How Dispersion Scales Response Variance

Introduction

A mean describes the centre of a response distribution, but it does not say how widely responses vary around that centre. In generalized linear models, a variance function describes how variance depends on the mean, while a dispersion parameter supplies a separate scale factor. Keeping these roles distinct makes it easier to interpret model assumptions and to calculate response variances correctly.

The basic relation is simple: response variance equals dispersion multiplied by the variance function evaluated at the mean. The important details are what each quantity represents, which quantities are held fixed in a comparison, and how the multiplier affects variance and standard deviation.

Mean, Variance Function, and Dispersion

Definition: Let \(Y\) be a response with finite mean \(\mu=E[Y]\), where \(\mu\) belongs to a specified mean domain \(\mathcal{D}\). A variance function is a nonnegative function \(V(\mu)\) on that domain. A mean-variance specification with dispersion parameter \(\phi>0\) states that

E[Y]=\mu,\qquad \operatorname{Var}(Y)=\phi V(\mu).

Here, \(\mu\) sets the mean, \(V(\mu)\) describes how the variance-function part depends on that mean, and \(\phi\) multiplies that part. The response must have a finite variance for this specification to apply.

This relation is a statement about the first two moments. It does not, by itself, specify a complete probability distribution for \(Y\). The function \(V\) and the allowed mean domain depend on the model under consideration. In this tutorial, they will be given explicitly when a numerical calculation is required.

The roles of \(\mu\) and \(\phi\) differ. Changing \(\mu\) changes the stated response mean and may also change \(V(\mu)\). Changing \(\phi\), while holding \(\mu\) and \(V\) fixed, multiplies the variance without changing the stated mean. Thus dispersion is not another name for the mean, nor is it the variance itself in every model.

Worked Example: A Linear Variance Function

Suppose a response has mean \(\mu=5\), variance function \(V(\mu)=\mu\) for positive \(\mu\), and dispersion \(\phi=0.4\). Find its variance and standard deviation.

First evaluate the variance function at the specified mean:

$$ V(5)=5. $$

Then multiply by the dispersion:

\operatorname{Var}(Y)=\phi V(\mu)=0.4(5)=2.

The standard deviation is the positive square root of the variance:

\operatorname{SD}(Y)=\sqrt{\operatorname{Var}(Y)}=\sqrt{2}\approx 1.414.

The mean remains \(5\). The value \(0.4\) is not the response variance; it scales the value \(V(5)=5\) to give variance \(2\).

What Changing Dispersion Does

At a fixed mean, \(V(\mu)\) is fixed. If \(V(\mu)>0\), the variance is directly proportional to \(\phi\): doubling \(\phi\) doubles the variance, and multiplying \(\phi\) by any positive factor multiplies the variance by that factor. Standard deviation behaves differently because it is the square root of variance. Multiplying dispersion by a factor \(c\) multiplies standard deviation by \(\sqrt{c}\), not by \(c\).

Theorem: Fix a mean \(\mu\) in the domain of \(V\), and suppose \(V(\mu)>0\). For two positive dispersion values \(\phi_1\) and \(\phi_2\), if each response variance is specified by \(\operatorname{Var}(Y_j)=\phi_j V(\mu)\), then

\frac{\operatorname{Var}(Y_2)}{\operatorname{Var}(Y_1)} =\frac{\phi_2}{\phi_1}, \qquad \frac{\operatorname{SD}(Y_2)}{\operatorname{SD}(Y_1)} =\sqrt{\frac{\phi_2}{\phi_1}}.

Proof. Since the mean is the same in both specifications, they use the same value \(V(\mu)\). By the variance relation,

\frac{\operatorname{Var}(Y_2)}{\operatorname{Var}(Y_1)} = \frac{\phi_2 V(\mu)}{\phi_1 V(\mu)} = \frac{\phi_2}{\phi_1},

where cancellation is valid because \(V(\mu)>0\) and \(\phi_1>0\). Both variances are positive, so taking their positive square roots gives

\frac{\operatorname{SD}(Y_2)}{\operatorname{SD}(Y_1)} = \sqrt{\frac{\operatorname{Var}(Y_2)}{\operatorname{Var}(Y_1)}} = \sqrt{\frac{\phi_2}{\phi_1}}.

This proves both claims. \(\square\)

The condition \(V(\mu)>0\) matters for the ratios: if \(V(\mu)=0\), both specified variances are zero, so their ratio is not defined. In that case, changing a positive dispersion value still leaves the variance at zero.

Worked Example: Comparing Two Dispersion Values

Let \(\mu=3\) and \(V(\mu)=\mu^2\), so \(V(3)=9\). Compare the response variances and standard deviations when \(\phi_1=0.2\) and \(\phi_2=0.8\).

For the first dispersion value,

\operatorname{Var}(Y_1)=0.2(9)=1.8, \qquad \operatorname{SD}(Y_1)=\sqrt{1.8}\approx 1.342.

For the second,

\operatorname{Var}(Y_2)=0.8(9)=7.2, \qquad \operatorname{SD}(Y_2)=\sqrt{7.2}\approx 2.683.

The dispersion ratio is \(0.8/0.2=4\), and the variance ratio is \(7.2/1.8=4\), as the theorem predicts. The standard-deviation ratio is approximately \(2.683/1.342=2\), which equals \(\sqrt{4}\). The mean is \(3\) under both specifications.

Calculating Variance for a Specified Mean

For a numerical calculation, evaluate the variance function first and multiply by dispersion second. This order helps prevent two common confusions: using \(\phi\) as if it were the variance, and evaluating the function at a value other than the stated mean. The following process makes the inputs and output explicit.

Check the mean domain.
Confirm that the specified \(\mu\) is in the domain where the given variance function is defined.

Evaluate the variance function.
Compute \(V(\mu)\) using the stated mean, not the dispersion value.

Multiply by dispersion.
Set \(\operatorname{Var}(Y)=\phi V(\mu)\). If requested, take the positive square root to obtain standard deviation.

Worked Example: A Constant Variance Function

Suppose the mean domain is all real numbers and the specified variance function is \(V(\mu)=9\) throughout that domain. For a response with mean \(\mu=-2\) and dispersion \(\phi=0.6\), calculate the variance and standard deviation.

The mean \(-2\) is in the stated domain. The variance function is constant, so

$$ V(-2)=9. $$

The dispersion-scaled variance is

\operatorname{Var}(Y)=0.6(9)=5.4.

Therefore,

\operatorname{SD}(Y)=\sqrt{5.4}\approx 2.324.

The negative mean does not cause a problem here because the stated domain includes it and the variance function is nonnegative there. A different variance function could have a more restricted domain; checking that domain is part of the calculation.

Dispersion and Relative Variability

When \(\mu\ne 0\), one can compare variance with the squared mean. This ratio is often useful for interpreting scale, but it is not the dispersion parameter alone unless the variance function has a particular form. Under the mean-variance specification,

\frac{\operatorname{Var}(Y)}{\mu^2} = \phi\frac{V(\mu)}{\mu^2}.

Thus, at a fixed nonzero mean and fixed variance function, the ratio of variance to squared mean is also multiplied by dispersion. Across different means, however, the factor \(V(\mu)/\mu^2\) may change. The next result records precisely what follows at a fixed mean.

Result: If \(\mu\ne 0\) and \(V(\mu)\) is finite, then

\frac{\operatorname{Var}(Y)}{\mu^2} =\phi\frac{V(\mu)}{\mu^2}.

For two positive dispersion values at that same mean, the ratio of these relative-variance quantities equals \(\phi_2/\phi_1\), provided \(V(\mu)>0\).

Derivation. Substitute the specified variance \(\operatorname{Var}(Y)=\phi V(\mu)\) into the ratio and divide by the nonzero value \(\mu^2\):

\frac{\operatorname{Var}(Y)}{\mu^2} = \frac{\phi V(\mu)}{\mu^2} = \phi\frac{V(\mu)}{\mu^2}.

At the same mean, \(V(\mu)/\mu^2\) is common to the two specifications. If it is positive, it cancels when their relative-variance quantities are divided, leaving \(\phi_2/\phi_1\). \(\square\)

The condition \(\mu\ne 0\) is essential because division by \(\mu^2\) is otherwise undefined. This ratio is a derived summary, not a replacement definition for dispersion. Its value depends both on \(\phi\) and on the shape of the variance function at the mean.

Common Mistakes and Interpretation

Treating dispersion as the response variance. The variance is \(\phi V(\mu)\), not generally \(\phi\). Evaluate \(V\) at the mean and multiply.
Changing the mean when only dispersion changes. In the stated specification, changing \(\phi\) alone changes the variance scale; the mean remains \(\mu\).
Scaling standard deviation by the full dispersion ratio. Variance scales by \(c\) when dispersion scales by \(c\); standard deviation scales by \(\sqrt{c}\).
Ignoring the domain of \(V\). A numerical input outside the function's permitted mean domain does not yield a valid calculation under that specification.
Assuming the same variance function in every model. The scaling rule uses the specified \(V\); the form and domain of that function must come from the model being used.

Dispersion is useful because it separates two features of the moment specification: how the mean is located and how variance is scaled around it. For a fixed variance function and mean, comparisons of dispersion translate directly into comparisons of variance. But a calculated mean and variance do not by themselves identify every feature of a response distribution; the variance relation should be interpreted as a moment statement unless further distributional assumptions have been supplied.

Key takeaway: Start with the specified mean \(\mu\), evaluate \(V(\mu)\), and then multiply by \(\phi\). Dispersion scales response variance, not the mean; standard deviation changes by the square root of the variance-scale factor.

Check Your Understanding

Use the stated mean-variance relation in each question.

If \(V(\mu)=2\mu\), \(\mu=4\), and \(\phi=0.5\), what are the response variance and standard deviation?
At a fixed mean with \(V(\mu)>0\), dispersion changes from \(0.3\) to \(1.2\). By what factors do variance and standard deviation change?
Why is \(\phi\) not generally equal to \(\operatorname{Var}(Y)\)?
If \(V(\mu)=\mu^2\), \(\mu=0\), and \(\phi>0\), what does the variance relation give, and why is a variance ratio between two dispersion values undefined?
For \(\mu\ne0\), write the ratio \(\operatorname{Var}(Y)/\mu^2\) in terms of \(\phi\) and \(V(\mu)\). Which quantities can change if the mean changes?