Why the Boundary Needs Careful Interpretation
A variance component can equal zero at the boundary of its parameter space. In a GLMM, that boundary has a clear model interpretation: a random effect with variance zero has no between-group variation under the model. But a prior that allows values arbitrarily close to zero does not necessarily assign positive probability to the value zero itself. This distinction matters when describing prior beliefs about negligible variation, inspecting prior densities, and interpreting posterior output.
A second complication is that a prior density’s appearance near zero depends on the parameter used to express it. A density on a random-effect standard deviation can be finite at zero, while the corresponding density on the variance diverges there. That divergence does not imply a point mass or an infinite probability. Probabilities over intervals are the comparable quantities.
For example, consider a random intercept \(b_j\sim N(0,v)\), where \(j\) indexes groups. The standard deviation \(\tau\) and variance \(v\) describe the same scale of group-level variation, but their numerical units differ: if \(b_j\) is measured in outcome-link units, \(\tau\) has those units and \(v\) has squared units. A density on \(v\) therefore has units inverse to squared link units; a density on \(\tau\) has units inverse to link units. Their density heights should not be compared as if they were probabilities.
Transforming a Prior Between Variance and Standard Deviation
Suppose the prior density on \(\tau\geq 0\) is \(\pi_\tau(\tau)\). Because \(v=\tau^2\), the inverse transformation is \(\tau=\sqrt{v}\), with derivative \(d\tau/dv=1/(2\sqrt{v})\) for \(v>0\). The change-of-variables formula gives the density on the variance scale.
This formula applies to continuous probability on the positive part of the parameter space; it does not create or remove a point mass at zero. If \(\pi_\tau(\tau)\) approaches a finite, positive value \(c\) as \(\tau\) approaches zero, then \(\pi_v(v)\) behaves approximately as \(c/(2\sqrt{v})\). The variance density diverges, but the divergence is integrable. In fact, the probability of a small interval satisfies
As \(\epsilon\) shrinks to zero, this probability also shrinks to zero. Thus a density that becomes arbitrarily high at the boundary can still describe a continuous prior with no probability atom there. The divergence arises because a short interval of variance corresponds to a different interval length on the standard-deviation scale.
Worked Example: A Half-Normal Prior on the Standard Deviation
Suppose the random-effect standard deviation has a half-normal prior with scale \(0.5\) link units: \(\tau\sim\text{Half-Normal}(0,0.5^2)\). Find the prior probability that the variance is at most \(0.01\) squared link units, and describe the induced variance density near zero.
Since \(v=\tau^2\), the event \(v\leq0.01\) is equivalent to \(\tau\leq\sqrt{0.01}=0.1\). A half-normal variable is the absolute value of a centered normal variable, so
Thus the prior assigns about 15.85% probability to variances between zero and \(0.01\) squared link units. The half-normal density on \(\tau\) is \(\pi_\tau(\tau)=\sqrt{2/\pi}(0.5)^{-1}\exp[-\tau^2/(2(0.5)^2)]\). Transforming it gives
The density tends to infinity as \(v\) approaches zero, but the calculated interval probability is finite, and \(P(v=0)=0\). The boundary height alone does not say how much probability lies in a specified neighborhood.
What Different Variance-Scale Priors Imply
A prior specified directly on \(v\) need not have the same boundary behavior as a prior specified on \(\tau\). For instance, an exponential prior on variance with rate \(\lambda>0\) has density \(\pi_v(v)=\lambda e^{-\lambda v}\) for \(v\geq0\). Its density is finite at zero, with \(\pi_v(0)=\lambda\), and its near-zero probability is \(1-e^{-\lambda\epsilon}\), approximately \(\lambda\epsilon\) for small \(\epsilon\). By contrast, an exponential prior on standard deviation induces a variance density proportional to \(v^{-1/2}\) near zero, when its standard-deviation density is positive at zero. Those are different prior distributions, not two descriptions of the same one.
The following example compares their probabilities on a fixed interval. The chosen rate values are illustrative; a useful prior scale must be calibrated to the model’s units and plausible group heterogeneity.
Worked Example: Exponential Priors on Two Different Scales
Compare (a) an exponential prior with rate \(2\) per link unit on \(\tau\), and (b) an exponential prior with rate \(4\) per squared link unit on \(v\). Calculate \(P(v\leq0.01)\) under each prior.
(a) Prior on \(\tau\). Its cumulative distribution function is \(P(\tau\leq t)=1-e^{-2t}\) for \(t\geq0\). Since \(v\leq0.01\) means \(\tau\leq0.1\),
The induced variance density is \(\pi_v(v)=e^{-2\sqrt{v}}/\sqrt{v}\) for \(v>0\), which diverges at zero but integrates to the stated probability.
(b) Prior on \(v\). An exponential variance prior with rate \(4\) has
The first prior places about 18.13% of its probability below \(0.01\); the second places about 3.921%. Both are continuous and assign zero probability to \(v=0\), but their near-zero probability differs substantially. Also note that the rates have different units because the priors are defined on different scales.
A useful general description is to focus on the cumulative probability near the boundary. If a variance density behaves as \(C v^{a-1}\) for small positive \(v\), where \(C>0\) and \(a>0\), then
The exponent determines how rapidly the near-zero probability falls as the interval contracts. A finite positive variance density at zero corresponds to \(a=1\), giving probability proportional to \(\epsilon\). A \(v^{-1/2}\) divergence corresponds to \(a=1/2\), giving probability proportional to \(\sqrt{\epsilon}\). In both cases, the probability tends to zero as \(\epsilon\) tends to zero; neither behavior is an atom.
Worked Example: Separating a Continuous Prior From a Point Mass
Suppose a prior is defined as follows: with probability \(0.20\), set \(v=0\); otherwise draw \(v\) from an exponential distribution with rate \(4\) per squared link unit. Find the probability that \(v\leq0.01\), and distinguish it from the probability of exactly zero.
The point mass contributes \(0.20\) to the interval event. Conditional on drawing from the continuous component, the probability of \(v\leq0.01\) is \(1-e^{-0.04}\approx0.0392106\). Therefore,
But the probability of exactly zero is \(P(v=0)=0.20\), not \(0.2314\). The remaining probability, about \(0.03137\), is continuous probability on positive variances up to \(0.01\). This example illustrates the distinction between near-zero probability and an atom; it does not address how to test competing variance-component models.
Assumptions, Checks, and Failure Modes
The transformations and calculations above assume that \(v\) is a nonnegative scalar variance component, \(\tau=\sqrt{v}\), and the stated prior is a probability distribution on that scale. The change-of-variables calculation applies for \(v>0\), with any mass at zero handled separately. When comparing probabilities, use the same interval and the same variance units. In a GLMM, whether that interval is substantively “near zero” depends on the link scale, the model, and the scientific setting.
Several checks help catch misleading interpretations:
- Identify the scale. Record whether a prior is specified on variance, standard deviation, or precision before interpreting its density. A density on one scale cannot be compared directly with a density on another.
- Check the Jacobian. For \(v=\tau^2\), the factor \(1/(2\sqrt{v})\) is essential. Omitting it gives an incorrect induced density.
- Compute interval probabilities. Evaluate \(P(v\leq\epsilon)\) for meaningful values of \(\epsilon\), rather than inferring probability from the density’s value at zero.
- Check normalization and the boundary separately. A density that diverges near zero can still integrate to one. A point mass, however, must be represented separately from the continuous density.
- Inspect the scale and units. A small numerical value of \(v\) may not imply negligible variation if the model’s link-scale units or prior scale make that value consequential.
Numerical plots can also mislead. A histogram or density curve may appear to show a spike at zero when it is depicting a continuous distribution concentrated near the boundary, or when the first bin includes a range of positive values. Conversely, a plot with a finite-looking boundary may conceal an actual atom if the plotting routine does not display discrete mass separately. Check how the prior is represented and calculate the probability of the exact point and of an interval as distinct quantities.
Why the Distinction Matters
In Bayesian GLMM analysis, the prior combines with the likelihood to produce a posterior distribution. If a prior is continuous on \(v\), the prior probability of exactly zero is zero. Under an ordinary likelihood with a finite density over positive \(v\), the posterior remains continuous with zero probability at the single point \(v=0\), even if substantial posterior probability lies near zero. This is different from a prior that explicitly assigns positive probability to zero. The prior choice can therefore affect whether inference describes a small positive variance or gives positive probability to the model with no random-effect variation.
These two questions should not be conflated: “How much prior probability lies below a practically chosen threshold?” is answered by an interval probability; “How much prior probability is assigned to exactly zero?” is answered by the mass at that point. A high density at zero answers neither question by itself.
| Prior description | Behavior at zero | Probability at exactly zero | Useful quantity to report |
|---|---|---|---|
| Continuous prior on variance with finite density | Finite boundary density | Zero | \(P(v\leq\epsilon)\) |
| Continuous prior on standard deviation with positive density at zero | Induced variance density diverges as \(v^{-1/2}\) | Zero | \(P(\tau\leq\sqrt{\epsilon})\) |
| Prior with an explicit atom at \(v=0\) | Discrete mass plus any continuous component | Positive by specification | Report the atom and continuous interval probability separately |
Check Your Understanding
Answer each question, then compare with the solutions below.
- A continuous prior on \(\tau\) has a finite positive density at zero. Does the induced density on \(v=\tau^2\) have to be finite at zero? Does the prior assign positive probability to \(v=0\)?
- For an exponential prior on \(v\) with rate \(3\) per variance unit, calculate \(P(v\leq0.02)\).
- A half-normal prior on \(\tau\) has scale \(1\) link unit. Express \(P(v\leq0.04)\) in terms of the standard normal cumulative distribution function \(\Phi\), and give its approximate value.
- Explain why a density that tends to infinity at zero does not establish a point mass there.
- A prior assigns probability \(0.10\) to \(v=0\) and otherwise uses a continuous distribution. What is known about \(P(v=0)\), and why can \(P(v\leq\epsilon)\) be larger?
Solutions. (1) No: the transformation gives \(\pi_v(v)=\pi_\tau(\sqrt v)/(2\sqrt v)\), which diverges as \(v\) approaches zero when \(\pi_\tau(0)>0\). Nevertheless, \(P(v=0)=0\) for a continuous prior. (2) \(P(v\leq0.02)=1-e^{-3(0.02)}=1-e^{-0.06}\approx0.05824\). (3) The event is \(\tau\leq0.2\); hence \(P(v\leq0.04)=2\Phi(0.2)-1\approx0.1585\). (4) A diverging density can be integrable, so the probability of a shrinking interval can tend to zero even as the density grows without bound; a point mass is a separate positive probability assigned to one value. (5) \(P(v=0)=0.10\). For \(\epsilon>0\), \(P(v\leq\epsilon)\) includes that \(0.10\) and may also include positive probability from the continuous component on \((0,\epsilon]\).