Transforming Random-Effect Predictions to Odds and Rates

Why Transform Random-Effect Predictions?

In a generalized linear mixed model, a group’s random effect is added to the linear predictor. In logit and log-rate models, that predictor is not itself an odds ratio or a rate ratio: it is on a logarithmic scale. Exponentiation converts a log-scale group deviation into a multiplicative comparison that is often easier to communicate. The interpretation depends on which random-effect terms are included and which covariates are held fixed.

There is a second issue. Exponentiating a predicted log-scale effect gives a transformed point prediction, but it does not generally give the average multiplicative effect when the predicted effect is uncertain. Keeping those two quantities separate is essential for both reporting and prediction.

Model and Interpretation

Definition: Let \(b_j\) denote the vector of random effects for group \(j\), and let \(\hat b_j\) denote its predicted value. In a logit model, the group-specific linear predictor is \(\eta_{ij}=x_{ij}^{\mathsf T}\beta+z_{ij}^{\mathsf T}b_j\), with \(P(Y_{ij}=1\mid b_j,x_{ij})=\operatorname{logit}^{-1}(\eta_{ij})\). In a log-rate model, \(\log(\mu_{ij})=\log(e_{ij})+x_{ij}^{\mathsf T}\beta+z_{ij}^{\mathsf T}b_j\), where \(e_{ij}\) is exposure and \(\mu_{ij}\) is the expected count. A predicted group deviation \(z_{ij}^{\mathsf T}\hat b_j\) is on the link scale; exponentiating it gives a multiplicative effect on the odds or rate, respectively.

The usual assumptions for these interpretations are that the chosen link and conditional outcome distribution are appropriate, the random-effect structure represents the intended group variation, and the fixed and random effects are combined as specified in the model. For a common Gaussian random-effects model, \(b_j\) is modeled as having mean zero and covariance matrix \(G\) across groups. The zero value is therefore the model’s reference deviation, not necessarily an observed group. Predictions such as \(\hat b_j\) are conditional on the observed data and fitted model; they are not the group’s known true effect.

For a random intercept \(b_{0j}\), the odds multiplier is \(\exp(\hat b_{0j})\). For a random intercept and slope, the deviation at covariate value \(x\) is \(\hat b_{0j}+x\hat b_{1j}\), so the odds multiplier at that value is \(\exp(\hat b_{0j}+x\hat b_{1j})\). It is not generally valid to exponentiate the intercept alone and call the result the group’s effect at every covariate value.

Worked Example: A Predicted Logit Deviation at Two Covariate Values

Suppose a hypothetical binary-outcome model has a random intercept and random slope for a covariate \(x\), and a particular group has predictions \(\hat b_{0j}=-0.40\) and \(\hat b_{1j}=0.25\). The covariate is measured in tens of a unit, and the fixed effects are held the same for the group comparison. Find the group’s odds multiplier relative to the zero-random-effect reference at \(x=0\) and \(x=2\).

At \(x=0\), the group deviation is \(-0.40+0(0.25)=-0.40\). Its conditional odds multiplier is

\exp(-0.40)=0.6703.

Thus, at \(x=0\), the group’s conditional odds are estimated to be about 0.670 times the reference odds, or about 33.0% lower. At \(x=2\), the deviation is \(-0.40+2(0.25)=0.10\), giving

\exp(0.10)=1.105.

At that covariate value, the group’s conditional odds are estimated to be about 1.105 times the reference odds, or about 10.5% higher. The direction changes because the predicted random slope offsets the negative intercept deviation as \(x\) increases. These are comparisons of odds at fixed covariates, not comparisons of probabilities and not effects of changing \(x\) by one unit.

For two groups \(j\) and \(k\), their conditional odds ratio at the same covariate vector is \(\exp\{z_{ij}^{\mathsf T}\hat b_j-z_{ik}^{\mathsf T}\hat b_k\}\), assuming the same fixed-effect contribution applies to both. In the random-intercept-only case this reduces to \(\exp(\hat b_j-\hat b_k)\). The exponential is always positive: a negative deviation yields a multiplier below one, zero yields one, and a positive deviation yields a multiplier above one.

Worked Example: From a Log-Rate Prediction to an Expected Count

Suppose a hypothetical Poisson log-rate model includes exposure \(e\) and a group random intercept. A group’s predicted random intercept is \(\hat b_j=-0.2231\). At the same covariates and exposure, the reference expected count is 12 events over 1,000 person-years. Find the group’s rate multiplier and expected count over that exposure.

The estimated rate multiplier is

\exp(\hat b_j)=\exp(-0.2231)\approx 0.8000.

The group’s rate is estimated to be 0.8000 times the reference rate, or 20.0% lower, at the same covariates. Because exposure is held fixed at 1,000 person-years, the corresponding fitted expected count is \(12(0.8000)=9.60\) events over 1,000 person-years. If exposure differed, the expected counts would also differ because of exposure; the rate multiplier alone does not account for that difference.

This result concerns a fitted conditional expected count, not a guarantee that exactly 9.60 events will occur. Counts are random, and the model also has uncertainty in the predicted random effect and fitted parameters.

Transforming a Point Prediction Versus Averaging

If the predicted random effect is represented by one value \(\hat b_j\), exponentiating it gives the plug-in multiplier \(\exp(\hat b_j)\). If instead there is a conditional distribution for the group effect, there are several possible summaries. In particular, the exponentiated conditional mean, \(\exp\{E(b_j\mid\text{data})\}\), is not generally the conditional mean multiplier, \(E\{\exp(b_j)\mid\text{data}\}\). This follows from the nonlinearity of the exponential function.

For a logit model, the same distinction arises when converting a linear predictor into a probability. Applying the inverse-logit to a mean linear predictor does not generally equal the average probability over uncertainty in the random effect. Odds multipliers, probabilities, and averages over uncertain effects are different targets and should be named explicitly.

Derivation: If the conditional distribution of a predicted random effect is approximately \(b_j\mid\text{data}\sim N(m,v)\), write \(b_j=m+\sqrt{v}Z\), where \(Z\sim N(0,1)\). The normal moment-generating function gives \(E\{\exp(tZ)\}=\exp(t^2/2)\). Therefore

E\{\exp(b_j)\mid\text{data}\} =\exp(m)E\{\exp(\sqrt{v}Z)\} =\exp(m+v/2).

So \(\exp(m)\) is the exponential of the conditional mean on the log scale and, under this normal model, the median multiplier. The conditional mean multiplier is larger when \(v>0\). This result concerns uncertainty in the random effect under the stated normal approximation; it does not by itself incorporate uncertainty in estimated fixed effects or variance components.

Worked Example: Accounting for Uncertainty in a Predicted Log Effect

Suppose a hypothetical group’s conditional distribution for a log-rate deviation is approximated by \(b_j\mid\text{data}\sim N(-0.20,0.16)\). Thus the conditional mean is \(-0.20\) and the conditional standard deviation is \(\sqrt{0.16}=0.40\). Compare the exponential of the mean with the conditional mean rate multiplier, and transform an approximate 95% interval.

The plug-in multiplier, which is also the median multiplier under this normal approximation, is

\exp(-0.20)=0.8187.

The conditional mean multiplier is

E\{\exp(b_j)\mid\text{data}\} =\exp(-0.20+0.16/2) =\exp(-0.12) =0.8869.

Thus the mean multiplier is about 0.8869, rather than 0.8187. This does not mean the plug-in calculation is wrong; it answers a different question. The first summarizes the conditional mean log deviation after exponentiation, while the second averages the multiplier over the conditional distribution.

An approximate normal 95% interval on the log scale is \(-0.20\pm1.96(0.40)\), or \([-0.984,0.584]\). Because exponentiation is strictly increasing, transforming both endpoints gives

[\exp(-0.984),\exp(0.584)]\approx[0.374,1.793].

Under the stated conditional normal approximation, this is an approximate 95% interval for the group’s rate multiplier. It is asymmetric around 0.8187 on the multiplier scale, as expected after exponentiating a symmetric interval. It includes one, corresponding to no multiplicative deviation from the reference.

How to Report and Check a Transformed Prediction

Name the target.
Specify whether the quantity is a conditional odds multiplier, a rate multiplier, a fitted expected count, or an average over uncertainty in the random effect.

Form the relevant log-scale deviation.
For an intercept-and-slope model, use the predicted random-effect combination at the stated covariate value. For a rate model, keep exposure and other covariates clear in the comparison.

Exponentiate and interpret in context.
Report the multiplier with a reference and, where helpful, its percent difference, calculated as \(100\{\exp(d)-1\}\%\) for deviation \(d\).

Handle uncertainty on the intended scale.
Transform interval endpoints for a monotone exponential transformation. If the target is an average multiplier, integrate or use an appropriate distributional calculation rather than simply exponentiating the mean log effect.

Several checks catch common failures. A rate ratio should not depend on a change in exposure when comparing rates at otherwise identical conditions; a predicted count will depend on exposure. For a random slope, verify the covariate value and its units before exponentiating. A one-unit change in a covariate recorded in tens is not the same as a one-unit change in the underlying measurement. Very large positive or negative predicted deviations can produce extreme multipliers, so check the fitted model, data support, and numerical calculation rather than silently reporting an implausible value.

Also check the comparison being claimed. In a logit mixed model, exponentiating a group deviation produces a conditional odds ratio relative to the zero-random-effect reference at fixed covariates. It does not directly give a probability difference, and it is not generally an odds ratio comparing a group with a population-averaged outcome. The inverse-logit must be used to obtain a conditional probability from a complete linear predictor, and averaging probabilities over random-effect uncertainty requires averaging the inverse-logit, not applying it just once to the mean predictor.

Common Pitfalls and Alternatives

A common mistake is to say, “The predicted logit effect is \(-0.40\), so the group’s probability is 40% lower.” This treats a log-odds deviation as a probability change. The correction is to exponentiate the deviation to obtain the conditional odds multiplier: \(\exp(-0.40)=0.6703\). To report a probability, specify the other terms in the linear predictor and apply the inverse-logit; the resulting probability change depends on the starting probability.

Another mistake is to report \(\exp(m)\) as the mean multiplier when \(m\) is the mean of an uncertain log effect. If the target is the mean multiplier and the conditional normal approximation is appropriate, use \(\exp(m+v/2)\). If the target is a plug-in multiplier based on the predicted mean effect, report \(\exp(m)\) and label it accordingly.

Reporting the predicted deviation on the log scale is a reasonable alternative when the multiplicative comparison is not the main goal or when several random-effect components need to be shown together. Reporting an odds or rate multiplier improves direct interpretation, but can obscure how the components combine. A probability-scale summary can be useful when probabilities are the substantive target, but must be calculated at specified covariates and with an explicit choice about whether to condition on or average over random effects. None of these summaries removes uncertainty in the fitted model. Intervals based only on conditional random-effect uncertainty may omit parameter-estimation uncertainty, so the uncertainty source should be stated.

Exponentiating an interval is justified because the exponential function preserves order: if \(L\le b_j\le U\), then \(\exp(L)\le\exp(b_j)\le\exp(U)\). But a transformed interval answers only the question represented by its original interval. An interval for a latent group deviation is not automatically an interval for a marginal probability, and an interval calculated with parameters treated as fixed does not automatically include their estimation uncertainty.

Key takeaway: Exponentiation converts a log-scale random-effect deviation into a conditional multiplicative odds or rate effect. State the reference, covariate values, and uncertainty target; do not confuse a transformed point prediction with an average over uncertainty or with a probability-scale effect.

Check Your Understanding

Answer each question, then compare with the solutions below.

A binary GLMM has a predicted random-intercept deviation of \(0.30\). What is the conditional odds multiplier relative to the zero-random-effect reference, and how should it be interpreted?
A log-rate model predicts a random-effect deviation of \(-0.10\). Calculate the rate multiplier and the percent difference from the reference rate.
If \(b\mid\text{data}\sim N(0.10,0.09)\), calculate \(\exp\{E(b\mid\text{data})\}\) and \(E\{\exp(b)\mid\text{data}\}\). Are they the same?
Why is a random-slope odds multiplier incomplete unless the covariate value is specified?

Solutions. (1) \(\exp(0.30)=1.350\). At fixed covariates, the group’s conditional odds are estimated to be 1.350 times the zero-random-effect reference odds, or 35.0% higher; this is not a probability increase of 35 percentage points. (2) \(\exp(-0.10)=0.9048\), so the rate is about \(100(0.9048-1)=-9.52\%\), or 9.52% lower, at the same covariates. (3) The exponential of the mean is \(\exp(0.10)=1.105\). The conditional mean multiplier is \(\exp(0.10+0.09/2)=\exp(0.145)=1.156\). They differ because exponentiation is nonlinear and the conditional variance is positive. (4) The group deviation is the sum of the random intercept and the covariate value multiplied by the random slope. Changing the covariate changes that sum and therefore changes its exponential, the odds multiplier.