Why Adjust the Empirical Sandwich?
The empirical sandwich covariance estimates variability by adding the outer products of cluster-level estimating contributions. With few independent clusters, this empirical “meat” can be too small because the fitted coefficients have already used information from those same clusters. A high-leverage cluster, in particular, tends to have a reduced fitted residual. The Mancl–DeRouen (MD) and Kauermann–Carroll (KC) corrections address this effect by adjusting residuals according to cluster leverage.
The corrections differ in how strongly they inflate residual components in high-leverage directions. Neither is a universal remedy: both are finite-sample adjustments to a sandwich estimate, not guarantees of unbiased covariance estimation or accurate coverage.
Setup and Definitions
Let \(Y_i\) be the response vector for cluster \(i\), for \(i=1,\ldots,K\), and let \(\hat\beta\) be the fitted GEE coefficient vector of dimension \(p\). At the fitted values, write \(D_i\) for the derivative of the marginal mean vector with respect to \(\beta\), and \(V_i\) for the working covariance matrix. Assume each \(V_i\) is symmetric positive definite and the estimating-equation bread is nonsingular. Define
Here \(r_i\) is the fitted residual vector and \(s_i\) is the cluster estimating contribution. The ordinary empirical meat and sandwich covariance are
To define leverage adjustments without ambiguity about matrix square roots, use whitened cluster coordinates. Let \(V_i^{-1/2}\) be the symmetric positive-definite inverse square root of \(V_i\), and set
Then \(A=\sum_i Z_i^T Z_i\), and \(s_i=Z_i^T e_i\). The full projection matrix for the stacked whitened design has cluster-diagonal block \(P_{ii}=Z_iA^{-1}Z_i^T\). This symmetric matrix is the leverage matrix used below.
\(\hat B_{\mathrm{MD}}=\sum_i Z_i^T(I-P_{ii})^{-1}e_i e_i^T(I-P_{ii})^{-1}Z_i\),
and
\(\hat B_{\mathrm{KC}}=\sum_i Z_i^T(I-P_{ii})^{-1/2}e_i e_i^T(I-P_{ii})^{-1/2}Z_i\).
The corrected sandwich estimates are \(A^{-1}\hat B_{\mathrm{MD}}A^{-1}\) and \(A^{-1}\hat B_{\mathrm{KC}}A^{-1}\).
These inverses require every eigenvalue of \(P_{ii}\) to be strictly less than one. The formulas are plug-in corrections evaluated at the fitted model quantities; they do not alter the fitted GEE coefficients.
What the Leverage Eigenvalues Show
The leverage matrices are contractions. To see this, stack all \(Z_i\) into a matrix \(Z\). Since \(A=Z^TZ\) is nonsingular, \(P=ZA^{-1}Z^T\) is symmetric and idempotent, hence an orthogonal projection. The matrix \(P_{ii}\) is its principal block for cluster \(i\). For any vector \(x\) of the appropriate dimension, embed \(x\) in the stacked space as \(u\), with zeros in the other cluster positions. Then
Thus the eigenvalues \(h\) of \(P_{ii}\) lie in \([0,1]\); the inverse adjustments additionally require \(h<1\). Because the adjustments are functions of the same symmetric matrix, they act separately along its eigenvectors.
Indeed, putting \(t=1-h\), we have \(0<t\leq1\), so \(1/t\geq1/\sqrt{t}\). By the spectral theorem, this also means \((I-P_{ii})^{-1}-(I-P_{ii})^{-1/2}\) is positive semidefinite. Consequently, for a fixed residual vector, the MD adjusted residual has at least as large a Euclidean norm as the KC adjusted residual. This comparison concerns the residual multipliers themselves. It does not establish an ordering of the corrected meat matrices.
Worked Example: Comparing the Corrections in a Scalar GEE
Suppose a one-coefficient fit has \(A=10\) and three scalar cluster leverages \(h_1=0.2\), \(h_2=0.3\), and \(h_3=0.5\). Suppose the unadjusted scalar cluster scores are \(s_1=2\), \(s_2=-1\), and \(s_3=1\). For a scalar cluster, the adjusted score is the original score multiplied by the corresponding residual multiplier.
The ordinary meat is \(2^2+(-1)^2+1^2=6\). The MD multipliers are \(1/0.8=1.25\), \(1/0.7=10/7\), and \(1/0.5=2\). Its corrected meat is therefore
The KC multipliers are \(1/\sqrt{0.8}\), \(1/\sqrt{0.7}\), and \(1/\sqrt{0.5}\). Squaring each multiplier in the scalar meat gives
Since \(A^{-1}=0.1\), the sandwich variances are \(\hat C_{\mathrm{MD}}=0.01(601/49)\approx0.12265\) and \(\hat C_{\mathrm{KC}}=0.01(59/7)\approx0.08429\). In this one-dimensional example, MD gives the larger meat and variance. That scalar ordering follows because each corrected squared score is a nonnegative number that increases with its multiplier; it does not establish a matrix ordering in a multivariate fit.
Why the Meat Matrices Need Not Be Ordered
A corrected meat contribution is an outer product of an adjusted score vector. Even if one adjustment matrix is larger in the positive-semidefinite ordering, that fact alone does not order the outer products of the resulting vectors. In particular, if two adjusted score vectors point in different directions, their outer products can differ by an indefinite matrix.
Worked Example: A Matrix Difference That Is Indefinite
Consider a cluster with \(P_{ii}=\operatorname{diag}(0.2,0.8)\) and whitened residual \(e_i=(1,2)^T\). The MD adjustment matrix is \(\operatorname{diag}(1.25,5)\), giving adjusted residual \((1.25,10)^T\). The KC adjustment matrix is \(\operatorname{diag}(1/\sqrt{0.8},1/\sqrt{0.2})\), giving adjusted residual \((1/\sqrt{0.8},2/\sqrt{0.2})^T\).
For this illustration, take \(Z_i=I\), so each corrected cluster-meat contribution is the outer product of its adjusted residual. The MD contribution is
The KC contribution is
The determinant of the difference is \(0.3125(80)-7.5^2=25-56.25=-31.25\). A positive-semidefinite matrix cannot have a negative determinant, so this difference is indefinite. Thus the larger MD multiplier in each leverage eigen-direction does not imply that the MD contribution, or the sum of corrected contributions, is greater in the positive-semidefinite ordering. The same warning applies to the sandwich covariance matrices.
Near-Unit Leverage and Practical Use
Both corrections become increasingly aggressive as leverage approaches one, but at different rates. The MD multiplier grows as \((1-h)^{-1}\), while the KC multiplier grows as \((1-h)^{-1/2}\). Since the meat uses an outer product, the corresponding multiplier on a residual component’s squared contribution is \((1-h)^{-2}\) for MD and \((1-h)^{-1}\) for KC.
Worked Example: A Cluster With Leverage 0.9
Suppose one scalar cluster has leverage \(h=0.9\), bread \(A=5\), and unadjusted cluster score \(s=0.5\). Its ordinary meat contribution is \(s^2=0.25\). The MD score multiplier is \(1/(1-0.9)=10\), so the corrected score is \(5\) and its meat contribution is \(25\). The KC score multiplier is \(1/\sqrt{0.1}=\sqrt{10}\), so its corrected score is \(0.5\sqrt{10}\) and its meat contribution is \(2.5\).
Because the scalar sandwich contribution is the meat contribution divided by \(A^2=25\), these contributions are \(1\) under MD and \(0.1\) under KC, compared with \(0.25/25=0.01\) without correction. A single high-leverage cluster can therefore dominate a corrected estimate. This calculation does not show that either estimate is closer to the true sampling variance; it shows why leverage and sensitivity to individual clusters should be examined.
In practice, MD is the more aggressive of the two adjustments in each leverage eigen-direction, while KC applies the square-root adjustment and is less aggressive near unit leverage. That comparison is useful when explaining why estimates differ, but it is not a universal ranking of their performance. Small-cluster behavior depends on the design, cluster-size pattern, working covariance, and mean-model adequacy. A correction cannot make a misspecified mean model correct, and a nearly singular \(I-P_{ii}\) can make either estimate unstable.
For a small-cluster analysis, report which correction was used and examine whether the resulting standard errors are sensitive to the choice. Also inspect leverage and the influence of individual clusters; a dramatic increase can be a warning about limited information rather than evidence that one formula has solved the small-sample problem. These covariance adjustments alone do not specify the reference distribution or degrees of freedom for a test.
Common Mistakes
- Claiming a matrix ordering from multiplier ordering. The MD residual multiplier is larger in each leverage eigen-direction, but this does not imply that its adjusted score outer product exceeds the KC outer product in every direction. The difference can be indefinite.
- Comparing only the residual multipliers. The meat uses outer products, so a score multiplier is squared in a one-dimensional contribution. In several dimensions, the direction of the adjusted score also matters.
- Treating either correction as a guarantee. These are finite-sample covariance adjustments, not guarantees of unbiasedness, correct coverage, or stability.
- Ignoring leverage near one. The inverse adjustments become large as an eigenvalue approaches one. Check whether the fit is sufficiently informative for the correction to be numerically and substantively credible.
Check Your Understanding
Use the eigenvalue formulas and the distinction between residual adjustment and matrix ordering.
- For a leverage eigenvalue \(h=0.36\), calculate the MD and KC residual multipliers and compare them.
- Why must every eigenvalue of \(P_{ii}\) be less than one for both corrections to be defined?
- Explain why a larger MD multiplier in each eigen-direction does not prove that \(\hat B_{\mathrm{MD}}-\hat B_{\mathrm{KC}}\) is positive semidefinite.
- If a scalar cluster has leverage \(0.75\), by what factors do MD and KC multiply its squared residual contribution?
- Name one reason a large corrected standard error in a small-cluster analysis should be interpreted cautiously.