The Inverse-Gaussian Cubic Variance Function

Why the Inverse-Gaussian Variance Is Cubic

For some response distributions, variability changes substantially as the mean changes. The inverse-Gaussian family gives a particularly clear example: with dispersion held fixed, multiplying the mean by a factor multiplies the variance by the cube of that factor. A mean increase of 50%, for instance, does not produce a 50% variance increase.

The relationship follows directly from the inverse-Gaussian exponential dispersion form. We will identify that form, differentiate its log-partition function to obtain the variance, and then use the resulting variance function for explicit scaling calculations.

The Inverse-Gaussian Exponential Dispersion Form

Let \(Y\) have an inverse-Gaussian distribution with mean parameter \(\mu>0\) and dispersion parameter \(\phi>0\). Its density, with respect to Lebesgue measure on \(y>0\), is

To see the exponential dispersion structure, expand the quadratic expression in the exponent:

Set \(\theta=-1/(2\mu^2)\). Since \(\mu>0\), this natural parameter satisfies \(\theta<0\), and the mapping between \(\mu\) and \(\theta\) is one-to-one. In particular, \(1/\mu=\sqrt{-2\theta}\). Define the log-partition function by \(\kappa(\theta)=-\sqrt{-2\theta}\). Then the density can be written as

The first two factors that depend on \(y\) but not on \(\theta\) make up the carrier term. The parameter-dependent part has the standard exponential dispersion form. For a regular exponential dispersion model with open natural-parameter space and twice differentiable \(\kappa\), the mean is \(\kappa'(\theta)\) and the variance is \(\phi\kappa''(\theta)\). Here the natural-parameter space is the open interval \((-\infty,0)\).

Deriving the Cubic Variance Function

Differentiate the log-partition function on its domain \(\theta<0\). Its first derivative is

Because \(\theta=-1/(2\mu^2)\), we have \(-2\theta=1/\mu^2\), so \(\kappa'(\theta)=\mu\). Differentiating once more gives

For \(\mu>0\), \((-2\theta)^{-3/2}=(1/\mu^2)^{-3/2}=\mu^3\). Applying the exponential dispersion mean-variance relationship therefore gives the following result.

What Happens When the Mean Is Rescaled?

Suppose the dispersion stays fixed and the mean changes from \(\mu\) to \(c\mu\), where \(c>0\). The new mean remains in the inverse-Gaussian parameter domain because \(c\mu>0\). Applying the variance function at the new mean gives

This yields an exact scaling rule, not an approximation.

Proof. At the original mean \(\mu\), the variance is \(\phi\mu^3\). At the rescaled mean \(c\mu\), the variance is \(\phi(c\mu)^3\). Since \(c>0\), \((c\mu)^3=c^3\mu^3\). Therefore the new variance is \(c^3\phi\mu^3\), which is \(c^3\) times the original variance. Subtracting the original variance gives \((c^3-1)\phi\mu^3\). \(\square\)

Interpreting the Cubic Relationship

The variance function \(V(\mu)=\mu^3\) describes how variance depends on the mean before the dispersion multiplier is applied. The actual variance is \(\phi V(\mu)\). Consequently, comparisons based on the cubic scaling rule hold when \(\phi\) is the same in both cases. If dispersion also changes, the variance ratio must include that change:

This equation follows by dividing \(\phi_{\text{new}}\mu_{\text{new}}^3\) by \(\phi_{\text{old}}\mu_{\text{old}}^3\), assuming both dispersions and means are positive. It makes the fixed-dispersion condition explicit: only when \(\phi_{\text{new}}=\phi_{\text{old}}\) does the ratio reduce to the cube of the mean ratio.

The relationship also distinguishes a change in the variance function from a change in the dispersion. The cubic power comes from the inverse-Gaussian family’s mean-variance structure; \(\phi\) scales the variance at any given mean. For example, doubling \(\phi\) doubles the variance if the mean is held fixed, while doubling \(\mu\) multiplies the variance by eight if \(\phi\) is held fixed.

Common Mistake

• Wrong reasoning: “The mean increases by 50%, so the variance increases by 50%.” This treats the variance as directly proportional to \(\mu\).
• Correction: Use \(\operatorname{Var}(Y)=\phi\mu^3\). With dispersion fixed, a 50% mean increase has \(c=1.5\), so the variance ratio is \(1.5^3=3.375\). The variance becomes 3.375 times its original value, a 237.5% increase.
• Another check: Keep the dispersion in the calculation. Writing the variance as just \(\mu^3\) omits \(\phi\) and gives the wrong numerical variance unless \(\phi=1\).