Random-Intercept Links for Continuous Outcomes and Dropout

Why Link the Outcome and Dropout Processes?

Suppose a study measures a continuous outcome repeatedly, but some participants stop providing measurements. If participants with systematically higher or lower outcome levels also tend to drop out at different rates, the observed outcome histories can be selected with respect to the outcome process. A shared-parameter model represents this dependence by giving both processes a common latent subject-specific effect.

Here the shared effect is a random intercept. It shifts a participant’s outcome trajectory up or down and also enters that participant’s dropout hazard. This tutorial specifies that particular link and explains its implications. It does not make the common effect a directly observed cause of dropout: it is a model for dependence between the processes after accounting for included covariates.

Model and Assumptions

Let $Y_{ij}$ be the continuous outcome for subject $i$ at scheduled occasion $j$, observed while the subject remains under follow-up. Let $t_{ij}$ denote the occasion time, and let $D_{ij}=1$ indicate dropout during the interval after occasion $j$, among subjects still at risk at its start. Set $D_{ij}=0$ if the subject remains under follow-up through that interval. For clarity, the model below omits measured covariates; they can be added to the linear predictor for each process.

Definition: A Gaussian random-intercept outcome model linked to a dropout hazard is $$ Y_{ij}=\beta_0+\beta_1t_{ij}+b_i+\epsilon_{ij}, \qquad b_i\sim N(0,\tau^2), \qquad \epsilon_{ij}\sim N(0,\sigma^2), $$ together with $$ \operatorname{logit}\{h_{ij}(b_i)\} =\log\!\left(\frac{h_{ij}(b_i)}{1-h_{ij}(b_i)}\right) =\gamma_0+\alpha b_i, \qquad h_{ij}(b_i)=P(D_{ij}=1\mid\text{at risk at }j,b_i). $$ The random intercept $b_i$ is shared by the two processes. The usual specification assumes independent outcome errors across occasions, independent of $b_i$, and conditionally independent dropout-interval outcomes given $b_i$ and the risk history. Outcomes are conditionally independent of the dropout process given $b_i$ and the included covariates.

The outcome distribution conditional on $b_i$ is Gaussian with mean $\beta_0+\beta_1t_{ij}+b_i$ and variance $\sigma^2$. The marginal outcome mean, before conditioning on dropout, is $\beta_0+\beta_1t_{ij}$, because $E(b_i)=0$. The variance at one occasion is $\tau^2+\sigma^2$; for two different occasions on the same subject, the covariance is $\tau^2$, under the independent-error assumption.

The hazard is conditional on being at risk at that interval. The logit intercept $\gamma_0$ is the log odds of dropout in an interval for a subject with $b_i=0$. The coefficient $\alpha$ is the change in log odds for a one-unit increase in the outcome-scale random intercept. Thus $\exp(\alpha)$ is the conditional odds ratio for dropout per one-unit increase in $b_i$, holding the interval and other modeled predictors fixed. When $\alpha>0$, subjects with higher latent outcome levels have higher interval dropout risks. When $\alpha<0$, the direction is reversed.

These interpretations are conditional on $b_i$, not marginal across subjects. The population hazard is obtained by averaging the conditional hazard over the distribution of $b_i$ among subjects still at risk. That distribution can change as subjects drop out; consequently, the average hazard need not equal the hazard evaluated at the average random intercept.

Worked Example: Interpreting the Two Linked Components

Consider a hypothetical study with outcome measured in points and time measured in months. Let $\beta_0=50$, $\beta_1=2$ points per month, $\tau=5$ points, $\sigma=3$ points, $\gamma_0=-1$, and $\alpha=1$ per point. Compare subjects with $b_i=-1$ and $b_i=1$ at month 3.

For $b_i=-1$, the conditional outcome mean is $50+2(3)-1=55$ points. For $b_i=1$, it is $50+2(3)+1=57$ points. The two latent intercept values therefore imply a 2-point difference in conditional mean.

The corresponding dropout logits are $-1+(-1)=-2$ and $-1+1=0$. The hazards are $$ h(-1)=\frac{1}{1+\exp(2)}\approx0.1192, \qquad h(1)=\frac{1}{1+\exp(0)}=0.5000. $$ Thus, among subjects at risk in a given interval, the model assigns a 11.9% dropout probability to the first intercept value and a 50.0% probability to the second. These are conditional interval probabilities, not probabilities of dropout over the entire study.

The odds ratio for a one-point increase in $b_i$ is $\exp(1)\approx2.718$. This is an odds ratio, not a risk ratio. Also, a one-point shift in the latent intercept is a model-based comparison, not necessarily a contrast between two observed groups.

How the Shared Effect Induces Association

Even though outcomes and dropout are assumed conditionally independent given $b_i$, they are generally associated after $b_i$ is integrated out. A subject with a high $b_i$ tends to have higher outcomes and, if $\alpha>0$, a higher chance of dropout. The dependence arises because both processes respond to the same latent quantity.

A short derivation makes the direction of this association precise for one dropout interval. At a fixed occasion, write $Y=\mu+B+\epsilon$, where $\mu=\beta_0+\beta_1t$, $B$ has mean zero, and $\epsilon$ is independent of $B$ and the dropout indicator $D$, with mean zero. Let $q(b)=P(D=1\mid B=b)$. Conditional expectation gives $$ \operatorname{Cov}(Y,D) =\operatorname{Cov}(B,D) =\operatorname{Cov}\{B,q(B)\}. $$ If $\alpha>0$, $q(b)=\operatorname{expit}(\gamma_0+\alpha b)$ is increasing. For an independent copy $B'$ of $B$, $$ \operatorname{Cov}\{B,q(B)\} =\frac{1}{2}E\!\left[(B-B')\{q(B)-q(B')\}\right]\geq0, $$ because the two differences inside the expectation have the same sign. The covariance is nonnegative, and is strictly positive for a non-degenerate $B$ when $q$ is strictly increasing. When $\alpha<0$, $q$ is decreasing and the covariance is nonpositive. This establishes the induced association under the stated assumptions; it does not establish a causal effect of outcome level on dropout.

Worked Example: From Interval Hazard to Cumulative Dropout

Suppose a hypothetical study has three equal follow-up intervals, with the same conditional hazard in each interval for a given subject. Let $\gamma_0=-1$ and $\alpha=1$. Calculate the probability of dropout by the end of the third interval for subjects with $b_i=-1$ and $b_i=1$.

For $b_i=-1$, the interval hazard is $\operatorname{expit}(-2)\approx0.119203$. Conditional on this fixed intercept, the probability of remaining in one interval is $1-0.119203=0.880797$. Therefore, $$ P(\text{dropout by interval 3}\mid b_i=-1) =1-(1-0.119203)^3 =1-(0.880797)^3 \approx0.3167. $$

For $b_i=1$, the interval hazard is $\operatorname{expit}(0)=0.5000$, so $$ P(\text{dropout by interval 3}\mid b_i=1) =1-(1-0.5000)^3 =1-(0.5000)^3 =0.8750. $$ The difference is $0.8750-0.3167=0.5583$, or about 55.8 percentage points. These calculations use a constant hazard conditional on $b_i$; they do not imply a constant hazard after averaging over subjects whose random-effect distribution among those still at risk can change.

Selection of the Observed Outcome Distribution

The same mechanism changes the distribution of $b_i$ among subjects who drop out or remain. In a single interval, if $f_B(b)$ is the population density of the random intercept, then the density among those who drop out is proportional to $q(b)f_B(b)$. Among those who remain, it is proportional to $\{1-q(b)\}f_B(b)$. In particular, $$ f(b\mid D=1)=\frac{q(b)f_B(b)}{P(D=1)}, \qquad f(b\mid D=0)=\frac{\{1-q(b)\}f_B(b)}{P(D=0)}. $$ These formulas apply to a single interval and condition on its risk set. Later risk sets can have a different distribution of $b_i$, even if the random-effect distribution at study entry is Gaussian.

Because the outcome mean includes $b_i$, this selection can make the average outcome among subjects who drop out differ from the average among those who remain. The following calculation uses a numerical integration approximation to show the mechanism. It is not an exact replacement for integrating the logistic hazard against a normal density.

Worked Example: Approximate Outcome Means by Dropout Status

Consider a hypothetical single-interval comparison with $B\sim N(0,1)$, $\gamma_0=-1.3$, and $\alpha=1$. At the outcome occasion, suppose the conditional mean is $10+B$ units. Approximate the mean random intercept among those who drop out and those who remain using three-point Gaussian quadrature.

For a standard normal random effect, the three quadrature nodes are $-\sqrt{3},0,\sqrt{3}$, with weights $1/6,2/3,1/6$. The corresponding hazards are approximately $0.04599$, $0.21417$, and $0.60634$, respectively: these are $\operatorname{expit}(-1.3-\sqrt{3})$, $\operatorname{expit}(-1.3)$, and $\operatorname{expit}(-1.3+\sqrt{3})$.

The approximate overall dropout probability is the weighted sum $$ P(D=1)\approx\frac{0.04599}{6}+\frac{2(0.21417)}{3}+\frac{0.60634}{6} \approx0.2515. $$ The weighted mean intercept among dropouts is $$ E(B\mid D=1)\approx \frac{(-\sqrt{3})(0.04599/6)+(\sqrt{3})(0.60634/6)} {0.2515} \approx0.643. $$ The zero node contributes nothing to the numerator. Among those who remain, the corresponding approximate probability is $1-0.2515=0.7485$, and $$ E(B\mid D=0)\approx \frac{(-\sqrt{3})(1-0.04599)/6+(\sqrt{3})(1-0.60634)/6} {0.7485} \approx-0.216. $$ Thus the approximate conditional outcome means are $10+0.643=10.643$ units among dropouts and $10-0.216=9.784$ units among those who remain. The difference is approximately $0.859$ outcome units. The quadrature values are approximate because the logistic-normal integrals are not exactly evaluated by a finite three-node rule. The direction of the difference follows from the positive shared-effect link; its magnitude depends on the model parameters and the integration accuracy.

Assumptions, Diagnostics, and Alternatives

The interpretation above relies on substantive assumptions, not merely on the Gaussian form of the outcome. The same scalar $b_i$ must plausibly represent a persistent subject-specific shift relevant to both processes. The outcome errors must be adequately described by the stated conditional distribution, and the hazard link must represent how dropout risk changes with that shift. Conditional independence asserts that no remaining association between outcome errors and dropout persists after conditioning on the shared effect and modeled covariates.

Useful diagnostics and warning signs include:

Outcome residual structure: systematic curvature over time, changing residual spread, or serial dependence beyond the random intercept can indicate that the Gaussian outcome component is inadequate.
Dropout patterns: observed dropout frequencies by interval and by available outcome history can be compared with model-based summaries. Large discrepancies suggest that the hazard form or its predictors may be inadequate.
Implausible fitted risks: very large $|\alpha|$, extreme fitted hazards, or estimates that change substantially under reasonable model adjustments warrant careful scrutiny.
Distributional sensitivity: the normal random-intercept assumption and the logit link are modeling choices. A fit that depends strongly on them should not be presented as definitive evidence about the unobserved outcomes.

These checks can reveal conflict between the model and observed data, but they cannot verify that the assumed relationship with unobserved outcomes is correct. In particular, similar observed-data fit can coexist with different assumptions about dropout. The earlier material on model fit and missingness assumptions is relevant: goodness of fit does not by itself validate the missingness mechanism.

The shared-random-intercept specification is more restrictive than allowing a flexible association between each outcome history and dropout: it uses one latent shift to connect both processes. A model with separate independent random effects does not induce this association through a common latent variable. Pattern-mixture models organize assumptions by missingness pattern, whereas selection models specify dropout dependence through outcome-related quantities; the present construction instead expresses dependence through a shared latent effect. These approaches are not interchangeable without additional assumptions.

The model also has limitations. It assumes the same random intercept acts across occasions and does not, by itself, represent subject-specific changes in outcome trajectory or a changing effect of $b_i$ on dropout risk. Those mechanisms require a different model structure. Further, the random effect is latent, and the relationship between it and dropout is informed through the joint model and its assumptions. A numerically estimated $\alpha$ should therefore be interpreted as a model-conditional association parameter, not as a directly observed or causal effect.

Common Pitfall

Common mistake: “The outcome and dropout are independent once the model is fitted, so the observed outcome means for dropouts and retained subjects should be equal.”

Correction: Conditional independence is asserted only after conditioning on $b_i$. Once that latent effect is averaged out, subjects with different $b_i$ values have different outcome means and different dropout risks. The resulting groups can have different distributions of $b_i$, and therefore different outcome means. Conditional independence is not marginal independence.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If $\alpha=0$, what does the model say about the dependence between the random intercept and the conditional dropout hazard?
With $\gamma_0=-2$, $\alpha=0.5$, and $b_i=2$, calculate the conditional interval hazard.
Under the assumptions of the covariance derivation, what is the sign of $\operatorname{Cov}(Y,D)$ when $\alpha<0$? Briefly justify it.
Why is the average hazard across the original population generally not the same as the hazard evaluated at $b_i=0$?

Solutions. (1) The hazard no longer depends on $b_i$, since $\operatorname{logit}(h_{ij})=\gamma_0$. The shared intercept still shifts outcomes, but this specified link does not connect it to dropout risk. (2) The logit is $-2+(0.5)(2)=-1$, so $h=\operatorname{expit}(-1)=1/(1+\exp(1))\approx0.2689$, or 26.9% per interval conditional on being at risk. (3) The covariance is nonpositive: when $\alpha<0$, the dropout probability $q(b)$ decreases as $b$ increases, so the product of differences in the covariance identity is nonpositive. (4) The hazard is a nonlinear function of $b_i$, so averaging $\operatorname{expit}(\gamma_0+\alpha b_i)$ is generally not equal to evaluating it at $E(b_i)=0$. Among later risk sets, the distribution of $b_i$ can also differ from its distribution at study entry because dropout has selectively removed subjects.