What Local Influence Measures in Selection Models

From a Fitted Model to a Local Question

A fitted selection model can be sensitive to small changes in the information or assumptions that enter its likelihood. Local influence makes that question precise: if a specified perturbation is introduced by a very small amount, how much does the fitted model move away from its baseline fit, as measured on the baseline likelihood scale? The answer is local to a chosen perturbation path. It is not a universal ranking of subjects, and it is not, by itself, evidence that a model is wrong.

The essential comparison is between a baseline fit and fits obtained after small perturbations. A path describes how the model is changed continuously from its baseline specification. Likelihood displacement then measures the loss in the original, unperturbed log likelihood when evaluated at a perturbed fit. This keeps the scale of comparison fixed while allowing the fit to move.

Definition: Let $\ell(\theta\mid\omega)$ be a log likelihood indexed by model parameter $\theta$ and perturbation setting $\omega$. Let $\omega_0$ denote the baseline setting, and let $\hat{\theta}$ maximize $\ell(\theta\mid\omega_0)$. For a nearby setting $\omega$, let $\hat{\theta}_{\omega}$ maximize $\ell(\theta\mid\omega)$. The baseline likelihood displacement is $$ LD(\omega)=2\{\ell(\hat{\theta}\mid\omega_0)-\ell(\hat{\theta}_{\omega}\mid\omega_0)\}. $$ It measures how much baseline log-likelihood is lost by evaluating the original likelihood at the perturbed estimate rather than at its maximum.

The same baseline likelihood is used at both parameter values in this definition. Since $\hat{\theta}$ maximizes that likelihood, $LD(\omega)\geq 0$, provided both parameter values are in its domain. This is a likelihood-scale measure of displacement, not a p-value, a probability that a subject is influential, or a test of a null hypothesis.

A perturbation path is a smooth one-dimensional route away from the baseline, written $\omega(t)$, where $\omega(0)=\omega_0$. The scalar $t$ controls the size of the change. A subject-level path might alter some aspect of one subject’s contribution by a small amount, while leaving the remaining model specification unchanged. The exact meaning and scale of that alteration must be stated; local influence is always relative to the path chosen.

What “Local” Means

Local influence concerns the behavior of the fitted model and its likelihood displacement for $t$ close to zero. It asks about the initial response to an infinitesimal change, not the effect of a large change. If the fitted parameter changes smoothly, write $\hat{\theta}(t)$ for the estimate along the path, with $\hat{\theta}(0)=\hat{\theta}$. The displacement compares $\hat{\theta}(t)$ with the baseline maximum using the baseline likelihood.

Regularity conditions: The baseline fit should be an interior local maximum of a twice-differentiable log likelihood, with a nonsingular negative-definite Hessian in the parameters being estimated. The perturbed likelihood and its maximizer should vary smoothly for small $t$, and the perturbation path must remain within the model’s valid domain. For a likelihood displacement interpreted around a unique fitted solution, numerical optimization and any numerical integration used in the likelihood must also be sufficiently accurate to distinguish genuine movement from computational error.

These conditions make the local description meaningful. They may fail, for example, when the fit is on a parameter boundary, the likelihood has multiple competing maxima, or the parameter estimates change abruptly as the perturbation changes. Selection-model likelihoods can exhibit such difficulties, so local influence should be calculated only after the baseline fit has been checked. The earlier material on likelihood instability and reproducible likelihood analyses is relevant to those checks.

Why the Displacement Starts at Second Order

At the baseline estimate, the first derivative of the baseline log likelihood with respect to $\theta$ is zero. Consequently, a small movement away from the baseline maximum has no first-order effect on the baseline log likelihood. Its leading change is quadratic in the parameter movement. This is why likelihood displacement is naturally suited to describing local movement, rather than simply reporting the sign of a parameter change.

Expand the baseline likelihood.
Write $\Delta(t)=\hat{\theta}(t)-\hat{\theta}$. A second-order Taylor expansion around the baseline maximum gives $\ell(\hat{\theta}(t)\mid\omega_0)=\ell(\hat{\theta}\mid\omega_0)+\frac{1}{2}\Delta(t)^{T}H_0\Delta(t)+o(\|\Delta(t)\|^2)$, because the first derivative is zero. Here $H_0$ is the baseline Hessian.

Use the local maximum condition.
At a regular maximum, $H_0$ is negative definite. Subtracting the expansion from the maximum log likelihood and multiplying by two yields $LD(\omega(t))=-\Delta(t)^T H_0\Delta(t)+o(\|\Delta(t)\|^2)$, which is nonnegative to leading order.

Relate displacement to path size.
If $\Delta(t)=t v+o(|t|)$ for some parameter-change vector $v$, then $LD(\omega(t))=-t^2v^TH_0v+o(t^2)$. Thus the first-order displacement is zero; the leading change is quadratic in the small path step.

This justification describes the local order of the displacement; it does not say that every small perturbation produces the same response. The vector $v$ depends on the model and the chosen path. Nor does a small displacement guarantee that a large perturbation would be harmless: local behavior need not extrapolate over a long path.

Worked Example: A Small Perturbation of a Fitted Contrast

Suppose a hypothetical selection-model analysis has a one-parameter profile log likelihood for a treatment contrast $\theta$, measured in log-response units. Use the following local quadratic profiles solely to illustrate the calculation: $$ \ell(\theta\mid 0)=-\frac{(\theta-1.20)^2}{2(0.25)},\qquad \ell(\theta\mid t)=-\frac{(\theta-(1.20+0.08t))^2}{2(0.25)}. $$ The baseline setting is $t=0$. Assume these profiles are valid near the displayed maxima and that no boundary or competing maximum is involved.

The baseline estimate is $\hat{\theta}=1.20$. At perturbation $t$, the perturbed estimate is $\hat{\theta}(t)=1.20+0.08t$. The displacement is evaluated using the baseline profile, not the perturbed one: $$ LD(t)=2\left\{0-\left[-\frac{(0.08t)^2}{2(0.25)}\right]\right\} =\frac{0.0064t^2}{0.25}=0.0256t^2. $$

For a small path step $t=0.10$, the estimate moves by $0.08(0.10)=0.008$ log-response units, from 1.20 to 1.208. The baseline likelihood displacement is $0.0256(0.10)^2=0.000256$. It is small on this illustrative likelihood scale. This calculation describes the specified path and step; it is not a significance test and does not establish that the subject or model is unimportant under other changes.

Infinitesimal Change Is Not Case Deletion

A subject-level infinitesimal perturbation changes the model continuously by a small amount. Deleting a subject is instead a finite, discrete operation: the subject is removed, the likelihood is changed accordingly, and the model is refitted. The deletion result can be useful as a benchmark, but it is not automatically the local influence of that subject. In particular, there is generally no reason to treat deletion as a small step on a given perturbation path or to assume that a local approximation remains accurate all the way to deletion.

Worked Example: A Local Change Versus Removing a Subject

Consider a hypothetical one-parameter profile for a selection-model fit, with $\theta$ measured in log-response units: $$ \ell_{\mathrm{full}}(\theta)=-5(\theta-0.50)^2. $$ Its maximum is at $\hat{\theta}=0.50$, and its curvature is negative. Suppose a small, continuous subject-level perturbation is represented for illustration by adding $2\varepsilon(\theta-0.50)$, where $\varepsilon=0$ is the baseline.

The perturbed log likelihood is $-5(\theta-0.50)^2+2\varepsilon(\theta-0.50)$. Differentiating with respect to $\theta$ and setting the derivative to zero gives $-10(\theta-0.50)+2\varepsilon=0$, so $\hat{\theta}(\varepsilon)=0.50+0.20\varepsilon$. At $\varepsilon=0.10$, the estimate is 0.52, a change of 0.02 log-response units. Evaluating the original profile there gives $$ LD(0.10)=2\{0-[-5(0.02)^2]\}=0.004. $$

Now suppose the actual likelihood after deleting that subject has the hypothetical profile $\ell_{-\!i}(\theta)=-4(\theta-0.40)^2$. Refitting gives $\hat{\theta}_{-\!i}=0.40$, a finite change of $-0.10$ log-response units. Its displacement on the original profile is $$ 2\{\ell_{\mathrm{full}}(0.50)-\ell_{\mathrm{full}}(0.40)\} =2\{0-[-5(0.10)^2]\}=0.10. $$ The two calculations differ because the continuous perturbation and deletion are different operations. The local result does not predict the deletion result without additional justification.

Scale, Diagnostics, and Alternatives

The numerical size of a local displacement depends on how the path is scaled. If the same physical change is labelled by a different path parameter, the coefficient of $t^2$ changes. Thus a displacement has no meaningful interpretation without the perturbation definition, direction, and scale. Comparisons across subjects or models require comparable perturbation scales; otherwise a larger value may merely reflect a larger unit of change.

Worked Example: Why Path Scaling Must Be Reported

Suppose a hypothetical baseline profile is $\ell_0(\theta)=-(25/2)(\theta-2.00)^2$, with $\hat{\theta}=2.00$, and a specified path gives $\hat{\theta}(t)=2.00+0.04t$. Assume the profile and path are smooth near zero and the estimate is an interior maximum.

The baseline displacement is $$ LD(t)=2\left\{0-\left[-\frac{25}{2}(0.04t)^2\right]\right\} =25(0.0016)t^2=0.04t^2. $$ At $t=0.10$, the estimate moves by 0.004 units and $LD=0.0004$.

Re-label the same physical path using $s=2t$. Then $\hat{\theta}=2.00+0.02s$, and the formula becomes $LD(s)=25(0.02s)^2=0.01s^2$. At the same physical change, $t=0.10$ corresponds to $s=0.20$, and both formulas give displacement 0.0004. Comparing the coefficients 0.04 and 0.01 without comparing the path scales would be misleading.

Useful diagnostics follow directly from the definition and its assumptions:

Check the baseline fit first. A parameter on a boundary, a nearly flat likelihood, or an alternative local maximum can make small perturbations appear disproportionately consequential or make the fitted path discontinuous.
Check numerical stability. If optimization error or inaccurate numerical integration is comparable to the apparent change, the reported displacement may reflect computation rather than perturbation. Refit from suitable starting values and verify the likelihood calculation.
State what changes and by how much. Name the subject-level feature or model component being perturbed, identify the baseline, and report the path scale. “Subject $i$ has high local influence” is incomplete without saying under which perturbation.
Do not extrapolate without checking. A small-step result describes behavior near the baseline. For a larger change, evaluate the fit along the path or use a finite-change analysis rather than assuming the local approximation still applies.

Local influence and its alternatives answer related but different questions. Exact case deletion asks what happens after removing a subject and refitting; it is a finite benchmark and can reveal a practically consequential change, but it does not isolate infinitesimal behavior. Broader sensitivity analysis asks how conclusions respond to a range of specified changes, which can be more informative when the range itself matters. A likelihood-ratio test instead addresses a specified model constraint and its calibration; it is not an influence measure. These methods can complement one another, but their targets and scales should not be conflated.

Key takeaway: Local influence is the response of a fitted likelihood to an explicitly defined, infinitesimal perturbation path, summarized through displacement on the baseline likelihood. It is distinct from deleting a subject, and its interpretation depends on regular fitting conditions, a stable computation, and a clearly stated perturbation scale.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Why is the first-order change in baseline log likelihood zero when the fitted parameter moves smoothly from an interior baseline maximum?
A baseline profile is $\ell_0(\theta)=-2(\theta-3)^2$. A perturbed fit is $\hat{\theta}_{\omega}=3.1$. Calculate its baseline likelihood displacement.
Why is deleting a subject not generally the same as taking a small subject-level perturbation to its largest possible value?
If the perturbation parameter is rescaled, what must be held fixed to make comparisons of displacement meaningful?
Name two conditions or diagnostics that matter before interpreting a local displacement from a selection-model fit.

Solutions. (1) The baseline score is zero at an interior maximum, so the linear term in the Taylor expansion vanishes; the leading likelihood change is quadratic in the parameter movement. (2) The baseline maximum is $\ell_0(3)=0$, while $\ell_0(3.1)=-2(0.1)^2=-0.02$. Therefore $LD=2[0-(-0.02)]=0.04$. (3) Deletion is a discrete change to the likelihood followed by refitting; it need not lie on the same smooth path, and a local approximation near zero need not remain accurate for a finite change. (4) Hold the physical perturbation definition and size fixed, and report the path scale; raw coefficients based on differently scaled path parameters are not directly comparable. (5) Examples include an interior fit with a nonsingular negative-definite Hessian, a smooth valid perturbation path, adequate optimization and integration accuracy, and checks for boundary estimates or competing maxima.