Variance Functions And Their Domains

Why the Domain Belongs to the Variance Function

A variance function describes how response variability changes with the response mean, after separating out a positive scale factor. Its formula alone is not the whole object: it is defined only for mean values that the model can actually attain. For example, a polynomial expression may be evaluable at negative numbers even when the response mean in the model must be positive. Those negative inputs do not thereby become part of the variance function’s domain.

This distinction matters when expressing a model in terms of its mean. The natural parameter may range over one set, while the mean ranges over another. The mean domain is the image of the allowable parameter values under the mean map. The earlier material on inverting the mean parameter supplies the key tool: where that map is one-to-one, it can be used to express variance as a function of the mean.

Definition: Consider a family of response distributions indexed by a mean $\mu$ and a positive scale factor $\phi$. If, for every allowable pair $(\mu,\phi)$, the variance can be written $$ \operatorname{Var}(Y\mid \mu,\phi)=\phi V(\mu), $$ where the same function $V$ works for all allowable $\phi$, then $V$ is the variance function. Its domain is the set $\mathcal{M}$ of allowable mean values: the values of $\mu$ for which the model specifies a response distribution. The factor $\phi$ is kept separate from $V(\mu)$.

In this definition, $\mathcal{M}$ is part of the specification of $V$, not a consequence of simplifying its formula. A variance function is therefore the pair of information “formula and domain.” The scale factor is positive; its precise interpretation depends on the response family and model. Some families have a fixed scale, which can be set to one when describing their variance function.

For a regular one-parameter exponential-dispersion family with density or mass function $$ f(y;\theta,\phi)=\exp\left\{\frac{y\theta-b(\theta)}{\phi}+c(y,\phi)\right\}, $$ the natural parameter $\theta$ belongs to an allowable set $\Theta$. Under the regularity conditions for differentiating the normalizing integral, the mean and variance are $b'(\theta)$ and $\phi b''(\theta)$. If the mean map is one-to-one, then $$ \mu=b'(\theta),\qquad \theta=(b')^{-1}(\mu),\qquad V(\mu)=b''\big((b')^{-1}(\mu)\big). $$ The mean domain is $\mathcal{M}=\{b'(\theta):\theta\in\Theta\}$. Thus the natural-parameter domain determines which inputs to the variance function are valid.

Recovering the Mean Domain

The following result gives a useful domain check for a regular scalar family. Assume that $\Theta=(\alpha,\beta)$ is an open interval, allowing infinite endpoints, and that $b$ is twice continuously differentiable on $\Theta$ with $b''(\theta)>0$ throughout. These conditions ensure that the mean map is continuous and strictly increasing.

Theorem: Under those assumptions, the mean domain is the open interval $$ \mathcal{M}=\bigl(\lim_{\theta\downarrow\alpha}b'(\theta),\ \lim_{\theta\uparrow\beta}b'(\theta)\bigr), $$ where either endpoint may be infinite. Every $\mu\in\mathcal{M}$ corresponds to exactly one $\theta\in\Theta$, and the variance function is $V(\mu)=b''(\theta)$ at that unique parameter value.

Proof. For any $\theta_1,\theta_2\in\Theta$ with $\theta_1<\theta_2$, the fundamental theorem of calculus gives $$ b'(\theta_2)-b'(\theta_1)=\int_{\theta_1}^{\theta_2}b''(t)\,dt>0. $$ The strict inequality follows because $b''(t)>0$ throughout the interval. Hence $b'$ is strictly increasing and therefore one-to-one. It is continuous because $b$ is twice continuously differentiable. By continuity and the intermediate value theorem, its image is an interval; strict increase identifies that image as the interval between its limits at the two endpoints. Since $\Theta$ is open and $b'$ is strictly increasing, neither finite endpoint value is attained inside $\Theta$. Thus the image is the stated open interval, and every point in it has exactly one preimage. Substituting that preimage into the variance expression $\phi b''(\theta)$ gives $V(\mu)=b''(\theta)$. $\square$

This result makes an important distinction. A limit of attainable means at the edge of $\mathcal{M}$ need not itself be attainable by any finite natural parameter. Sometimes a distribution can be added at that boundary, often as a degenerate limiting case; sometimes it is not part of the model under consideration. The boundary must be checked against the family’s definition rather than included automatically.

Worked Example: Bernoulli Mean Domain and Boundary Values

Let $Y$ have a Bernoulli distribution with success probability $p$, where $0<p<1$ in the regular family. The response takes the values 0 and 1. Directly, $$ E[Y]=0(1-p)+1p=p, $$ and, because $Y^2=Y$, $$ \operatorname{Var}(Y)=E[Y^2]-E[Y]^2=p-p^2=p(1-p). $$ With the fixed scale factor set to one, the mean is $\mu=p$, so $V(\mu)=\mu(1-\mu)$.

The regular mean domain is $\mathcal{M}=(0,1)$, since the regular Bernoulli family has $0<p<1$. The formula also evaluates to zero at $\mu=0$ and $\mu=1$, but these values are boundary extensions, not members of that regular mean domain. If the model is explicitly enlarged to include the degenerate distributions at $p=0$ and $p=1$, its mean domain becomes $[0,1]$. In either version, the variance formula does not authorize values such as $\mu=1.2$: those do not correspond to Bernoulli distributions.

Examples of Mean Domains and Variance Functions

The examples below emphasize how the distribution determines the domain. They also show that an integer-valued response does not force its mean to be integer-valued: the mean changes continuously as the distribution’s parameter changes.

Worked Example: Poisson Mean Domain

Suppose $Y\sim\operatorname{Poisson}(\lambda)$ with $\lambda>0$. Its probability mass function is $P(Y=y)=e^{-\lambda}\lambda^y/y!$ for $y=0,1,2,\ldots$. The mean can be calculated from the series: $$ E[Y]=\sum_{y=1}^{\infty}y e^{-\lambda}\frac{\lambda^y}{y!} =\lambda e^{-\lambda}\sum_{y=1}^{\infty}\frac{\lambda^{y-1}}{(y-1)!} =\lambda. $$ For the second factorial moment, $$ E[Y(Y-1)]=\sum_{y=2}^{\infty}y(y-1)e^{-\lambda}\frac{\lambda^y}{y!} =\lambda^2 e^{-\lambda}\sum_{y=2}^{\infty}\frac{\lambda^{y-2}}{(y-2)!} =\lambda^2. $$ Since $Y^2=Y(Y-1)+Y$, it follows that $E[Y^2]=\lambda^2+\lambda$, and therefore $$ \operatorname{Var}(Y)=E[Y^2]-E[Y]^2=(\lambda^2+\lambda)-\lambda^2=\lambda. $$ Thus $\mu=\lambda$, $V(\mu)=\mu$, and the regular mean domain is $(0,\infty)$. The response values are nonnegative integers, but the mean can be any positive real number. A Poisson distribution with $\lambda=0$ is a degenerate boundary case, which may be included only if the model is defined to include it.

Worked Example: Normal Responses Have an Unrestricted Mean

Let $Y\sim N(\mu,\phi)$, with $\phi>0$, using the convention that the second argument is the variance. By the definition of this normal distribution, $E[Y]=\mu$ and $\operatorname{Var}(Y)=\phi$. Hence the scale-separated representation is $$ \operatorname{Var}(Y)=\phi V(\mu),\qquad V(\mu)=1. $$ For every real $\mu$ and every positive $\phi$, this specifies a normal distribution. Therefore the mean domain is $\mathcal{M}=\mathbb{R}$, and the variance function is constant on all of $\mathbb{R}$.

The fact that a response value is real-valued does not alone establish that its mean domain is all real numbers; that conclusion here follows from the normal model’s parameterization. Conversely, the constant formula $V(\mu)=1$ can be evaluated at any real input, and every such input is valid because the model permits every real mean.

Together, these examples show why a variance formula should always be reported with its domain. The Bernoulli formula is naturally attached to means between zero and one; the Poisson formula to positive means; and the normal constant function to all real means. A formula may have an algebraic extension beyond a model’s domain, but that extension does not describe variance for distributions in the specified family.

Two Useful Consequences

The separation of scale and mean-dependent variance also gives a simple uniqueness property. Suppose a family has the same allowable mean domain $\mathcal{M}$ for its positive scale values and admits a representation $\operatorname{Var}(Y\mid\mu,\phi)=\phi V(\mu)$. Then $V$ is fixed by the model’s variances; it cannot be changed independently while preserving the same scale convention.

Result: If two functions $V_1$ and $V_2$ both satisfy $\operatorname{Var}(Y\mid\mu,\phi)=\phi V_j(\mu)$ for every $\mu\in\mathcal{M}$ and every allowed $\phi>0$, then $V_1(\mu)=V_2(\mu)$ for all $\mu\in\mathcal{M}$.

Proof. Fix any $\mu\in\mathcal{M}$ and choose any allowed $\phi>0$. The two representations imply $$ \phi V_1(\mu)=\operatorname{Var}(Y\mid\mu,\phi)=\phi V_2(\mu). $$ Dividing by the positive number $\phi$ gives $V_1(\mu)=V_2(\mu)$. Since $\mu$ was arbitrary, the functions agree everywhere on $\mathcal{M}$. $\square$

A second basic check follows from the definition of variance: $V(\mu)\geq 0$ at every allowable mean, since $\phi>0$ and variance cannot be negative. For a nondegenerate response distribution, its variance is positive, so $V(\mu)>0$ there. Zero values can occur at allowable means only when the response is degenerate at those means. These are checks on the values of the function, not a way to infer the domain: nonnegativity of a formula at some input does not prove that a distribution exists there.

Common Mistake: Using the Formula to Invent Its Domain

A common error is to find an algebraic expression for $V(\mu)$, then declare its domain to be every real number at which that expression can be evaluated. For example, the expression $V(\mu)=\mu$ is mathematically defined for negative $\mu$, but in the Poisson example the allowable means are positive. A negative value would also make $\phi V(\mu)$ negative, which cannot be a variance.

The correction is to establish the model’s allowable parameter values first, map them through the mean function, and state the resulting mean domain alongside the variance formula. When a finite endpoint is only a limit of the mean map, say so; include it only if the family explicitly includes a corresponding distribution. This procedure prevents algebraic extensions, limiting cases, and actual model parameters from being confused.

Key takeaway: A variance function is defined on the model’s allowable mean domain, not on the largest set where its formula happens to make algebraic sense. In a regular scalar exponential-dispersion family, obtain that domain by mapping the natural-parameter space through $\mu=b'(\theta)$, then express $b''(\theta)$ in terms of $\mu$.

Check Your Understanding

For each question, keep the allowable mean domain separate from the expression for the variance function.

In the regular Bernoulli family, what is the mean domain, and why does the variance formula’s value at $\mu=0$ not by itself determine whether that endpoint is included?
For a Poisson response with positive rate, state the mean domain and variance function. Can its integer-valued outcomes imply that the mean must be an integer?
For $Y\sim N(\mu,\phi)$, identify the scale factor, variance function, and mean domain.
Under what assumptions in the theorem above is the mean map $b'(\theta)$ strictly increasing?
If two variance functions use the same scale convention and give the same variance for every allowed mean and scale, explain why they must agree on the entire mean domain.