Projecting Mean Ellipsoids Onto Individual Contrasts

Why Project a Mean Ellipsoid?

A multivariate mean confidence region describes plausible values of an entire mean vector. Often, however, the scientific question concerns just one linear combination of its coordinates: a difference between two means, a weighted sum, or another prespecified contrast. Projecting the region onto that one quantity gives an interval that follows directly from the joint confidence statement.

The key calculation is geometric. The interval’s center is the contrast evaluated at the sample mean. Its half-width depends on the contrast’s direction and the estimated covariance matrix, as well as on the size of the full mean ellipsoid. Thus, this interval is not generally the same as an interval calibrated only for one scalar parameter.

The Gaussian Mean Ellipsoid

Suppose $X_1,\ldots,X_n$ are independent observations from a $p$-variate normal distribution $N_p(\mu,\Sigma)$, where $\mu$ is unknown and $\Sigma$ is positive definite. Let $\bar X$ be the sample mean and let $S$ be the usual sample covariance matrix, calculated with divisor $n-1$. Assume $n>p$, so $S$ is nonsingular with probability one under this model.

For $0<\alpha<1$, write $F_{p,n-p}(1-\alpha)$ for the $1-\alpha$ quantile of the central $F$ distribution with $p$ and $n-p$ degrees of freedom. The exact one-sample Hotelling mean ellipsoid has the following form.

Definition: The $1-\alpha$ Gaussian mean confidence ellipsoid is $$ \mathcal E_{1-\alpha} = \left\{\theta\in\mathbb R^p: n(\bar X-\theta)'S^{-1}(\bar X-\theta) \le \frac{p(n-1)}{n-p}F_{p,n-p}(1-\alpha) \right\}. $$ Under the stated independent multivariate-normal model, this region contains $\mu$ with probability $1-\alpha$. This calibration follows from the one-sample Hotelling $T^2$ distribution.

Fix a nonzero vector $c\in\mathbb R^p$. The scalar quantity of interest is $c'\mu$. A contrast is often such a linear combination, with additional constraints on its coefficients determined by the scientific question; the projection calculation itself only requires that $c$ be fixed and nonzero.

Deriving the Projected Interval

The projection of $\mathcal E_{1-\alpha}$ onto the quantity $c'\theta$ is the set of all values $c'\theta$ as $\theta$ ranges over the ellipsoid. To find its endpoints, maximize and minimize $c'\theta$ subject to the ellipsoid constraint.

Theorem: The projection of $\mathcal E_{1-\alpha}$ onto $c'\theta$ is the interval $$ \left[ c'\bar X-h_c,\;c'\bar X+h_c \right], \qquad h_c= \sqrt{ \frac{p(n-1)}{n(n-p)} F_{p,n-p}(1-\alpha)\,c'Sc }. $$

To derive the endpoints, set $\delta=\theta-\bar X$ and $K=\frac{p(n-1)}{n-p}F_{p,n-p}(1-\alpha)$. The ellipsoid constraint becomes $n\delta'S^{-1}\delta\le K$. Since $S$ is positive definite, it has a positive definite square root. Put $z=S^{-1/2}\delta$; then $\delta=S^{1/2}z$ and the constraint is $z'z\le K/n$. Also, $c'\delta=(S^{1/2}c)'z$. By the Cauchy–Schwarz inequality,

|c'\delta| = |(S^{1/2}c)'z| \le \sqrt{c'Sc}\sqrt{z'z} \le \sqrt{\frac{K}{n}c'Sc}.

This bound is attained: for the maximum, choose $z=\sqrt{K/n}\,S^{1/2}c/\sqrt{c'Sc}$; for the minimum, use its negative. Therefore the possible values of $c'\theta=c'\bar X+c'\delta$ run from $c'\bar X-\sqrt{Kc'Sc/n}$ to $c'\bar X+\sqrt{Kc'Sc/n}$. Substituting the definition of $K$ gives the stated half-width and endpoints. $\square$

The same argument shows why the center and half-width have different roles: $c'\bar X$ locates the interval, while $c'Sc$ measures sample variability in the direction $c$. In particular, the off-diagonal entries of $S$ can matter just as much as its variances.

Worked Calculations

Worked Example: A Difference Between Two Coordinates

Suppose $p=2$, $n=10$, $\bar X=(4,7)'$, and $$ S=\begin{pmatrix}4&1\\1&9\end{pmatrix}. $$ For the contrast $c=(1,-1)'$, find the 95% interval obtained by projecting the mean ellipsoid.

The center is $c'\bar X=4-7=-3$. The covariance contribution is $$ c'Sc = (1,-1)\begin{pmatrix}4&1\\1&9\end{pmatrix} \begin{pmatrix}1\\-1\end{pmatrix} = (1,-1)\begin{pmatrix}3\\-8\end{pmatrix} =11. $$ For $p=2,n=10$, $F_{2,8}(0.95)\approx4.459$. Thus $$ K=\frac{2(9)}{8}(4.459)\approx10.0328, \qquad h_c=\sqrt{\frac{10.0328(11)}{10}}\approx3.322. $$ The projected interval is $-3\pm3.322$, or approximately $[-6.322,\;0.322]$.

The covariance term includes the cross-product contribution: $c'Sc=4+9-2(1)=11$. Omitting the covariance would incorrectly give 13.

Worked Example: A Weighted Contrast in Three Dimensions

Let $p=3$, $n=12$, $\bar X=(2,5,1)'$, and $$ S=\begin{pmatrix}4&1&0\\1&3&1\\0&1&2\end{pmatrix}. $$ For $c=(1,1,-2)'$, calculate the 95% projected interval.

The center is $c'\bar X=2+5-2=5$. Multiplying the covariance matrix by $c$ gives $Sc=(5,2,-3)'$, so $$ c'Sc=(1,1,-2)\begin{pmatrix}5\\2\\-3\end{pmatrix}=5+2+6=13. $$ The matrix is positive definite: its leading principal minors are $4$, $11$, and $18$, all positive. The calculation therefore uses a valid nonsingular covariance matrix.

Using $F_{3,9}(0.95)\approx3.863$, $$ K=\frac{3(11)}{9}(3.863)\approx14.164, \qquad h_c=\sqrt{\frac{14.164(13)}{12}}\approx3.917. $$ The interval is $5\pm3.917$, approximately $[1.083,\;8.917]$.

Worked Example: Rescaling a Contrast

Use the two-dimensional data in the first example, but take $d=(2,1)'$. The center is $d'\bar X=2(4)+7=15$. Since $$ Sd=\begin{pmatrix}4&1\\1&9\end{pmatrix} \begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}9\\11\end{pmatrix}, \qquad d'Sd=(2,1)\begin{pmatrix}9\\11\end{pmatrix}=29, $$ the half-width is $$ h_d=\sqrt{\frac{10.0328(29)}{10}}\approx5.394. $$ The projected interval is approximately $[9.606,\;20.394]$.

This example also illustrates scale: if a contrast vector is multiplied by a constant $a$, its center is multiplied by $a$, and its half-width by $|a|$. Indeed, $(ac)'S(ac)=a^2c'Sc$. For $a<0$, the endpoints reverse order before being written from smallest to largest.

What Coverage Does Projection Provide?

The confidence ellipsoid is calibrated for the whole vector $\mu$, while the projected interval concerns one scalar function of that vector. The interval inherits a coverage guarantee because whenever the ellipsoid contains $\mu$, its projection must contain $c'\mu$.

Coverage consequence: Under the Gaussian model and conditions stated above, $$ P\!\left(c'\mu\in[c'\bar X-h_c,\;c'\bar X+h_c]\right) \ge 1-\alpha. $$ The projection is not generally calibrated to have exactly $1-\alpha$ coverage for this single scalar.

For completeness, the inclusion behind this guarantee is $\{\mu\in\mathcal E_{1-\alpha}\}\subseteq \{c'\mu\in[c'\bar X-h_c,c'\bar X+h_c]\}$, by the definition of projection. Taking probabilities gives the inequality, since the ellipsoid has coverage $1-\alpha$. The projection can include the true contrast even when the full mean vector lies outside the ellipsoid. That is why its coverage can exceed the ellipsoid’s coverage. $\square$

The distinction matters in practice. This interval uses the critical value for a $p$-dimensional mean region, even though only one contrast is being reported. Its guarantee comes from the joint region, not from a separate calibration targeted specifically at that one contrast.

Common Mistakes

Dropping the off-diagonal covariances. The correct variance factor is $c'Sc$, not a sum of squared coefficients times diagonal variances unless the omitted covariance terms are actually zero.
Using the wrong ellipsoid multiplier. With $S$ defined using divisor $n-1$, the half-width uses $\frac{p(n-1)}{n(n-p)}F_{p,n-p}(1-\alpha)c'Sc$ under the square root. Changing the covariance convention requires changing the formula consistently.
Calling the projected interval an exact single-contrast interval. Projection guarantees coverage at least as large as the ellipsoid’s level; it does not assert exact marginal calibration.
Treating one projection as protection for every possible contrast. The result applies to the specified $c$. A guarantee over a collection of contrasts is a different inferential question.
Ignoring the regularity conditions. The displayed finite-sample ellipsoid relies on independent multivariate-normal observations and $n>p$, with a nonsingular covariance estimate.

Key takeaway: To project a Gaussian mean ellipsoid onto one fixed contrast, compute the center $c'\bar X$ and the half-width $\sqrt{\frac{p(n-1)}{n(n-p)}F_{p,n-p}(1-\alpha)c'Sc}$. The covariance matrix determines how uncertainty is oriented in the contrast’s direction.

Check Your Understanding

Use the ellipsoid projection argument to answer these questions.

For a fixed nonzero $c$, which quantity determines the center of the projected interval, and which determines its half-width?
Why can off-diagonal entries of $S$ affect the width even when the contrast coefficients are unchanged?
What Gaussian-model and sample-size conditions are used for the stated finite-sample ellipsoid?
Explain why projection guarantees at least $1-\alpha$ coverage for $c'\mu$, rather than necessarily exactly $1-\alpha$.
If a contrast vector is multiplied by $-3$, how do the projected interval’s center and half-width change?