Half-Student-t Priors for Random-Effect Standard Deviations

Why Use a Half-Student-t Prior?

A random-effect standard deviation must be nonnegative, but the data may not determine its plausible range precisely, especially when there are few groups. A prior for that standard deviation therefore matters: it helps define which amounts of group-to-group variation receive substantial prior support. The half-Student-t family provides a flexible choice. Its degrees of freedom control tail heaviness, while its scale sets the units and broad magnitude of the prior.

Consider a GLMM with a normally distributed random intercept $b_j$ for group $j$, conditional on its standard deviation $\sigma$:

b_j\mid \sigma \sim N(0,\sigma^2),\qquad \sigma>0.

The prior discussed here is assigned to $\sigma$, not directly to the individual $b_j$. For a random intercept in a logistic model, $\sigma$ is measured in log-odds units. For a random slope, it is measured in the units of that slope coefficient. The interpretation and numerical scale must therefore match the model parameter.

Definition: Let $T$ have a standard Student-t distribution with $\nu>0$ degrees of freedom, with density $f_\nu$ and cumulative distribution function $T_\nu$. For scale $s>0$, a half-Student-t prior for a standard deviation is the distribution of $\sigma=s|T|$. Its density is $$ \pi(\sigma\mid \nu,s)=\frac{2}{s}f_\nu\left(\frac{\sigma}{s}\right),\qquad \sigma>0. $$ Here $\nu$ is the degrees of freedom and $s$ is the scale parameter.

The factor of 2 folds the symmetric Student-t density onto the nonnegative half-line. Thus the density integrates to one over $\sigma>0$. The prior is proper for every $\nu>0$ and $s>0$. It assigns zero probability to any exact value, including $\sigma=0$; it is a continuous prior, not a model that assigns a separate probability to zero variance.

Tail Probabilities and Prior Interpretation

A useful way to understand the prior is to calculate the probability it assigns to standard deviations above a substantively meaningful threshold $c>0$. Because $\sigma=s|T|$, the event $\sigma>c$ is the event $|T|>c/s$. By symmetry of the standard Student-t distribution,

P(\sigma>c) =P\left(|T|>\frac{c}{s}\right) =2\left\{1-T_\nu\left(\frac{c}{s}\right)\right\}.

This formula separates the two controls. The ratio $c/s$ compares the threshold to the prior scale. At fixed $s$, decreasing $\nu$ makes the tails heavier, so sufficiently large standard deviations are more plausible. At fixed $\nu$, increasing $s$ increases the probability of exceeding a fixed positive threshold. These are statements about prior probabilities, not posterior conclusions about the data.

Worked Example: A Large Random-Intercept Standard Deviation

Suppose a hypothetical logistic random-intercept model uses $\sigma\sim\text{half-}t(\nu=3,s=0.5)$, with both $\sigma$ and $s$ measured in log-odds units. Find the prior probability that $\sigma$ exceeds $1.5$ log-odds units.

The standardized threshold is $1.5/0.5=3$. The tail formula gives

P(\sigma>1.5) =2\{1-T_3(3)\} =2(1-0.9711655) \approx 0.05767.

Thus this prior assigns about $5.767\%$ probability to a random-intercept standard deviation above $1.5$ log-odds units. This is a statement about the prior distribution. It does not say there is a $5.767\%$ probability that a fitted model’s unknown standard deviation exceeds that threshold after observing data.

The scale $s$ is not generally the prior mean or standard deviation of $\sigma$. For $\nu>1$, its prior mean is

E(\sigma) =s\frac{\sqrt{\nu}\,\Gamma\left((\nu-1)/2\right)} {\sqrt{\pi}\,\Gamma(\nu/2)}.

For $\nu>2$, its second moment is $E(\sigma^2)=s^2\nu/(\nu-2)$; for $\nu\leq 2$, that second moment is infinite. The scale therefore gives a reference magnitude, while the tail behavior can make the mean or variance much larger than the scale suggests—or make a moment fail to exist altogether. In practice, threshold probabilities and quantiles are often more directly interpretable than moments.

How Degrees of Freedom and Scale Change the Tails

For a fixed scale, smaller $\nu$ gives a heavier tail than larger $\nu$. As $\nu$ increases without bound, the standard Student-t distribution approaches the standard normal distribution. Accordingly, a half-Student-t prior with fixed scale approaches a half-normal prior with that same scale. The comparison is especially informative for thresholds several scale units from zero.

Worked Example: Comparing Degrees of Freedom and Scale

Suppose $\sigma$ is measured in log-odds units and the threshold of interest is $1.5$. First compare $\nu=1$ and $\nu=3$ at the same scale $s=0.5$. The standardized threshold is 3 for both priors. For $\nu=1$, the standard Student-t distribution is Cauchy, so

P(\sigma>1.5) =2\left\{1-\left(\frac{1}{2}+\frac{\arctan(3)}{\pi}\right)\right\} =1-\frac{2\arctan(3)}{\pi} \approx 0.2048.

For $\nu=3$, the calculation in the previous example gives $P(\sigma>1.5)\approx0.05767$. At this threshold, the $\nu=1$ prior assigns substantially more probability to large standard deviations than the $\nu=3$ prior, even though both have the same scale.

Now hold $\nu=3$ fixed and increase the scale to $s=0.75$. The standardized threshold becomes $1.5/0.75=2$. Using $T_3(2)\approx0.93036$,

P(\sigma>1.5) =2\{1-T_3(2)\} \approx 2(1-0.93036) \approx 0.1393.

Increasing the scale from $0.5$ to $0.75$ log-odds units raises the prior probability of exceeding $1.5$ from about $5.767\%$ to about $13.93\%$. The two comparisons isolate different choices: $\nu$ changes tail heaviness, and $s$ changes the scale relative to the threshold.

A half-normal prior is a lighter-tailed alternative. If it is formed as the absolute value of a $N(0,s^2)$ variable, its tail probability is $2\Phi(-c/s)$, where $\Phi$ is the standard normal cumulative distribution function. At $\nu=3$ and $s=0.5$, the same threshold $c=1.5$ gives

P(\sigma>1.5)=2\Phi(-3)\approx0.002700.

That is about $0.2700\%$, compared with $5.767\%$ for the half-Student-t prior with $\nu=3$ and $s=0.5$. Matching scale parameters does not make the two priors equivalent: the tail probability can differ sharply. More generally, the Student-t density has polynomially decaying tails, whereas the normal density decays exponentially in the squared standardized value.

Worked Example: Finding a Prior’s 95th Percentile

Suppose a hypothetical prior for a random-intercept standard deviation is half-Student-t with $\nu=3$ and $s=0.5$ log-odds units. Find its 95th percentile.

For $q>0$, the half-Student-t cumulative distribution function is $P(\sigma\leq q)=2T_\nu(q/s)-1$. Set this equal to $0.95$:

2T_3(q/0.5)-1=0.95 \quad\Longrightarrow\quad T_3(q/0.5)=0.975.

The 97.5th percentile of a standard Student-t distribution with 3 degrees of freedom is approximately $3.182$. Therefore,

q=0.5(3.182)\approx1.591.

The prior’s 95th percentile is about $1.591$ log-odds units: under this prior, 95% of its probability lies below that value and 5% lies above it. This is a prior quantile, not a 95% posterior credible interval.

Justification of the Tail Formula

The tail calculation follows directly from the definition and symmetry; it does not require an approximation. Since $s>0$, dividing the event $\sigma>c$ by $s$ preserves the inequality. Since $\sigma=s|T|$, the event becomes $|T|>c/s$. The two disjoint possibilities are $T>c/s$ and $T<-c/s$. Symmetry makes their probabilities equal, and each is $1-T_\nu(c/s)$. Hence

\begin{aligned} P(\sigma>c) &=P\left(T>\frac{c}{s}\right)+P\left(T<-\frac{c}{s}\right)\\ &=1-T_\nu\left(\frac{c}{s}\right) +1-T_\nu\left(\frac{c}{s}\right)\\ &=2\left\{1-T_\nu\left(\frac{c}{s}\right)\right\}. \end{aligned}

The same symmetry gives the cumulative distribution function for $q\geq0$: $P(\sigma\leq q)=2T_\nu(q/s)-1$. The factor of 2 in both expressions is essential: a half-Student-t prior uses both tails of the original symmetric distribution, folded onto the positive axis.

Specification Checks, Comparisons, and Limitations

Before using a half-Student-t prior, state the random-effect coefficient and its units, then assess thresholds in those same units. Useful checks include calculating $P(\sigma>c)$ for one or more values of $c$, and calculating a quantile such as the 95th percentile. Those summaries make it easier to see whether the chosen tail and scale imply plausible amounts of group heterogeneity.

The main lighter-tailed comparison is the half-normal prior. A half-Student-t prior can be useful when it is desirable to retain appreciable prior support for large standard deviations without making the prior improper. However, a heavy tail is not automatically safer: it can place more prior probability on implausibly large variation. The model, measurement scale, and scientific context still determine whether that support is reasonable. A prior on the variance rather than the standard deviation is another possible parameterization; the induced distribution on $\sigma$ must be considered before comparing its implications.

A continuous half-Student-t prior also cannot represent a prior hypothesis that the variance is exactly zero with positive probability. Its density is defined over positive values, and $P(\sigma=0)=0$. In addition, prior probabilities for large scales do not show whether the observed data can estimate $\sigma$ precisely. Sparse groups or weak information can leave posterior conclusions sensitive to the prior. These are reasons to examine the model’s implications and uncertainty, not reasons to interpret a prior tail probability as a data-based finding.

Common pitfall: Saying “the scale is $0.5$, so standard deviations above $1.5$ are negligible” ignores the degrees of freedom. With $s=0.5$ and $\nu=3$, the probability above $1.5$ is about $0.05767$; the half-normal prior with the same scale gives about $0.002700$. Calculate the tail probability for the specified family instead of inferring it from the scale alone.

Check Your Understanding

Answer each question using the half-Student-t tail and cumulative distribution formulas. Solutions follow the questions.

For $\nu=3$, $s=0.5$, and threshold $c=1.0$, what is $P(\sigma>c)$? Use $T_3(2)\approx0.93036$.
If $s$ increases while $\nu$ and a positive threshold $c$ stay fixed, what happens to $P(\sigma>c)$, and why?
For a half-Student-t prior with $\nu=3$ and $s=0.5$, express the 90th percentile in terms of a standard Student-t quantile.
Why is it incorrect to interpret $s$ as the prior standard deviation of $\sigma$ for every value of $\nu$?

Solutions. (1) The standardized threshold is $1.0/0.5=2$, so $P(\sigma>1.0)=2\{1-T_3(2)\}\approx2(1-0.93036)=0.1393$. (2) The ratio $c/s$ decreases as $s$ increases. The Student-t upper-tail probability at that smaller positive threshold increases, so the probability of exceeding $c$ increases. (3) Set $2T_3(q/0.5)-1=0.90$, giving $q=0.5T_{3,0.95}$, where $T_{3,0.95}$ denotes the 95th percentile of a standard Student-t distribution with 3 degrees of freedom. (4) $s$ is the distribution’s scale parameter. The second moment of $\sigma$ exists only for $\nu>2$ and then is $s^2\nu/(\nu-2)$, not $s^2$; for $\nu\leq2$, it is infinite.