Why Local Curvature Matters
A Wald interval summarizes uncertainty using curvature right at the fitted maximum. That summary is useful when the likelihood is well approximated by a quadratic over the region relevant to inference. Near a likelihood ridge, however, the surface can be nearly flat in one direction and can bend as parameters move. The local quadratic approximation may then describe only a small neighborhood of the estimate, while the reported interval reaches much farther.
This issue is especially important in selection models, where outcome parameters and response parameters can compensate for each other. A fitted model may reproduce the observed data well while allowing substantially different values of a missingness-related parameter. A small standard error computed from an unstable curvature calculation does not resolve that ambiguity; it can conceal it.
The inverse information is used because the local quadratic approximation has the form \(\ell(\boldsymbol{\theta})\approx\ell(\hat{\boldsymbol{\theta}})-\frac12(\boldsymbol{\theta}-\hat{\boldsymbol{\theta}})^T I(\hat{\boldsymbol{\theta}})(\boldsymbol{\theta}-\hat{\boldsymbol{\theta}})\). In that approximation, \(I^{-1}\) is the covariance matrix. This is a local argument: it does not show that the actual likelihood remains quadratic far from its maximum, or that the parameter is well identified by the observed data.
A Selection-Model Ridge in the Observed Data
Consider a hypothetical study with a binary outcome \(Y\) and response indicator \(R\), where \(R=1\) means \(Y\) is observed. Let \(p=P(Y=1)\), \(q_1=P(R=1\mid Y=1)\), and \(q_0=P(R=1\mid Y=0)\). A selection model specifies response probabilities through a logistic regression:
Thus \(\alpha=\operatorname{logit}(q_0)\), and \(\gamma=\operatorname{logit}(q_1)-\operatorname{logit}(q_0)\). The observed data distinguish the probabilities of a responding outcome \(Y=1\), a responding outcome \(Y=0\), and a missing outcome. They do not directly reveal the outcome distribution among those with \(R=0\).
Worked Example: An Exact Ridge from Three Observed Categories
Suppose a hypothetical sample of 100 people contains 36 observed \(Y=1\) outcomes, 24 observed \(Y=0\) outcomes, and 40 missing outcomes. The corresponding observed category probabilities are \(pq_1=0.36\), \((1-p)q_0=0.24\), and \(1-pq_1-(1-p)q_0=0.40\).
Likelihood and ridge. Apart from the multinomial coefficient, the observed-data likelihood is
For any \(p\) strictly between \(0.36\) and \(0.76\), set \(q_1=0.36/p\) and \(q_0=0.24/(1-p)\). Then \(q_1\) and \(q_0\) are both strictly between zero and one, and the three category probabilities remain \(0.36\), \(0.24\), and \(0.40\). The likelihood is therefore unchanged as \(p\) varies over this interval. The endpoints are excluded for finite logistic parameters: at \(p=0.36\), \(q_1=1\); at \(p=0.76\), \(q_0=1\). They are only limiting cases if infinite parameter values are admitted.
Different selection effects on the same ridge. At \(p=0.50\), the values are \(q_1=0.72\) and \(q_0=0.48\), so \(\gamma=\operatorname{logit}(0.72)-\operatorname{logit}(0.48)\approx0.9445-(-0.0800)=1.025\). At \(p=0.60\), \(q_1=q_0=0.60\), giving \(\gamma=0\). Both parameter choices give exactly the same observed category probabilities and likelihood. Thus the data do not uniquely identify \(\gamma\) in this example.
Inference consequence. Along the ridge, the likelihood has no curvature in the direction that changes \(p\) while preserving those observed probabilities. The information matrix is singular in the idealized model, so its ordinary inverse—and hence the usual Wald standard error—is unavailable. A numerical algorithm that returns a finite standard error after rounding, regularization, or other approximations has not created information that the observed data lack.
Why the Quadratic Approximation Can Mislead
An exact ridge is an extreme case. More commonly, the likelihood has some curvature along the ridge, but it is weak, and the ridge bends. A Wald interval uses the curvature at one point and projects it onto a parameter coordinate. A profile likelihood instead considers the best-fitting nuisance parameter values at each fixed value of the parameter of interest. When the ridge curves, these summaries need not agree.
The contrast can be seen by writing the local quadratic approximation at the maximum. Along a direction with small information eigenvalue \(\lambda\), a displacement \(t\) changes the approximate log likelihood by only \(-\lambda t^2/2\). If the actual surface bends or becomes steeper as \(t\) increases, the quadratic approximation can give an inaccurate account of how much likelihood support remains at the interval endpoints. If curvature is so small that numerical error is comparable to it, the estimated inverse information can also vary greatly with small computational perturbations.
Worked Example: A Curved Profile and a Wald Interval
Consider a hypothetical log likelihood for a selection parameter \(\gamma\) and nuisance parameter \(\eta\):
Here \(C\) is constant. For each fixed \(\gamma\), the likelihood is maximized over \(\eta\) at \(\hat{\eta}(\gamma)=\gamma^2\). The profile log likelihood is therefore \(C-\frac12\{\gamma^2/0.1225+8\gamma^4\}\), with maximum at \(\hat{\gamma}=0\). At that maximum, the second derivative with respect to \(\gamma\) is \(-1/0.1225\), so the local standard error is \(\sqrt{0.1225}=0.3500\).
Wald calculation. Using \(z_{0.975}=1.96\), the nominal 95% Wald interval is
This computation uses only the curvature at \(\gamma=0\). As a comparison, the usual one-parameter profile-likelihood cutoff uses a log-likelihood drop of \(3.841/2\), where \(3.841\) is the 95th percentile of a chi-squared distribution with one degree of freedom. The resulting profile endpoints satisfy
At \(d=0.591\), \(d^2=0.349281\) and \(d^4=(0.349281)^2\approx0.1219972\). Thus the left side is \(0.349281/0.1225+8(0.1219972)\approx2.8513+0.9760=3.8273\). At \(d=0.592\), the left side is about \(3.8435\), so the positive root is approximately \(0.5919\). By symmetry, the profile interval is approximately \([-0.5919,0.5919]\), narrower than the Wald interval here. This constructed likelihood demonstrates that the extra quartic curvature changes the interval endpoints even though it does not change the Hessian at the maximum. It does not establish that either method has correct coverage in a particular finite sample.
Reading the Hessian Without Overreading It
A Hessian diagnostic helps identify whether a Wald calculation deserves scrutiny. If the information matrix has a very small eigenvalue, the likelihood is locally flat along the corresponding eigenvector, which is a particular combination of parameters. A large condition number—the ratio of the largest to the smallest eigenvalue—signals that inversion may be sensitive to small numerical errors. The eigenvector matters: a weak direction can involve both an outcome parameter and a response parameter, even if neither coordinate alone looks obviously unstable.
Worked Example: A Nearly Flat Direction and a Large Wald Error
Suppose a hypothetical fitted selection model has two parameters, \(\gamma\) and \(\eta\), with estimate \((0.4,0.2)\) and observed information matrix
The eigenvalues are \(2.00\) and \(0.0200\), with condition number \(2/0.02=100\). The small eigenvalue corresponds to the direction \((1,1)\), so simultaneous movement in \(\gamma\) and \(\eta\) has little local likelihood penalty. The determinant is \(1.01^2-0.99^2=0.0400\), giving
The estimated standard error for \(\gamma\) is \(\sqrt{25.25}\approx5.025\). Its nominal 95% Wald interval is \(0.4\pm1.96(5.025)\), or approximately \([-9.449,10.249]\). The calculation is arithmetically consistent, but it is a warning rather than reassurance: the interval is extremely broad relative to the estimate, and the weak direction couples the two parameters. Inspecting the profile for \(\gamma\), the parameter’s allowed range, and the stability of the Hessian is necessary before treating this local interval as informative.
Diagnostics, Alternatives, and Limits
Warning signs include a singular or nearly singular information matrix, a large condition number, substantial changes in standard errors under small optimizer or numerical-integration changes, and a Wald interval that extends into regions where the likelihood is visibly nonquadratic. In selection models, also check whether different combinations of outcome and response parameters produce similar observed-data fits. The earlier material on Hessian diagnostics provides the relevant local curvature checks; those checks diagnose instability but do not by themselves repair weak identification.
Profile-likelihood intervals are a useful alternative when nuisance parameters can be optimized reliably at each fixed parameter value. They follow the likelihood along a curved ridge rather than using only its tangent quadratic approximation. They still depend on the specified model and on reliable optimization, and their usual calibration can be questionable in weakly identified or nonstandard settings. The introductory material on profile-likelihood confidence intervals for MNAR parameters provides the basic construction. A Wald interval can remain reasonable when the likelihood is smooth, well identified, and close to quadratic over the relevant range; its failure is not automatic merely because a model is a selection model.
A particularly common mistake is to interpret a positive-definite numerical Hessian as proof that the parameter is well identified. Positive definiteness says that the computed local approximation has positive curvature in every direction. It does not guarantee that the smallest curvature is large enough to be estimated stably, that the likelihood stays quadratic over the interval, or that the model identifies the parameter from the observed data. In an exact ridge, rounding can even turn a theoretically zero eigenvalue into a small positive one.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In the binary selection model, why must \(p\) lie strictly between \(0.36\) and \(0.76\) for the ridge parameterization to correspond to finite logistic parameters?
- What does a small eigenvalue of the observed information indicate, and why can it make a Wald standard error unstable?
- In the curved-profile example, what is the local Wald standard error, and why does the profile interval differ?
- Does a positive-definite numerical Hessian establish that a selection parameter is strongly identified? Explain.
Solutions. (1) The choices \(q_1=0.36/p\) and \(q_0=0.24/(1-p)\) must both be strictly between zero and one. This requires \(p>0.36\) and \(p<0.76\); at either endpoint one response probability equals one, which requires an infinite logistic linear predictor. (2) A small eigenvalue indicates little local curvature in its eigenvector direction. Inverting the information magnifies uncertainty and numerical error in that direction, making standard errors sensitive to small perturbations. (3) The local standard error is \(\sqrt{0.1225}=0.3500\). The profile likelihood includes the quartic term, which is zero in its second derivative at the maximum but changes the likelihood away from it; its endpoints are approximately \(\pm0.5919\), rather than the Wald endpoints \(\pm0.6860\). (4) No. Positive definiteness describes the computed local quadratic surface, not the stability of its smallest curvature, the global shape, or how well the observed data distinguish parameter combinations.