Scaling Perturbations for Comparable Subject Diagnostics

Why Perturbation Scale Matters

Local influence diagnostics compare how the fitted model responds to small changes. But a change of one unit does not mean the same thing for every perturbation: one unit might be one outcome point, one year of a selection covariate, or a full unit of case weight. Curvature calculated per raw unit can therefore rank subjects in a way that reflects the chosen measurement units as much as the model’s response.

The remedy is to state what size of change is being compared. A subject’s diagnostic should answer a concrete question such as, “How much curvature results from a change of one prespecified, scientifically meaningful size?” Standardization makes those sizes explicit. It does not make every perturbation scientifically equivalent, nor does it remove the need to inspect the model and the chosen scale.

Local Curvature and a Change of Coordinates

Let $\ell(\theta;\omega)$ be a differentiable log-likelihood, where $\theta$ contains the fitted model parameters and $\omega$ is a vector of perturbation coordinates. At the unperturbed value $\omega_0$, let $\hat{\theta}$ be an interior maximum. Define the log-likelihood Hessian $H$, the observed information $I$, and the parameter-score/perturbation cross-derivative matrix $\Delta$ by

H=\left.\frac{\partial^2\ell}{\partial\theta\,\partial\theta^\mathsf{T}}\right|_{(\hat{\theta},\omega_0)},\qquad I=-H,\qquad \Delta=\left.\frac{\partial^2\ell}{\partial\theta\,\partial\omega^\mathsf{T}}\right|_{(\hat{\theta},\omega_0)}.

For the regular local calculation below, the likelihood is twice differentiable near the fit, the score is zero at $(\hat{\theta},\omega_0)$, and $I$ is positive definite. In particular, at a regular local maximum $H$ is negative definite; the positive-definite matrix used in the curvature calculation is the observed information $I=-H$. If the model includes nuisance parameters, they are included in $\theta$ and in $I$. A nearly singular $I$ makes its inverse unstable and can make computed curvature unreliable.

Definition: Under the stated regularity conditions, define the local likelihood-displacement curvature in a perturbation direction $d$ by $$ C(d)=\frac{2d^\mathsf{T}\Delta^\mathsf{T}I^{-1}\Delta d}{d^\mathsf{T}d},\qquad d\ne 0. $$ The factor of 2 is the conventional likelihood-displacement scaling used here. The direction’s length is normalized, so $C(d)$ describes curvature per unit direction in the coordinates being used.

To compare the coordinate-wise, one-at-a-time perturbations, the curvature matrix is

K=2\Delta^\mathsf{T}I^{-1}\Delta.

For coordinate $j$, the unit-direction curvature is $K_{jj}$. That value is not an intrinsic subject ranking until the unit of $\omega_j$ has been specified. If one coordinate is an outcome change in points and another is a covariate change in years, their raw unit curvatures refer to different-sized scientific changes.

Derivation of the Standardized Diagnostic

Let $S$ be a diagonal matrix of positive, prespecified perturbation scales $s_j$. Write a perturbation in dimensionless coordinates $u$ as $\omega=\omega_0+Su$. Thus, $u_j=1$ means a change of $s_j$ in the original coordinate $\omega_j$. The score equation for the perturbed estimate is

\frac{\partial\ell(\hat{\theta}(\omega);\omega)}{\partial\theta}=0.

Differentiate this equation at the baseline fit. The chain rule gives $H\,d\hat{\theta}+\Delta\,d\omega=0$. Since $H=-I$, the first-order parameter response is $d\hat{\theta}=I^{-1}\Delta\,d\omega$. For a change $d\omega=Sd u$, the response is $I^{-1}\Delta S d u$. Substituting the transformed cross-derivative $\Delta S$ into the curvature expression yields the standardized curvature matrix.

K_{\mathrm{std}}=2S^\mathsf{T}\Delta^\mathsf{T}I^{-1}\Delta S=S^\mathsf{T}KS.

When $S$ is diagonal, the standardized one-at-a-time curvature for coordinate $j$ is $s_j^2K_{jj}$. The square appears because curvature is second order: reversing the sign of a perturbation leaves its one-coordinate curvature unchanged. This standardization changes the perturbation scale, not the fitted model. A coherent re-expression of the same perturbation in new units gives the same standardized result when its scale is converted consistently.

Worked Example: Standardizing Outcome Perturbations

Suppose a fitted hypothetical selection model has observed information $I=25$ for the parameter direction under consideration. For two subjects, the cross-derivatives for additive outcome perturbations, measured in outcome points, are $\Delta_1=0.60$ and $\Delta_2=0.30$. Assume the fit is an interior regular maximum and that this information value is positive and well-conditioned.

Per one outcome point, the two curvatures are $2\Delta_i^2/I$. Therefore,

K_{11}=\frac{2(0.60)^2}{25}=0.0288,\qquad K_{22}=\frac{2(0.30)^2}{25}=0.00720.

On this raw scale, subject 1 has four times the curvature of subject 2. Now suppose a one-at-a-time change of five outcome points is the prespecified scale for each subject. Then $s_1=s_2=5$, so the standardized values are

K_{\mathrm{std},11}=5^2(0.0288)=0.720,\qquad K_{\mathrm{std},22}=5^2(0.00720)=0.180.

The ranking is unchanged because both coordinates received the same scale multiplier. The values now describe curvature for five-point changes, not one-point changes. If the scientific scales differed by subject, the ranking could change; that would answer a different, explicitly scale-dependent comparison. The five-point calculation is useful only if five points remains within the region where the local quadratic approximation is adequate.

Choosing Scales for Different Perturbations

A scale should describe a meaningful small change in the perturbation coordinate, not be chosen after inspecting which subjects appear influential. Depending on the diagnostic, a defensible choice might be a clinically interpretable outcome increment, a standard deviation of a selection covariate, or a specified fractional change in a case weight. State the units and rationale. A standard deviation is convenient but not automatically scientifically meaningful, especially for a skewed or highly constrained variable.

For a selection-equation covariate, be precise about the coordinate being perturbed. If a logistic selection model uses $\operatorname{logit}(\pi_i)=\alpha+\gamma x_i$, perturbing $x_i$ is different from perturbing the linear predictor or the selection probability $\pi_i$. These changes have different derivatives and units. A useful diagnostic names the actual perturbed quantity and the size of the change. If the choice is to perturb $\pi_i$, care is also needed near zero and one, where an equal probability change can behave differently from an equal logit change.

Worked Example: Converting a Selection-Covariate Scale

In a hypothetical fitted selection model, suppose the curvature information for one subject’s covariate perturbation is $I=16$, and the cross-derivative is $\Delta_x=0.080$ per year. Assume the fitted information is positive and nonsingular. A one-year change has curvature

K_x=2\frac{(0.080)^2}{16}=0.000800.

Suppose the same covariate is recorded in decades, $z=x/10$. A one-unit change in $z$ represents ten years, so the corresponding cross-derivative is $\Delta_z=10(0.080)=0.800$. The curvature per one decade is therefore

K_z=2\frac{(0.800)^2}{16}=0.0800.

The tenfold larger change produces one hundred times the curvature, as expected for a second-order quantity: $0.0800=10^2(0.000800)$. Now take a standardized change of two years. In years, $s_x=2$; in decades, the same change is $s_z=0.2$. Both calculations agree:

s_x^2K_x=2^2(0.000800)=0.00320,\qquad s_z^2K_z=0.2^2(0.0800)=0.00320.

Thus, changing from years to decades changes the raw unit curvature but not the curvature for the same two-year perturbation. This comparison is valid because the scale conversion was applied consistently and the model represents the same covariate change in either unit.

Subject-Level Outcome, Covariate, and Weight Diagnostics

For subject-level comparisons, construct one perturbation coordinate per subject, or clearly identify which subjects share a coordinate. Choose a scale $s_i$ for each coordinate and calculate $s_i^2K_{ii}$. For example, outcome perturbations could be expressed as a common clinically relevant change; selection-covariate perturbations could use a common reference increment, such as one standard deviation; and case-weight perturbations could use a prespecified fractional change. These are separate analyses unless a joint perturbation scheme and a common scientific interpretation have been defined.

Case-weight coordinates illustrate why scale must be explicit. In a weighted log-likelihood $\ell(\theta;\omega)=\sum_i\omega_i\ell_i(\theta)$, the baseline weights are often $\omega_i=1$. A perturbation of $0.1$ means a ten-percent weight increase, while a perturbation of $-1$ corresponds to removing that case from a unit baseline weight. Local curvature for a small weight change does not, by itself, give the exact effect of deletion. Deletion is a finite change and can require a separate approximation or refit.

Worked Example: A Ten-Percent Case-Weight Change

Consider a hypothetical fitted model with observed information $I=100$ and a subject’s cross-derivative $\Delta_i=2.50$ per unit change in that subject’s case weight. Assume the likelihood is smooth near baseline weight $1$, and $I$ is positive and adequately conditioned. Curvature per full weight unit is

K_{ii}=2\frac{(2.50)^2}{100}=0.125.

If the diagnostic is intended to represent a ten-percent weight change, take $s_i=0.10$. The standardized curvature is

K_{\mathrm{std},ii}=(0.10)^2(0.125)=0.00125.

This value concerns a local increase of $0.10$ in the weight, around its baseline value of $1$. It is not the curvature for deletion: deletion changes the weight by $-1$, ten times the size in absolute value. Simply multiplying by one hundred would be an extrapolation of the local quadratic approximation to a potentially nonlocal change, and need not match a case-deletion result.

Diagnostics, Comparisons, and Limits

Before interpreting a standardized ranking, check both the model calculation and the scale choice. At minimum, verify that the baseline is a genuine interior maximum, the observed information $I$ is positive definite and not nearly singular, and the perturbation derivatives are computed in the stated units. Compare results under a couple of defensible nearby scales: because curvature is quadratic, doubling a scale multiplies the reported curvature by four. The ranking can also change when scale multipliers differ across subjects.

Raw versus standardized curvature: raw curvature is per unit of the recorded coordinate; standardized curvature is per prespecified change $s_i$. Neither is meaningful without its unit.
One-at-a-time versus joint changes: $K_{ii}$ describes a coordinate direction by itself. For simultaneous changes, off-diagonal terms matter through $d^\mathsf{T}Kd$; adding separate diagonal curvatures does not generally give the joint result.
Local curvature versus finite deletion: curvature describes second-order behavior near the baseline. It does not guarantee accurate prediction for a large perturbation or case removal.
Subject ranking versus evidence about missing outcomes: high curvature flags local sensitivity of the fitted likelihood calculation. It does not establish that a subject’s missing outcome is unusual or that the assumed selection model is correct.

A common mistake is to rank subjects by unstandardized values from different perturbation types and call the largest value “most influential.” That reasoning treats one outcome point, one covariate year, and one case-weight unit as comparable changes. The correction is to define a scientifically interpretable scale for each coordinate, compute $S^\mathsf{T}KS$, and report the units and scale alongside the ranking. If no defensible scales make the changes comparable, present separate diagnostics rather than forcing a single ranking.

Standardization also does not cure weak identification, poor numerical differentiation, model misspecification, or a perturbation that is scientifically implausible. If the information matrix is nearly singular, its inverse can greatly amplify small derivative errors. If a scale is too large, the local approximation can fail; if it is too small, numerical precision may dominate. These conditions call for caution and additional numerical checks, not stronger claims from the curvature value.

Key takeaway: A subject-level curvature is interpretable only relative to a stated perturbation coordinate and scale. Use the observed information $I=-H$, transform the curvature as $K_{\mathrm{std}}=S^\mathsf{T}KS$, and describe the result as local sensitivity for the chosen change—not as a scale-free measure or an exact deletion effect.

Check Your Understanding

Answer each question, then compare with the brief solutions.

A perturbation coordinate has raw curvature $0.020$ per unit. What is the standardized curvature for a scale of $3$ units?
A covariate is converted from years to decades. What happens to its raw one-unit curvature, and what happens to curvature for the same two-year change?
Why is a ten-percent case-weight perturbation not equivalent to case deletion?
What matrix must be positive definite at a regular local maximum in the curvature calculation: the log-likelihood Hessian $H$, or the observed information $I=-H$?

Solutions. (1) $3^2(0.020)=0.180$. (2) One decade is ten years, so curvature per decade is $10^2$ times curvature per year; for the same two-year change, the scale is converted consistently and standardized curvature is unchanged. (3) A ten-percent increase changes a baseline weight of $1$ by $0.10$, whereas deletion changes it by $-1$; the latter is a much larger finite perturbation and is not generally covered by the local approximation. (4) $I=-H$ is positive definite; $H$ is negative definite at a regular local maximum.