Worked Influence Analysis of a Longitudinal Trial Participant

Why Follow One Participant Through the Calculation?

A subject-level influence result is only as trustworthy as the likelihood and derivatives behind it. In a longitudinal selection model, a participant’s contribution must account for both the outcomes observed and the probability of the observed response pattern. An outcome missing at an intermediate visit is integrated out; outcomes after dropout are also unobserved and integrated out. Treating the observed outcomes as if they were the whole trajectory can therefore change both the likelihood and the influence diagnostic.

This tutorial follows one hypothetical trial participant from the observed likelihood to a subject-weight perturbation, a curvature calculation, and an exact deletion refit. The goal is not to decide automatically that an unusual participant should be removed. It is to determine how sensitive the fitted analysis is to that participant and whether the local approximation is informative at the finite change corresponding to deletion.

Observed Likelihood for an Intermittent Pattern and Dropout

Definition: Let $Y_i=(Y_{i1},\ldots,Y_{iT})$ denote participant $i$'s full outcome trajectory, $R_i$ its observed response pattern, and $A_i$ the treatment and other observed covariates. A selection model specifies an outcome distribution $f_Y(y_i\mid A_i;\theta_Y)$ and a response-pattern probability $\pi(R_i\mid y_i,A_i;\theta_R)$. The participant's observed-data likelihood contribution is $$ L_i^{\mathrm{obs}}(\theta)= \sum_{y_{i,\mathrm{mis}}} f_Y(y_{i,\mathrm{obs}},y_{i,\mathrm{mis}}\mid A_i;\theta_Y) \pi(R_i\mid y_{i,\mathrm{obs}},y_{i,\mathrm{mis}},A_i;\theta_R), $$ where the sum is replaced by an integral for continuous missing outcomes and $\theta=(\theta_Y,\theta_R)$. The observed log-likelihood is $\ell(\theta)=\sum_i\ell_i(\theta)$, where $\ell_i(\theta)=\log L_i^{\mathrm{obs}}(\theta)$.

This expression is a joint likelihood for outcomes and the response pattern. In a selection model, the response probability may depend on outcomes that are themselves unobserved, so those outcomes remain inside the sum or integral. The model must specify normalized outcome and response distributions; participants are treated as independent conditional on their observed covariates. Inference also relies on the chosen outcome and selection models and their identification assumptions. The observed data alone do not generally determine the distribution of unobserved outcomes under an MNAR model.

For a trial with four scheduled visits, consider the pattern $R_i=(1,0,1,0)$: the participant has an intermittent missing outcome at visit 2, returns at visit 3, and then drops out before visit 4. The visit-2 and visit-4 outcomes are both missing, but their roles in the response history differ. The selection model assigns a probability to this full pattern conditional on the full outcome trajectory.

Worked Example: Constructing One Observed Contribution

Suppose a hypothetical fitted model has binary outcomes and, for a participant with observed $Y_{i1}=1$ and $Y_{i3}=0$, gives the following joint outcome probabilities and response-pattern probabilities. The response pattern is $R_i=(1,0,1,0)$. Each row represents one possible pair of missing outcomes $(Y_{i2},Y_{i4})$.

$(Y_{i2},Y_{i4})$	Joint outcome probability	$\pi(R_i\mid Y_i,A_i)$	Product
(0, 0)	0.06	0.50	0.030
(0, 1)	0.04	0.25	0.010
(1, 0)	0.08	0.25	0.020
(1, 1)	0.02	0.50	0.010

The observed likelihood contribution sums over both missing outcomes: $$ L_i^{\mathrm{obs}}=0.06(0.50)+0.04(0.25)+0.08(0.25)+0.02(0.50) =0.030+0.010+0.020+0.010=0.070. $$ Thus $\ell_i=\log(0.070)\approx -2.659$. This is a single contribution under the specified hypothetical model, not an estimate of a real trial probability. In particular, using only the first row would incorrectly treat the missing outcomes as known to be $(0,0)$.

Subject-Weight Perturbation and Curvature

Let $\hat\theta$ maximize the unweighted observed log-likelihood, and define the observed information at that fit by $J=-\ell_{\theta\theta}(\hat\theta)$. Assume the estimate is an interior, locally identifiable maximum, the relevant derivatives exist, and $J$ is positive definite. These conditions make the local inverse-information calculation well-defined. Write $u_i=\ell_{i,\theta}(\hat\theta)$ for participant $i$'s observed-data score. At the unweighted maximum, $\sum_i u_i=0$.

Give participant $i$ weight $1+\epsilon$, leaving all other weights at one. The perturbed objective is $\ell(\theta;\epsilon)=\ell(\theta)+\epsilon\ell_i(\theta)$, with maximizer $\hat\theta(\epsilon)$. For influence displacement, use the unperturbed likelihood evaluated at that perturbed estimate:

LD(\epsilon)=2\{\ell(\hat\theta)-\ell(\hat\theta(\epsilon))\}.

This definition compares fitted parameter values on the original likelihood surface. It is not the difference between the unperturbed maximum and the perturbed objective value. That distinction determines the sign of the curvature.

Result: For the unit subject-weight direction, let $q_i=u_i^{\mathsf T}J^{-1}u_i$. Under the regularity conditions above, $$ \hat\theta'(0)=J^{-1}u_i,\qquad LD(\epsilon)=\epsilon^2q_i+o(\epsilon^2),\qquad LD''(0)=2q_i. $$ The value $2q_i$ is the normal curvature for this direction under the stated baseline-likelihood displacement.

To justify the result, differentiate the perturbed score equation $\ell_\theta(\hat\theta(\epsilon))+\epsilon\ell_{i,\theta}(\hat\theta(\epsilon))=0$. At $\epsilon=0$, its derivative is $-J\hat\theta'(0)+u_i=0$, so $\hat\theta'(0)=J^{-1}u_i$. Taylor expansion of the unperturbed log-likelihood about its maximum gives $$ \ell(\hat\theta(\epsilon)) =\ell(\hat\theta)-\frac{1}{2}\epsilon^2 \hat\theta'(0)^{\mathsf T}J\hat\theta'(0)+o(\epsilon^2). $$ Substitution yields $LD(\epsilon)=\epsilon^2u_i^{\mathsf T}J^{-1}u_i+o(\epsilon^2)$, hence the stated second derivative. Because $J$ is positive definite, $q_i\geq 0$, as expected for a displacement from a local maximum.

Worked Example: Calculating a Participant’s Local Curvature

Suppose a hypothetical fitted selection model has parameter vector $\theta=(\beta,\gamma)^{\mathsf T}$, where $\beta$ is a treatment contrast in standardized outcome units and $\gamma$ is a selection coefficient per standardized outcome unit. For one participant, suppose $$ J=\begin{pmatrix}4&1\\1&3\end{pmatrix}, \qquad u_i=\begin{pmatrix}1\\2\end{pmatrix}. $$ The matrix is positive definite: its leading diagonal entry is $4>0$, and its determinant is $4(3)-1(1)=11>0$. The information condition is therefore satisfied for this calculation.

First invert $J$: $$ J^{-1}=\frac{1}{11}\begin{pmatrix}3&-1\\-1&4\end{pmatrix}, \qquad J^{-1}u_i=\frac{1}{11}\begin{pmatrix}1\\7\end{pmatrix}. $$ The first-order change in the fitted parameters per unit increase in this participant’s weight is therefore approximately $(1/11,7/11)^{\mathsf T}$, or $(0.09091,0.6364)^{\mathsf T}$ in the stated parameter units. The curvature calculation is $$ q_i=u_i^{\mathsf T}J^{-1}u_i =(1,2)\frac{1}{11}\begin{pmatrix}1\\7\end{pmatrix} =\frac{1+14}{11}=\frac{15}{11}\approx1.364. $$ Thus $LD''(0)=2q_i=30/11\approx2.727$. This describes local sensitivity near weight one; it does not, by itself, give the exact result of deleting the participant.

Exact Deletion: A Separate Refit

Deleting participant $i$ sets that participant’s weight to zero, or $\epsilon=-1$. This is a finite change, not an infinitesimal one. The local derivative predicts the initial movement as the weight changes near one, but it does not account fully for changing curvature or for a different likelihood shape farther from the original fit. Exact deletion requires maximizing the likelihood with that participant’s contribution removed and checking that the reduced-data fit is itself well-defined.

Worked Example: Comparing Local and Exact Deletion

Continue with the matrices from the previous example. For a transparent hypothetical calculation, suppose the full log-likelihood is exactly quadratic in $\delta=\theta-\hat\theta$, with information $J$. Suppose the participant’s contribution has local score $u_i=(1,2)^{\mathsf T}$ and information $H_i=I$, the $2$ by $2$ identity matrix. The remaining participants then have information $J-H_i=\begin{pmatrix}3&1\\1&2\end{pmatrix}$, which is positive definite because its determinant is $5>0$.

At deletion, the remaining-data score at $\delta=0$ is $-u_i$, since the full-data score is zero. The remaining-data score equation is therefore $-u_i-(J-H_i)\delta=0$. Solving gives $$ \delta_{\mathrm{delete}}=-(J-H_i)^{-1}u_i =-\frac{1}{5}\begin{pmatrix}2&-1\\-1&3\end{pmatrix} \begin{pmatrix}1\\2\end{pmatrix} =-\begin{pmatrix}0\\1\end{pmatrix} =\begin{pmatrix}0\\-1\end{pmatrix}. $$ In this constructed quadratic example, the exact deletion refit leaves $\beta$ unchanged and moves $\gamma$ down by one unit. The linear local prediction at $\epsilon=-1$ would instead be $-J^{-1}u_i=(-1/11,-7/11)^{\mathsf T}$, approximately $(-0.09091,-0.6364)^{\mathsf T}$. The difference arises because the information after deletion is $J-H_i$, not $J$.

The displacement measured on the original, full-data likelihood is also directly calculable here: $$ LD(-1)=\delta_{\mathrm{delete}}^{\mathsf T}J\delta_{\mathrm{delete}} =(0,-1)\begin{pmatrix}4&1\\1&3\end{pmatrix} \begin{pmatrix}0\\-1\end{pmatrix}=3. $$ The local second-order approximation is $q_i\epsilon^2=15/11\approx1.364$. Its difference from 3 illustrates why exact deletion and local curvature answer related but distinct questions.

Diagnostics, Alternatives, and Limitations

A reliable influence calculation should be checked against the likelihood actually fitted. Useful diagnostics include verifying that the unweighted score is close to zero, checking that $J$ is symmetric and positive definite, and comparing analytic or automatic derivatives with finite differences. For the observed likelihood, also check that the integration or summation over missing outcomes is implemented consistently with the response pattern. In particular, a returning participant with an intermittent missing visit must not be coded as having dropped out at that visit.

If $J$ is nearly singular, the inverse can amplify small score errors and produce unstable curvature. If the fitted estimate is on a boundary, or the likelihood is nonsmooth, the derivative argument may not apply. Large curvature can also reflect a scaling choice: the parameter units and any scaling of perturbations should be reported when comparing subjects. A strong diagnostic is to refit after deleting a participant with high local curvature and compare the actual parameter change with the local prediction. Large disagreement signals that the local approximation is inadequate for that finite deletion, not necessarily that either calculation is erroneous.

The main alternatives serve different purposes. A subject-weight curvature is a local, derivative-based screening diagnostic and is efficient to compute for many participants. Exact deletion is more direct for the question “what changes if this participant is removed?” but requires repeated optimization and may expose convergence or identification problems. Prespecified directional perturbations can target a scientifically defined change in several weights, while a target-parameter diagnostic focuses on a particular contrast rather than all parameters jointly. None of these procedures repairs a misspecified outcome or selection model, and a large influence measure is not evidence that the participant is invalid or should be excluded.

Common Mistake

Pitfall: Do not define $LD(\epsilon)=2\{\ell(\hat\theta)-\ell(\hat\theta(\epsilon);\epsilon)\}$ and then use the positive curvature $2u_i^{\mathsf T}J^{-1}u_i$. That expression evaluates the perturbed objective and includes its direct weight change; it is not the baseline-likelihood displacement defined here. For the definition used in this tutorial, evaluate the unperturbed $\ell$ at $\hat\theta(\epsilon)$, giving $LD''(0)=2u_i^{\mathsf T}J^{-1}u_i$. Keep the objective and displacement definition aligned throughout.

The practical lesson is to report what was perturbed, how the displacement was defined, and whether a result is local or from an exact refit. A curvature value is a model-based sensitivity diagnostic, not a test of the participant’s data quality and not a substitute for examining the response and outcome histories.

Check Your Understanding

Answer each question, then compare with the solutions below.

For pattern $R=(1,0,1,0)$, which outcomes are summed or integrated out in the observed likelihood, and what does the response probability condition on in a selection model?
If $J$ is positive definite and $u_i$ is a subject score, explain why $u_i^{\mathsf T}J^{-1}u_i$ cannot be negative.
For $J=\begin{pmatrix}2&0\\0&5\end{pmatrix}$ and $u_i=(2,5)^{\mathsf T}$, calculate the local curvature $LD''(0)$.
Why can the exact deletion estimate differ from the first-order local prediction at $\epsilon=-1$?

Solutions. (1) The visit-2 intermittent outcome and the post-dropout visit-4 outcome are unobserved, so both are summed or integrated out. The selection probability is conditional on the full outcome trajectory, including those missing outcomes, and on observed covariates. (2) Positive definiteness of $J$ implies positive definiteness of $J^{-1}$; hence the quadratic form is nonnegative, with equality only when $u_i=0$. (3) $J^{-1}=\operatorname{diag}(1/2,1/5)$, so $q_i=(2,5)J^{-1}(2,5)^{\mathsf T}=2+5=7$, and $LD''(0)=2q_i=14$. (4) The local derivative describes movement near weight one using the full-data information. At deletion the weight change is finite, the likelihood curvature can change, and the reduced-data maximum must be found by refitting.