gaussian model, the maximum-likelihood estimate divides the residual sum of squares by $n$; the unbiased estimate divides it by the residual degrees of freedom, $n-p$.

Here $n=8$, $p=3$, and the residual degrees of freedom are

$$ n-p=8-3=5. $$

Therefore, the maximum-likelihood estimate is

\widehat{\phi}_{\mathrm{ML}}=\frac{72}{8}=9\text{ dollars}^2.

The unbiased estimate of the error variance is

s^2=\frac{72}{5}=14.4\text{ dollars}^2.

The corresponding estimated standard deviation from the unbiased variance estimate is $\sqrt{14.4}\approx 3.7947$ dollars. The two variance estimates differ because their denominators differ; they are not two calculations of the same estimator. Check: multiplying $9$ by $8$ gives the original residual sum of squares, $72$, and multiplying $14.4$ by $5$ also gives $72$. The common mistake avoided is using $n-p$ for the Gaussian maximum-likelihood estimate or using $n$ for the usual unbiased estimate.

Worked Example: A Poisson Model Has Fixed Dispersion

Problem. In a hypothetical model, the number $N$ of emergency calls received by a service desk during one shift is Poisson with mean $\mu=2.4$. Find the dispersion under the standard Poisson model, the variance, and the standard deviation. Explain whether a separate dispersion estimate is part of this model.

Method and assumptions. Assume the shift count follows the standard Poisson distribution, with no extra scale parameter. For a Poisson response, $\phi=1$ and $V(\mu)=\mu$. Hence

\operatorname{Var}(N)=\phi V(\mu)=1(2.4)=2.4.

The standard deviation is

\operatorname{SD}(N)=\sqrt{2.4}\approx1.5492\text{ calls}.

The dispersion is fixed at one in this standard model; a separate dispersion estimate is not calculated. Check: the Poisson mean and variance are both $2.4$, as required. The common mistake avoided is treating $\mu$ as a dispersion estimate: here $\mu$ determines both the mean and the variance, while $\phi$ remains fixed.

Worked Example: Gamma Dispersion and Claim-Size Variance

Problem. A hypothetical model for positive repair costs uses a gamma response with conditional mean $\mu=\$500$ and dispersion $\phi=0.04$. Calculate its variance, standard deviation, and coefficient of variation. Under the gamma parameterization with shape $1/\phi$ and scale $\phi\mu$, verify the variance independently.

Method and assumptions. Assume the repair cost follows this gamma model and that the stated $\phi$ is the dispersion parameter, not the gamma scale parameter. The gamma variance function is $V(\mu)=\mu^2$, so

\operatorname{Var}(Y)=\phi\mu^2 =0.04(500)^2 =0.04(250{,}000) =\$10{,}000.

The standard deviation is $\sqrt{10{,}000}=\$100$, and the coefficient of variation is the standard deviation divided by the mean:

\operatorname{CV}=\frac{100}{500}=0.20.

For the independent parameterization check, the gamma shape is $1/\phi=1/0.04=25$, and the gamma scale is $\phi\mu=0.04(500)=\$20$. A gamma distribution with shape $k$ and scale $s$ has mean $ks$ and variance $ks^2$. Thus its mean is $25(20)=\$500$, while its variance is $25(20)^2=25(400)=\$10{,}000$, matching the variance-function calculation. Check: the coefficient of variation is also $1/\sqrt{25}=0.20$. The common mistake avoided is reporting the gamma scale, $\$20$, as the variance or as the dispersion; the variance is $\$10{,}000$, and the dispersion is $0.04$.

Interpreting Dispersion Carefully

Dispersion is not always a free parameter that can be estimated in the same way. In a Gaussian model, estimating the common error variance is a standard part of fitting the model. In many other GLMs, a Pearson-based estimate of dispersion may be used as a diagnostic or as part of a quasi-likelihood approach. Such an estimate does not change the fact that the ordinary binomial or Poisson likelihood fixes dispersion at one; it reflects a different modeling choice about variation.

For example, a standard Poisson model specifies $\operatorname{Var}(Y\mid X)=\mu$. If a proposed model instead allows $\operatorname{Var}(Y\mid X)=\phi\mu$ with a free $\phi$, then the variance is no longer tied to the mean in precisely the standard Poisson way. Keep the stated model in view before calculating or interpreting a dispersion value.

Exam trap: Do not confuse “fixed at one” with “variance equals one.” In a standard Poisson model, $\phi=1$, but $\operatorname{Var}(Y)=V(\mu)=\mu$, which need not equal one. Likewise, in a gamma model the variance is $\phi\mu^2$, not simply $\phi$.

A reliable solution sequence is to identify the distributional model first, write its variance as dispersion times variance function, then determine whether dispersion is fixed, known, or estimated. Only after those steps should a numerical variance or standard deviation be calculated.

Name the model.
For example, distinguish standard Poisson from a model that allows additional dispersion.

Separate the roles.
Use the natural parameter to determine the mean and the dispersion parameter to scale the variance.

Apply the variance function.
Calculate $\operatorname{Var}(Y)=\phi V(\mu)$ under the convention $a(\phi)=\phi$.

Check the dispersion treatment.
Determine whether the stated model fixes dispersion, supplies it as known, or requires an estimate.

Check Your Understanding

For each problem, identify the dispersion treatment before calculating any requested variance.

A standard Bernoulli response has mean $0.3$. State $\phi$ and calculate its variance. (Answer: $\phi=1$; variance $=0.3(0.7)=0.21$.)
A standard Poisson count has mean $4.5$. Find its variance and standard deviation to four decimal places. (Answer: variance $=4.5$; standard deviation $=\sqrt{4.5}=2.1213$.)
A Gaussian regression has $n=10$, $p=2$, and residual sum of squares $96$. Find the maximum-likelihood estimate of $\phi$ and the usual unbiased estimate of error variance. (Answer: $\widehat{\phi}_{\mathrm{ML}}=96/10=9.6$; unbiased estimate $=96/(10-2)=12$.)
A gamma response has mean $\$300$ and dispersion $0.09$. Find its variance and standard deviation. (Answer: variance $=0.09(300)^2=\$8{,}100$; standard deviation $=\$90$.)
A model states $\operatorname{Var}(Y\mid X)=\phi\mu$, with $\mu=6$ and a specified $\phi=1.5$. Calculate the variance and explain why it is not the standard Poisson variance. (Answer: variance $=1.5(6)=9$; standard Poisson variance would be $6$, since its dispersion is fixed at one.)

Response model	Typical variance statement	Usual dispersion treatment
Bernoulli with mean \(\mu\)	\(\mu(1-\mu)\)	Fixed at \(\phi=1\)
Binomial count with \(n\) trials and success probability \(p\)	\(np(1-p)\)	Fixed at \(\phi=1\)
Poisson count with mean \(\mu\)	\(\mu\)	Fixed at \(\phi=1\)
Gaussian response with mean \(\mu\)	\(\phi\)	Usually estimated unless known
Gamma response with mean \(\mu\)	\(\phi\mu^2\)	Usually estimated unless known

What You'll Learn