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Inference on Mean Vectors · Tutorial 193 of 975

Geometry of Mean Confidence Ellipsoids

Use the covariance eigendecomposition to interpret an ellipsoid’s orientation, axis lengths, radial distances, projections, and volume.

Intermediate 10 min read

What You'll Learn

  • Identify the principal directions and semi-axis lengths of a known-covariance mean confidence ellipsoid
  • Explain how covariance eigenvalues determine the ellipsoid’s shape and scale
  • Distinguish radial distance in a direction from projected half-width in that direction
  • Calculate the volume of a confidence ellipsoid from the covariance determinant
  • Interpret repeated eigenvalues and the resulting non-unique orientation

From a Confidence Region to Its Geometry

A confidence ellipsoid is more informative than a list of simultaneous intervals: its shape shows which combinations of mean components are estimated more or less precisely. The covariance matrix determines that shape. Its eigenvectors identify the principal directions, and its eigenvalues determine the lengths of the corresponding axes.

Let \(X_1,\ldots,X_n\) be independent observations from a \(p\)-variate normal distribution with mean vector \(\mu\) and known positive-definite covariance matrix \(\Sigma\). Write \(\bar X\) for the sample mean, and let \(c\) be the \(1-\alpha\) quantile of a chi-square distribution with \(p\) degrees of freedom. The known-covariance confidence ellipsoid is

$$ \mathcal C = \left\{\mu\in\mathbb R^p: n(\bar X-\mu)^\mathsf T\Sigma^{-1}(\bar X-\mu)\le c \right\}. $$

Under the stated model, \(n(\bar X-\mu)^\mathsf T\Sigma^{-1}(\bar X-\mu)\) has a chi-square distribution with \(p\) degrees of freedom. Thus the region has coverage probability \(1-\alpha\). Here the geometry is considered for the observed value of \(\bar X\); the ellipsoid is centered at that value.

Definition: A confidence ellipsoid’s principal directions are the eigenvector directions of \(\Sigma\). Its semi-axis length in a principal direction is the distance from its center to the boundary along that direction. The full axis length is twice the semi-axis length.

Eigenvectors Set the Orientation; Eigenvalues Set the Axes

Because \(\Sigma\) is symmetric and positive definite, it has an orthonormal eigendecomposition

$$ \Sigma=Q\Lambda Q^\mathsf T, \qquad \Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_p), \qquad \lambda_i>0, $$

where the columns \(q_1,\ldots,q_p\) of \(Q\) are unit eigenvectors. Consider a displacement \(d=\mu-\bar X\), expressed in this eigenvector basis as \(d=Qa\). Substituting into the ellipsoid inequality gives

$$ n d^\mathsf T\Sigma^{-1}d = n a^\mathsf T\Lambda^{-1}a = n\sum_{i=1}^p\frac{a_i^2}{\lambda_i} \le c. $$

This is the equation of an ellipsoid in coordinates \(a_i\). If all coordinates except \(a_i\) are zero, its boundary satisfies \(n a_i^2/\lambda_i=c\). The semi-axis in direction \(q_i\) therefore has length \(\sqrt{c\lambda_i/n}\). A larger eigenvalue means a longer confidence axis in that principal direction. The eigenvalue is the variance of an observation projected onto that unit eigenvector; the factor \(c/n\) converts that variance scale into the confidence-region scale for the mean.

Theorem: If \(\lambda_i\) and \(q_i\) are the eigenvalues and corresponding unit eigenvectors of \(\Sigma\), then the known-covariance mean confidence ellipsoid has center \(\bar X\), principal directions \(q_i\), and semi-axis lengths \(\sqrt{c\lambda_i/n}\). Its full axis lengths are \(2\sqrt{c\lambda_i/n}\).

The calculation above proves the result: in eigenvector coordinates, the inequality is \(\sum_i a_i^2/(\lambda_i c/n)\le 1\), which has semi-axis lengths \(\sqrt{c\lambda_i/n}\) along the coordinate axes. Transforming back by \(Q\) maps those coordinate axes to the eigenvector directions \(q_i\). \(\square\)

If an eigenvalue is repeated, there is no unique choice of individual eigenvectors within its eigenspace. The ellipsoid is nevertheless unambiguous: in the corresponding subspace, all principal semi-axes have the same length. When all eigenvalues are equal, the region is a sphere, and its orientation has no geometric effect.

Worked Example: An Ellipse With Coordinate-Aligned Axes

Suppose \(p=2\), \(n=25\), and \(\Sigma=\begin{pmatrix}9&0\\0&1\end{pmatrix}\). Use a 95% confidence level, for which \(c=\chi^2_{2,0.95}\approx5.991\). Find the principal directions and semi-axis lengths.

The coordinate vectors \(q_1=(1,0)^\mathsf T\) and \(q_2=(0,1)^\mathsf T\) are eigenvectors, with eigenvalues \(\lambda_1=9\) and \(\lambda_2=1\). The semi-axis lengths are

$$ \sqrt{\frac{5.991(9)}{25}} =\sqrt{2.15676} \approx1.469, \qquad \sqrt{\frac{5.991(1)}{25}} =\sqrt{0.23964} \approx0.490. $$

The longer axis is aligned with the first coordinate and has full length about \(2.938\); the shorter is aligned with the second coordinate and has full length about \(0.980\). The shape is three times as long along the first direction because the ratio of the semi-axis lengths is \(\sqrt{9/1}=3\), not \(9\). The eigenvalue ratio describes squared lengths.

Radial Distance Is Not the Same as Projected Width

In a principal direction, the distance from the center to the boundary is the corresponding semi-axis length. For a direction that is not a principal direction, two different geometric quantities can be useful: the distance to the boundary along a ray, and the maximum projection of the ellipsoid onto that direction. They generally are not equal.

Theorem: For any unit vector \(u\), the distance from \(\bar X\) to the ellipsoid boundary along the ray \(\bar X+tu\), \(t\ge0\), is \(\sqrt{c/(n\,u^\mathsf T\Sigma^{-1}u)}\). The maximum absolute projection of a displacement in the ellipsoid onto \(u\) is \(\sqrt{(c/n)u^\mathsf T\Sigma u}\).

To obtain the radial distance, substitute \(d=tu\) into the defining inequality. Since \(u\) has unit length and \(\Sigma^{-1}\) is positive definite, the boundary occurs at the positive solution of

$$ nt^2u^\mathsf T\Sigma^{-1}u=c, \qquad t=\sqrt{\frac{c}{n\,u^\mathsf T\Sigma^{-1}u}}. $$

For the projection, write any displacement in the ellipsoid as \(d=\sqrt{c/n}\,\Sigma^{1/2}z\), where \(\|z\|\le1\). Then

$$ |u^\mathsf T d| = \sqrt{\frac cn}\, |(\Sigma^{1/2}u)^\mathsf Tz| \le \sqrt{\frac cn}\,\|\Sigma^{1/2}u\|\,\|z\| \le \sqrt{\frac cn\,u^\mathsf T\Sigma u}. $$

The first inequality is Cauchy–Schwarz. Equality is attained by choosing \(z\) to be a unit vector in the direction of \(\Sigma^{1/2}u\), so the bound is the actual maximum. This proves both formulas. \(\square\)

The projection formula gives the half-width of the ellipsoid in direction \(u\); its full projected width is twice that value. The radial formula instead restricts attention to points lying on the ray in direction \(u\). Only in a principal direction do the two half-lengths necessarily coincide.

Worked Example: A Correlated Ellipse and a Coordinate Direction

Let \(p=2\), \(n=10\), \(c=5.991\), and \(\Sigma=\begin{pmatrix}4&3\\3&4\end{pmatrix}\). First find the principal axes, then compare the radial distance and projected half-width in direction \(u=(1,0)^\mathsf T\).

The vectors \(q_1=(1,1)^\mathsf T/\sqrt2\) and \(q_2=(1,-1)^\mathsf T/\sqrt2\) are orthonormal eigenvectors. Direct multiplication gives \(\Sigma q_1=7q_1\) and \(\Sigma q_2=q_2\), so the eigenvalues are \(7\) and \(1\). The ellipse is oriented at 45 degrees to the coordinate axes, and its semi-axis lengths are

$$ \sqrt{\frac{5.991(7)}{10}} =\sqrt{4.1937} \approx2.048, \qquad \sqrt{\frac{5.991(1)}{10}} =\sqrt{0.5991} \approx0.774. $$

The inverse covariance matrix is \(\Sigma^{-1}=\frac17\begin{pmatrix}4&-3\\-3&4\end{pmatrix}\), so \(u^\mathsf T\Sigma^{-1}u=4/7\). The radial distance along the first coordinate ray is therefore

$$ \sqrt{\frac{5.991}{10(4/7)}} = \sqrt{\frac{7(5.991)}{40}} \approx1.024. $$

By contrast, \(u^\mathsf T\Sigma u=4\), so the projected half-width in that direction is \(\sqrt{(5.991/10)4}=\sqrt{2.3964}\approx1.548\). These values differ because the coordinate direction is not a principal direction. The radial boundary point is constrained to lie on that ray; the point giving the largest coordinate projection need not lie on the ray.

Volume and Overall Scale

The volume combines all axis lengths. The unit ball in \(\mathbb R^p\) has volume \(\pi^{p/2}/\Gamma(p/2+1)\), where \(\Gamma\) is the gamma function. The ellipsoid is the image of that ball under the linear map \(z\mapsto\sqrt{c/n}\,\Sigma^{1/2}z\), followed by a translation to \(\bar X\). A linear map scales volume by the absolute value of its determinant. Since \(\Sigma\) is positive definite, \(\det(\Sigma^{1/2})=\sqrt{\det\Sigma}\). Thus

$$ \operatorname{Vol}(\mathcal C) = \frac{\pi^{p/2}}{\Gamma(p/2+1)} \left(\frac cn\right)^{p/2} \sqrt{\det\Sigma}. $$

Equivalently, multiplying the semi-axis lengths gives \((c/n)^{p/2}\sqrt{\prod_i\lambda_i}\), and \(\prod_i\lambda_i=\det\Sigma\). This confirms that the determinant summarizes the ellipsoid’s overall volume scale, while the eigenvectors and individual eigenvalues describe its orientation and axis-specific shape.

Worked Example: Volume of a Three-Dimensional Ellipsoid

Suppose \(p=3\), \(n=20\), \(c=7.815\), and the covariance eigenvalues are \(4,1,\) and \(0.25\). Find the semi-axis lengths and the volume.

The semi-axis lengths are

$$ \sqrt{\frac{7.815(4)}{20}}=\sqrt{1.563}\approx1.250, \qquad \sqrt{\frac{7.815}{20}}=\sqrt{0.39075}\approx0.625, \qquad \sqrt{\frac{7.815(0.25)}{20}}=\sqrt{0.0976875}\approx0.313. $$

The determinant is the product of the eigenvalues: \(\det\Sigma=4(1)(0.25)=1\). The volume of a three-dimensional ellipsoid is \(4\pi/3\) times the product of its semi-axis lengths. Hence

$$ \operatorname{Vol}(\mathcal C) = \frac{4\pi}{3} \sqrt{\frac{c}{n}\frac{c}{n}\frac{c(0.25)}{n}} = \frac{4\pi}{3} \left(\frac{7.815}{20}\right)^{3/2} \sqrt{1} \approx1.02. $$

Using the rounded axis lengths gives \((4\pi/3)(1.250)(0.625)(0.313)\approx1.02\) as well. The agreement checks the volume calculation.

Common Pitfall and Interpretation

A common mistake is to use \(\sqrt{(c/n)u^\mathsf T\Sigma u}\) as the distance to the boundary along every unit direction \(u\). That expression is the projected half-width, not generally the radial distance. Substitution of \(tu\) into the ellipsoid equation gives the radial distance \(\sqrt{c/(n\,u^\mathsf T\Sigma^{-1}u)}\). In the correlated example, the two quantities in the first coordinate direction were approximately \(1.024\) and \(1.548\), respectively; treating them as interchangeable would misdescribe the ellipse.

For interpretation, begin with the principal axes rather than the diagonal entries of \(\Sigma\). With nonzero correlations, the axes can be rotated relative to the measured variables, as the second example shows. Eigenvalues describe squared spread along those rotated directions, while the ellipsoid’s semi-axis lengths scale as their square roots. Volume provides a single summary of overall size, but by itself it does not reveal orientation or how that size is distributed across directions.

Key takeaway: The eigenvectors of the known covariance matrix give the ellipsoid’s principal directions, and its eigenvalues determine the squared semi-axis lengths through \(c\lambda_i/n\). For a non-principal direction, distinguish the radial boundary distance from the projection half-width.

Check Your Understanding

Use the definitions and formulas in this tutorial to answer the following.

  1. If a covariance eigenvalue is multiplied by four while \(c\) and \(n\) stay fixed, what happens to the corresponding semi-axis length?
  2. What does a repeated covariance eigenvalue imply about the ellipsoid’s axes within its eigenspace?
  3. For a unit direction \(u\), which formula gives the distance to the boundary along the ray from the center: the one involving \(u^\mathsf T\Sigma u\), or the one involving \(u^\mathsf T\Sigma^{-1}u\)?
  4. Why can the projected half-width in a direction differ from the radial distance in that direction?
  5. Which feature of \(\Sigma\) appears in the volume formula, and what geometric information does it summarize?
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