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Large-Sample Likelihood Theory · Tutorial 256 of 981

Information Equality and Sandwich Covariance

Learn why misspecification separates score variability from likelihood curvature, and how their combination determines asymptotic covariance.

Intermediate 12 min read

What You'll Learn

  • Define the curvature and score-variance matrices at a pseudo-true parameter
  • Derive the sandwich asymptotic covariance from a likelihood score expansion
  • State conditions that justify centering the score by differentiating under expectation
  • Prove information equality for a correctly specified regular likelihood
  • Compare sandwich and model-based covariance in explicit examples
  • Interpret an empirical sandwich covariance estimate

Why Likelihood Curvature May Not Describe Uncertainty

A likelihood fit summarizes the data using a specified model. If the data-generating distribution is not in that model, the fitted parameter typically converges to the parameter that maximizes expected log-likelihood, not to a true parameter of the fitted family. The curvature of that expected log-likelihood still describes how sharply the criterion is maximized. But the actual variation of the score across observations need not match that curvature.

The distinction matters for standard errors. In a correctly specified regular likelihood, expected curvature and score variance are equal, giving the familiar inverse-information covariance. Under misspecification they can differ. The asymptotic covariance then combines both matrices in a sandwich: curvature on the outside, score variance in the middle.

Definitions and the Information Equality

Let \(Z_1,\ldots,Z_n\) be independent and identically distributed observations from a distribution \(P_0\), which need not belong to the fitted model. For one observation, write the fitted log-likelihood contribution as \(q(Z;\theta)\), where \(\theta\in\mathbb{R}^p\). Define its score by \(s(Z;\theta)=\nabla_\theta q(Z;\theta)\), and its Hessian by \(H(Z;\theta)=\nabla_\theta^2q(Z;\theta)\). The expected log-likelihood is \(M(\theta)=E_0[q(Z;\theta)]\).

Definition: A pseudo-true parameter \(\theta^*\) is an interior maximizer of \(M(\theta)\). At \(\theta^*\), define the curvature matrix \(A=-E_0[H(Z;\theta^*)]\) and the score-variance matrix \(B=\operatorname{Var}_0(s(Z;\theta^*))\). When \(A\) is nonsingular and the score has mean zero, the sandwich covariance per observation is \(A^{-1}BA^{-T}\). In a regular likelihood \(A\) is symmetric, so this is \(A^{-1}BA^{-1}\).

The expectation of the score is not automatically the derivative of the expected log-likelihood. To equate them, differentiation must be allowed to pass under the expectation. One sufficient condition is that \(q(Z;\theta)\) is differentiable near \(\theta^*\), and that each score component is dominated there by an integrable random variable. Under such a condition, differentiation under the integral or expectation gives \(\nabla M(\theta^*)=E_0[s(Z;\theta^*)]\). Since \(\theta^*\) is an interior maximizer, \(\nabla M(\theta^*)=0\). Thus the score is centered, and \(B=E_0[s(Z;\theta^*)s(Z;\theta^*)^T]\).

Under correct specification, the expected curvature and score variance agree, subject to regularity conditions. To see the source of the equality, let \(f_\theta(z)\) be a density or mass function relative to a fixed dominating measure, with support not depending on \(\theta\). Suppose it is twice differentiable, its first and second derivatives can be passed under the integral, and the resulting score and Hessian expectations are finite. Since \(\int f_\theta(z)\,dz=1\), differentiation once gives \(E_\theta[s(Z;\theta)]=0\). Also,

$$ \nabla_\theta^2 f_\theta(z) =f_\theta(z)\left\{H(z;\theta)+s(z;\theta)s(z;\theta)^T\right\}. $$

Integrating the identity, the left side is \(\nabla_\theta^2 1=0\), by the assumed interchange. Therefore \(E_\theta[H]+E_\theta[ss^T]=0\). The score has mean zero, so its second moment is its covariance, and consequently \(A=B\). This is the information equality. Under misspecification, the expectation is taken under \(P_0\), not the fitted density \(f_\theta\); the normalization argument for \(f_\theta\) no longer implies \(A=B\).

Deriving the Sandwich Limit

The next result states the usual independent-observation form. Its assumptions separate the statistical argument from the technical conditions that justify it.

Theorem: Suppose \(Z_1,\ldots,Z_n\) are i.i.d. from \(P_0\), and \(\theta^*\) is an interior maximizer of \(M(\theta)=E_0[q(Z;\theta)]\). Assume: (1) \(M\) is differentiable at \(\theta^*\), differentiation under expectation is valid there, and \(E_0[s(Z;\theta^*)]=0\); (2) \(E_0[\|s(Z;\theta^*)\|^2]<\infty\); (3) \(A=-E_0[H(Z;\theta^*)]\) exists and is nonsingular; (4) the likelihood is twice differentiable locally, and a uniform law of large numbers for its Hessian implies that the average Hessian along the line segment from \(\theta^*\) to any consistent estimator converges in probability to \(-A\); and (5) a consistent interior maximum-likelihood estimator \(\hat\theta\) satisfies the score equation with probability tending to one. Then \( \sqrt{n}(\hat\theta-\theta^*)\xrightarrow{d}N(0,A^{-1}BA^{-T})\), where \(B=\operatorname{Var}_0(s(Z;\theta^*))\).

The centering condition in assumption (1) is important: an interior maximum of the expected log-likelihood makes its derivative zero, but only valid differentiation under expectation identifies that derivative with the expected score. The derivation uses the multivariate central limit theorem for the centered scores and a Taylor expansion of the sample score.

1
Expand the score.
Write \(S_n(\theta)=\sum_{i=1}^n s(Z_i;\theta)\). On the event that \(\hat\theta\) is an interior maximum, \(S_n(\hat\theta)=0\). Taylor expansion along the segment joining \(\theta^*\) and \(\hat\theta\) gives \(0=S_n(\theta^*)+J_n(\hat\theta-\theta^*)\), where \(J_n=\int_0^1\sum_{i=1}^n H(Z_i;\theta^*+t(\hat\theta-\theta^*))\,dt\).
2
Use the Hessian limit.
By the assumed uniform law of large numbers and consistency, \(J_n/n\xrightarrow{p}-A\). Since \(A\) is nonsingular, \(J_n/n\) is invertible with probability tending to one, and its inverse converges in probability to \(-A^{-1}\).
3
Apply the score central limit theorem.
The scores \(s(Z_i;\theta^*)\) are i.i.d., have mean zero, and have finite covariance \(B\). Therefore \(S_n(\theta^*)/\sqrt n\xrightarrow{d}N(0,B)\).
4
Combine the limits.
Rearranging the expansion gives \(\sqrt n(\hat\theta-\theta^*)=-(J_n/n)^{-1}S_n(\theta^*)/\sqrt n\). The matrix multiplier converges to \(A^{-1}\). Slutsky's theorem gives a centered normal limit with covariance \(A^{-1}BA^{-T}\). The minus sign does not change the covariance. \(\square\)

The inverse-curvature approximation \(A^{-1}/n\) is therefore not generally the covariance of \(\hat\theta\). It is correct when \(B=A\), as under the information equality, but otherwise it omits the variability of the score under the actual distribution. In practice, the matrices are often estimated by

$$ \hat A=-\frac{1}{n}\sum_{i=1}^n H(Z_i;\hat\theta), \qquad \hat B=\frac{1}{n}\sum_{i=1}^n s(Z_i;\hat\theta)s(Z_i;\hat\theta)^T, \qquad \widehat{\operatorname{Var}}(\hat\theta) =\frac{1}{n}\hat A^{-1}\hat B\hat A^{-T}. $$

With an interior maximum, the sample scores sum to zero, so the displayed outer-product estimate is also the empirical covariance of the fitted scores using divisor \(n\). Under the theorem's regularity conditions and corresponding laws of large numbers, these estimates converge to \(A\) and \(B\). The factor \(1/n\) is essential: \(A\) and \(B\) are per-observation matrices.

Worked Examples

Worked Example: A Correctly Specified Bernoulli Intercept Model

Let \(Y_i\) be i.i.d. Bernoulli with success probability \(p=0.3\), and fit the intercept-only Bernoulli model with parameter \(\theta=\operatorname{logit}(p)\). The fitted probability is \(p_\theta=e^\theta/(1+e^\theta)\). For one observation, the log-likelihood is \(q(y;\theta)=y\theta-\log(1+e^\theta)\), so \(s(y;\theta)=y-p_\theta\) and \(H(y;\theta)=-p_\theta(1-p_\theta)\).

The true parameter is \(\theta^*=\log(0.3/0.7)\), and at this value the score has mean \(E[Y-p]=0\). Its variance is \(B=\operatorname{Var}(Y)=p(1-p)=0.3(0.7)=0.21\). The negative expected Hessian is \(A=p(1-p)=0.21\). Thus information equality holds, and the sandwich variance is \(A^{-1}BA^{-1}=1/0.21\). The asymptotic variance of \(\hat\theta\) is \(1/(0.21n)\), exactly the inverse-information variance.

Worked Example: A Poisson Likelihood for Overdispersed Counts

Suppose the actual observations are i.i.d. with \(P_0(Y=0)=P_0(Y=4)=1/2\). Then \(E_0[Y]=2\) and \(\operatorname{Var}_0(Y)=E_0[Y^2]-E_0[Y]^2=8-4=4\). Fit an intercept-only Poisson model with log mean \(\theta\), even though the actual distribution is not Poisson. The one-observation log-likelihood, score, and Hessian are \(q(y;\theta)=y\theta-e^\theta-\log(y!)\), \(s(y;\theta)=y-e^\theta\), and \(H(y;\theta)=-e^\theta\).

The expected log-likelihood has derivative \(E_0[Y]-e^\theta=2-e^\theta\), so its unique maximizer is \(\theta^*=\log 2\). Here the differentiation is legitimate directly: the score is \(Y-e^\theta\), and \(Y\) is bounded by 4, so the score is dominated on every bounded neighborhood of \(\theta^*\) by an integrable constant. At \(\theta^*\), the score is \(Y-2\), which has mean zero and variance \(B=4\). The curvature is \(A=-E_0[H(Y;\theta^*)]=2\). Hence the sandwich variance per observation is \(A^{-1}BA^{-1}=(1/2)(4)(1/2)=1\), giving asymptotic variance \(1/n\) for \(\hat\theta\). The model-based inverse-information variance would instead be \(1/(2n)\), half as large. The discrepancy comes from the actual count variation, not from the fitted Poisson curvature.

Worked Example: A Gaussian Location Likelihood with the Wrong Variance

Let the actual data be i.i.d. with \(P_0(Y=-1)=P_0(Y=1)=1/2\), so \(E_0[Y]=0\) and \(\operatorname{Var}_0(Y)=1\). Fit a normal location likelihood with known variance 4: \(Y\) is modeled as \(N(\theta,4)\). Ignoring terms independent of \(\theta\), \(q(y;\theta)=-(y-\theta)^2/8\). Thus \(s(y;\theta)=(y-\theta)/4\) and \(H(y;\theta)=-1/4\).

The expected log-likelihood is maximized at \(\theta^*=E_0[Y]=0\). The score at this value has mean zero and variance \(B=\operatorname{Var}_0(Y/4)=1/16\). The curvature is \(A=1/4\). Therefore the sandwich variance per observation is \((4)(1/16)(4)=1\), and the asymptotic variance of \(\hat\theta\) is \(1/n\). The model-based variance is \(A^{-1}/n=4/n\). In this example, the fitted likelihood assumes much more observation variability than the data actually have, so its model-based uncertainty is too large.

Interpretation and Common Pitfall

The matrices have distinct roles. \(A\) measures how quickly the expected fitted log-likelihood declines away from its pseudo-true maximizer; \(B\) measures how much the individual fitted scores vary under the actual data distribution. A steep criterion does not by itself guarantee a precise estimator if the scores are highly variable, and a shallow criterion does not by itself imply large actual variance if the scores vary little.

  • Common mistake: using inverse curvature as the covariance under misspecification. The reasoning “the estimator maximizes a likelihood, so its covariance is inverse information” silently assumes information equality. Under misspecification, the curvature \(A\) and score variance \(B\) need not agree. Use \(A^{-1}BA^{-T}/n\), and use \(A^{-1}/n\) only when the equality is justified.
  • Common mistake: declaring the expected score zero without checking differentiation. An interior maximum gives \(\nabla M(\theta^*)=0\). It gives \(E_0[s(Z;\theta^*)]=0\) only when differentiation under expectation is valid. A suitable integrable dominating function for the score near \(\theta^*\) is one standard sufficient condition.
  • Common mistake: mixing total and per-observation information. The theorem uses \(A\) and \(B\) for one-observation quantities. The covariance of the estimator includes the additional factor \(1/n\).

In GLM work, the sandwich covariance provides asymptotic uncertainty for the likelihood estimator even when the specified response distribution is not exactly correct, provided the regularity conditions for the estimator and sandwich limit hold. It does not repair bias from an unsuitable mean structure, dependence between observations, or failure of the required regularity conditions; those require separate analysis.

Check Your Understanding

Use the definitions and derivations above to answer the following questions.

  1. In the sandwich formula, what does \(A\) measure, and what does \(B\) measure?
  2. Why does an interior maximizer of expected log-likelihood not, by itself, prove that the expected score is zero?
  3. For the two-point count distribution in the Poisson example, calculate \(A\), \(B\), and the model-based variance for a sample of size \(n\).
  4. State the information equality and give the key normalization argument behind it.
  5. Under what condition does the sandwich covariance reduce to the inverse-curvature covariance?
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