Coordinatewise Intervals From Ellipsoidal Regions

From a Joint Region to Componentwise Statements

A confidence ellipsoid for a mean vector gives a joint statement: it describes a region of plausible vectors, not separate limits for each component. In many applications, however, the questions are coordinate-specific: what values of the first mean, the second mean, and so on are compatible with that region? The answer is to project the ellipsoid onto each coordinate axis.

Projection preserves a useful guarantee. If the true mean vector lies in the ellipsoid, then every one of its coordinates lies within the corresponding projected interval. Thus the intervals provide simultaneous coverage, although they can cover more often than the ellipsoid itself. Their widths also have a simple form: each depends on the estimated variance of that coordinate and on the common critical value for the entire ellipsoid.

The Ellipsoid and Its Coordinate Projections

Suppose $X_1,\ldots,X_n$ are independent observations from a $p$-variate normal distribution $N_p(\mu,\Sigma)$, where $\Sigma$ is positive definite, and $n>p$. Write $\bar X$ for the sample mean vector and $S$ for the usual unbiased sample covariance matrix. Under these assumptions, $S$ is positive definite with probability one. For a confidence level $1-\alpha$, let $F_{p,n-p}(1-\alpha)$ denote the $1-\alpha$ quantile of the $F$ distribution with $p$ and $n-p$ degrees of freedom, and set

c_\alpha=\frac{p(n-1)}{n-p}F_{p,n-p}(1-\alpha).

The exact Hotelling confidence ellipsoid for $\mu$ is

\mathcal C=\left\{x\in\mathbb R^p: n(x-\bar X)^\mathsf{T}S^{-1}(x-\bar X)\le c_\alpha \right\}.

The exact Hotelling $T^2$ distribution gives $P(\mu\in\mathcal C)=1-\alpha$ under the stated model and regularity conditions. This joint region is the starting point; the coordinate intervals below come from projecting this region, not from applying separate univariate procedures.

Definition: The projection of $\mathcal C$ onto coordinate $j$ is the set of values $x_j$ for which there is at least one vector $x\in\mathcal C$ having that $j$th coordinate. Its endpoints are the minimum and maximum possible $j$th coordinates among vectors in the ellipsoid.

Write $S_{jj}$ for the $j$th diagonal entry of $S$, and $e_j$ for the vector with a 1 in coordinate $j$ and 0 elsewhere. Then the projection is the interval

I_j= \left[ \bar X_j-\sqrt{\frac{c_\alpha S_{jj}}{n}}, \ \bar X_j+\sqrt{\frac{c_\alpha S_{jj}}{n}} \right], \qquad j=1,\ldots,p.

The half-width is $h_j=\sqrt{c_\alpha S_{jj}/n}$, and the full width is $2h_j$. The intervals share the same joint critical value, but their widths can differ because the estimated marginal variances $S_{jj}$ differ.

Why These Endpoints Are the Projection

The key geometric calculation is the largest possible change in one coordinate under a quadratic constraint. For any vector $a$, the maximum of $a^\mathsf{T}(x-\bar X)$ over the ellipsoid is $\sqrt{(c_\alpha/n)a^\mathsf{T}Sa}$. Taking $a=e_j$ gives the upper endpoint; taking $-e_j$ gives the lower endpoint.

Result: For positive definite $S$ and $c_\alpha>0$, the projection of $\mathcal C$ onto the $j$th coordinate axis is exactly $I_j$, not merely a subset of it.

To prove this, let $y=x-\bar X$. The ellipsoid constraint is $y^\mathsf{T}S^{-1}y\le c_\alpha/n$. For any $a$, the Cauchy–Schwarz inequality applied after factoring the quadratic form gives

a^\mathsf{T}y =(S^{1/2}a)^\mathsf{T}(S^{-1/2}y) \le \sqrt{a^\mathsf{T}Sa}\sqrt{y^\mathsf{T}S^{-1}y} \le \sqrt{\frac{c_\alpha}{n}a^\mathsf{T}Sa}.

The bound is attained by

y= \sqrt{\frac{c_\alpha/n}{a^\mathsf{T}Sa}}\,Sa,

when $a\ne 0$: this vector satisfies $y^\mathsf{T}S^{-1}y=c_\alpha/n$, and its value of $a^\mathsf{T}y$ is the displayed upper bound. Replacing $a$ with $-a$ gives the matching lower bound. With $a=e_j$, $a^\mathsf{T}Sa=S_{jj}$, so the extremes of coordinate $j$ are $\bar X_j\pm\sqrt{c_\alpha S_{jj}/n}$. The ellipsoid is convex, so every value between these two extremes also occurs in the projection. This proves the claimed interval. $\square$

This derivation also explains an important detail: the diagonal entry of $S$, not the diagonal entry of $S^{-1}$, determines the projected coordinate width. The inverse matrix defines the ellipsoid’s quadratic constraint; maximizing a coordinate over that constraint produces $S_{jj}$.

Simultaneous Coverage and Widths

Collect the projected intervals into a rectangle $R=I_1\times\cdots\times I_p$. Every point in $\mathcal C$ has each coordinate inside its projection, so $\mathcal C\subseteq R$. Consequently, if $\mu\in\mathcal C$, then $\mu_j\in I_j$ for every coordinate. The joint coverage guarantee follows:

Theorem: Under the independent multivariate normal model with positive definite $\Sigma$ and $n>p$, the coordinatewise intervals satisfy $$ P\left(\mu_j\in I_j\text{ for every }j=1,\ldots,p\right)\ge 1-\alpha. $$

Indeed, the event $\{\mu\in\mathcal C\}$ is contained in the event that every coordinate is covered. Taking probabilities and using the exact ellipsoid coverage gives the inequality. The bound is generally conservative when $p\ge2$: the projection rectangle extends beyond the ellipsoid, so the rectangle can cover mean vectors that the ellipsoid does not. The intervals are simultaneous because their coverage event concerns all coordinates together; this does not mean each interval has exactly $1-\alpha$ marginal coverage.

The widths also have a useful aggregate identity. Since $h_j^2=c_\alpha S_{jj}/n$, summing over coordinates yields

\sum_{j=1}^p h_j^2 =\frac{c_\alpha}{n}\sum_{j=1}^p S_{jj} =\frac{c_\alpha}{n}\operatorname{tr}(S).

Thus the sum of squared half-widths depends on the trace of the sample covariance matrix, while the common multiplier $c_\alpha/n$ accounts for the joint confidence requirement. This identity does not say that every component has the same width: a coordinate with larger sample variance still has a wider projected interval.

Worked Example: Projecting a Two-Dimensional Ellipsoid

Let $p=2$, $n=6$, and use a 95% confidence level. Suppose $\bar X=(10,20)^\mathsf{T}$ and $S=\begin{pmatrix}4&1\\1&9\end{pmatrix}$. The covariance estimate is positive definite because its determinant is $4(9)-1^2=35>0$.

For these degrees of freedom, $F_{2,4}$ has distribution function $1-\{4/(4+2x)\}^2$. Setting this equal to $0.95$ gives $4/(4+2x)=1/\sqrt{20}$, hence $F_{2,4}(0.95)=2\sqrt{20}-2\approx6.9443$. Therefore

$c_{0.05}=\{2(6-1)/(6-2)\}(6.9443)=2.5(6.9443)\approx17.3607$.

For the first coordinate, $S_{11}=4$, so its half-width is $\sqrt{17.3607(4)/6}=\sqrt{11.5738}\approx3.402$. The interval is approximately $[6.598,13.402]$.

For the second coordinate, $S_{22}=9$, so its half-width is $\sqrt{17.3607(9)/6}=\sqrt{26.0411}\approx5.103$. Its interval is approximately $[14.897,25.103]$. Both intervals use the same ellipsoid-based critical value, and together they cover both mean components with probability at least $0.95$.

Worked Example: Changing Correlation Without Changing Projected Widths

Keep $p=2$, $n=6$, the 95% confidence level, and $\bar X=(10,20)^\mathsf{T}$, but change the covariance estimate to $S=\begin{pmatrix}4&5\\5&9\end{pmatrix}$. This matrix is positive definite because its determinant is $4(9)-5^2=11>0$.

The diagonal entries are still 4 and 9. The same critical value $c_{0.05}\approx17.3607$ therefore gives the same projected half-widths: $\sqrt{17.3607(4)/6}\approx3.402$ and $\sqrt{17.3607(9)/6}\approx5.103$. The coordinate intervals remain approximately $[6.598,13.402]$ and $[14.897,25.103]$.

The two ellipses do not have the same orientation: their off-diagonal covariance entries differ. But projection onto an individual coordinate axis depends only on that coordinate’s variance. The changed correlation affects the joint shape, not these projected endpoints.

Worked Example: Calculating Widths with a Simple Critical Value

Suppose $p=2$, $n=4$, and the desired confidence level is 90%. Let $\bar X=(3,8)^\mathsf{T}$ and $S=\begin{pmatrix}1&0\\0&4\end{pmatrix}$. For an $F_{2,2}$ random variable, the distribution function is $x/(1+x)$, so its 90th percentile solves $x/(1+x)=0.90$, giving $F_{2,2}(0.90)=9$.

The ellipsoid multiplier is $c_{0.10}=\{2(4-1)/(4-2)\}(9)=3(9)=27$. Thus $h_1=\sqrt{27(1)/4}=\sqrt{6.75}\approx2.598$, and the first interval is approximately $[0.402,5.598]$, with full width $2(2.598)\approx5.196$.

For the second coordinate, $h_2=\sqrt{27(4)/4}=\sqrt{27}\approx5.196$. Its interval is approximately $[2.804,13.196]$, with full width $2(5.196)\approx10.392$. As a check on the aggregate identity, $h_1^2+h_2^2=6.75+27=33.75$, while $(c_{0.10}/n)\operatorname{tr}(S)=(27/4)(1+4)=33.75$.

Common Pitfalls and Interpretation

Using $S^{-1}_{jj}$ in the width. This confuses the ellipsoid’s defining quadratic form with its coordinate projection. The support calculation gives $e_j^\mathsf{T}Se_j=S_{jj}$, so the half-width is $\sqrt{c_\alpha S_{jj}/n}$.
Calling each interval a 95% marginal interval. The guarantee is joint coverage of all $p$ components at least $1-\alpha$. It does not assert that each interval individually has exactly that coverage.
Expecting the intervals to reproduce the ellipsoid. The intervals are projections, and their Cartesian product is a rectangle containing the ellipsoid. The rectangle discards information about which combinations of coordinate values are jointly plausible.
Ignoring the common critical value. These are not ordinary componentwise intervals with separately chosen critical values. Their widths use the multivariate Hotelling cutoff, which reflects the dimension and degrees of freedom of the joint region.

These intervals are useful when results must be reported component by component while retaining a joint confidence guarantee. Their simplicity comes with a tradeoff: projection provides transparent coordinate limits, but the resulting rectangle can be wider as a joint set than the original ellipsoid. If the scientific target is a contrast or another linear combination rather than a coordinate, its own projection is the relevant calculation.

Check Your Understanding

Answer each question using the definitions and projection argument above.

Under what model and sample-size conditions is the stated Hotelling ellipsoid exact?
For coordinate $j$, which entry of $S$ determines the projected half-width, and why?
Why do the projected intervals have joint coverage at least $1-\alpha$, rather than necessarily exactly $1-\alpha$?
If two covariance estimates have the same diagonal entries but different off-diagonal entries, what happens to the coordinatewise widths when the other inputs are fixed?
For $p=2$, $c_\alpha=12$, $n=6$, and $S_{22}=3$, calculate the second interval’s half-width and full width.