Why compare influence by dropout history?
In a longitudinal study, subjects can contribute different observed records because they drop out at different times. A subject with an unusual record may have substantial influence on the fitted selection model, but a large dropout history can also produce a large total influence simply because it contains many subjects. These are different phenomena: influence concentrated in particular records is not the same as influence associated with a numerous response pattern.
This tutorial compares subject-level curvature across observed dropout histories. The histories are used to group subjects after their individual contributions have been assessed. The comparison describes local sensitivity of the fitted observed selection likelihood; it does not establish that a particular history caused an outcome or that the fitted model is correct.
Definition and assumptions
Let \(y_i^{\mathrm{obs}}\) denote the observed record for subject \(i\), including the subject’s observed dropout history, and let \(\ell_i(\theta)\) be that subject’s contribution to the observed selection log likelihood. The full log likelihood is \(\ell(\theta)=\sum_{i=1}^n \ell_i(\theta)\), and \(\hat{\theta}\) is an interior maximum-likelihood estimate. Define the subject score \(u_i=\partial \ell_i(\theta)/\partial\theta\) evaluated at \(\hat{\theta}\). Under a differentiable likelihood and an interior maximum, the scores satisfy \(\sum_i u_i=0\).
The calculations below assume that the observed selection likelihood is additive over subjects, that the likelihood is twice differentiable near \(\hat{\theta}\), and that the observed information matrix is positive definite at the fitted value. These conditions give a stable local quadratic approximation. If parameters are on a boundary, the likelihood is not smooth, or the information matrix is singular or nearly singular, the curvature calculations can be invalid or unstable.
To see the meaning of this definition, give each subject a weight \(\omega_i\) and write \(\ell(\theta;\omega)=\sum_i\omega_i\ell_i(\theta)\). At the fitted weights \(\omega_i=1\), perturb in direction \(d\) by setting \(\omega_i=1+t d_i\). Differentiating the weighted score equation at \(t=0\) gives the local change in the estimate. In particular, for a single subject direction, this change is \(J^{-1}u_i\). Its squared length measured using \(J\) is \(C_i\).
Here \(\dot{\theta}\) is the derivative of the estimate with respect to \(t\) at zero. The sign in the second equation follows because the derivative of the unweighted score with respect to \(\theta\) is \(-J\). Setting \(d\) to the unit vector for subject \(i\) yields \(C_i\). For a pattern \(g\) containing \(n_g\) subjects, a useful common-weight comparison takes \(d_i=1/\sqrt{n_g}\) for subjects in that pattern and zero otherwise. This direction has squared length one, and its curvature is \(C_g^{\mathrm{common}}=(\sum_{i\in g}u_i)^TJ^{-1}(\sum_{i\in g}u_i)/n_g\). It measures the response to a common change spread across the pattern, on this specified unit scale.
Worked Example: A smaller history contains the most influential subjects
Worked Example: A smaller history contains the most influential subjects
Consider six hypothetical subjects in two observed dropout histories. For illustration, suppose the fitted information matrix is the two-dimensional identity matrix, \(J=I\), so each subject’s curvature is the squared length of the score vector. The score vectors at the fitted value are:
History A has four subjects, each with score \((-0.25,\,0.25)^T\). History B has two subjects with scores \((2,\,0)^T\) and \((-1,\,-1)^T\).
Check the fitted-score condition. The four History A scores sum to \((-1,\,1)^T\). The two History B scores sum to \((1,\,-1)^T\). Thus the total score is \((0,\,0)^T\), as required at the fitted value.
Calculate individual curvature. For each A subject, \(C_i=(-0.25)^2+(0.25)^2=0.125\). For the two B subjects, the values are \(C_i=2^2+0^2=4\) and \(C_i=(-1)^2+(-1)^2=2\). The smaller history contains both subjects with substantially larger individual curvature.
Compare common-weight pattern curvature. For A, the summed score is \((-1,1)^T\), so \(C_A^{\mathrm{common}}=[(-1)^2+1^2]/4=0.500\). For B, the summed score is \((1,-1)^T\), so \(C_B^{\mathrm{common}}=[1^2+(-1)^2]/2=1.00\). On the stated unit perturbation scale, B also has greater common-weight curvature.
The result does not follow from pattern size: B has fewer subjects. It follows from the score contributions and the information metric. The score check is important: assigning subject scores that do not sum to zero would contradict the assumed fitted likelihood maximum.
How to summarize influence across histories
No single pattern summary answers every question. For each observed history, useful descriptive summaries include the number of subjects, the mean and median of \(C_i\), the maximum \(C_i\), and the proportion above a prespecified review threshold. The mean and median describe typical subject curvature; the maximum and upper tail help identify concentration in a few records. Report the number of subjects alongside these quantities, since a median based on two records is less stable as a description of a broader pattern than one based on many.
The common-weight pattern curvature answers a different question: whether a unit-length perturbation that changes the weights of all subjects in the history together produces a large local displacement. It depends on the sum of their score vectors, so cross-subject cancellation matters. It is not generally equal to the sum or average of the individual \(C_i\) values. Thus compare individual summaries and, when a coherent pattern-wide perturbation is substantively relevant, report its curvature separately.
Worked Example: Pattern size does not determine typical influence
Suppose two hypothetical histories have \(J=I\). History A contains eight subjects, each with score \((-0.25,\,0)^T\). History B contains two subjects, each with score \((1,\,0)^T\). These scores sum to \((-2,0)^T\) for A and \((2,0)^T\) for B, so the total score is zero.
For each A subject, \(C_i=(-0.25)^2=0.0625\); for each B subject, \(C_i=1^2=1\). The mean curvature is therefore \(0.0625\) in A and \(1.00\) in B. The sum of individual curvatures is \(8(0.0625)=0.500\) in A and \(2(1)=2.00\) in B. Both the typical and summed individual curvature are larger in the smaller history, even though A has four times as many subjects.
For common-weight perturbations, A has curvature \([(-2)^2]/8=0.500\), while B has curvature \([2^2]/2=2.00\). In this example there is no within-history score cancellation, so the common-weight comparison also shows greater curvature for B. The arithmetic illustrates why pattern counts alone cannot rank influence.
Cancellation and interpretation
Worked Example: Large individual curvature can cancel within a history
Consider four hypothetical subjects in one history with scores \((1,0)^T,(-1,0)^T,(1,0)^T,(-1,0)^T\), and two subjects in another history with scores \((0,1)^T,(0,-1)^T\). Let \(J=I\). The scores in each history sum to zero, so the total fitted score is also zero.
Every subject has \(C_i=1\), since each score vector has squared length one. Yet the common-weight curvature is zero for both histories: each history’s summed score is zero, and hence \(C_g^{\mathrm{common}}=0\). The pattern-wide common change causes no first-order displacement because opposing subject scores cancel. This does not mean the individual records have zero curvature; it means the selected group perturbation direction has zero curvature.
This distinction is central to pattern-specific interpretation. A pattern may contain several individually influential records whose effects point in opposing directions. Conversely, a common shift in a pattern’s weights may align contributions and produce a substantial displacement. Always state which summary and perturbation direction are being compared.
Derivation and practical checks
The local result follows directly from the weighted score equation. Write \(U=\sum_i d_i u_i\), and let \(\theta(t)\) solve the weighted score equation at \(\omega_i=1+td_i\), with \(\theta(0)=\hat{\theta}\). Differentiating that equation with respect to \(t\) at zero gives \(0=-J\dot{\theta}+U\). Since \(J\) is invertible, \(\dot{\theta}=J^{-1}U\). Substitution into the information-metric squared displacement gives \(C(d)=U^TJ^{-1}U\). This is a local, first-derivative calculation; it does not claim that a finite weight change has exactly this effect.
A practical comparison should check that the score contributions sum approximately to zero at the fitted estimate, subject to numerical tolerance. A large discrepancy suggests a fitting or score-evaluation problem. Check that \(J\) is symmetric and positive definite and inspect its conditioning: if it is nearly singular, small numerical changes can greatly alter \(J^{-1}\) and therefore the curvature rankings. Also inspect whether a high pattern mean is driven by one extreme subject, and whether small patterns make their summaries imprecise.
Group membership must be defined from the observed dropout history consistently. If subjects have different observed response sequences, do not silently combine them merely to obtain larger groups. A history can be rare, and a high-curvature subject in a rare history merits inspection, but curvature alone cannot distinguish data error, model inadequacy, or a genuinely unusual record.
How this comparison differs from alternatives
| Approach | Question answered | Important distinction |
|---|---|---|
| Subject-level curvature summarized by history | Which records or histories have large local curvature? | Local and dependent on the fitted likelihood’s information metric. |
| Exact case deletion | How does the refitted estimate change after removing a subject? | Finite refitting can differ from the local approximation. |
| Pattern common-weight curvature | How does a specified equal-direction weight change within a history act locally? | Score cancellation and the chosen direction affect the result. |
| Largest-eigenvalue direction | Which combined perturbation direction has greatest curvature? | It need not correspond to a single history or individual subject. |
These summaries complement rather than replace one another. Exact deletion is a useful benchmark when a record appears influential, while local subject curvature provides a computationally convenient screen. Pattern summaries locate where high individual curvatures occur; they do not by themselves diagnose why. Nor should the history with the largest total or maximum curvature automatically be called the most problematic: counts, within-pattern dispersion, cancellation, and the parameter direction of displacement all matter.
Common pitfalls and limitations
- Ranking histories by size. A larger history may have more total curvature simply because it contains more records. Report subject-level distributions and counts, not only totals.
- Equating zero group curvature with no influential records. A common-weight score sum can cancel. Inspect individual \(C_i\) values as well.
- Adding subject curvatures to obtain group curvature. The group formula contains cross-products among scores; it is not generally the sum of the individual values.
- Treating local curvature as an exact deletion effect. The derivation is local around the fitted estimate. Large changes or nonlinear likelihood behavior can make the approximation inadequate.
- Ignoring the information matrix. The same score vector can produce different curvature under different fitted information. Near-singularity makes rankings especially fragile.
These diagnostics describe sensitivity under the specified observed selection likelihood and perturbation. They do not resolve whether the missingness model is scientifically plausible, nor do they establish that a high-curvature record is erroneous. Those judgments require substantive review and complementary analyses.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For a one-parameter model with \(J=4\), what is the subject curvature when \(u_i=2\)?
- Two histories have respective mean individual curvatures 0.2 and 0.8. The first has 20 subjects and the second has 3. Which has higher typical individual curvature, and why is this not a comparison of total curvature?
- Why can a history have zero common-weight curvature even when every subject in it has positive individual curvature?
- State one condition required for the local derivative calculation and one diagnostic for a failure of that condition.
Solutions. (1) \(C_i=u_i^2/J=2^2/4=1\). (2) The second history has higher typical curvature because its mean is 0.8 rather than 0.2. Mean curvature is not total curvature; the subject counts differ, and the sums would be \(20(0.2)=4\) and \(3(0.8)=2.4\). (3) The score vectors can sum to zero within the history, so the common-weight perturbation has zero first-order score displacement even though each squared individual contribution is positive. (4) The information matrix must be invertible (and in the stated setup positive definite); a singular or nearly singular \(J\), detected through its eigenvalues or condition number, makes the inverse-based calculation undefined or unstable.