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Mean Variance and Dispersion · Tutorial 54 of 981

Dispersion Conventions Across Response Families

Learn how the dispersion convention differs across common GLM response families, and how it affects the interpretation of their variances.

Beginner 10 min read

What You'll Learn

  • Distinguish a fixed unit dispersion from a response variance of one
  • Identify the conventional dispersion for Bernoulli, binomial, and Poisson responses
  • Identify the common dispersion parameter for normal, gamma, and inverse-Gaussian responses
  • Translate between binomial counts and proportions without mistaking trial counts for estimated dispersion
  • Check family variance formulas using explicit parameter values

Why Dispersion Conventions Matter

A response variance in a generalized linear model has two parts: a family-specific variance function and a dispersion multiplier. The multiplier describes the scale of variation around the mean, but its treatment differs across response families. In the conventional GLM specification, Bernoulli, binomial, and Poisson responses have dispersion fixed at one. Normal, gamma, and inverse-Gaussian responses instead have a common dispersion parameter that is ordinarily estimated from data.

“Fixed at one” does not mean that the response variance equals one. It means that the family’s variance function is used without an additional estimated multiplier. For example, a Bernoulli response with mean \(0.3\) has variance \(0.3(1-0.3)=0.21\), despite having unit dispersion. Keeping this distinction clear prevents confusion between a variance function and a dispersion convention.

Definition: In the mean-variance representation \(\operatorname{Var}(Y)=\phi V(\mu)\), \(\mu=E[Y]\) is the response mean, \(V(\mu)\) is the family’s variance function, and \(\phi\) is the dispersion parameter. A fixed unit dispersion convention sets \(\phi=1\) as part of the response-family specification. An estimated common dispersion convention treats \(\phi\) as an unknown shared parameter for responses in that model.

The word “common” refers to a single dispersion value shared across the response observations in the model; it does not mean that their variances are identical. If their means differ, their variance functions can differ too. For instance, with a common \(\phi\), a gamma response with mean \(\mu_i\) has variance \(\phi\mu_i^2\), which varies with \(\mu_i\).

The Conventional Family-by-Family Comparison

The table gives the standard mean-variance forms and dispersion conventions. For a binomial count, \(m\) is its known number of trials. A Bernoulli response is the special case \(m=1\).

Response familyMean-variance formConventional dispersion
Bernoulli\(\operatorname{Var}(Y)=\mu(1-\mu)\)Fixed at \(\phi=1\)
Binomial count with \(m\) trials\(\operatorname{Var}(Y)=\mu(1-\mu/m)\)Fixed at \(\phi=1\)
Poisson\(\operatorname{Var}(Y)=\mu\)Fixed at \(\phi=1\)
Normal\(\operatorname{Var}(Y)=\phi\)Common \(\phi\), ordinarily estimated
Gamma\(\operatorname{Var}(Y)=\phi\mu^2\)Common \(\phi\), ordinarily estimated
Inverse-Gaussian\(\operatorname{Var}(Y)=\phi\mu^3\)Common \(\phi\), ordinarily estimated

In each row, the displayed variance can be read as \(\phi V(\mu)\). For the first three families the conventional value is \(\phi=1\), so the variance function itself gives the variance. For the last three, \(\phi\) remains in the variance expression. These are standard modeling conventions for these response families, not a claim that no alternative model can ever introduce additional variation.

Worked Example: Bernoulli and Binomial Dispersion

Suppose a Bernoulli response has success probability \(p=0.3\). Its mean is \(\mu=p=0.3\), and its variance is \[ \operatorname{Var}(Y)=\mu(1-\mu)=0.3(0.7)=0.21. \] The conventional dispersion is still \(\phi=1\); the variance is \(0.21\), not one.

Now let \(Y\) be a binomial count from \(m=8\) trials with success probability \(p=0.25\). The mean is \(\mu=mp=8(0.25)=2\). Its variance is \[ \operatorname{Var}(Y)=mp(1-p)=8(0.25)(0.75)=1.5. \] Using the mean-based form gives the same result: \[ \mu(1-\mu/m)=2(1-2/8)=2(0.75)=1.5. \] Here too, the conventional dispersion is one. The known number of trials affects the variance function; it is not an estimated common dispersion.

Worked Example: Poisson Dispersion

Let \(Y\) have a Poisson distribution with mean \(\mu=4\). The Poisson mean-variance relationship is \(\operatorname{Var}(Y)=\mu\), so \[ E[Y]=4,\qquad \operatorname{Var}(Y)=4. \] Writing the variance with an explicit dispersion multiplier gives \(\operatorname{Var}(Y)=\phi\mu\). Under the conventional Poisson GLM specification, \(\phi=1\), and therefore \(\operatorname{Var}(Y)=1(4)=4\). It would be incorrect to infer that the variance is one from the statement that the dispersion is one.

Deriving the Discrete-Family Conventions

The unit-dispersion entries follow directly from the usual probability models. A Bernoulli variable with success probability \(p\) has \(E[Y]=p\) and \(\operatorname{Var}(Y)=p(1-p)\). For a binomial count, write \(Y=\sum_{j=1}^{m}X_j\), where the \(X_j\) are independent Bernoulli variables with success probability \(p\). Linearity of expectation and independence give

$$ E[Y]=\sum_{j=1}^{m}E[X_j]=mp,\qquad \operatorname{Var}(Y)=\sum_{j=1}^{m}\operatorname{Var}(X_j)=mp(1-p). $$

Since \(\mu=mp\), we have \(p=\mu/m\), and thus

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

This has the listed binomial variance function with multiplier one. The Poisson distribution has \(E[Y]=\mu\) and \(\operatorname{Var}(Y)=\mu\), so its listed variance function also has multiplier one. These calculations explain the conventional entries; they do not estimate a separate dispersion parameter for any of the three models.

For a binomial proportion \(Y/m\), the mean is \(p\) and the variance is \(p(1-p)/m\). The factor \(1/m\) reflects averaging \(m\) trials. Whether the response is recorded as a count or as a proportion, the trial count is known and belongs in the variance structure; it is not a free common \(\phi\) to estimate.

Common Dispersion in Continuous Response Families

For the normal, gamma, and inverse-Gaussian families, the usual GLM parameterizations retain a scale parameter in the variance. One can see the role of that parameter by comparing standard distribution parameters with the mean-variance forms.

Result: Under the standard GLM parameterizations, a normal response has variance \(\phi\), a gamma response with mean \(\mu\) has variance \(\phi\mu^2\), and an inverse-Gaussian response with mean \(\mu\) has variance \(\phi\mu^3\). In these three families, \(\phi\) is a common scale parameter that is ordinarily estimated rather than fixed at one.

For a normal response, the usual notation is \(Y\sim N(\mu,\sigma^2)\). Its variance is \(\sigma^2\), so identifying \(\phi=\sigma^2\) and \(V(\mu)=1\) yields \(\operatorname{Var}(Y)=\phi V(\mu)=\phi\). For the gamma family, use a shape parameter \(\alpha=1/\phi\) and a scale parameter \(\beta=\phi\mu\). The gamma mean and variance in shape-scale form are \(E[Y]=\alpha\beta\) and \(\operatorname{Var}(Y)=\alpha\beta^2\). Substitution gives

$$ E[Y]=\frac{1}{\phi}(\phi\mu)=\mu,\qquad \operatorname{Var}(Y)=\frac{1}{\phi}(\phi\mu)^2=\phi\mu^2. $$

For the inverse-Gaussian family, use mean \(\mu\) and shape parameter \(\lambda=1/\phi\). Its variance is \(\mu^3/\lambda\), and therefore

$$ \operatorname{Var}(Y)=\frac{\mu^3}{\lambda}=\phi\mu^3. $$

Thus the scale parameter multiplies the family-specific variance function in all three cases. The variance-function shapes \(1\), \(\mu^2\), and \(\mu^3\) are distinct from the shared multiplier \(\phi\). Estimating \(\phi\) does not mean estimating a separate dispersion for each response.

Worked Example: A Shared Dispersion for Normal Responses

Suppose four normal responses have known means \(10,10,10,10\), and the observed values are \(11,9,12,8\). The residuals from those means are \(1,-1,2,-2\), so the squared residuals are \(1,1,4,4\), with sum \(10\). For independent normal observations with known means and common variance \(\phi\), the likelihood as a function of \(\phi>0\) is proportional to \[ \phi^{-4/2}\exp\left(-\frac{10}{2\phi}\right). \] Its log-likelihood, up to a constant, is \(\ell(\phi)=-2\log\phi-5/\phi\). Differentiating gives \[ \frac{d\ell}{d\phi}=-\frac{2}{\phi}+\frac{5}{\phi^2} =\frac{5-2\phi}{\phi^2}. \] The derivative is positive for \(\phi<2.5\) and negative for \(\phi>2.5\), so the likelihood is maximized at \(\hat{\phi}=2.5\). This is one common estimated variance for the four responses, not four separate variances. The known means make this a direct illustration of estimating the common dispersion without involving estimation of the means.

Worked Example: Gamma and Inverse-Gaussian Variances

For a gamma response with mean \(\mu=5\) and common dispersion \(\phi=0.2\), the variance function is \(V(\mu)=\mu^2=25\). Hence \[ \operatorname{Var}(Y)=\phi V(\mu)=0.2(25)=5. \] In the shape-scale parameterization, \(\alpha=1/\phi=5\) and \(\beta=\phi\mu=1\). The mean is \(\alpha\beta=5(1)=5\), and the variance is \(\alpha\beta^2=5(1)^2=5\), checking the result.

For an inverse-Gaussian response with mean \(\mu=2\) and dispersion \(\phi=0.1\), the variance function is \(V(\mu)=\mu^3=8\). Therefore \[ \operatorname{Var}(Y)=\phi V(\mu)=0.1(8)=0.8. \] Equivalently, the shape parameter is \(\lambda=1/\phi=10\), and the standard variance formula gives \(\mu^3/\lambda=8/10=0.8\). In both calculations the dispersion is shared and multiplies the variance function.

How to Read the Convention in Practice

When a model specification names a response family, separate three questions: what is the mean, what is the family’s variance function, and how is dispersion treated? For Bernoulli, binomial, and Poisson responses, the conventional answer to the third question is that dispersion is fixed at one. For normal, gamma, and inverse-Gaussian responses, the conventional answer is that a common \(\phi\) is estimated. The value of \(\phi\) is not itself the response variance unless the variance function equals one.

For example, a gamma model with \(\phi=0.2\) and mean \(5\) has variance \(5\), not \(0.2\), because its variance function is \(5^2\). Conversely, a Poisson model with mean \(4\) has variance \(4\), not one, because its unit dispersion multiplies the variance function \(V(4)=4\).

Common Mistakes

  • Equating unit dispersion with unit variance. The incorrect reasoning is “the Bernoulli dispersion is one, so its variance is one.” The correction is to multiply the unit dispersion by the variance function: at mean \(0.3\), the variance is \(1(0.3)(0.7)=0.21\).
  • Treating a binomial trial count as an estimated dispersion. The trial count \(m\) is part of the known binomial variance function. For a count, \(\operatorname{Var}(Y)=\mu(1-\mu/m)\); it is not an estimate of \(\phi\).
  • Using the same variance function for all families. A common multiplier does not imply a common function: normal, gamma, and inverse-Gaussian variances use \(1\), \(\mu^2\), and \(\mu^3\), respectively.
  • Assuming every response family estimates dispersion. Under the conventional GLM specification, Bernoulli, binomial, and Poisson dispersion is fixed at one. Adding a separately estimated scale is a change to the model specification, not the standard convention.
Key takeaway: First identify the family’s variance function, then apply its dispersion convention. Bernoulli, binomial, and Poisson use fixed unit dispersion; normal, gamma, and inverse-Gaussian use a common estimated dispersion.

Check Your Understanding

Use the variance function and dispersion convention together to answer each question.

  1. A Bernoulli response has mean \(0.6\). What are its variance and conventional dispersion?
  2. A binomial count has \(m=10\) trials and mean \(\mu=3\). Calculate its variance and state whether the trial count is an estimated dispersion.
  3. A Poisson response has mean \(7\). What variance follows from the conventional dispersion specification?
  4. A gamma response has mean \(4\) and common dispersion \(0.5\). Calculate its variance.
  5. Explain why a normal response with dispersion \(2\) need not have variance \(2\) if its variance function is not one. Which variance function does the standard normal GLM use?
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