Conditional and Population-Averaged Effects in NLME

Why Conditional and Population-Averaged Effects Differ

A coefficient in a nonlinear mixed-effects model describes a contrast on the model’s link scale while holding the random effects fixed. That is a conditional effect: it compares responses for individuals with the same random-effect values. A population-averaged effect instead compares expected responses after averaging over the random-effects distribution. These answer different questions, and in nonlinear models the second is generally not obtained by simply applying the inverse link to the average linear predictor.

This distinction matters when a model is used to communicate treatment effects or predict response rates for a population. The conditional coefficient describes how the modeled response changes for a specified latent individual profile. The population-averaged prediction describes the response expected across individuals with the modeled distribution of latent profiles. Even when the random effects have mean zero, the two may differ because the response function is nonlinear.

Definition: Let $Y_{ij}$ be the response for individual $i$ at observation $j$, let $x_{ij}$ denote covariates, and let $b_i$ be that individual’s random-effects vector with distribution $f(b;\Omega)$. A conditional mean is $E(Y_{ij}\mid x_{ij},b_i)$. The population-averaged mean at covariate value $x$ is $$ \mu_{\mathrm{PA}}(x)=E_b\{E(Y\mid x,b)\}=\int E(Y\mid x,b)f(b;\Omega)\,db. $$ If covariates and random effects are associated, the relevant distribution for a conditional-on-$x$ prediction is $f(b\mid x)$, not automatically the unconditional $f(b)$.

For a model with link function $g$, the conditional mean is specified by $g\{E(Y\mid x,b)\}=\eta(x,b)$, where $\eta(x,b)$ is the linear predictor. The marginal mean is therefore the expectation of $g^{-1}\{\eta(x,b)\}$, not generally $g^{-1}\{E_b[\eta(x,b)]\}$. These quantities coincide for some linear functions, but not generally for nonlinear inverse links.

Conditional Effects in a Binary-Response Model

For a binary response, a common conditional mean is a probability $p(x,b)=P(Y=1\mid x,b)$, modeled using the logistic inverse link: $$ p(x,b)=\frac{\exp\{\eta(x,b)\}}{1+\exp\{\eta(x,b)\}}. $$ Suppose a binary covariate $x$ has coefficient $\beta$ and the predictor includes a random intercept $b$: $\eta(x,b)=\alpha+\beta x+b$. At any fixed $b$, the conditional odds are $\exp(\alpha+\beta x+b)$. Thus the conditional odds ratio comparing $x=1$ with $x=0$ is $\exp(\beta)$. It does not depend on $b$, under this model with no random slope for $x$.

The population-averaged probability at each covariate value must instead be integrated over the random-intercept distribution. The population-averaged odds ratio is calculated from those two averaged probabilities. In general, it is not $\exp(\beta)$: averaging probabilities and then converting them to odds is not the same operation as averaging conditional odds.

Worked Example: Conditional and Population-Averaged Odds Ratios

Consider a hypothetical binary-response model with $\alpha=0$, $\beta=1$, and a random intercept that equals $-1$ or $1$, each with probability $0.5$. Assume this same random-effects distribution applies at both values of $x$. The conditional odds ratio is $\exp(1)=2.718$.

For $x=0$, the conditional probabilities at $b=-1$ and $b=1$ are $\operatorname{expit}(-1)=0.2689$ and $\operatorname{expit}(1)=0.7311$. Therefore, $$ p_{\mathrm{PA}}(0)=0.5(0.2689)+0.5(0.7311)=0.5000. $$

For $x=1$, the corresponding probabilities are $\operatorname{expit}(0)=0.5000$ and $\operatorname{expit}(2)=0.8808$. Thus, $$ p_{\mathrm{PA}}(1)=0.5(0.5000)+0.5(0.8808)=0.6904. $$

The population-averaged odds ratio is $$ \frac{0.6904/(1-0.6904)}{0.5000/(1-0.5000)} =\frac{2.230}{1.000}=2.230. $$ It differs from the conditional odds ratio of 2.718. The conditional contrast holds $b$ fixed; the population contrast averages predicted probabilities over the stated random-intercept distribution before forming odds.

Why Averaging a Nonlinear Prediction Changes It

The reason can be seen directly from the logistic function. For the same random-intercept distribution as above, set the fixed part of the predictor to $1$. The random intercept has mean zero, so the inverse link evaluated at the mean predictor is $\operatorname{expit}(1)=0.7311$. But the average of the inverse-link predictions is different:

Worked Example: Integrating Probabilities Rather Than Predictors

Let the fixed part of the predictor be $1$, and let $b=-1$ or $b=1$ with equal probability. The two conditional probabilities are $\operatorname{expit}(1-1)=0.5000$ and $\operatorname{expit}(1+1)=0.8808$. Hence the population-averaged probability is $$ p_{\mathrm{PA}}=0.5(0.5000)+0.5(0.8808)=0.6904. $$

The alternative plug-in calculation uses the mean random effect, $E(b)=0$, and gives $\operatorname{expit}(1+0)=0.7311$. The difference, $0.7311-0.6904=0.0407$, shows that a zero-mean random effect does not make the two calculations equivalent. For this example, the sigmoid curve is curved across the range of predictors; averaging its values is not the same as evaluating it at the average predictor.

This is a consequence of integrating the response function, not a special property of the two-point distribution. With a continuous random-effects distribution, the same expectation is an integral. It may require numerical quadrature or simulation, depending on the model and the dimension of $b$. The distribution, its estimated parameters, and the covariate values all matter to the resulting averaged prediction.

Justification: For a nonlinear inverse link $g^{-1}$, the population mean is $E_b[g^{-1}\{\eta(x,b)\}]$. Replacing this by $g^{-1}\{E_b[\eta(x,b)]\}$ moves the expectation through a nonlinear function. There is no general identity permitting that step. It is valid only under particular conditions, such as a constant predictor with respect to $b$, or a linear inverse link.

Rate Ratios: A Useful Special Case and an Important Exception

For a log-link count model, write the conditional rate as $\lambda(x,b)=\lambda_0\exp(\beta x+b_0+b_1x)$, where $b_0$ is a random intercept and $b_1$ a random slope for the covariate $x$. A rate here is an expected count per unit of observation time. Holding both random effects fixed, the rate ratio for increasing $x$ from 0 to 1 is $\exp(\beta+b_1)$. It is not necessarily the same for every individual if $b_1$ varies.

The population-averaged rates and their ratio are $$ \lambda_{\mathrm{PA}}(x)=\lambda_0\exp(\beta x)E_b[\exp(b_0+b_1x)], \qquad RR_{\mathrm{PA}}=\exp(\beta)\frac{E_b[\exp(b_0+b_1)]}{E_b[\exp(b_0)]}. $$

If there is only a random intercept, so $b_1=0$, and the same random-intercept distribution applies at both covariate values, the expectation $E_b[\exp(b_0)]$ cancels in the ratio. In that special case, $RR_{\mathrm{PA}}=\exp(\beta)$. This differs from the logistic example: the log link makes a random intercept a multiplicative factor in the rate, and that common factor cancels. With random slopes, the exponential expectation in the numerator can differ from the one in the denominator, so the population-averaged rate ratio generally changes.

Worked Example: A Random Slope Changes the Averaged Rate Ratio

Suppose a hypothetical count model has a baseline rate $\lambda_0=4$ events per person-year and a fixed covariate coefficient $\beta=\log(2)$. Let $b_0=0$, and let the random slope $b_1$ equal $-0.5$ or $0.5$ with equal probability. At fixed $b_1$, the conditional rate ratio for $x=1$ versus $x=0$ is $\exp(\log(2)+b_1)$.

For individuals with $b_1=-0.5$, this ratio is $2\exp(-0.5)=1.213$. For those with $b_1=0.5$, it is $2\exp(0.5)=3.297$. The population-averaged control rate is $4$ events per person-year. The treated rate is $$ 4\left\{0.5[2\exp(-0.5)]+0.5[2\exp(0.5)]\right\} =4(2.255)=9.021 $$ events per person-year. The population-averaged rate ratio is $9.021/4=2.255$. It is neither the fixed-effect ratio $\exp(\beta)=2$ nor either individual conditional ratio. These calculations use the stated distribution of random slopes and average rates before taking their ratio.

Assumptions, Interpretation, and Practical Checks

An averaged effect is meaningful only after specifying what population is being averaged over. The calculations above assume the stated random-effects distribution is appropriate and applies comparably to the covariate groups being contrasted. When random effects are associated with covariates, averaging over the same unconditional distribution may not describe the covariate-specific population. For a prediction in a target population, the covariate distribution may also need to be incorporated: first calculate the appropriate conditional predictions, then average them over the target population’s covariates.

In practice, inspect the model’s random-effects structure and the target of the reported effect. For a logistic model, determine whether the reported quantity is $\beta$, $\exp(\beta)$, or an odds ratio formed from integrated probabilities; these are not interchangeable. For a log-link model, check whether random slopes are present before assuming a rate ratio is preserved. For continuous random effects, compare integration-based predictions with plug-in predictions as a check, and verify that integration is performed over the fitted distribution rather than only over estimated individual random effects.

A frequent failure is to use empirical Bayes estimates of $b_i$ for observed individuals, average their fitted predictions, and describe the result as the model-implied population average. Those estimates are individual-level predictions informed by each individual’s data and are typically shrunk toward the population distribution. They do not, by themselves, replace integration over the fitted random-effects distribution for a new or target population. Another warning sign is reporting a conditional coefficient as a population effect without stating the conditioning.

Quantity	What it compares	Typical calculation
Conditional effect	Responses at fixed random effects	Contrast of conditional means or link-scale predictors
Population-averaged effect	Responses averaged over a specified random-effects distribution	Integrate conditional means, then form the contrast
Covariate-standardized effect	Responses averaged over random effects and a target covariate distribution	Integrate predictions over both distributions

A conditional mixed-effects model and a marginal model, such as one that specifies population means directly, can target different estimands. The conditional model represents latent individual heterogeneity through random effects; a marginal approach specifies population-level response relationships directly. Neither is universally preferable. Choose and report the estimand that matches the scientific question, and do not assume a population-average contrast can be recovered from a single fixed-effect coefficient without calculation.

Common Mistakes and Limitations

Applying the inverse link to the mean predictor. This usually gives a plug-in prediction, not the integrated population mean. Average conditional means over the random-effects distribution instead.
Calling $\exp(\beta)$ a population odds ratio. In a logistic model with random intercepts, it is a conditional odds ratio under the stated model. Calculate the odds ratio from population-averaged probabilities for a marginal odds ratio.
Assuming every nonlinear-link effect changes in the same way. A log-link model with only a shared random intercept preserves the rate ratio under the stated conditions; random slopes or different random-effects distributions can alter it.
Leaving the target population unspecified. The averaged prediction depends on the random-effects distribution and may also depend on how covariates are distributed or associated with random effects.

Integration does not remove model limitations. If the random-effects distribution is misspecified, the averaged predictions can be misleading. Numerical integration can also be difficult with many correlated random effects; the method used should be checked for adequate accuracy. Finally, a population-averaged association is not automatically a causal effect. Causal interpretation requires appropriate design and assumptions beyond the act of averaging.

Check Your Understanding

Answer each question by identifying the estimand and showing the relevant calculation.

A logistic mixed model has a fixed coefficient $\beta=0.7$ for a binary covariate. What is the conditional odds ratio, and is it automatically the population-averaged odds ratio?
For a random-intercept logistic model, why is averaging $\operatorname{expit}(\alpha+\beta x+b)$ generally different from applying expit to $\alpha+\beta x+E(b)$?
A log-link rate model has a random intercept but no random slope, and the same random-intercept distribution at both exposure levels. What happens to the population-averaged rate ratio?
In a log-link model, the conditional rate ratio is $\exp(\beta+b_1)$, with $b_1=-0.2$ or $0.2$ equally likely and $\exp(\beta)=1.5$. Calculate both conditional rate ratios and the population-averaged rate ratio, assuming the baseline rate does not depend on $b_1$.

Solutions. (1) The conditional odds ratio is $\exp(0.7)=2.014$. It is not automatically the population-averaged odds ratio, which must be formed from probabilities integrated over the random-effects distribution. (2) The inverse logistic function is nonlinear, so expectation does not generally commute with applying the inverse link: $E_b[\operatorname{expit}(\eta+b)]\ne\operatorname{expit}(\eta+E[b])$. (3) The common factor $E[\exp(b)]$ cancels from the two averaged rates, so the population-averaged rate ratio is $\exp(\beta)$. (4) The conditional ratios are $1.5\exp(-0.2)=1.228$ and $1.5\exp(0.2)=1.832$. With equal probabilities and a baseline rate independent of $b_1$, the ratio of averaged rates is their mean: $0.5(1.228)+0.5(1.832)=1.530$, approximately.