Skewness and Shape Across Inverse-Gaussian Means

Why Mean Matters for Distributional Shape

An inverse-Gaussian response is positive and typically right-skewed, but the amount of skewness is not fixed across different means. In the mean-dispersion parameterization used for inverse-Gaussian generalized linear models, the mean and dispersion together determine both the variance and the standardized shape. This tutorial derives the exact formulas for skewness and excess kurtosis, then examines what happens when the mean changes while dispersion is held fixed.

The formulas are useful for understanding how a fitted mean can affect the shape of the response distribution. They describe standardized shape, not the response’s spread in its original units: variance and standardized skewness answer different questions.

Parameterization and Shape Measures

Definition: Let $X$ have an inverse-Gaussian distribution with mean $\mu>0$ and dispersion $\phi>0$, written $X\sim\operatorname{IG}(\mu,\phi)$, with density $$ f(x)=\frac{1}{\sqrt{2\pi\phi x^3}}\exp\left\{-\frac{(x-\mu)^2}{2\phi\mu^2x}\right\},\qquad x>0. $$ In this parameterization, $E[X]=\mu$ and $\operatorname{Var}(X)=\phi\mu^3$. The skewness is $E[(X-\mu)^3]/\operatorname{Var}(X)^{3/2}$. The kurtosis is $E[(X-\mu)^4]/\operatorname{Var}(X)^2$, and the excess kurtosis is the kurtosis minus 3.

The parameterization matters: in another convention the second parameter may be called a shape parameter rather than dispersion. Here $\phi$ is the GLM dispersion, so increasing $\phi$ at a fixed $\mu$ increases the variance. The density above also makes clear that both parameters must be positive and that the distribution has positive support.

We will obtain the shape measures through cumulants. The cumulant-generating function is $K(t)=\log E[e^{tX}]$, wherever this expectation is finite in a neighborhood of zero. Its $r$th derivative at zero is the $r$th cumulant, denoted $\kappa_r$. In particular, $\kappa_2$ is the variance, $\kappa_3$ is the third central moment, and $\kappa_4=E[(X-\mu)^4]-3\operatorname{Var}(X)^2$.

Deriving the Cumulants

For $\phi>0$, $\mu>0$, and $t<1/(2\phi\mu^2)$, the moment-generating function is finite. To calculate it from the density, first expand the exponent: $$ -\frac{(x-\mu)^2}{2\phi\mu^2x} =-\frac{x}{2\phi\mu^2}+\frac{1}{\phi\mu}-\frac{1}{2\phi x}. $$

The integral can then be evaluated using, for $a,b>0$, the identity $$ \int_0^\infty x^{-3/2}e^{-a/x-bx}\,dx =\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}. $$

For completeness, set $x=a/u^2$ in the integral. It becomes $(2/\sqrt{a})I(\sqrt{ab})$, where $I(c)=\int_0^\infty e^{-u^2-c^2/u^2}\,du$. For $c>0$, differentiating under the integral is valid on any compact interval of positive $c$, since the exponential controls the integrand at both endpoints. Substituting $v=c/u$ in the resulting integral gives $I'(c)=-2I(c)$. Also, dominated convergence gives $I(c)\to\sqrt{\pi}/2$ as $c\downarrow0$. Therefore $I(c)=(\sqrt{\pi}/2)e^{-2c}$, which proves the identity.

Applying the identity with $a=1/(2\phi)$ and $b=1/(2\phi\mu^2)-t>0$, and cancelling the density’s normalizing constant, gives $$ E[e^{tX}] =\exp\left\{\frac{1}{\phi\mu}\left(1-\sqrt{1-2\phi\mu^2t}\right)\right\}. $$

Thus the cumulant-generating function is $$ K(t)=\frac{1}{\phi\mu}\left(1-\sqrt{1-2\phi\mu^2t}\right), \qquad t<\frac{1}{2\phi\mu^2}. $$

The next result extracts the cumulants needed for the shape calculations.

Theorem: For $X\sim\operatorname{IG}(\mu,\phi)$, the first four cumulants are $$ \kappa_1=\mu,\qquad \kappa_2=\phi\mu^3,\qquad \kappa_3=3\phi^2\mu^5,\qquad \kappa_4=15\phi^3\mu^7. $$

Proof. The function $K(t)$ is analytic for $t$ in an open interval containing zero, so its Taylor expansion there has coefficients $\kappa_r/r!$. Let $z=2\phi\mu^2t$. The Taylor expansion of the square root around zero gives $$ 1-\sqrt{1-z}=\frac{z}{2}+\frac{z^2}{8}+\frac{z^3}{16}+\frac{5z^4}{128}+O(z^5). $$

Substituting $z=2\phi\mu^2t$ into $K(t)$ yields $$ K(t)=\mu t+\frac{\phi\mu^3}{2}t^2+\frac{\phi^2\mu^5}{2}t^3+\frac{5\phi^3\mu^7}{8}t^4+O(t^5). $$

Matching the coefficient of $t^r$ to $\kappa_r/r!$ gives $\kappa_1=\mu$, $\kappa_2=2!(\phi\mu^3/2)=\phi\mu^3$, $\kappa_3=3!(\phi^2\mu^5/2)=3\phi^2\mu^5$, and $\kappa_4=4!(5\phi^3\mu^7/8)=15\phi^3\mu^7$. $\square$

Skewness and Excess Kurtosis

The cumulants give the standardized shape directly. Skewness divides the third central moment by the variance to the power $3/2$. Excess kurtosis divides the fourth cumulant by the variance squared; subtracting 3 from ordinary kurtosis is exactly what removes the Gaussian reference value.

Theorem: For $X\sim\operatorname{IG}(\mu,\phi)$, the skewness $\gamma_1$, excess kurtosis $\gamma_2$, and ordinary kurtosis are $$ \gamma_1=3\sqrt{\frac{\phi}{\mu}},\qquad \gamma_2=15\frac{\phi}{\mu},\qquad \text{kurtosis}=3+15\frac{\phi}{\mu}. $$

Derivation. From the cumulants, the variance is $\kappa_2=\phi\mu^3$ and the third central moment is $\kappa_3=3\phi^2\mu^5$. Since $\phi,\mu>0$, $$ \gamma_1=\frac{\kappa_3}{\kappa_2^{3/2}} =\frac{3\phi^2\mu^5}{(\phi\mu^3)^{3/2}} =3\sqrt{\frac{\phi}{\mu}}. $$

The fourth central moment is $\kappa_4+3\kappa_2^2$, by the definition of the fourth cumulant. Consequently ordinary kurtosis is $3+\kappa_4/\kappa_2^2$, and excess kurtosis is $$ \gamma_2=\frac{\kappa_4}{\kappa_2^2} =\frac{15\phi^3\mu^7}{(\phi\mu^3)^2} =15\frac{\phi}{\mu}. $$

This also shows the exact relationship $\gamma_2=(5/3)\gamma_1^2$. The inverse-Gaussian skewness is positive for all permitted $\mu,\phi$, and its excess kurtosis is also positive. Ordinary kurtosis therefore exceeds 3 for every finite positive mean and dispersion.

Worked Shape Calculations

Worked Example: A Mean Increase at Fixed Dispersion

Suppose dispersion is $\phi=0.25$. Calculate skewness and excess kurtosis at $\mu=4$, then at $\mu=16$.

At $\mu=4$, $\phi/\mu=0.25/4=1/16$. Therefore $$ \gamma_1=3\sqrt{\frac{1}{16}}=\frac{3}{4}=0.75,\qquad \gamma_2=15\left(\frac{1}{16}\right)=\frac{15}{16}=0.9375. $$

At $\mu=16$, $\phi/\mu=0.25/16=1/64$. Thus $$ \gamma_1=3\sqrt{\frac{1}{64}}=\frac{3}{8}=0.375,\qquad \gamma_2=15\left(\frac{1}{64}\right)=\frac{15}{64}=0.234375. $$

The mean has quadrupled while dispersion stayed fixed. Skewness is halved, because it is proportional to $1/\sqrt{\mu}$, while excess kurtosis is quartered, because it is proportional to $1/\mu$. The corresponding ordinary kurtoses are $3.9375$ and $3.234375$.

Worked Example: Calculating Shape at a Given Mean and Dispersion

Let $\mu=9$ and $\phi=0.36$. Find the variance, skewness, excess kurtosis, and ordinary kurtosis.

First, $\phi/\mu=0.36/9=0.04$. The variance is $\phi\mu^3=0.36(9^3)=0.36(729)=262.44$. The skewness is $$ \gamma_1=3\sqrt{0.04}=3(0.2)=0.6. $$

The excess kurtosis is $\gamma_2=15(0.04)=0.6$, so ordinary kurtosis is $3+0.6=3.6$. As a check, $(5/3)\gamma_1^2=(5/3)(0.36)=0.6$, agreeing with the calculated excess kurtosis.

Worked Example: Finding the Mean for a Target Skewness

At fixed dispersion $\phi=0.5$, what mean gives skewness $0.6$, and what is the excess kurtosis at that mean?

Set the skewness formula equal to the target and solve for the positive mean: $$ 0.6=3\sqrt{\frac{0.5}{\mu}} \quad\Longrightarrow\quad 0.2=\sqrt{\frac{0.5}{\mu}} \quad\Longrightarrow\quad 0.04=\frac{0.5}{\mu} \quad\Longrightarrow\quad \mu=12.5. $$

Check the result: $3\sqrt{0.5/12.5}=3\sqrt{0.04}=0.6$. The excess kurtosis is $$ \gamma_2=15\frac{0.5}{12.5}=15(0.04)=0.6. $$

The equality of the numerical skewness and excess kurtosis here is not a general rule; it follows from this particular target and the relationship $\gamma_2=(5/3)\gamma_1^2$.

How Shape Changes as the Mean Varies

At fixed $\phi>0$, both shape measures are functions of $\mu>0$. Their dependence can be stated precisely.

Result: For fixed $\phi>0$, skewness $3\sqrt{\phi/\mu}$ and excess kurtosis $15\phi/\mu$ are strictly decreasing in $\mu$, and both approach zero as $\mu\to\infty$.

Proof. Differentiating with respect to $\mu$ while holding $\phi$ constant gives $$ \frac{d\gamma_1}{d\mu} =-\frac{3\sqrt{\phi}}{2\mu^{3/2}}<0, \qquad \frac{d\gamma_2}{d\mu} =-\frac{15\phi}{\mu^2}<0, \qquad \mu>0. $$

The derivatives are strictly negative, so both measures decrease strictly with the mean. Also, $\mu^{-1/2}\to0$ and $\mu^{-1}\to0$ as $\mu\to\infty$. Multiplying by the fixed constants in the formulas shows that both measures tend to zero. $\square$

This limiting statement concerns these standardized moments. It does not say that the distribution is symmetric at any finite mean: the skewness is strictly positive for every $\mu,\phi>0$. Nor does it say that the raw variance decreases as the mean increases; at fixed dispersion, the variance is $\phi\mu^3$, which increases with $\mu$. The spread on the original measurement scale and standardized shape are different features.

Common Mistake

Confusing kurtosis with excess kurtosis. Using $15\phi/\mu$ as the ordinary kurtosis omits the Gaussian reference value of 3. The correction is to report excess kurtosis as $15\phi/\mu$, and ordinary kurtosis as $3+15\phi/\mu$.
Changing the mean and dispersion together unintentionally. The decrease in shape established above assumes $\phi$ is fixed. If $\phi$ also changes, the relevant quantity is the ratio $\phi/\mu$, so the mean alone does not determine the direction of change.
Calling the response symmetric because skewness gets small. For every finite positive $\mu$ and $\phi$, skewness remains positive. A value approaching zero as the mean grows is a limiting statement, not exact symmetry at a finite mean.

In inverse-Gaussian regression, the mean may vary with predictors while dispersion is treated as common across observations. The formulas then show how the model’s standardized shape changes along with its fitted mean. In particular, observations with larger fitted means have lower skewness and lower excess kurtosis when the common dispersion is fixed. These are distributional implications of the mean-dispersion parameterization, not separate coefficients to interpret as regression effects.

Check Your Understanding

Use the mean-dispersion parameterization and show the calculation for each answer.

For $\mu=25$ and $\phi=1$, calculate the skewness and excess kurtosis.
At fixed dispersion, by what factor does skewness change if the mean is multiplied by 9? By what factor does excess kurtosis change?
If skewness is $1.5$ and dispersion is $0.25$, solve for the mean and then calculate excess kurtosis.
Explain why $15\phi/\mu$ is excess kurtosis rather than ordinary kurtosis.
Can an inverse-Gaussian distribution with positive finite mean and dispersion have exactly zero skewness? Explain using the formula.