Duality Between Mean Tests and Confidence Regions

One Procedure, Two Ways to Report Its Result

A test of a proposed mean vector gives a decision: reject or do not reject. A confidence region gives a set of plausible mean vectors. When both are built from the same exact test, these are not separate analyses. The test rejects a particular vector precisely when the confidence region excludes it.

We develop this equivalence for the one-sample multivariate normal model with unknown covariance. Its practical value is immediate: a single confidence ellipsoid summarizes the test decision for every possible proposed mean, while a test of one specified mean is a membership check against that ellipsoid.

The Exact Hotelling Test and Its Region

Suppose \(X_1,\ldots,X_n\) are independent and identically distributed \(p\)-dimensional random vectors with distribution \(N_p(\mu,\Sigma)\), where \(\Sigma\) is positive definite. Write \(\bar X=n^{-1}\sum_{i=1}^n X_i\), and let \(S=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)'\) be the sample covariance matrix. Assume \(n>p\), so \(S\) is positive definite with probability one under this model.

Definition: For a candidate mean vector \(\mu_0\), the one-sample Hotelling statistic is \(T^2(\mu_0)=n(\bar X-\mu_0)'S^{-1}(\bar X-\mu_0)\). For a test of \(H_0:\mu=\mu_0\) against \(H_1:\mu\ne\mu_0\), reject when this statistic is larger than its critical value.

Under \(H_0\), the exact Hotelling calibration is

\frac{n-p}{p(n-1)}T^2(\mu_0)\sim F_{p,n-p}.

Here \(F_{p,n-p}\) denotes the central F distribution with \(p\) and \(n-p\) degrees of freedom. Let \(q_{1-\alpha}\) be its \(1-\alpha\) quantile, where \(0<\alpha<1\). Thus \(P(F_{p,n-p}\le q_{1-\alpha})=1-\alpha\). The level-\(\alpha\) test rejects when \(T^2(\mu_0)>c_\alpha\), with

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

Inverting this test gives the region

C_{1-\alpha}= \left\{\mu\in\mathbb{R}^p: n(\bar X-\mu)'S^{-1}(\bar X-\mu)\le c_\alpha \right\}.

It is an ellipsoid centered at \(\bar X\). Its size and orientation depend on \(S\), and its cutoff uses the same \(c_\alpha\) as the test. That shared cutoff is essential: using a different calibration would break the claimed correspondence.

The Duality, Precisely

Theorem: Under the stated multivariate normal model with \(n>p\), for every observed sample and every candidate vector \(\mu_0\), \(\mu_0\notin C_{1-\alpha}\) if and only if the level-\(\alpha\) Hotelling test of \(H_0:\mu=\mu_0\) rejects. Moreover, \(C_{1-\alpha}\) has exact coverage \(1-\alpha\): for every true mean vector \(\mu\), \(P_\mu(\mu\in C_{1-\alpha})=1-\alpha\).

Proof of the decision equivalence. By the definition of the region, \(\mu_0\in C_{1-\alpha}\) if and only if \(n(\bar X-\mu_0)'S^{-1}(\bar X-\mu_0)\le c_\alpha\). The expression on the left is exactly \(T^2(\mu_0)\). The test rejects exactly when \(T^2(\mu_0)>c_\alpha\). These conditions are logical complements, so the test rejects if and only if \(\mu_0\notin C_{1-\alpha}\). The strict rejection rule corresponds to a closed region: equality is included in the region and is not a rejection. \(\square\)

Proof of exact coverage. Fix any true mean vector \(\mu\). The event that the confidence region contains the true vector is

\{\mu\in C_{1-\alpha}\} = \left\{ \frac{n-p}{p(n-1)} n(\bar X-\mu)'S^{-1}(\bar X-\mu) \le \frac{n-p}{p(n-1)}c_\alpha \right\}.

Under \(P_\mu\), the statistic inside the event has distribution \(F_{p,n-p}\), and the right-hand cutoff is \(q_{1-\alpha}\). Consequently, the event has probability \(P(F_{p,n-p}\le q_{1-\alpha})=1-\alpha\). This holds for every \(\mu\), which proves exact coverage. \(\square\)

The theorem is an instance of test inversion. More generally, suppose a test of each point null value \(\theta=\theta_0\) rejects when a statistic \(R(\theta_0)\) exceeds a cutoff \(k\). Define the nonrejection set as \(\{\theta:R(\theta)\le k\}\). For any observed data, a point is outside that set exactly when its test rejects. If each point-null test has rejection probability \(\alpha\) when that point is true, the set has coverage \(1-\alpha\). In the Hotelling case, the exact F calibration supplies that pointwise rejection probability and the resulting set is an ellipsoid.

Worked Examples

Worked Example: A Proposed Mean Outside the Ellipsoid

Suppose \(p=2\), \(n=10\), \(\bar x=(1,2)'\), and \(S=\begin{pmatrix}4&1\\1&2\end{pmatrix}\). Test \(\mu_0=(0,0)'\) at level \(\alpha=0.05\). For these degrees of freedom, \(q_{0.95}\) for \(F_{2,8}\) is approximately \(4.459\), so

c_{0.05}=\frac{2(9)}{8}(4.459)\approx10.033.

The determinant of \(S\) is \(4(2)-1(1)=7\), hence \(S^{-1}=\frac17\begin{pmatrix}2&-1\\-1&4\end{pmatrix}\). For \(d=\bar x-\mu_0=(1,2)'\),

S^{-1}d =\frac17\begin{pmatrix}2&-1\\-1&4\end{pmatrix} \begin{pmatrix}1\\2\end{pmatrix} =\begin{pmatrix}0\\1\end{pmatrix}, \qquad d'S^{-1}d=(1,2)\begin{pmatrix}0\\1\end{pmatrix}=2.

Therefore \(T^2(\mu_0)=10(2)=20\), which is greater than \(10.033\); the test rejects. The same calculation says that \(\mu_0\) does not belong to the confidence ellipsoid. As a check using the F scale, the observed statistic is \((8/18)(20)=8.889\), larger than \(4.459\).

Worked Example: A Proposed Mean Inside the Ellipsoid

Let \(p=2\), \(n=10\), \(\bar x=(0.8,0.4)'\), \(S=I_2\), and test \(\mu_0=(0,0)'\) at level \(0.05\). The same cutoff is \(c_{0.05}\approx10.033\). Since \(S^{-1}=I_2\),

T^2(\mu_0) =10\left[(0.8)^2+(0.4)^2\right] =10(0.64+0.16) =8.

Because \(8<10.033\), the test does not reject, and \(\mu_0\) lies in the ellipsoid. In F units, the statistic is \((8/18)(8)=32/9\approx3.556\), below the \(0.95\) quantile \(4.459\). Both comparisons give the same decision.

Worked Example: A Candidate Exactly on the Boundary

Take \(p=2\), \(n=10\), \(S=I_2\), and the level-\(0.05\) cutoff \(c_{0.05}\approx10.033\). For any observed \(\bar x\), choose a candidate mean \(\mu_0=\bar x-(\sqrt{2c_{0.05}/5},0)'\). The difference vector has squared length \(2c_{0.05}/5\), so

T^2(\mu_0) =10\left(\frac{2c_{0.05}}5\right) =4c_{0.05} \qquad\text{is not the correct substitution if the first coordinate is unscaled only when }S=I_2.

With \(S=I_2\), the quadratic form is the squared length, and the multiplier \(n=10\) gives \(T^2=4c_{0.05}\), not \(c_{0.05}\). To place the candidate on the boundary, instead take \(\mu_0=\bar x-(\sqrt{c_{0.05}/10},0)'\). Then

T^2(\mu_0)=10\left(\frac{c_{0.05}}{10}\right)=c_{0.05}.

Thus the candidate belongs to the closed confidence ellipsoid and is not rejected by the strict rule \(T^2>c_{0.05}\). This example illustrates why the boundary convention must be stated, even though under the continuous null distribution an exact-boundary observation has probability zero.

Why the Equivalence Matters

The duality makes inference easier to communicate and audit. A reported confidence ellipsoid answers all point-null questions at once: to assess any specified \(\mu_0\), check whether it lies inside. Conversely, a collection of level-\(\alpha\) tests over all candidate vectors determines the confidence region through its nonrejection set. The coverage statement is pointwise in the true vector: each fixed \(\mu\) is included with probability \(1-\alpha\) under the model.

It does not mean that every possible assertion about a vector-valued parameter has simultaneous coverage beyond the specified region, nor does it turn the ellipsoid into a probability distribution over \(\mu\). The confidence statement concerns the repeated-sampling behavior of the procedure. Exactness also depends on the model and calibration: the F cutoff is exact for independent multivariate normal observations with positive definite covariance and \(n>p\), not automatically for arbitrary data.

Common Mistakes

Reversing the inequality. Large Mahalanobis distance is evidence against the proposed mean. The test rejects above the cutoff; the region contains vectors at or below it.
Using different cutoffs. A test and region are dual only when they use the same statistic and critical value. For unknown covariance, the exact Hotelling F calibration is not replaced by a chi-square cutoff.
Treating the boundary as a rejection. With the convention \(T^2>c_\alpha\), equality is nonrejection and belongs to the closed region. A different boundary convention should be matched consistently.
Calling the ellipsoid a probability statement about the parameter. Exact coverage is a repeated-sampling probability, not a posterior probability that the fixed mean lies in the observed region.

Key takeaway: The exact Hotelling confidence ellipsoid is the set of all mean vectors not rejected by the corresponding level-\(\alpha\) test. Its \(1-\alpha\) coverage follows from the same F calibration that determines the test cutoff.

Check Your Understanding

Use the shared cutoff and the strict rejection rule to answer these questions.

Why does the nonrejection set use a less-than-or-equal inequality when the test rejects using a strict greater-than inequality?
For \(p=2\), \(n=10\), and \(\alpha=0.05\), what is the formula for the Hotelling critical value in terms of the \(F_{2,8}\) quantile?
If a candidate mean has \(T^2(\mu_0)=c_\alpha\), is it inside or outside the confidence region, and is it rejected?
What model assumptions justify the exact F calibration used to prove coverage?
Explain why a confidence ellipsoid with a different cutoff from the test would not satisfy the stated duality.