Common Error: Replacing the Missing Current Outcome

Why the Missing Current Outcome Matters

In a longitudinal study, dropout at a visit can depend on the outcome a participant would have had at that visit. For someone who drops out before the measurement, that current outcome is unobserved. It is tempting to replace it with the last recorded outcome because that value is available. But availability does not make the two outcomes interchangeable: the replacement changes which variable the selection model conditions on, and generally changes the predicted dropout probability.

The issue is not whether the last observed value is useful. It can be part of a model for the current outcome or a predictor in a dropout model. The error is treating it as though it were the missing current value without stating and justifying that modeling choice.

Definition: Let $R_{ij}=1$ if participant $i$'s outcome at visit $j$ is observed, and $R_{ij}=0$ if it is missing. For monotone dropout, a participant at risk of dropout at visit $j$ has $R_{i,j-1}=1$. Write $Y_{ij}$ for the visit-$j$ outcome and $H_{i,j-1}$ for the observed history before that visit, including earlier outcomes and relevant covariates. A current-outcome selection model specifies $$ \operatorname{logit}\{P(R_{ij}=0\mid R_{i,j-1}=1,H_{i,j-1},Y_{ij})\} =\alpha_j+\gamma Y_{ij}+\boldsymbol{\beta}^{\mathsf T}X_{ij}, $$ where $X_{ij}$ denotes included covariates. The probability is conditional on the current outcome, even when that outcome is missing for a participant who drops out.

This model describes a hypothetical selection mechanism: for each possible value of $Y_{ij}$, it gives the dropout probability conditional on that value and the specified history. It does not mean the analyst observes $Y_{ij}$ for dropouts. The outcome distribution conditional on history, $f(y_{ij}\mid H_{i,j-1})$, is needed to connect this conditional model to observable dropout rates.

What the Observed Dropout Probability Requires

For a participant who drops out at visit $j$, the current outcome is unavailable. The dropout probability given the observed history is therefore obtained by averaging the conditional selection probability over the possible current outcomes:

P(R_{ij}=0\mid R_{i,j-1}=1,H_{i,j-1}) = \int P(R_{ij}=0\mid R_{i,j-1}=1,H_{i,j-1},y) f(y\mid H_{i,j-1})\,dy.

For a discrete outcome, the integral is a sum. This expression follows from conditioning on the unobserved outcome: each possible value contributes its conditional dropout probability, weighted by how likely that value is under the outcome model. Replacing $y$ with $Y_{i,j-1}$ instead evaluates the selection probability at one observed value. That is not the same average, except in special cases—for example, if the selection probability does not depend on the current outcome or if the current outcome is known to equal the last observed value.

Worked Example: Exact Calculation With a Binary Current Outcome

Suppose the current outcome is binary and, for a particular observed history, $P(Y_{ij}=1\mid H_{i,j-1})=P(Y_{ij}=0\mid H_{i,j-1})=0.5$. Consider the current-outcome selection model $$ \operatorname{logit}\{P(R_{ij}=0\mid Y_{ij})\}=-2+4Y_{ij}. $$ The last observed outcome is $Y_{i,j-1}=0$. Compare the model’s dropout probability given the history with the result of substituting that last value.

For $Y_{ij}=0$, the linear predictor is $-2$, so the conditional dropout probability is $$ \operatorname{expit}(-2)=\frac{1}{1+e^2}=0.1192. $$ For $Y_{ij}=1$, the linear predictor is $2$, giving $$ \operatorname{expit}(2)=\frac{1}{1+e^{-2}}=0.8808. $$ Averaging over the current outcome distribution gives $$ P(R_{ij}=0\mid H_{i,j-1}) =(0.5)(0.1192)+(0.5)(0.8808)=0.5000. $$ By contrast, replacing the missing current outcome by the last observed value $0$ yields a predicted dropout probability of $0.1192$. It is not a close approximation here: the correct probability under the specified model is $0.5000$.

The difference also changes the implied composition of dropouts. Under the current-outcome model, $$ P(Y_{ij}=1\mid R_{ij}=0,H_{i,j-1}) =\frac{P(R_{ij}=0\mid Y_{ij}=1)P(Y_{ij}=1\mid H_{i,j-1})} {P(R_{ij}=0\mid H_{i,j-1})} =\frac{(0.8808)(0.5)}{0.5000}=0.8808. $$ Thus, among those who drop out in this hypothetical example, most would have had $Y_{ij}=1$. A model that assigns every participant with this history the dropout probability $0.1192$ does not represent that current-outcome selection mechanism.

Continuous Outcomes: Integrate, Do Not Impute by Substitution

With a continuous outcome, the same principle applies, but the average is an integral rather than a finite sum. In particular, the dropout probability at the mean current outcome is generally not the average dropout probability. The logistic function is nonlinear, so evaluating it at a single representative value need not preserve the expectation.

Worked Example: A Gaussian Current Outcome

Suppose, conditional on the observed history, $Y_{ij}\sim N(0,1)$, and suppose $$ \operatorname{logit}\{P(R_{ij}=0\mid Y_{ij})\}=-1+2Y_{ij}. $$ For illustration, take the last observed value to be $0$. The actual dropout probability given the history is $$ \int \operatorname{expit}(-1+2y)\,\phi(y;0,1)\,dy, $$ where $\phi(y;0,1)$ is the standard normal density.

A five-point Gauss–Hermite calculation approximates this expectation using standard-normal nodes $0,\ \pm1.3556,\ \pm2.8570$, with weights $0.53333,\ 0.22208$ for each of the first pair, and $0.01126$ for each of the second pair. Substitution gives $$ (0.53333)\operatorname{expit}(-1) +(0.22208)\{\operatorname{expit}(1.7112)+\operatorname{expit}(-3.7112)\} +(0.01126)\{\operatorname{expit}(4.7139)+\operatorname{expit}(-6.7139)\} \approx 0.3480. $$ The last-value substitution instead evaluates $\operatorname{expit}(-1+2(0))=\operatorname{expit}(-1)=0.2689$. Therefore, under the stated Gaussian outcome distribution and current-outcome selection model, the approximate dropout probability is $0.3480$, not $0.2689$. The calculation uses the distribution of the missing outcome; it does not pretend that the missing outcome was observed.

Same Last Value, Different Current-Outcome Distributions

Substitution can also erase differences between participants’ histories. Two people can have the same last recorded outcome but different earlier trajectories or covariates, and consequently different distributions for their current outcomes. A rule based only on the last recorded value assigns them the same selection probability, even when integrating the current-outcome model gives different probabilities.

Worked Example: A Shared Last Value Does Not Fix the Current Outcome

Consider two hypothetical histories with the same last observed outcome, $Y_{i,j-1}=4$. Under history A, suppose the current outcome is equally likely to be 3 or 5. Under history B, suppose it is equally likely to be 5 or 7. Use the selection model $$ \operatorname{logit}\{P(R_{ij}=0\mid Y_{ij})\}=-8+2Y_{ij}. $$

For history A, the dropout probabilities at current outcomes 3 and 5 are $\operatorname{expit}(-2)=0.1192$ and $\operatorname{expit}(2)=0.8808$. Thus its history-specific dropout probability is $$ (0.5)(0.1192)+(0.5)(0.8808)=0.5000. $$ For history B, the probabilities at 5 and 7 are $0.8808$ and $\operatorname{expit}(6)=0.9975$, so $$ (0.5)(0.8808)+(0.5)(0.9975)=0.9392. $$ A rule that substitutes $Y_{i,j-1}=4$ into the selection equation gives $\operatorname{expit}(0)=0.5000$ for both histories. It happens to match the integrated probability for history A, but it misses history B’s probability by $0.4392$. Agreement in one case would not validate the substitution; it is the averaging under each history’s current-outcome distribution that defines the current-outcome model.

Why Substitution Changes the Model

The distinction can be stated directly in terms of conditioning sets. A current-outcome model specifies dropout conditional on $Y_{ij}$; a last-observed-outcome model specifies dropout conditional on $Y_{i,j-1}$. These are different variables, generally with different values and distributions. If a model is explicitly defined to depend on the lagged outcome, that may be a legitimate alternative specification. It is not a computational shortcut for the current-outcome model.

For a continuous current outcome, the observed-data contribution for dropout at visit $j$, conditional on the observed history, contains the integrated probability $$ \int f(y\mid H_{i,j-1})\, P(R_{ij}=0\mid H_{i,j-1},y)\,dy. $$ For a participant observed at visit $j$, the corresponding joint contribution includes the observed outcome density and the probability of remaining observed: $$ f(Y_{ij}\mid H_{i,j-1}) \{1-P(R_{ij}=0\mid H_{i,j-1},Y_{ij})\}. $$ The second expression can use the observed current outcome. The first cannot, which is why it averages across possible values. Substituting the last value in the first expression replaces the averaging step with a different conditional probability.

A lagged-outcome model, a model based on observed history alone, and a pattern-mixture model are alternative ways to describe missingness and outcomes. They answer related but distinct modeling questions. No one is obtained automatically by filling in the missing current outcome with the last observation.

Assumptions, Diagnostics, and Limits

The calculations above rely on two specified components: a conditional outcome distribution $f(y\mid H_{i,j-1})$ and a conditional selection probability given the current outcome and history. The integration also assumes the stated conditioning history and visit structure are appropriate. In the Gaussian example, the normal distribution and its parameters were assumed for illustration; changing the outcome distribution can change the integrated dropout probability. The quadrature value is an approximation to the integral, not an observed dropout rate.

A practical check is to write down the model’s conditioning set before calculating any probability. For every dropout probability, ask whether the model calls for the missing $Y_{ij}$, the observed $Y_{i,j-1}$, or a broader observed history. Then verify that the outcome distribution being averaged over matches that choice. When numerical integration is used, compare the calculation at increased quadrature accuracy or by simulation to ensure the estimate is stable.

Observed data alone do not reveal the current outcomes of participants who dropped out. Consequently, the dependence of dropout on those unobserved outcomes cannot generally be checked directly from their missing values. Sensitivity to the assumed outcome and selection models matters; a precise integration under a misspecified model does not repair the misspecification.

Common pitfall: “The current outcome is missing, so I will put the last observed value into the current-outcome dropout equation.” This changes the equation’s predictor and discards uncertainty about the current outcome. The correction is to average the current-outcome selection probability over the specified conditional distribution of $Y_{ij}$, or to state clearly that a different, lagged-outcome model is being used.

Check Your Understanding

Answer each question, then compare with the worked solutions below.

If the current outcome is binary with equal probabilities and its dropout probabilities are 0.2 at 0 and 0.8 at 1, what is the dropout probability given history?
In the Gaussian example, why is evaluating the selection probability at the mean current outcome not generally the correct calculation?
Two participants have the same last observed outcome. Must a current-outcome model assign them the same dropout probability given history? Explain.
When would replacing the current outcome by the last observed outcome make no difference to the selection probability?

Solutions. (1) The required average is $(0.5)(0.2)+(0.5)(0.8)=0.5000$. (2) The selection probability is a nonlinear function of the outcome, so its average over the outcome distribution generally differs from its value at the mean. (3) No. Their current-outcome distributions can differ because their histories differ, producing different averages of the conditional selection probability. (4) For example, the selection probability may not depend on the current outcome, or the current outcome may be known to equal the last observed value. Without such a condition, substitution defines a different model.