Serial Dependence Beyond Random Intercepts

Why Random Effects May Not Explain All Dependence

Repeated measurements from one person are related. A person-level random intercept represents one source of that relationship: measurements share a stable person-specific deviation from the population mean. But it does not, by itself, represent every pattern over time. After accounting for stable differences and any modeled trajectory, adjacent measurements may still have more similar residual deviations than measurements far apart. That remaining pattern is residual serial correlation.

This distinction matters in chained-equation imputation because the model for a missing measurement determines how information from other times contributes to its imputed value and its uncertainty. A model that accounts for a shared intercept but assumes independent residuals can miss local temporal information. Conversely, adding an autoregressive structure without evidence or adequate data can make imputations unstable.

Definition: Let $Y_{it}$ be a repeated outcome for person $i$ at equally spaced occasion $t$. A random-intercept model with AR(1) residuals can be written $$ Y_{it}=X_{it}^{\mathsf T}\beta+b_i+e_{it}, $$ where $X_{it}^{\mathsf T}\beta$ is the fixed-effects mean, $b_i$ is a person-level random intercept with $E(b_i)=0$ and $\operatorname{Var}(b_i)=\tau_{00}$, and $e_{it}$ is a within-person residual. The AR(1) assumption is $e_{it}=\rho e_{i,t-1}+\eta_{it}$, with $|\rho|<1$, stationary residual variance $\operatorname{Var}(e_{it})=\sigma_e^2$, and independent innovations $\eta_{it}\sim N(0,\sigma_e^2(1-\rho^2))$. Different people are independent, and the random intercepts are independent of the residual processes.

The stationary AR(1) specification implies $\operatorname{Cov}(e_{it},e_{is})=\sigma_e^2\rho^{|t-s|}$. Thus, for two occasions $t$ and $s$, the outcome covariance for the same person is

\operatorname{Cov}(Y_{it},Y_{is}\mid X)=\tau_{00}+\sigma_e^2\rho^{|t-s|}.

This covariance formula describes the model after accounting for the fixed mean. If the model also includes random slopes, their covariance contribution must be added; the expression above isolates the random-intercept and AR(1) distinction. With $\rho=0$, measurements still covary through $\tau_{00}$, but residuals at different occasions are uncorrelated. With $\rho\ne0$, the residual covariance declines with the time lag, while the random-intercept contribution does not.

What the Added AR Structure Changes

In longitudinal MICE, a target variable’s conditional imputation model can include person-level effects and, when justified, a residual covariance structure across occasions. For a continuous outcome under a Gaussian model, this may be represented by a multivariate conditional distribution for a person’s repeated measurements. The missing components are drawn conditional on the observed components, predictors, and model parameters. A draw should reflect uncertainty in the missing value, not merely insert a fitted conditional mean.

The AR term changes how measurements near the missing occasion inform its imputation. A random intercept shares information across all occasions through a stable person effect. An AR residual term additionally lets a nearby residual deviation predict the missing residual, with the contribution typically weakening as the lag increases. These are different information channels, not competing names for the same correlation.

Worked Example: Separating the Two Covariance Sources

Suppose, hypothetically, a fitted model for a continuous outcome has random-intercept variance $\tau_{00}=9$, residual variance $\sigma_e^2=4$, and AR(1) parameter $\rho=0.50$. The fixed mean has already been accounted for. Calculate the variance at one occasion, the covariance at lag one, and the resulting correlation.

The variance at an occasion is the random-intercept variance plus residual variance:

\operatorname{Var}(Y_{it}\mid X)=9+4=13.

At lag one, the covariance is:

\operatorname{Cov}(Y_{it},Y_{i,t+1}\mid X)=9+4(0.50)=11.

The two occasion-specific variances are both 13, so their correlation is $11/\sqrt{13\times13}=11/13=0.8462$. If the residuals were independent, the covariance would instead be 9 and the correlation would be $9/13=0.6923$. The added covariance of 2 is the lag-one residual contribution; it is not an increase in the random-intercept variance.

This calculation assumes the stated variance components apply to the same population and that occasions are equally spaced as specified by the AR(1) model. It illustrates a model-implied covariance, not proof that a real data set has $\rho=0.50$.

Deriving a Conditional Imputation From Neighboring Residuals

A simple derivation shows how an AR(1) model uses observations on both sides of a missing occasion. Suppose the mean and person-specific intercept are known, so the residuals can be considered directly. Let $e_{i2}$ be missing, with $e_{i1}$ and $e_{i3}$ observed. Under a stationary Gaussian AR(1) process, the covariance matrix of the three residuals is $\sigma_e^2$ times the correlation matrix with entries $\rho^{|t-s|}$. The Gaussian conditional distribution of the middle residual given the two endpoints has mean and variance

E(e_{i2}\mid e_{i1},e_{i3})= \frac{\rho}{1+\rho^2}(e_{i1}+e_{i3}),\qquad \operatorname{Var}(e_{i2}\mid e_{i1},e_{i3})= \sigma_e^2\frac{1-\rho^2}{1+\rho^2}.

For justification, the covariance between the middle residual and the two endpoints is $\sigma_e^2(\rho,\rho)$, while the endpoints’ covariance matrix is $\sigma_e^2\begin{pmatrix}1&\rho^2\\\rho^2&1\end{pmatrix}$. The Gaussian conditioning formula gives the conditional mean by multiplying the middle-to-endpoint covariance row by the inverse endpoint covariance matrix. The inverse is $\{\sigma_e^2(1-\rho^4)\}^{-1}\begin{pmatrix}1&-\rho^2\\-\rho^2&1\end{pmatrix}$. The resulting two weights are both $\rho/(1+\rho^2)$. Subtracting the corresponding covariance reduction from the marginal variance gives $\sigma_e^2-\sigma_e^2\,2\rho^2/(1+\rho^2)=\sigma_e^2(1-\rho^2)/(1+\rho^2)$, as stated.

Worked Example: Imputing a Missing Middle Residual

Suppose, hypothetically, the fitted residual model has $\rho=0.60$ and $\sigma_e^2=4$. After subtracting the fixed mean and person intercept, the observed residuals immediately before and after a missing value are $e_{i1}=3$ and $e_{i3}=-1$. Calculate the conditional mean and variance of the missing residual.

The conditional mean is

\frac{0.60}{1+0.60^2}(3+(-1)) =\frac{0.60}{1.36}(2) =0.8824.

The conditional variance is

4\frac{1-0.60^2}{1+0.60^2} =4\frac{0.64}{1.36} =1.882.

Thus the conditional distribution is normal with mean 0.8824 and variance 1.882, conditional on the model parameters, the known mean and intercept, and these two residuals. Its standard deviation is $\sqrt{1.882}=1.372$. A proper stochastic imputation draws from this distribution rather than always imputing 0.8824. In an actual analysis, the mean, intercept, and covariance parameters are uncertain and should also be handled by the imputation procedure.

For comparison, under independent residuals with $\rho=0$, the conditional mean of $e_{i2}$ given these endpoint residuals is zero and its conditional variance remains 4. The random intercept can still induce dependence in the original outcomes, but once the intercept and mean are conditioned on, independent residuals do not let the neighboring residuals predict the missing residual. This contrast is why an intercept alone cannot reproduce local residual persistence.

Choosing and Checking an Imputation Structure

The random-intercept-only and random-intercept-plus-AR(1) models make different assumptions about residuals. The first assumes that, conditional on the modeled mean and person effect, residuals at distinct occasions are independent (or at least uncorrelated in the Gaussian model). The second assumes a stationary, geometrically decaying residual correlation for equally spaced occasions. Selection should be guided by the study design, substantive knowledge, residual diagnostics, and whether there is enough repeated information to estimate the extra dependence parameter.

Feature	Random intercept, independent residuals	Random intercept plus AR(1) residuals
Dependence represented	Shared stable person deviation	Shared deviation plus lag-decaying residual association
Residual correlation after conditioning on the intercept	Zero across occasions under the model	$\rho^{\|t-s\|}$
Information for a missing occasion	Shared person effect and predictors	Those sources plus local residual information
Additional requirement	No serial parameter to estimate	Equally spaced ordering and an estimable, appropriate $\rho$

Useful checks focus on the residual process after accounting for the fixed mean and person effects. Examine residual correlations or autocorrelation by lag, preferably using observed portions of the data and patterns that are informative about within-person dependence. Compare observed and imputed distributions of within-person changes and lagged differences, and check whether imputed series show implausible smoothness or excessive roughness. In an imputation analysis, also inspect trace behavior of model parameters and summaries across iterations and imputations, and compare substantive estimates under defensible alternative covariance structures.

These checks are not mechanical tests that establish one model as true. Sparse observations per person, short series, substantial missingness, or a narrow range of observed lags can make $\rho$ weakly identified. A fitted value near a boundary, unstable estimates across chains, or residual dependence left at short lags can signal a poor fit or insufficient information. Imputed values themselves should not be treated as independent evidence for the structure that generated them; assess diagnostics using observed data and model-based checks with that limitation in mind.

Worked Example: Comparing Two Candidate Imputations

Consider a hypothetical continuous-outcome series with equally spaced occasions. After conditioning on a known fixed mean and person intercept, the residuals at occasions 1 and 3 are 3 and $-1$, and occasion 2 is missing. For a transparent comparison, take $\sigma_e^2=4$, and compare an independent-residual model with an AR(1) model having $\rho=0.60$. These parameters and the conditioned effects are treated as known for this calculation.

Under the independent-residual model, conditioning on the endpoint residuals gives $E(e_{i2}\mid e_{i1},e_{i3})=0$ and $\operatorname{Var}(e_{i2}\mid e_{i1},e_{i3})=4$. Under AR(1), the calculation in the preceding derivation gives mean 0.8824 and variance 1.882.

The AR(1) conditional mean is positive because the positive preceding residual contributes more than the negative following residual in this example: the two are added to give 2, then multiplied by $0.60/1.36$. The conditional variance is also smaller because both neighboring values provide information about the missing residual under the assumed correlation. Neither result means the missing value is known: the AR(1) model still assigns a standard deviation of 1.372, and an imputation must preserve that uncertainty.

The comparison is valid only for the stated equally spaced Gaussian model and conditioning assumptions. It is not a general claim that AR imputations are always closer to neighbors or have smaller uncertainty; the direction and size depend on the observed residuals and fitted parameters.

Assumptions, Failure Modes, and Scope

The AR(1) residual structure is most defensible when occasion spacing is regular and residual association is approximately stationary and decays geometrically with lag. The Gaussian conditional calculation additionally assumes a continuous outcome and a Gaussian residual process. The missingness mechanism must also be ignorable for the specified imputation model—for example, missingness at a time may depend on observed history included in the model, but not on the missing value itself after conditioning on that history and the other modeled variables. The imputation equations must include the variables and relationships needed by the planned analysis.

Several departures require care. A residual correlation that oscillates, changes over time, or decays at a different rate may not be represented by AR(1). Unequal time gaps are not handled by simply treating each row as one lag; the relevant time-distance structure requires a different model, which is outside the equally spaced setup here. For binary or count outcomes, a Gaussian AR residual model on the observed scale is generally not automatically appropriate; one may need a model on a latent or otherwise suitable scale. If chained conditional models with separately specified covariance structures do not correspond to a coherent joint distribution, the resulting imputations can have compatibility problems. Parameter uncertainty and random-effect uncertainty also matter: using only fitted values can understate imputation uncertainty.

A common mistake is to reason, “The random intercept makes measurements within a person correlated, so no serial term is needed.” The correction is to ask whether residuals remain associated by lag after accounting for the modeled person effect and trajectory. Another mistake is to add AR(1) solely because measurements are ordered. Ordering alone does not establish residual autocorrelation, and a more elaborate model can be less reliable when the data provide little information about its parameters.

Key takeaway: A random intercept creates a common covariance component across occasions; an AR(1) residual creates an additional covariance that depends on lag. Include the latter when its assumptions are plausible and the data can support its estimation, and verify that the imputation model preserves both the dependence structure and uncertainty relevant to the analysis.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $\tau_{00}=5$, $\sigma_e^2=3$, and $\rho=0.40$, what are the model-implied variance at one occasion and covariance at lag two?
Under the Gaussian AR(1) model, why can an adjacent observed residual help predict a missing residual even after a random intercept is included?
For the worked middle-residual setup with $e_{i1}=2$, $e_{i3}=0$, $\rho=0.50$, and $\sigma_e^2=4$, calculate the conditional mean and variance.
Name two diagnostics or warning signs relevant to deciding whether an AR(1) residual structure is adequate.

Solutions. (1) The variance is $5+3=8$. The lag-two covariance is $5+3(0.40^2)=5+0.48=5.48$; the residual contribution is smaller at lag two than at lag one. (2) The intercept accounts for a shared stable deviation, while AR(1) specifies correlated residual deviations that decay with lag; a neighboring residual therefore carries information about the missing residual conditional on the intercept. (3) The mean is $\{0.50/(1+0.50^2)\}(2+0)=1/1.25=0.8000$. The variance is $4\{(1-0.25)/(1+0.25)\}=4(0.60)=2.400$. (4) Examples include residual autocorrelation by lag after accounting for person effects, implausible imputed lagged differences, unstable or boundary estimates of $\rho$, and poor parameter trace behavior.