Matching Variance Growth to a Family

Why Compare Variance Growth?

For a positive response, a plausible mean is not the only feature a model must represent. The amount of variation around that mean matters too. In particular, does the response variance stay roughly constant as the mean changes, or does it rise with the mean? If it rises, does it grow at a rate closer to the square or the cube of the mean? These questions distinguish three familiar continuous families.

The comparison is about the relationship between a response’s mean and variance, not simply about whether the response is positive or skewed. A variance pattern can help screen candidate models, but it cannot by itself establish a full probability distribution. The shape of the distribution, its support, and the scientific setting also matter.

Here \(\mu>0\) for the gamma and inverse-Gaussian models, whose responses have positive support. The Gaussian response can take any real value. Its constant variance may still be a useful approximation for a positive outcome in some settings, but positivity alone does not make a Gaussian model appropriate. The dispersion parameter scales each variance function; the key comparison is how the variance changes with \(\mu\) when \(\phi\) is held fixed.

The Three Growth Patterns

The Gaussian variance does not depend on the mean. The gamma variance is proportional to \(\mu^2\), so multiplying the mean by a factor \(c\) multiplies the variance by \(c^2\). The inverse-Gaussian variance is proportional to \(\mu^3\), so the same change multiplies the variance by \(c^3\). Thus the inverse-Gaussian variance grows fastest as the mean increases.

A useful comparison is the ratio of variances at two means. If both observations follow the same variance function with the same dispersion, then for positive means \(\mu_1\) and \(\mu_2\), the variance ratio is \(1\), \((\mu_2/\mu_1)^2\), or \((\mu_2/\mu_1)^3\), respectively. The ratio removes the shared dispersion parameter. This makes the rate of growth easier to compare, although real data provide estimates, not exact population variances.

Proof. Substituting the assumed variance at each mean gives \[ \frac{\operatorname{Var}(Y\mid\mu_2)}{\operatorname{Var}(Y\mid\mu_1)} =\frac{\phi\mu_2^p}{\phi\mu_1^p} =\left(\frac{\mu_2}{\mu_1}\right)^p, \] because \(\phi\) is positive and cancels. For a constant variance, the ratio is \(\phi/\phi=1\). \(\square\)

The same idea can be expressed as a local growth rate. For a power variance, the logarithm of the variance is a straight-line function of the logarithm of the mean, with slope equal to the power. This is a concise way to describe why “squared” and “cubed” refer to distinctly different rates.

Proof. Since \(\log V(\mu)=\log\phi+p\log\mu\), differentiating with respect to \(\log\mu\) gives \(p\). Equivalently, using the chain rule, \(\frac{d\log V}{d\log\mu}=\frac{\mu}{V(\mu)}\frac{dV(\mu)}{d\mu}=\frac{\mu}{\phi\mu^p}p\phi\mu^{p-1}=p\). \(\square\)

Where the Gamma and Inverse-Gaussian Variances Come From

The powers in the variance functions are properties of the distributions’ parameterizations, not labels chosen from a visual impression. For the gamma model, take shape \(a=1/\phi\) and scale \(b=\phi\mu\). A gamma random variable with shape \(a\) and scale \(b\) has mean \(ab\) and variance \(ab^2\). 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 model, use mean parameter \(\mu>0\) and shape parameter \(\lambda=1/\phi>0\). Its density for \(y>0\) is \[ f(y)=\left(\frac{\lambda}{2\pi y^3}\right)^{1/2} \exp\left\{-\frac{\lambda(y-\mu)^2}{2\mu^2y}\right\}. \] Its moment-generating function, for \(t<\lambda/(2\mu^2)\), is \[ M(t)=\exp\left\{\frac{\lambda}{\mu}\left(1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right)\right\}. \] Let \(g(t)=\log M(t)\). Direct differentiation gives \[ g'(t)=\mu\left(1-\frac{2\mu^2t}{\lambda}\right)^{-1/2}, \qquad g''(t)=\frac{\mu^3}{\lambda}\left(1-\frac{2\mu^2t}{\lambda}\right)^{-3/2}. \] Since \(M(0)=1\), \(M'(0)=g'(0)=\mu\), and \(M''(0)=g''(0)+[g'(0)]^2=\mu^3/\lambda+\mu^2\). Therefore \(\operatorname{Var}(Y)=M''(0)-[M'(0)]^2=\mu^3/\lambda=\phi\mu^3\).

Worked Comparisons

Using the Pattern Without Overclaiming

A practical comparison asks whether the observed variation changes little, approximately with the square of the mean, or more steeply in a way compatible with cubic growth. Means and variances are most informative when they summarize comparable observations under similar conditions. If observations within a group have different underlying means, the group variance can reflect that extra variation as well as the response’s conditional variance. Thus a rough match in summaries should be treated as a clue, not a definitive test.

The dispersion parameter also matters. Two families may have very different values of \(\phi\), yet each can match the variance at one particular mean. The distinction appears in how variance then changes across means. This is why comparing growth across a range of means is more informative than comparing a single variance in isolation.

Finally, variance growth is only one feature of a distribution. A Gaussian model has support on the whole real line, whereas gamma and inverse-Gaussian distributions are positive. Even among positive families, similar variance functions do not guarantee the same tail behavior or distributional shape. A proposed family should be assessed using the response’s support and other relevant features as well as its variance pattern.

Common Pitfall

• Wrong reasoning: “The variance is larger in the group with the larger mean, so the inverse-Gaussian model must be correct.”
• Correction: A larger variance alone does not identify cubic growth. Compare the variance ratio with the predictions from each candidate power, and account for uncertainty in estimated means and variances. Even a clear match to cubic growth supports a variance relationship, not a proof of the full inverse-Gaussian distribution.
• Wrong reasoning: “The variance function has a squared mean, so the variance must equal \(\mu^2\).”
• Correction: In this parameterization, the variance is \(\phi\mu^2\). The dispersion \(\phi\) scales the variance; the exponent describes its growth with the mean.