The Positive-Support Pitfall in Inverse-Gaussian Models

Why Support Matters

An inverse-Gaussian generalized linear model is not merely a model with a positive fitted mean. Its conditional response distribution assigns outcomes only to the positive half-line. This distinction matters whenever a data set contains zeros or negative values: changing the link or improving the numerical optimizer cannot make an ordinary inverse-Gaussian likelihood describe observations outside its response support.

The issue is often practical rather than mathematical. A zero may be a genuine outcome, a rounded measurement, a value below an assay's detection limit, or a data-entry convention. Those cases call for different likelihood contributions. A negative response, by contrast, is outside the inverse-Gaussian support and requires a model for a different response scale or a different data-generating process.

Definition: For mean parameter $\mu>0$ and shape parameter $\lambda>0$, an inverse-Gaussian response $Y$ has density $$ f(y;\mu,\lambda)=\sqrt{\frac{\lambda}{2\pi y^3}}\exp\!\left\{-\frac{\lambda(y-\mu)^2}{2\mu^2y}\right\}, \qquad y>0, $$ and density zero for $y<0$. It has no point mass at zero: $P(Y=0)=0$. In the dispersion parameterization, $\phi=1/\lambda>0$, so $\operatorname{Var}(Y)=\phi\mu^3$. In an ordinary inverse-Gaussian GLM, covariates may determine $\mu_i$, but each conditional response still has this positive support.

The boundary $y=0$ deserves care. The density formula is given for $y>0$, and an isolated density value can in general be changed without changing a continuous distribution. For the standard inverse-Gaussian density, its continuous extension to zero is zero. This is the usual density convention for evaluating the likelihood at an observed boundary value. More importantly, the distribution itself has no atom at zero. A zero is not an outcome that this model generates with positive probability.

The Boundary Density and the Likelihood Consequence

The boundary behavior follows directly from the density. Rewriting its exponent gives $$ -\frac{\lambda(y-\mu)^2}{2\mu^2y} =-\frac{\lambda y}{2\mu^2}+\frac{\lambda}{\mu}-\frac{\lambda}{2y}. $$ As $y$ approaches zero from above, the factor $y^{-3/2}$ grows, but the exponential factor $\exp\{-\lambda/(2y)\}$ shrinks faster than any power of $1/y$.

Result: For every fixed $\mu>0$ and $\lambda>0$, the standard inverse-Gaussian density satisfies $\lim_{y\downarrow 0}f(y;\mu,\lambda)=0$. Consequently, under the ordinary density convention, an exact zero contributes zero likelihood.

Proof. Set $t=1/y$. The terms in the density that depend on $y$, apart from constants, become $t^{3/2}\exp\{-\lambda t/2\}\exp\{-\lambda/(2\mu^2t)\}$. The last factor converges to 1 as $t$ increases without bound. For any $c>0$, $t^{3/2}e^{-ct}\to0$: for example, its logarithm is $(3/2)\log t-ct$, which tends to $-\infty$ because $\log t/t\to0$. Taking $c=\lambda/2>0$ proves the claimed limit. $\square$

Result: Suppose $Y_1,\ldots,Y_n$ are conditionally independent inverse-Gaussian responses with parameters $\mu_i>0$ and $\lambda_i>0$. If any observed $y_i<0$, the ordinary inverse-Gaussian likelihood is zero. If any observed $y_i=0$, its standard density likelihood is also zero. Thus, for data containing either kind of value, the log-likelihood is $-\infty$ for every admissible regression parameter value.

Proof. Conditional independence gives the likelihood $L=\prod_{i=1}^n f(y_i;\mu_i,\lambda_i)$. If $y_i<0$, the $i$th density is zero by the support definition. If $y_i=0$, it is zero under the continuous extension just established. In either case the product is zero, regardless of the values of the admissible $\mu_i$ and $\lambda_i$; taking its logarithm gives $-\infty$. $\square$

This result is about the response distribution, not the linear predictor. A link function and its predictor constraints determine which positive means can be represented. They do not add a mass at zero or extend the distribution to negative responses. An optimizer reporting convergence cannot remove this incompatibility.

Worked Example: An Exact Zero Makes the Ordinary Likelihood Zero

Suppose two independent observations are modeled with the same inverse-Gaussian parameters, $\mu=1$ and $\lambda=2$, and the recorded data are $y_1=0$ and $y_2=1$. For the positive observation, $$ f(1;1,2)=\sqrt{\frac{2}{2\pi}}\exp\!\left\{-\frac{2(1-1)^2}{2}\right\}=\frac{1}{\sqrt{\pi}}\approx0.5642. $$

For the zero, the density's boundary extension is $f(0;1,2)=0$. Therefore the joint likelihood is $$ L=f(0;1,2)f(1;1,2)=0\times\frac{1}{\sqrt{\pi}}=0, \qquad \ell=\log L=-\infty. $$

The positive observation has an ordinary, finite density contribution; the exact zero makes the full likelihood vanish. No alternative positive value of $\mu$ or $\lambda$ changes that conclusion.

When a Recorded Zero Represents an Interval

An instrument may report zero whenever a positive latent measurement is too small to resolve. In that case, zero is not the exact value of the latent response. It means that the response fell in a specified interval, such as $0<Y<L$, where $L$ is the reporting threshold. The appropriate likelihood contribution is the probability of that interval, not the density evaluated at zero.

Result: If a recorded zero means that the unobserved response satisfies $0<Y<L$, its likelihood contribution under an inverse-Gaussian model is $$ P(0<Y<L\mid\mu,\lambda)=\int_0^L f(y;\mu,\lambda)\,dy. $$ If the observation is instead recorded as “at or below $L$,” the contribution is $P(Y\le L)$; these probabilities agree here because the inverse-Gaussian distribution has no point mass at $L$.

Justification. The observed information is the event that the latent response lies in the reporting interval. A likelihood for coarsened data must assign that event its model probability. Since the response is continuous, this probability is the integral of its density over the interval. It is not $f(0;\mu,\lambda)$: a density describes probability per unit of response near a value, whereas the recorded value here identifies an interval.

Worked Example: A Zero Recorded Below a Threshold

Suppose an assay reports zero whenever a positive response is below one measurement unit. For a particular covariate pattern, assume $Y\sim\operatorname{IG}(\mu=1,\lambda=1)$. A reported zero then contributes $$ \int_0^1 \sqrt{\frac{1}{2\pi y^3}}\exp\!\left\{-\frac{(y-1)^2}{2y}\right\}\,dy $$ to the likelihood.

The upper limit is $1$, the assay threshold, rather than the recorded number $0$. The integral is a positive probability: the inverse-Gaussian density is positive for every $y$ strictly between zero and one, and the interval has positive length. In a sample with other exact positive observations $y_2,\ldots,y_n$, the full likelihood includes this interval probability multiplied by $\prod_{i=2}^n f(y_i;\mu_i,\lambda_i)$.

If the assay uses a different threshold, the upper limit must change accordingly. The observation mechanism therefore supplies essential likelihood information; treating the report as an exact zero discards that information and produces the wrong likelihood.

Genuine Zeros and Model Choice

If zero is a genuine possible outcome rather than a reporting code, a model needs to assign it positive probability. One option is a two-part model: a Bernoulli component describes whether the response is zero, and a positive-response component describes the response conditional on being nonzero. For example, let $P(Y=0\mid x)=\pi(x)$, and let $Y\mid(Y>0,x)$ have an inverse-Gaussian distribution with mean $\mu(x)>0$ and shape $\lambda>0$. This model has a point mass at zero and a continuous positive component.

For an observation $y$, its likelihood contribution is $\pi(x)$ if $y=0$, and $(1-\pi(x))f(y;\mu(x),\lambda)$ if $y>0$. This is not an ordinary inverse-Gaussian GLM: it is a model with an additional probability mechanism. It still does not accommodate negative responses. If negative values are scientifically meaningful, the response scale and distribution must be reconsidered rather than forcing those values into a positive-support model.

Worked Example: A Separate Probability for Structural Zeros

Consider three independent responses: $y_1=0$, $y_2=1$, and $y_3=2$. Use the two-part model with $\pi=0.2$ and a conditional positive-response distribution $\operatorname{IG}(\mu=1,\lambda=2)$. The zero contributes $0.2$. For $y=1$, $$ f(1;1,2)=\frac{1}{\sqrt{\pi}}\approx0.56419. $$ For $y=2$, $$ f(2;1,2) =\sqrt{\frac{2}{2\pi(2^3)}}\exp\!\left\{-\frac{2(2-1)^2}{2(1)^2(2)}\right\} =\frac{1}{\sqrt{8\pi}}e^{-1/2} \approx0.12099. $$

Each positive response also contributes its probability of belonging to the positive part, $1-\pi=0.8$. The joint likelihood is therefore $$ L=0.2\,[0.8f(1;1,2)]\,[0.8f(2;1,2)] =0.2(0.8)^2(0.56419)(0.12099)\approx0.00874. $$

Unlike the ordinary inverse-Gaussian likelihood, this model gives the observed zero a positive likelihood contribution. Whether a two-part model is suitable depends on the scientific meaning of the zeros and the study design; its extra component should represent a plausible process, not merely a numerical workaround.

Common Mistakes and Practical Checks

A useful first step is to determine what each recorded zero means before fitting any model. The following checks separate the main cases.

Exact zero treated as an ordinary inverse-Gaussian value. The model has no atom at zero, and its standard boundary density is zero. An exact zero makes the ordinary likelihood zero.
Below-threshold value treated as exact zero. If the instrument reports an interval, use the interval probability determined by the threshold and the measurement protocol.
Small constant substituted for every zero. Replacing zero by $\varepsilon>0$ changes the recorded data. Moreover, for fixed $\mu,\lambda>0$, $f(\varepsilon;\mu,\lambda)\to0$ as $\varepsilon\downarrow0$. The result therefore depends on an arbitrary choice of $\varepsilon$, unless that value has a defensible measurement-based meaning.
Zero-valued rows deleted without justification. Deleting them changes the analyzed sample and can alter the target population or the data-generating process being modeled. It is not a likelihood correction for a support mismatch.
Negative values handled by changing the link. A link changes the relation between predictors and the mean; it does not change the response support. A scientifically meaningful negative response requires a compatible response model.

A zero's origin is often more important than its numeric appearance. A true structural zero, a rounded value, and a censored value may all be printed as “0,” yet imply different likelihoods. Recording the measurement threshold, rounding rule, and reason a zero can occur makes the modeling decision explicit and reproducible.

Key takeaway: An ordinary inverse-Gaussian GLM models strictly positive responses. Exact zeros and negative values are support violations; zeros standing for intervals require interval probabilities, while genuine zero outcomes require a model with positive probability at zero.

Check Your Understanding

Use the response mechanism, not just the printed value, to answer each question.

For an inverse-Gaussian response with $\mu>0$ and $\lambda>0$, what happens to the standard density as $y\downarrow0$, and what does that imply for an exact zero in the ordinary likelihood?
A negative response is observed. Why can changing from a log link to another inverse-Gaussian link not make the ordinary model assign it positive likelihood?
An instrument reports zero for every response below a known threshold $L$. What likelihood contribution should replace evaluation of the density at zero?
Why can replacing zero by a small positive constant produce an analysis that depends on an arbitrary modeling choice?
In a two-part model with a separate zero probability, which component contributes to a zero observation, and which components contribute to a positive observation?