Deriving Inverse-Gaussian Moments

Why Derive the Moments?

The inverse-Gaussian distribution is positive and generally right-skewed, so its mean alone does not describe how much individual responses vary around it. Its variance depends on both the mean and a dispersion parameter: at the same mean, a larger dispersion produces more variability. Deriving the moments makes that relationship precise and gives useful measures of relative variation and shape.

Use the parameterization with mean parameter \(\mu>0\) and dispersion parameter \(\phi>0\). Its density is

f(y;\mu,\phi)=\frac{1}{\sqrt{2\pi\phi y^3}} \exp\left\{-\frac{(y-\mu)^2}{2\phi\mu^2y}\right\},\qquad y>0.

Some references instead write the shape parameter \(\lambda=1/\phi\). Keeping these parameterizations distinct matters: the variance is \(\mu^3/\lambda\), which is \(\phi\mu^3\), not \(\lambda\mu^3\), in the notation used here.

Deriving the Moment-Generating Function

The moment-generating function is \(M(t)=E[e^{tX}]\), when the expectation is finite. To derive it from the density, first rewrite the exponent:

-\frac{(y-\mu)^2}{2\phi\mu^2y} =-\frac{y}{2\phi\mu^2}+\frac{1}{\phi\mu}-\frac{1}{2\phi y}.

For \(t<1/(2\phi\mu^2)\), set \(a=1/(2\phi)\) and \(b=1/(2\phi\mu^2)-t\), both positive. The density integral then uses the identity

\int_0^\infty x^{-3/2}e^{-a/x-bx}\,dx =\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}},\qquad a,b>0.

For completeness, substitute \(u=\sqrt{a/x}\). The integral becomes \(2J(\sqrt{ab})/\sqrt a\), where \(J(c)=\int_0^\infty e^{-u^2-c^2/u^2}\,du\). For \(c>0\), differentiation under the integral is valid locally in \(c\): on any compact interval of positive \(c\), the differentiated integrand is bounded near zero by a constant times \(u^{-2}e^{-c_0^2/u^2}\), and near infinity by a constant times \(e^{-u^2}\). Substitution \(v=c/u\) in the derivative gives \(J'(c)=-2J(c)\). Also \(J(c)\to\sqrt{\pi}/2\) as \(c\downarrow0\) by dominated convergence, since the integrand is bounded by \(e^{-u^2}\). Thus \(J(c)=(\sqrt{\pi}/2)e^{-2c}\), proving the identity.

Applying the identity to the density integral yields the moment-generating function:

\begin{aligned} M(t) &=\frac{e^{1/(\phi\mu)}}{\sqrt{2\pi\phi}} \int_0^\infty y^{-3/2}e^{-a/y-by}\,dy\\ &=\exp\left\{\frac{1}{\phi\mu} -\frac{\sqrt{1-2\phi\mu^2t}}{\phi\mu}\right\}\\ &=\exp\left\{\frac{1-\sqrt{1-2\phi\mu^2t}}{\phi\mu}\right\}, \qquad t<\frac{1}{2\phi\mu^2}. \end{aligned}

In particular, this function is finite on an open interval containing zero. Consequently, its derivatives at zero give the moments of every nonnegative integer order.

Definition: The cumulant-generating function is \(K(t)=\log M(t)\) wherever \(M(t)\) is finite. The cumulant of order \(r\) is \(\kappa_r=K^{(r)}(0)\), for integer \(r\geq1\). In particular, \(\kappa_1=E[X]\) and \(\kappa_2=\operatorname{Var}(X)\).

Cumulants and Moments

The cumulant-generating function here is

K(t)=\frac{1-\sqrt{1-2\phi\mu^2t}}{\phi\mu}.

Expanding the square root around zero gives a formula for every cumulant. For \(|z|<1\), \(1-\sqrt{1-z}=\sum_{r=1}^{\infty}(2r-3)!!\,z^r/(2^r r!)\), where \((-1)!!=1\). Substituting \(z=2\phi\mu^2t\) and multiplying by \(1/(\phi\mu)\) gives the following result.

Theorem: For every integer \(r\geq1\), the inverse-Gaussian cumulant of order \(r\) is

\kappa_r=(2r-3)!!\,\phi^{r-1}\mu^{2r-1}.

Derivation: The coefficient of \(t^r\) in \(K(t)\) is \((2r-3)!!\,\phi^{r-1}\mu^{2r-1}/r!\). Since \(K\) has a convergent power series in a neighborhood of zero, its \(r\)th derivative at zero is \(r!\) times that coefficient. This proves the stated formula. \(\square\)

The first four cumulants are therefore

\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.

The first two establish the mean and variance. Higher cumulants also give central moments: the third central moment equals \(\kappa_3\), while the fourth central moment equals \(\kappa_4+3\kappa_2^2\). Raw moments are obtained by expanding \(M(t)=e^{K(t)}\), or by the general recurrence below.

Result: Write \(m_r=E[X^r]\) and \(m_0=1\). For every integer \(n\geq0\), raw moments satisfy

m_{n+1}=\sum_{j=0}^{n}\binom{n}{j}\kappa_{j+1}m_{n-j}.

Derivation: Differentiate \(M(t)=e^{K(t)}\) to obtain \(M'(t)=K'(t)M(t)\). Expand \(M(t)=\sum_{r\geq0}m_rt^r/r!\) and \(K'(t)=\sum_{j\geq0}\kappa_{j+1}t^j/j!\). Equating the coefficients of \(t^n/n!\) in \(M'(t)=K'(t)M(t)\) gives the displayed recurrence. The expansions are valid near zero because the moment-generating function is finite there. \(\square\)

For example, applying the recurrence for \(n=0,1,2,3\) gives

\begin{aligned} E[X]&=\mu,\\ E[X^2]&=\mu^2+\phi\mu^3,\\ E[X^3]&=\mu^3+3\phi\mu^4+3\phi^2\mu^5,\\ E[X^4]&=\mu^4+6\phi\mu^5+15\phi^2\mu^6+15\phi^3\mu^7. \end{aligned}

For the central moments, subtract the mean before taking powers. The resulting variance, third central moment, and fourth central moment are

\begin{aligned} \operatorname{Var}(X)&=\phi\mu^3,\\ E[(X-\mu)^3]&=3\phi^2\mu^5,\\ E[(X-\mu)^4]&=3\phi^2\mu^6+15\phi^3\mu^7. \end{aligned}

The standardized skewness is \(3\sqrt{\phi\mu}\), and the excess kurtosis is \(15\phi\mu\). Thus, at a fixed value of \(\phi\), both standardized measures increase with \(\mu\). These formulas describe the distribution’s shape as well as its absolute variability.

Coefficient of Variation

The coefficient of variation (CV) is the standard deviation divided by the mean. It measures spread relative to the scale of the mean, and is defined here because \(\mu>0\). From the derived mean and variance,

\operatorname{CV}(X) =\frac{\sqrt{\operatorname{Var}(X)}}{E[X]} =\frac{\sqrt{\phi\mu^3}}{\mu} =\sqrt{\phi\mu}.

The CV is not \(\phi\) alone: it depends on the product \(\phi\mu\). In particular, changing the mean while holding dispersion fixed changes the relative variability.

Worked Example: Mean Four and Dispersion 0.05

Let \(X\) have an inverse-Gaussian distribution with \(\mu=4\) and \(\phi=0.05\). Calculate its mean, variance, standard deviation, CV, and skewness.

The mean is \(E[X]=\mu=4\). Since \(4^3=64\), the variance is \(\operatorname{Var}(X)=\phi\mu^3=0.05(64)=3.2\), and the standard deviation is \(\sqrt{3.2}\approx1.789\).

The coefficient of variation is \(\sqrt{\phi\mu}=\sqrt{0.05(4)}=\sqrt{0.2}\approx0.447\). Equivalently, \(1.789/4\approx0.447\). The skewness is \(3\sqrt{\phi\mu}=3\sqrt{0.2}\approx1.342\).

Worked Example: Converting From the Shape Parameter

A reference gives the inverse-Gaussian shape parameter \(\lambda=8\) and mean \(\mu=2\). Find the variance and CV using the dispersion parameterization.

Since \(\phi=1/\lambda\), the dispersion is \(\phi=1/8=0.125\). The variance is \(\phi\mu^3=(1/8)(2^3)=1\), so the standard deviation is \(1\). The CV is \(1/2=0.5\); the formula in the course notation gives the same result, \(\sqrt{(1/8)(2)}=\sqrt{1/4}=0.5\).

Worked Example: Third and Fourth Moments

Let \(\mu=0.8\) and \(\phi=0.2\). Calculate the variance and the raw moments of orders two and three, then check the third central moment.

First, \(0.8^2=0.64\), \(0.8^3=0.512\), \(0.8^4=0.4096\), and \(0.8^5=0.32768\). Thus the variance is \(0.2(0.512)=0.1024\). The second raw moment is \(0.64+0.1024=0.7424\).

Using the raw-moment formula, \(E[X^3]=0.512+3(0.2)(0.4096)+3(0.2^2)(0.32768)\). The two added terms are \(0.24576\) and \(0.0393216\), respectively, giving \(E[X^3]=0.7970816\).

The third central moment formula gives \(3(0.2^2)(0.8^5)=3(0.04)(0.32768)=0.0393216\). Directly, the same quantity is \(E[X^3]-3\mu E[X^2]+2\mu^3=0.7970816-3(0.8)(0.7424)+2(0.512)=0.0393216\), as required.

Interpretation and a Common Pitfall

The variance \(\phi\mu^3\) grows cubically with the mean if \(\phi\) is held fixed. The CV, by contrast, grows as the square root of \(\mu\). These are different statements: variance is measured in squared response units, whereas CV is dimensionless. The higher-moment formulas likewise show how dispersion and mean jointly determine asymmetry and tail shape.

Pitfall: “The CV is \(\phi\), because \(\phi\) is the dispersion.” This confuses a model parameter with relative standard deviation. The correction is to divide the standard deviation by the mean: \(\sqrt{\phi\mu^3}/\mu=\sqrt{\phi\mu}\).
Pitfall: “The variance is \(\mu^3/\phi\).” That formula uses the shape parameter \(\lambda\) in place of dispersion. If \(\lambda=1/\phi\), then \(\mu^3/\lambda=\phi\mu^3\).
Pitfall: “The third cumulant is the third raw moment.” The third cumulant is the third central moment, \(E[(X-\mu)^3]\). The raw third moment also includes terms involving the mean and variance.

A reliable calculation therefore begins by confirming which parameterization is in use, then derives the requested quantity from the mean and variance in that parameterization. For standardized quantities such as the CV, keep the mean in the calculation rather than treating dispersion by itself as relative variability.

Check Your Understanding

Use the density parameterization and moment formulas developed here.

For \(\mu=3\) and \(\phi=0.04\), calculate the variance and coefficient of variation.
If a source reports \(\lambda=5\), what is the corresponding dispersion \(\phi\)?
Write the cumulant of order four as a function of \(\mu\) and \(\phi\).
Explain why the third cumulant is not generally equal to the third raw moment.
For fixed \(\phi\), how does the CV change when \(\mu\) is multiplied by four?