Dispersion Parameters and Variance Functions

Why Separate Dispersion From the Variance Function?

A model’s mean does not by itself determine how variable its responses are. Two responses with different means may also have different variances, and the rule linking mean to variance depends on the response distribution. In a generalized linear model, it is useful to distinguish that rule from a separate scale factor: the variance function describes how variance changes with the mean, while the dispersion parameter scales that variance.

This distinction is especially useful when comparing response families. A Normal response has variance that does not change with its mean, whereas a Poisson response has variance equal to its mean. A Binomial count has variance determined by both its mean and its number of trials. The earlier material on deriving the variance from the log-partition function supplies the key result: in exponential-family form, variance is the scale factor times the second derivative of the log-partition function.

Definition: For a response $Y$ with mean $\mu$, a variance function $V(\mu)$ and dispersion parameter $\phi$ are defined by $$ \operatorname{Var}(Y)=\phi V(\mu). $$ In the common exponential-family parameterization with scale $a(\phi)=\phi$, $V(\mu)$ gives the mean-dependent part of the variance and $\phi$ multiplies it. Some response families fix $\phi$ at 1; in other families it may be an unknown parameter. For a grouped Binomial response, the number of trials is also part of the variance function.

The notation $V(\mu)$ does not mean that every variance changes with the mean. A constant variance is represented by a constant variance function. Nor does $\phi$ always mean an extra quantity to estimate: in the standard Binomial and Poisson likelihoods, it is fixed at 1. The value of $\phi$ and the definition of $V$ must be read together, since only their product is the response variance.

From the Exponential-Family Variance to a Variance Function

Write a one-parameter exponential-family density or mass function in the form

f(y;\theta,\phi) = \exp\left\{ \frac{y\theta-b(\theta)}{a(\phi)}+c(y,\phi) \right\}.

The mean and variance results for this form are $E(Y)=b'(\theta)$ and $\operatorname{Var}(Y)=a(\phi)b''(\theta)$. When $a(\phi)=\phi$, the mean relation gives $\mu=b'(\theta)$. If this relation can be inverted to express $\theta$ in terms of $\mu$, define

V(\mu)=b''\bigl(\theta(\mu)\bigr).

Substituting into the variance result gives the desired decomposition:

\operatorname{Var}(Y) = \phi b''(\theta) = \phi V(\mu).

This is a re-expression of the variance result, not an additional assumption about a new distribution. It separates the family’s mean-to-variance relationship from a scale factor. If a family has $a(\phi)$ different from $\phi$, the same idea applies with $a(\phi)$ as the multiplying scale; the convention here uses $a(\phi)=\phi$ when displaying the familiar form.

Three Standard Variance Structures

The following forms use the standard Normal, Binomial, and Poisson likelihoods. Unless stated otherwise, a response is considered conditional on the covariates that determine its modeled mean. The formulas specify its variance under the stated family; they do not claim that every observed sample will display exactly that amount of scatter.

Response family	Mean	Variance function	Dispersion in the standard model
Normal $Y\sim N(\mu,\sigma^2)$	$\mu$	$V(\mu)=1$	$\phi=\sigma^2$
Binomial count $Y\sim\operatorname{Binomial}(m,p)$	$\mu=mp$	$V(\mu)=\mu(1-\mu/m)$	$\phi=1$
Poisson $Y\sim\operatorname{Poisson}(\mu)$	$\mu$	$V(\mu)=\mu$	$\phi=1$

Normal Responses: Constant Variance

For a Normal response $Y\sim N(\mu,\sigma^2)$, the variance is $\sigma^2$, regardless of the value of $\mu$. Thus the variance function is $V(\mu)=1$, and the dispersion parameter is $\phi=\sigma^2$. The decomposition is therefore $\operatorname{Var}(Y)=\sigma^2(1)$. The value $\sigma^2$ is measured in squared response units; the standard deviation $\sigma$ is in the original response units.

Worked Example: Normal Variance in Physical Units

Suppose a hypothetical model for package fill weight, measured in grams, specifies $Y\sim N(500,2.5^2)$. The model assumes a Normal response with mean 500 grams and standard deviation 2.5 grams.

The dispersion is the variance, $\phi=\sigma^2=(2.5\text{ g})^2=6.25\text{ g}^2$, and the variance function is $V(\mu)=1$. Therefore,

\operatorname{Var}(Y) = \phi V(\mu) = 6.25\text{ g}^2\times 1 = 6.25\text{ g}^2.

The mean of 500 grams does not enter the variance calculation. Under this model, a different specified mean would still have variance 6.25 grams squared, unless the model also changed its dispersion.

Binomial Responses: Trial Count Matters

For a Binomial count $Y\sim\operatorname{Binomial}(m,p)$, $m$ is the number of trials and $p$ is the success probability on each trial. The mean is $\mu=mp$, and the variance is $mp(1-p)$. Since $p=\mu/m$, substitute this expression into the variance:

\operatorname{Var}(Y) = mp(1-p) = \mu\left(1-\frac{\mu}{m}\right).

Thus, for a fixed trial count $m$, one may write $V(\mu)=\mu(1-\mu/m)$ and $\phi=1$. The factor $1-\mu/m$ reflects the fact that the expected count cannot exceed $m$. For fixed $m$, the variance is zero at $p=0$ and $p=1$, and is largest at $p=0.5$, where it equals $m/4$.

If the response is instead the sample proportion $\hat{p}=Y/m$, its mean is $p$ and its variance is $p(1-p)/m$. This follows by scaling the count: $\operatorname{Var}(Y/m)=\operatorname{Var}(Y)/m^2$. Count and proportion formulas describe different response scales, so the variance units and formula must match the response being modeled.

Worked Example: Variance of a Binomial Count and Proportion

In a hypothetical set of 12 independent trials, suppose each trial has success probability $p=0.25$, and let $Y$ be the number of successes. The Binomial model assumes the same success probability on each trial and independence of the trial outcomes.

The expected count is $\mu=mp=12(0.25)=3$. With $\phi=1$, the variance of the count is

\operatorname{Var}(Y) = \mu\left(1-\frac{\mu}{m}\right) = 3\left(1-\frac{3}{12}\right) = 3(0.75) = 2.25\text{ successes}^2.

Equivalently, $mp(1-p)=12(0.25)(0.75)=2.25$. For the sample proportion $Y/12$, the variance is $2.25/12^2=0.015625$, and its standard deviation is $\sqrt{0.015625}=0.125$, or 12.5 percentage points. These are variances of different response scales, not competing answers.

Poisson Responses: Variance Equals the Mean

For a Poisson response with mean $\mu$, the standard Poisson model has $\operatorname{Var}(Y)=\mu$. Its dispersion is fixed at $\phi=1$, and its variance function is $V(\mu)=\mu$. As the modeled mean increases, so does the model-implied variance. Because a Poisson response is a count, the variance is expressed in squared count units, while its standard deviation is in count units.

Worked Example: Poisson Variance for an Event Count

Suppose a hypothetical model specifies that the number $Y$ of events in a fixed interval follows a Poisson distribution with mean $\mu=4.2$ events. The Poisson assumption is the stated model for this calculation; it fixes the dispersion at 1.

The variance is

\operatorname{Var}(Y) = \phi V(\mu) = 1(4.2) = 4.2\text{ events}^2.

The corresponding standard deviation is $\sqrt{4.2}\approx 2.049$ events. The standard deviation is not 4.2 events: 4.2 is the variance in squared count units and is also the mean count.

Comparing the Roles of Mean and Dispersion

The Normal, Binomial, and Poisson models differ in which part of their variance changes. In the Normal model, the variance function is constant, so changes in variance come through changes in $\phi$. In the standard Binomial and Poisson models, $\phi$ is fixed at 1, and the variance is governed by the mean together with the family-specific variance function. This comparison is about the variance implied by each stated model, not a claim that the same model is suitable for every data set.

Identify the response and family.
Determine whether the modeled response is a Normal measurement, a Binomial count or proportion, or a Poisson count.

State the response scale and mean.
For a Binomial response, distinguish a count $Y$ from the proportion $Y/m$, and record the trial count $m$ when the response is a count.

Apply the family’s variance function and dispersion.
Use $\operatorname{Var}(Y)=\phi V(\mu)$, substituting the appropriate $\mu$, $m$, and $\phi$.

Check units and range.
A variance has squared response units. For a Binomial count, also check $0\leq\mu\leq m$, as required by its possible range.

Diagnostics, Limitations, and a Common Mistake

A variance function is a model specification, not a guarantee that the model fits. After fitting a model, assess whether residual variation is broadly compatible with the chosen mean-variance relationship, using appropriate residual diagnostics and the data’s structure. A single observed response cannot establish that its variance equals a theoretical value; variance is a distributional property assessed across replicated or comparable observations. In particular, an apparent mismatch may reflect a poorly specified mean, dependence between observations, or a response distribution that does not capture the observed variation.

The standard Poisson likelihood fixes $\phi=1$; the standard Binomial likelihood also fixes $\phi=1$. If observations show more variation than those models imply, simply changing the dispersion while continuing to describe the model as the same standard likelihood is not justified. Alternative approaches exist, but choosing among them requires additional modeling assumptions. Conversely, the Normal dispersion is generally not fixed at 1: it is the response variance, $\sigma^2$, in the stated parameterization.

Mistake: “The dispersion is the variance, so every response variance is $\phi$.”
Correction: The variance is $\phi V(\mu)$. This equals $\phi$ only when $V(\mu)=1$, as in the Normal example.
Mistake: “For a Binomial count, use $p(1-p)$ as its variance.”
Correction: For $m$ trials, the count variance is $mp(1-p)$. The formula $p(1-p)/m$ is the variance of the sample proportion, not the count.
Mistake: “A Poisson model has an estimated dispersion parameter that can be any value.”
Correction: In the standard Poisson likelihood, $\phi=1$. The variance is the mean because $V(\mu)=\mu$.

Key takeaway: First identify the response scale and family; then use the matching variance function and dispersion. Normal variance is $\sigma^2$, Binomial count variance is $mp(1-p)$, and standard Poisson variance is $\mu$.

Check Your Understanding

Answer each question, then compare with the solutions below.

A hypothetical Normal response has standard deviation 3 units. State its dispersion, variance function, and variance.
A Binomial count has $m=20$ trials and $p=0.10$. Calculate its mean and variance.
A Poisson response has mean 7.5 counts. State its dispersion and variance.
For the Binomial setting in question 2, calculate the variance of the sample proportion.
Explain why a variance of 2.25 for a Binomial count with mean 3 is not obtained by using $p(1-p)$ alone.

Solutions. (1) $\phi=\sigma^2=3^2=9$ units squared, $V(\mu)=1$, and $\operatorname{Var}(Y)=9(1)=9$ units squared. (2) $\mu=mp=20(0.10)=2$; $\operatorname{Var}(Y)=mp(1-p)=20(0.10)(0.90)=1.8$ counts squared. (3) The standard Poisson model fixes $\phi=1$, so $\operatorname{Var}(Y)=1(7.5)=7.5$ counts squared. (4) The proportion is $Y/20$, so its variance is $1.8/20^2=0.0045$. (5) $p(1-p)=0.25(0.75)=0.1875$ is the variance of one Bernoulli trial. The count sums 12 such trials, so its variance is $12(0.1875)=2.25$ successes squared.

Response family	Mean	Variance function	Dispersion in the standard model
Normal \(Y\sim N(\mu,\sigma^2)\)	\(\mu\)	\(V(\mu)=1\)	\(\phi=\sigma^2\)
Binomial count \(Y\sim\operatorname{Binomial}(m,p)\)	\(\mu=mp\)	\(V(\mu)=\mu(1-\mu/m)\)	\(\phi=1\)
Poisson \(Y\sim\operatorname{Poisson}(\mu)\)	\(\mu\)	\(V(\mu)=\mu\)	\(\phi=1\)