Why the Contrast Subspace Matters
Scheffé inference protects a whole family of linear contrasts at once. But a family may be described by many coefficient vectors even when those vectors contain only a few independent directions. For example, ten requested contrasts may all be linear combinations of just two basic contrasts. The relevant dimension for the Scheffé multiplier is the number of independent directions, not the length of the list.
This reduction is useful when the scientific question concerns a structured set of comparisons, such as differences among group means subject to a constraint. It can avoid paying the larger multiplicity cost associated with protecting every contrast in the full mean-vector space. The protection remains simultaneous over every contrast in the specified subspace, including contrasts not explicitly listed.
We work in the classical one-sample Gaussian setting. Let \(X_1,\ldots,X_n\) be independent \(p\)-dimensional observations with \(X_i\sim N_p(\mu,\Sigma)\), where \(\Sigma\) is positive definite and \(n>p\). Write \(\bar X=n^{-1}\sum_{i=1}^n X_i\) and \(S=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)'\). Thus \(S\) estimates \(\Sigma\). Let \(C\) be a \(p\times q\) matrix of full column rank whose columns form a basis for \(L\). Then every \(c\in L\) can be written as \(c=Ca\) for some \(a\in\mathbb{R}^q\).
The Supremum Over a Subspace
For a fixed contrast \(c\), the estimated contrast is \(c'\bar X\), and its estimated standard error is \(\sqrt{c'Sc/n}\). The simultaneous Scheffé question is: how large can the absolute standardized estimation error become as \(c\) ranges over the subspace?
To derive the identity, set \(e=\bar X-\mu\), \(d=C'e\), and \(A=C'SC\). Since \(C\) has full column rank and \(S\) is positive definite, \(A\) is positive definite. Substituting \(c=Ca\) into the squared ratio gives $$ n\frac{(a'd)^2}{a'Aa}. $$ Write \(a'Aa=\|A^{1/2}a\|^2\) and \(a'd=(A^{1/2}a)'(A^{-1/2}d)\). By the Cauchy–Schwarz inequality, $$ (a'd)^2\leq (a'Aa)(d'A^{-1}d). $$ Equality is attained when \(A^{1/2}a\) is proportional to \(A^{-1/2}d\), provided \(d\ne0\). If \(d=0\), both sides of the claimed supremum identity are zero. Taking square roots proves the result. \(\square\)
Worked Example: Four Listed Contrasts, Two Independent Directions
Suppose \(p=3\) and the requested contrasts are \(c_1=(1,-1,0)'\), \(c_2=(0,1,-1)'\), \(c_1+c_2\), and \(2c_1-c_2\). There are four listed vectors, but the last two are combinations of the first two. The contrast space is \(L=\operatorname{span}(c_1,c_2)\), so its rank is \(q=2\).
Take \(C=(c_1\ c_2)\). For any \(a=(a_1,a_2)'\), the contrast \(Ca\) equals \(a_1c_1+a_2c_2\). Thus varying \(a\) over \(\mathbb{R}^2\) covers every contrast in the span, not merely the four listed vectors. The supremum result therefore uses a two-dimensional statistic.
As a numerical illustration, let \(n=15\), \(S=I_3\), and \(e=\bar X-\mu=(0.2,0,-0.2)'\). Then $$ C'C=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}, \qquad (C'C)^{-1}=\frac{1}{3}\begin{pmatrix}2&1\\1&2\end{pmatrix}, \qquad C'e=\begin{pmatrix}0.2\\0.2\end{pmatrix}. $$ Multiplying the inverse by \(C'e\) gives \((0.2,0.2)'\), so \((C'e)'(C'C)^{-1}(C'e)=0.08\). The observed supremum squared is therefore \(15(0.08)=1.20\), and the supremum is \(\sqrt{1.20}\approx1.095\).
Exact Critical Value for Rank \(q\)
For a confidence level \(1-\alpha\), denote the \(1-\alpha\) quantile of an \(F\) distribution with degrees of freedom \(r,s\) by \(F_{r,s;1-\alpha}\). The exact subspace Scheffé multiplier is $$ k_{q,\alpha}=\sqrt{\frac{q(n-1)}{n-q}F_{q,n-q;1-\alpha}}. $$ Its degrees of freedom depend on \(q\), the dimension of the contrast subspace.
To prove the distribution statement, consider the projected observations \(Y_i=C'X_i\). They are independent \(q\)-variate normal vectors with mean \(C'\mu\) and positive-definite covariance \(C'\Sigma C\). Their sample mean is \(C'\bar X\), and their sample covariance is \(C'SC\). The statistic in the theorem is therefore the one-sample Hotelling statistic for these \(q\)-dimensional observations.
For completeness, its exact distribution follows from the normal and Wishart representation. Let \(m=n-1\), and standardize using a positive-definite square root of \(C'\Sigma C\). Then $$ Z=\sqrt{n}(C'\Sigma C)^{-1/2}(C'\bar X-C'\mu)\sim N_q(0,I_q), $$ and $$ W=(C'\Sigma C)^{-1/2}(n-1)C'SC(C'\Sigma C)^{-1/2}\sim \operatorname{Wishart}_q(I_q,m). $$ The normal sample theory gives independence of \(Z\) and \(W\). Direct substitution shows that the quadratic statistic in the theorem is \((n-1)Z'W^{-1}Z\).
Here is the distribution of that expression. The squared length \(Z'Z\) has a \(\chi^2_q\) distribution. Conditional on the direction of \(Z\), rotational invariance of \(W\) implies that the corresponding diagonal element of \(W^{-1}\) has the same distribution as \((W^{-1})_{11}\), independent of \(Z'Z\). For a Wishart matrix with \(m\) degrees of freedom, the Schur complement representation gives \(1/(W^{-1})_{11}\sim\chi^2_{m-q+1}\): it is the squared length of the residual of one standard normal column after projection off the other \(q-1\) columns, a projection of rank \(m-q+1\). Thus $$ Z'W^{-1}Z\ \overset{d}{=}\ \frac{\chi^2_q}{\chi^2_{m-q+1}}, $$ where the two chi-square variables are independent. Since \(m-q+1=n-q\), multiplying this ratio by \(n-1=m\) yields $$ (n-1)\frac{\chi^2_q}{\chi^2_{n-q}} = \frac{q(n-1)}{n-q}F_{q,n-q}. $$ This proves the theorem's distribution assertion. Applying its \(1-\alpha\) quantile and the supremum identity gives the stated simultaneous coverage. \(\square\)
The result is also unchanged by a change of basis. If \(C_*=CA\), where \(A\) is any invertible \(q\times q\) matrix, then $$ C_*(C_*'SC_*)^{-1}C' _* =CA(A'C'SCA)^{-1}A'C' =C(C'SC)^{-1}C'. $$ The middle equality follows from \((A'BA)^{-1}=A^{-1}B^{-1}(A')^{-1}\) for invertible \(A\) and \(B\). Hence the statistic and its critical value depend on the subspace itself, not on how its basis is represented.
Worked Example: A 95% Scheffé Family With Rank Two
Suppose \(p=3\), \(n=15\), and the requested contrasts span a two-dimensional subspace. For a 95% family, the multiplier uses \(F_{2,13;0.95}\approx3.8056\). Therefore, $$ k_{2,0.05}^2 =\frac{2(15-1)}{15-2}(3.8056) =\frac{28}{13}(3.8056) \approx 8.1966, \qquad k_{2,0.05}\approx 2.863. $$
For the data summaries in the preceding example, the observed supremum is approximately \(1.095\), below the critical value \(2.863\). Equivalently, its squared value \(1.20\) is below the squared cutoff \(8.1966\). The confidence family therefore contains the true contrasts in this illustration. The four listed contrasts do not call for a rank-four multiplier: their span has rank two.
Worked Example: Rank One and the Usual Two-Sided \(t\) Critical Value
Let \(n=12\), and consider all nonzero scalar multiples of \(c=(1,-1,0)'\). This family has rank \(q=1\), regardless of how many multiples are written down. Suppose \(c'\bar X=1.4\) and \(c'Sc=4\). The estimated standard error is $$ \sqrt{\frac{c'Sc}{n}}=\sqrt{\frac{4}{12}}=\frac{1}{\sqrt{3}}\approx0.57735. $$
At 95% confidence, \(F_{1,11;0.95}=t_{11;0.975}^2\), so the Scheffé multiplier is the ordinary two-sided \(t\) critical value, approximately \(2.201\). The half-width is \(2.201(0.57735)\approx1.271\), giving the interval \(1.4\pm1.271\), or approximately \([0.129,2.671]\). Every nonzero multiple of \(c\) expresses the same one-dimensional direction, so it does not increase the rank or change this critical value.
Worked Example: A Rank-Two Subspace in Four Dimensions
Let \(p=4\), \(n=20\), and let \(L\) contain contrasts in the first two coordinate directions. Choose $$ C=\begin{pmatrix}1&0\\0&1\\0&0\\0&0\end{pmatrix}, \qquad S=\operatorname{diag}(1,4,9,16), \qquad \bar X-\mu=(0.3,-0.4,0,0)'. $$ Then \(C'SC=\operatorname{diag}(1,4)\), and \(C'(\bar X-\mu)=(0.3,-0.4)'\). The quadratic form is \(0.3^2/1+(-0.4)^2/4=0.09+0.04=0.13\), so the observed supremum squared is \(20(0.13)=2.6\), and the supremum is approximately \(1.612\).
For a 95% family, \(F_{2,18;0.95}\approx3.555\), giving $$ k_{2,0.05}^2=\frac{2(19)}{18}(3.555)\approx7.505, \qquad k_{2,0.05}\approx2.740. $$ The observed supremum is below this cutoff. Although the mean vector has four coordinates, the requested contrast space has rank two, which determines the multiplier.
Using the Result and Avoiding a Common Error
The theorem yields simultaneous intervals for all contrasts in \(L\): $$ c'\bar X\ \pm\ k_{q,\alpha}\sqrt{\frac{c'Sc}{n}}, \qquad c\in L,\ c\ne0. $$ The intervals are simultaneous in the sense that the probability that every one of them covers its corresponding \(c'\mu\) is \(1-\alpha\). A finite list of requested contrasts receives at least this coverage because it is contained in the full subspace family. If the list spans a larger space than intended, however, the rank and multiplier must be based on that actual span.
Common mistake: counting listed contrasts and using that count as the numerator degrees of freedom. Suppose five vectors are listed, but each is a combination of two basis contrasts. Choosing \(F_{5,n-5}\) because there are five entries treats the family as five-dimensional and gives the wrong calibration for the subspace. The correction is to form the span, find its rank \(q\), and use \(F_{q,n-q}\). Conversely, using a rank smaller than the true span fails to protect every requested direction.
A reduced-rank Scheffé family is especially useful when the inferential target is naturally restricted to a subspace. It offers exact finite-sample familywise coverage under the stated Gaussian model while avoiding unnecessary protection over directions that are not part of the question. The gain comes from a smaller family dimension, not from pretending that redundant vectors are independent tests.
Check Your Understanding
Answer each question by identifying the contrast space first.
- A list contains six contrast vectors, and their span has dimension three. Which degrees of freedom determine the Scheffé critical value?
- Why does replacing a basis matrix \(C\) by \(CA\), for invertible \(A\), leave the subspace statistic unchanged?
- For rank \(q=1\), how is the two-sided Scheffé multiplier related to a \(t\) critical value?
- In the exact distribution result, what assumptions ensure that the projected sample covariance is nonsingular?
- If a requested list has rank two but contains seven vectors, what does the rank-two guarantee protect, beyond the seven listed contrasts?