Inverse-Gaussian Density and Parameterizations

Introduction

Inverse-Gaussian models are used for positive continuous responses, including measurements such as waiting times and biological concentrations. Their notation can be confusing because the same distribution is commonly written with either a dispersion parameter or a shape parameter. The formulas are equivalent, but the two parameters are reciprocals, not interchangeable labels.

This tutorial defines the density using the mean $\mu$ and dispersion $\phi$, makes its strictly positive response domain explicit, and shows how to translate calculations to the shape convention $\lambda=1/\phi$. The focus is on the density and parameter conversion, not on deriving its moments.

Density and Parameter Meaning

Definition: For $\mu>0$ and $\phi>0$, the inverse-Gaussian distribution with mean parameter $\mu$ and dispersion parameter $\phi$ has density $$ f(y;\mu,\phi) = \frac{1}{\sqrt{2\pi\phi y^3}} \exp\!\left( -\frac{(y-\mu)^2}{2\phi\mu^2y} \right), \qquad y>0. $$ The density is zero for $y\leq 0$. The parameter $\mu$ is the mean, and $\phi$ is the dispersion parameter.

Both parameters must be strictly positive. The response is also strictly positive with probability one: $P(Y>0)=1$, and $P(Y\leq 0)=0$. The density is positive at every $y>0$. At the boundary $y=0$, the density is defined to be zero; there is no point mass there. In the topological definition of support, the closure of the positive density region includes zero, so the support is sometimes written $[0,\infty)$. For the response values permitted by this model, the important condition is $y>0$.

The dispersion parameter occurs in both the prefactor and the exponent. It is not itself a standard deviation. In particular, replacing $\phi$ by its square root in the density changes the distribution. The shape convention instead uses a parameter $\lambda>0$, related to dispersion by $\lambda=1/\phi$.

Two Equivalent Parameterizations

In the shape convention, the same density is written

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

Here $\lambda$ is often called the shape parameter. To translate a distribution specified with dispersion $\phi$, use $\lambda=1/\phi$; to translate back, use $\phi=1/\lambda$. The mean parameter $\mu$ stays the same in either convention.

Theorem: For every $y>0$, $\mu>0$, and $\phi>0$, setting $\lambda=1/\phi$ makes the dispersion and shape formulas identical: $$ f(y;\mu,\phi)=f(y;\mu,\lambda). $$ Conversely, setting $\phi=1/\lambda$ recovers the dispersion formula from the shape formula.

Proof. Substitute $\lambda=1/\phi$ into the shape density. Its prefactor becomes

\sqrt{\frac{\lambda}{2\pi y^3}} = \sqrt{\frac{1}{2\pi\phi y^3}},

and its exponent becomes

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

These are exactly the prefactor and exponent in the dispersion formula, so the densities agree for every $y>0$. Since $\phi=1/\lambda$ is the same reciprocal relation solved in the other direction, the reverse conversion also recovers the original formula. $\square$

The density formula must integrate to one. The following calculation checks its normalization and also confirms that the reciprocal parameterization is a probability density when its parameters are positive.

Result: For $a>0$ and $b>0$, $$ \int_0^\infty y^{-3/2}e^{-a/y-by}\,dy = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}. $$

Derivation. Make the substitution $t=\sqrt{a/y}$, so $y=a/t^2$. Reversing the limits after substituting gives

\int_0^\infty y^{-3/2}e^{-a/y-by}\,dy = \frac{2}{\sqrt a} \int_0^\infty e^{-t^2-ab/t^2}\,dt.

Let $c=\sqrt{ab}>0$ and define $J(c)=\int_0^\infty e^{-t^2-c^2/t^2}\,dt$. Differentiating under the integral is valid on any compact interval of positive $c$: near zero the differentiated integrand is bounded by a constant times $t^{-2}e^{-c_0^2/t^2}$ for some $c_0>0$, and near infinity it is bounded by a constant times $t^{-2}e^{-t^2}$. Both bounds are integrable. Thus

J'(c)=-2c\int_0^\infty t^{-2}e^{-t^2-c^2/t^2}\,dt.

In the integral on the right, substitute $u=c/t$. This gives

\int_0^\infty t^{-2}e^{-t^2-c^2/t^2}\,dt = \frac{1}{c}\int_0^\infty e^{-u^2-c^2/u^2}\,du = \frac{J(c)}{c}.

Therefore $J'(c)=-2J(c)$. Also, as $c$ decreases to zero, the integrand defining $J(c)$ converges pointwise on $t>0$ to $e^{-t^2}$ and is bounded by $e^{-t^2}$. The dominated convergence theorem gives $J(c)\to\sqrt{\pi}/2$. Solving the differential equation with this limit yields $J(c)=(\sqrt{\pi}/2)e^{-2c}$. Substitution into the transformed integral proves the result. $\square$

To apply the result to the inverse-Gaussian density, expand the quadratic term:

-\frac{\lambda(y-\mu)^2}{2\mu^2y} = \frac{\lambda}{\mu} -\frac{\lambda}{2y} -\frac{\lambda y}{2\mu^2}.

The density integral is therefore

\sqrt{\frac{\lambda}{2\pi}}e^{\lambda/\mu} \int_0^\infty y^{-3/2} \exp\!\left(-\frac{\lambda}{2y}-\frac{\lambda y}{2\mu^2}\right)\,dy.

Use the proved integral with $a=\lambda/2$ and $b=\lambda/(2\mu^2)$. Since $\mu,\lambda>0$, $\sqrt{ab}=\lambda/(2\mu)$, so the integral equals $\sqrt{2\pi/\lambda}\,e^{-\lambda/\mu}$. Multiplying by the factor outside it gives one. Thus the density is normalized.

Worked Parameter Conversions

Worked Example: From Dispersion to Shape

Suppose an inverse-Gaussian response has $\mu=4$ and $\phi=0.25$. Convert to the shape convention and calculate the density at $y=2$.

Convert the parameter. Since $\lambda=1/\phi$,

\lambda=\frac{1}{0.25}=4.

Evaluate the density. Using the shape formula,

f(2;4,4) = \sqrt{\frac{4}{2\pi(2)^3}} \exp\!\left(-\frac{4(2-4)^2}{2(4)^2(2)}\right).

The prefactor is $\sqrt{4/(16\pi)}=1/(2\sqrt{\pi})$. The exponent is $-16/64=-1/4$. Hence

f(2;4,4)=\frac{e^{-1/4}}{2\sqrt{\pi}}\approx 0.2197.

The same result follows by substituting $\phi=0.25$ directly into the dispersion formula.

Worked Example: From Shape to Dispersion

A distribution is specified by $\mu=3$ and $\lambda=12$. Express it using dispersion, then evaluate its density at $y=6$.

Convert the parameter. The reciprocal conversion gives

\phi=\frac{1}{\lambda}=\frac{1}{12}.

Calculate the density. In the shape convention,

f(6;3,12) = \sqrt{\frac{12}{2\pi(6)^3}} \exp\!\left(-\frac{12(6-3)^2}{2(3)^2(6)}\right).

The prefactor is $\sqrt{12/(432\pi)}=1/(6\sqrt{\pi})$. The exponent is $-108/108=-1$. Therefore

f(6;3,12)=\frac{e^{-1}}{6\sqrt{\pi}}\approx 0.0346.

Using $\phi=1/12$ in the dispersion formula gives the same prefactor and exponent.

Worked Example: Checking Both Forms at the Same Response

Take $\mu=2$ and $\phi=0.5$. First convert to shape and then evaluate the density at $y=1$ using each form.

The converted shape is $\lambda=1/0.5=2$. In the dispersion form,

f(1;2,0.5) = \frac{1}{\sqrt{2\pi(0.5)(1)^3}} \exp\!\left(-\frac{(1-2)^2}{2(0.5)(2)^2(1)}\right) = \frac{1}{\sqrt{\pi}}e^{-1/4}.

In the shape form,

f(1;2,2) = \sqrt{\frac{2}{2\pi(1)^3}} \exp\!\left(-\frac{2(1-2)^2}{2(2)^2(1)}\right) = \frac{1}{\sqrt{\pi}}e^{-1/4} \approx 0.4394.

The matching results verify the conversion for this numerical case.

Common Mistakes and Practical Checks

Before evaluating or reporting an inverse-Gaussian model, check which convention is being used. A parameter named $\lambda$ in one source may be the reciprocal of a parameter named $\phi$ in another. Carrying the name across without converting its value changes the density.

Using the same numerical value for $\phi$ and $\lambda$. If $\phi=0.25$, then $\lambda=4$, not $0.25$. Apply the reciprocal conversion before using a formula in the other convention.
Putting zero or a negative response into the density. The model assigns density only for $y>0$; zero and negative response values are outside its response domain.
Calling dispersion a standard deviation. The symbol $\phi$ appears directly in the density. It is not replaced by $\sqrt{\phi}$.
Changing the mean during conversion. Converting from $\phi$ to $\lambda$ changes only the second parameter. The mean parameter $\mu$ is unchanged.

A useful dimensional check reinforces the distinction: if $y$ is measured in units of a response, then $\mu$ has those same units, while $\lambda$ has response units and $\phi$ has inverse-response units. Their reciprocal relationship is consistent with the exponent being dimensionless.

Key takeaway: The inverse-Gaussian density is defined for strictly positive responses and positive parameters. Its mean parameter is $\mu$, and its two common second-parameter conventions are linked by exactly $\lambda=1/\phi$.

Check Your Understanding

Use the density definition and reciprocal conversion to answer each question.

If $\phi=0.2$, what is the corresponding shape parameter $\lambda$?
If a source reports $\lambda=5$, what dispersion value gives the same distribution?
Which response values are assigned positive density by the inverse-Gaussian model?
For $\mu=2$, $\phi=0.5$, and $y=1$, write the density expression before simplifying it.
Explain why using $\phi$ unchanged in the shape formula generally gives a different density.