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