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Asymptotic Confidence Intervals · Tutorial 907 of 970

Joint Wald Ellipsoids for Vector Parameters

Learn how to construct, interpret, and calculate a joint Wald confidence ellipsoid for a vector parameter.

Intermediate 10 min read

What You'll Learn

  • State the conditions for a joint Wald ellipsoid based on a multivariate normal limit.
  • Use an estimated covariance matrix and a chi-square critical value to construct the ellipsoid.
  • Interpret confidence level as coverage of the entire parameter vector.
  • Relate the ellipsoid’s shape and orientation to the covariance matrix.
  • Check parameter vectors and linear reparameterizations using the Wald quadratic form.

From a Vector Limit to a Joint Confidence Set

An estimate of a vector parameter reports several quantities at once. For example, a model may estimate two regression coefficients, or a study may estimate two population means. A confidence set for the vector should account for their joint sampling variation: it should describe a region that covers the whole parameter vector, not separate uncertainty statements considered one coordinate at a time.

When an estimator has a multivariate normal large-sample limit, its standardized distance from the true vector has an asymptotic chi-square distribution. Estimating the covariance matrix then gives a practical confidence set: an ellipsoid centered at the estimated vector. Its size is set by a chi-square critical value, and its shape and orientation reflect the estimated covariance.

Definition: Let \(\theta\in\mathbb{R}^p\) be a fixed parameter vector and let \(\hat{\theta}\) estimate it from a sample of size \(n\). Suppose \(\sqrt{n}(\hat{\theta}-\theta)\xrightarrow{d}N_p(0,V)\), where \(V\) is a positive-definite \(p\times p\) matrix. Let \(\hat V\) be a symmetric estimator satisfying \(\hat V\xrightarrow{p}V\). The level-\(1-\alpha\) joint Wald ellipsoid is \[ \mathcal C_{1-\alpha} = \left\{\vartheta\in\mathbb{R}^p: n(\hat{\theta}-\vartheta)^\mathsf{T}\hat V^{-1}(\hat{\theta}-\vartheta) \leq \chi^2_{p,1-\alpha} \right\}, \] when \(\hat V\) is invertible. Here \(\chi^2_{p,1-\alpha}\) is the \(1-\alpha\) quantile of a chi-square distribution with \(p\) degrees of freedom. Under the stated conditions, \(\hat V\) is positive definite with probability tending to one.

The matrix \(\hat V\) estimates the covariance of \(\sqrt{n}(\hat{\theta}-\theta)\), not the covariance of \(\hat{\theta}\) itself. Thus the estimated covariance matrix on the original estimator scale is \(\hat V/n\). Equivalently, the defining inequality can be written \((\hat{\theta}-\vartheta)^\mathsf{T}(\hat V/n)^{-1}(\hat{\theta}-\vartheta)\leq\chi^2_{p,1-\alpha}\).

Why the Chi-Square Cutoff Gives Joint Coverage

The key result turns the multivariate normal approximation into a coverage statement for the entire parameter vector. It requires both asymptotic normality and consistent estimation of the limiting covariance matrix. Positive definiteness matters: it makes the quadratic form well-defined and prevents any direction from having zero limiting variance.

Theorem: Suppose \(\sqrt{n}(\hat{\theta}-\theta)\xrightarrow{d}N_p(0,V)\), where \(V\) is positive definite, and suppose \(\hat V\xrightarrow{p}V\), with \(\hat V\) symmetric and positive definite with probability tending to one. For fixed \(0<\alpha<1\), the ellipsoid above satisfies \(P(\theta\in\mathcal C_{1-\alpha})\to 1-\alpha\).

Proof. Write \(Z_n=\sqrt{n}(\hat{\theta}-\theta)\). The event that the true vector belongs to the ellipsoid is exactly \[ \left\{Z_n^\mathsf{T}\hat V^{-1}Z_n\leq\chi^2_{p,1-\alpha}\right\}. \] Because \(V\) is positive definite, matrix inversion is continuous at \(V\). Therefore \(\hat V^{-1}\xrightarrow{p}V^{-1}\). By the assumed limit and Slutsky’s theorem, the quadratic form \(Z_n^\mathsf{T}\hat V^{-1}Z_n\) converges in distribution to \(Z^\mathsf{T}V^{-1}Z\), where \(Z\sim N_p(0,V)\).

To identify this limit, write \(Z=V^{1/2}U\), where \(U\sim N_p(0,I_p)\) and \(V^{1/2}\) is a symmetric square root of \(V\). Then \[ Z^\mathsf{T}V^{-1}Z = U^\mathsf{T}V^{1/2}V^{-1}V^{1/2}U = U^\mathsf{T}U = \sum_{j=1}^p U_j^2. \] The final sum has a chi-square distribution with \(p\) degrees of freedom. Its distribution is continuous at \(\chi^2_{p,1-\alpha}\), so convergence in distribution implies \[ P\!\left(Z_n^\mathsf{T}\hat V^{-1}Z_n\leq\chi^2_{p,1-\alpha}\right) \longrightarrow P\!\left(\chi^2_p\leq\chi^2_{p,1-\alpha}\right) = 1-\alpha. \] This is the claimed simultaneous coverage of the vector. \(\square\)

The coverage statement concerns one event: the true vector lies inside the random ellipsoid. The chi-square degrees of freedom are \(p\), the dimension of the vector. This is different from taking a one-dimensional normal cutoff and applying it separately to each coordinate.

Geometry of the Ellipsoid

The defining quadratic form measures distance from the center after accounting for covariance. A displacement along a direction with large estimated variance contributes less to the quadratic form than an equally long displacement along a direction with small estimated variance. Correlation also rotates the ellipsoid’s principal axes away from the coordinate axes.

Result: If \(\hat V\) is positive definite with eigenvalues \(\lambda_1,\ldots,\lambda_p\) and corresponding orthonormal eigenvectors \(q_1,\ldots,q_p\), the ellipsoid’s principal axes point in the directions \(q_j\). Its semi-axis length in direction \(q_j\) is \(\sqrt{\chi^2_{p,1-\alpha}\lambda_j/n}\).

Derivation. The spectral decomposition gives \(\hat V=Q\Lambda Q^\mathsf{T}\), where the columns of \(Q\) are the eigenvectors and \(\Lambda\) is diagonal with entries \(\lambda_j\). Let \(d=\hat{\theta}-\vartheta\), and write \(d=Qy\). Since \(Q^\mathsf{T}Q=I_p\), \[ n d^\mathsf{T}\hat V^{-1}d = n y^\mathsf{T}\Lambda^{-1}y = n\sum_{j=1}^p\frac{y_j^2}{\lambda_j}. \] The ellipsoid condition is therefore \(\sum_{j=1}^p y_j^2/(\chi^2_{p,1-\alpha}\lambda_j/n)\leq1\). In the \(j\)th eigenvector direction, the boundary occurs at \(|y_j|=\sqrt{\chi^2_{p,1-\alpha}\lambda_j/n}\), as claimed. \(\square\)

Worked Example: Constructing and Checking a Two-Parameter Ellipsoid

Suppose \(p=2\), \(n=100\), \(\hat{\theta}=(1.2,-0.4)^\mathsf{T}\), and the estimated covariance matrix for \(\sqrt{n}(\hat{\theta}-\theta)\) is \[ \hat V=\begin{pmatrix}4&1\\1&2\end{pmatrix}. \] For 95% confidence, \(\chi^2_{2,0.95}=5.991\). The determinant of \(\hat V\) is \(4(2)-1(1)=7\), so \[ \hat V^{-1}=\frac{1}{7}\begin{pmatrix}2&-1\\-1&4\end{pmatrix}. \] The confidence set consists of all \(\vartheta\) satisfying \[ 100(\hat{\theta}-\vartheta)^\mathsf{T} \frac{1}{7}\begin{pmatrix}2&-1\\-1&4\end{pmatrix} (\hat{\theta}-\vartheta)\leq5.991. \]

To check whether \(\vartheta=(1.0,-0.3)^\mathsf{T}\) is inside, use \(\hat{\theta}-\vartheta=(0.2,-0.1)^\mathsf{T}\). The quadratic form before multiplication by \(n\) is \[ \frac{1}{7}\left(2(0.2)^2-2(0.2)(-0.1)+4(-0.1)^2\right) = \frac{0.08+0.04+0.04}{7} = \frac{0.16}{7}. \] Thus the left side of the ellipsoid inequality is \(100(0.16/7)\approx2.286\), which is less than \(5.991\). The candidate vector is inside the ellipsoid.

For its shape, the eigenvalues of \(\hat V\) solve \(\lambda^2-6\lambda+7=0\), giving \(\lambda=3\pm\sqrt{2}\). The semi-axis lengths are therefore \(\sqrt{5.991(3+\sqrt{2})/100}\approx0.514\) and \(\sqrt{5.991(3-\sqrt{2})/100}\approx0.308\).

Worked Example: Exact Ellipsoid for a Normal Mean

Let \(X_1,\ldots,X_{25}\) be independent observations from \(N_2(\theta,\Sigma)\), where \[ \Sigma=\begin{pmatrix}1&0.6\\0.6&4\end{pmatrix}. \] The sample mean \(\bar X\) has the exact distribution \(N_2(\theta,\Sigma/25)\). Here \(\hat V=\Sigma\) is known, so the quadratic form has an exact chi-square distribution with two degrees of freedom. The 95% ellipsoid is \[ \left\{\vartheta: 25(\bar X-\vartheta)^\mathsf{T} \Sigma^{-1}(\bar X-\vartheta)\leq5.991 \right\}. \] The determinant of \(\Sigma\) is \(4-0.36=3.64\), and \[ \Sigma^{-1}=\frac{1}{3.64}\begin{pmatrix}4&-0.6\\-0.6&1\end{pmatrix}. \]

Suppose the observed mean is \(\bar X=(10,20)^\mathsf{T}\). To check the candidate \(\vartheta=(9.8,19.6)^\mathsf{T}\), the displacement is \((0.2,0.4)^\mathsf{T}\). Its quadratic form is \[ (0.2,0.4)\Sigma^{-1}(0.2,0.4)^\mathsf{T} = \frac{4(0.2)^2-2(0.6)(0.2)(0.4)+(0.4)^2}{3.64} = \frac{0.224}{3.64} \approx0.06154. \] Multiplication by \(25\) gives approximately \(1.538\), below \(5.991\); the candidate is inside. The exact normal model illustrates the chi-square calibration directly. In general, replacing \(\Sigma\) with an estimated covariance matrix gives the asymptotic construction, not this exact finite-sample guarantee.

Worked Example: Re-expressing the Ellipsoid After a Linear Transformation

Suppose \(n=40\), \(\hat{\theta}=(2,1)^\mathsf{T}\), and \(\hat V=\begin{pmatrix}4&2\\2&2\end{pmatrix}\). Consider the transformed parameter \(\phi=B\theta\), where \(B=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\). The transformed estimate is \(B\hat{\theta}=(3,1)^\mathsf{T}\), and its covariance estimate is \[ \hat V_\phi=B\hat V B^\mathsf{T} = \begin{pmatrix}10&2\\2&2\end{pmatrix}. \]

Check the original candidate \(\vartheta=(1.8,1.1)^\mathsf{T}\). Its displacement from the estimate is \(d=(0.2,-0.1)^\mathsf{T}\). Since \[ \hat V^{-1}=\begin{pmatrix}0.5&-0.5\\-0.5&1\end{pmatrix}, \] we obtain \(d^\mathsf{T}\hat V^{-1}d=0.020+0.020+0.010=0.050\). The Wald statistic is \(40(0.050)=2.00\).

In transformed coordinates, the displacement is \(Bd=(0.1,0.3)^\mathsf{T}\). The inverse of \(\hat V_\phi\), whose determinant is \(20-4=16\), is \[ \hat V_\phi^{-1}=\frac{1}{16}\begin{pmatrix}2&-2\\-2&10\end{pmatrix}. \] The transformed quadratic form is \[ (Bd)^\mathsf{T}\hat V_\phi^{-1}(Bd) = \frac{2(0.1)^2-4(0.1)(0.3)+10(0.3)^2}{16} = \frac{0.80}{16} = 0.050. \] Thus the Wald statistic remains \(2.00\). An invertible linear re-expression changes the coordinates of the ellipsoid but not the membership decision.

Common Mistake and Practical Interpretation

A common mistake is to use a separate one-dimensional normal cutoff for each coordinate and then call the resulting rectangular region a joint Wald ellipsoid. That changes both the shape of the region and the calibration of the joint event. The ellipsoid uses one quadratic form that includes the off-diagonal covariance terms, and its cutoff is a chi-square quantile with \(p\) degrees of freedom.

A second frequent error is to use \(\hat V\) as though it were already the covariance matrix of \(\hat{\theta}\). Under the stated convention, \(\hat V\) estimates the covariance on the \(\sqrt n\) scale; the covariance on the estimator scale is \(\hat V/n\). Omitting the factor \(n\) in the quadratic form therefore gives the wrong size of the confidence set.

The confidence level describes repeated-sampling coverage of the full vector: over repeated samples, the constructed ellipsoid contains the true vector with probability approaching \(1-\alpha\). It is not a probability assigned to the fixed parameter after the data have been observed. The ellipsoid also makes the dependence among estimated coordinates visible through its orientation and shape, while retaining a single joint coverage statement.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

  1. What covariance matrix does \(\hat V\) estimate, and how does it relate to the covariance of \(\hat{\theta}\)?
  2. Why does the quadratic form for the true parameter converge to a chi-square distribution with \(p\) degrees of freedom?
  3. What determines the direction and length of each principal semi-axis of a Wald ellipsoid?
  4. In the first worked example, why must the off-diagonal entries of \(\hat V^{-1}\) be included when checking a candidate vector?
  5. What is the difference between asymptotic coverage of a joint ellipsoid and exact finite-sample coverage?
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