Exact Hotelling Tests for Planned Group Contrasts

Why a Planned Contrast Has an Exact Test

A planned multivariate contrast asks about one specified linear combination of group mean vectors. For example, a study may compare one treatment group with the average of several control groups, where each patient contributes a vector of outcomes. The question is not whether all group means are equal; it is whether the particular vector-valued contrast is zero.

Under independent multivariate normal sampling with a common covariance matrix, the contrast estimate has a covariance that can be estimated from the pooled within-group covariance. Its standardized squared length has an exact Hotelling $T^2$ calibration. The result depends on accounting for the contrast coefficients and group sample sizes together.

Definition: Suppose there are $g$ independent groups. In group $i$, $X_{ij}$ is a $p$-dimensional observation, $j=1,\ldots,n_i$, with mean vector $\mu_i$ and common covariance matrix $\Sigma$. A single planned group contrast is $$ \psi=\sum_{i=1}^g c_i\mu_i, \qquad \sum_{i=1}^g c_i=0, $$ where the coefficients $c_i$ are fixed in advance and are not all zero. Its estimator is $\hat\psi=\sum_{i=1}^g c_i\bar X_i$, where $\bar X_i=n_i^{-1}\sum_{j=1}^{n_i}X_{ij}$. The zero-sum condition makes this a contrast of group means.

Write $N=\sum_{i=1}^g n_i$ and $\nu=N-g$, the total within-group residual degrees of freedom. The pooled within-group covariance estimate is

S_p=\frac{1}{\nu}\sum_{i=1}^g\sum_{j=1}^{n_i} (X_{ij}-\bar X_i)(X_{ij}-\bar X_i)'.

The subscript $p$ here labels the pooled estimate; the vectors and covariance matrices are $p$-dimensional. If $\nu\ge p$, this estimate is positive definite with probability one under the model, so its inverse exists.

The Contrast Covariance and Test Statistic

Independence across groups gives $\operatorname{Cov}(\bar X_i)=\Sigma/n_i$, and the group sample means are independent. Therefore,

\operatorname{Cov}(\hat\psi) =\sum_{i=1}^g c_i^2\operatorname{Cov}(\bar X_i) =q\Sigma, \qquad q=\sum_{i=1}^g\frac{c_i^2}{n_i}.

The scalar $q$ is determined by both the contrast coefficients and the group sample sizes. Replacing $\Sigma$ by $S_p$ estimates the covariance of the contrast by $qS_p$.

Theorem: Assume the observations are independent and $X_{ij}\sim N_p(\mu_i,\Sigma)$, with the same positive-definite $\Sigma$ in every group. For a fixed, nonzero contrast $c_1,\ldots,c_g$ summing to zero, suppose $\nu=N-g\ge p$. Under $H_0:\psi=0$, define $$ T^2=\hat\psi'(qS_p)^{-1}\hat\psi. $$ Then $$ \frac{\nu-p+1}{\nu p}T^2\sim F_{p,\,\nu-p+1}. $$

Worked Example: Comparing One Group With the Average of Two Others

Three groups have sample sizes $10,12,8$, and the planned contrast is $\psi=(\mu_1+\mu_2)/2-\mu_3$. Thus $c=(1/2,1/2,-1)$, which sums to zero. Suppose the observed sample means are $(10,5)'$, $(12,7)'$, and $(8,4)'$, and the pooled covariance estimate is $$ S_p=\begin{pmatrix}4&1\\1&3\end{pmatrix}. $$

The estimated contrast is $$ \hat\psi=\tfrac12(10,5)'+\tfrac12(12,7)'-(8,4)'=(3,2)'. $$ The coefficient-and-size factor is $$ q=\frac{(1/2)^2}{10}+\frac{(1/2)^2}{12}+\frac{(-1)^2}{8} =\frac1{40}+\frac1{48}+\frac18=\frac{41}{240}. $$ The determinant of $S_p$ is $4(3)-1(1)=11>0$, and $$ S_p^{-1}=\frac1{11}\begin{pmatrix}3&-1\\-1&4\end{pmatrix}. $$ Consequently, $$ \hat\psi'S_p^{-1}\hat\psi =\frac1{11}(3,2)\begin{pmatrix}3&-1\\-1&4\end{pmatrix}\begin{pmatrix}3\\2\end{pmatrix} =\frac1{11}(3,2)\begin{pmatrix}7\\5\end{pmatrix} =\frac{31}{11}. $$ Since $(qS_p)^{-1}=q^{-1}S_p^{-1}$, $$ T^2=\frac{240}{41}\frac{31}{11}=\frac{7440}{451}\approx16.50. $$ Here $N=30$, $g=3$, and $\nu=27$. The null calibration is $$ \frac{27-2+1}{27(2)}T^2=\frac{13}{27}T^2\sim F_{2,26}. $$ The observed transformed statistic is approximately $7.94$. A test at level $\alpha$ rejects when this value exceeds the $1-\alpha$ quantile of $F_{2,26}$.

Why the F Calibration Is Exact

The proof uses two facts supplied by multivariate normal sampling: the contrast estimate is normal, and it is independent of the pooled within-group covariance estimate. The distinction between centering by a population mean and centering by a sample mean matters in establishing this independence.

Transform each group’s observations.
For group $i$, apply an orthogonal transformation to the $n_i$ observations, with first row $n_i^{-1/2}(1,\ldots,1)$. Apply it to the observations centered by the population mean $\mu_i$. The first transformed vector is $\sqrt{n_i}(\bar X_i-\mu_i)$; the other $n_i-1$ transformed vectors are residual coordinates.

Use normality and orthogonality.
The transformed vectors are jointly normal. Their cross-covariances are zero because the transformation is orthogonal, so they are independent $N_p(0,\Sigma)$ vectors. In particular, the first coordinate is not zero: the observations were centered by $\mu_i$, not by $\bar X_i$.

Identify the within-group scatter.
Orthogonality preserves sums of outer products. Removing the first transformed coordinate leaves exactly $\sum_j(X_{ij}-\bar X_i)(X_{ij}-\bar X_i)'$. Across groups, the remaining $\sum_i(n_i-1)=\nu$ independent normal residual coordinates imply $\nu S_p\sim W_p(\Sigma,\nu)$. They are independent of all group sample means, and hence of $\hat\psi$.

Standardize the contrast.
Under $H_0$, $Z=(q\Sigma)^{-1/2}\hat\psi\sim N_p(0,I_p)$. It is independent of $W=\Sigma^{-1/2}(\nu S_p)\Sigma^{-1/2}\sim W_p(I_p,\nu)$, and $T^2=\nu Z'W^{-1}Z$.

For completeness, the normal-Wishart quadratic-form result used in the last step can be seen directly. If $Z\sim N_p(0,I_p)$ and $W\sim W_p(I_p,\nu)$ are independent, then $$ Z'W^{-1}Z\ \stackrel{d}{=}\ \frac{U}{V}, \qquad U\sim\chi^2_p,\quad V\sim\chi^2_{\nu-p+1}, $$ with $U$ and $V$ independent. To verify this, condition on $Z$ and rotate its direction to the first coordinate; the identity covariance of $Z$ and the identity-scale Wishart distribution are invariant under this rotation. The squared length $Z'Z$ has a $\chi^2_p$ distribution. Represent $W=A'A$, with $A$ a $\nu\times p$ matrix of independent standard normal entries, and partition its columns as $A=(a_1,A_2)$. The reciprocal of the first diagonal entry of $W^{-1}$ is the Schur complement $$ a_1'\{I-A_2(A_2'A_2)^{-1}A_2'\}a_1. $$ Conditional on $A_2$, the matrix in braces is an orthogonal projection of rank $\nu-p+1$. Since $a_1$ is an independent standard normal vector, this quadratic form has a $\chi^2_{\nu-p+1}$ distribution. This denominator is independent of the length of $Z$, giving the stated ratio. For $p=1$, the same conclusion follows directly with denominator degrees of freedom $\nu$.

By the definition of the $F$ distribution, $(U/p)/(V/(\nu-p+1))\sim F_{p,\nu-p+1}$. Since $T^2=\nu U/V$, multiplying by $(\nu-p+1)/(\nu p)$ gives the theorem. $\square$

Worked Example: A Treatment Versus the Average of Three Controls

Suppose four groups have sizes $12,9,15,12$, with group 1 the treatment group. The planned contrast compares it with the unweighted average of groups 2–4: $$ c=(1,-1/3,-1/3,-1/3). $$ Suppose the estimated group means are $(5,2)'$, $(2,1)'$, $(4,0)'$, and $(3,2)'$, respectively, and $$ S_p=\begin{pmatrix}2&0\\0&1\end{pmatrix}. $$ The coefficients sum to zero. The average of the three comparison-group means is $(3,1)'$, so $\hat\psi=(2,1)'$. The contrast variance factor is $$ q=\frac1{12}+\frac{1/9}{9}+\frac{1/9}{15}+\frac{1/9}{12} =\frac1{12}+\frac1{81}+\frac1{135}+\frac1{108} =\frac{91}{810}. $$ Because $S_p^{-1}=\operatorname{diag}(1/2,1)$, the squared length before dividing by $q$ is $$ \hat\psi'S_p^{-1}\hat\psi=(2,1)\begin{pmatrix}1/2&0\\0&1\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}=3. $$ Thus $T^2=3/(91/810)=2430/91\approx26.70$. Here $N=48$, $g=4$, and $\nu=44$, so the null-calibrated value is $$ \frac{44-2+1}{44(2)}T^2=\frac{43}{88}\frac{2430}{91}\approx13.05, $$ which is compared with $F_{2,43}$. The treatment-versus-average question is one contrast; this calculation alone does not provide simultaneous inference for a collection of contrasts.

Confidence Region and Scaling

The same exact calibration gives a confidence region for the unknown contrast vector. Let $F_{p,\nu-p+1;1-\alpha}$ denote the $1-\alpha$ quantile of the indicated $F$ distribution. Under the model conditions in the theorem, a $1-\alpha$ confidence region for $\psi$ is

\left\{\psi: (\hat\psi-\psi)'(qS_p)^{-1}(\hat\psi-\psi) \le \frac{p\nu}{\nu-p+1}F_{p,\nu-p+1;1-\alpha} \right\}.

Indeed, for each fixed true $\psi$, the centered estimator $\hat\psi-\psi$ has covariance $q\Sigma$, is independent of $S_p$, and the theorem’s derivation applies to its quadratic form. The region therefore has coverage $1-\alpha$ exactly under the stated Gaussian common-covariance model.

Worked Example: Rescaling Does Not Change the Test

For two groups of sizes $8$ and $12$, take the contrast coefficients $(1,-1)$. Suppose $\hat\psi=(1,2)'$ and $S_p=\operatorname{diag}(2,1)$. Then $q=1/8+1/12=5/24$, and $$ \hat\psi'S_p^{-1}\hat\psi=\frac12+4=\frac92, \qquad T^2=\frac{24}{5}\frac92=\frac{108}{5}=21.6. $$ Now multiply both contrast coefficients by $2$. The estimated contrast becomes $2\hat\psi$, and the variance factor becomes $4q$. Therefore the new statistic is $$ (2\hat\psi)'(4qS_p)^{-1}(2\hat\psi) =4\hat\psi'\{4qS_p\}^{-1}\hat\psi =\hat\psi'(qS_p)^{-1}\hat\psi =21.6. $$ The contrast’s units change, but the test statistic does not. Here $\nu=8+12-2=18$, and the exact null calibration is $(17/36)T^2\sim F_{2,17}$.

Common Mistakes and Practical Interpretation

Using the wrong factor for the estimated covariance. The covariance estimate is $qS_p$, where $q=\sum_i c_i^2/n_i$. Omitting the sample sizes or using $\sum_i c_i/n_i$ does not estimate the contrast covariance.
Using between-group scatter as the covariance estimate. The exact result uses pooled within-group residual scatter. Between-group variation reflects differences among the means being studied and is not a substitute for $S_p$.
Using the wrong degrees of freedom. The covariance estimate has $\nu=N-g$ residual degrees of freedom, and the transformed statistic has denominator degrees of freedom $\nu-p+1$. The numerator degrees of freedom are the outcome dimension $p$.
Centering the transformed observations by their sample means too early. After that centering, the transformed mean coordinate is zero. To establish independence, first transform observations centered by the population mean; the mean coordinate is then $\sqrt{n_i}(\bar X_i-\mu_i)$, independent of the residual coordinates.
Assuming exactness under unequal covariance matrices or nonnormal sampling. The stated F calibration relies on independent multivariate normal observations and one common positive-definite covariance matrix. It is not automatically exact when these assumptions fail.

The practical value of this result is that a single planned multivariate question can be answered without estimating each group covariance separately: the pooled residual covariance supplies the scale, and the contrast-specific factor $q$ accounts for the design. If $\nu<p$, the pooled covariance cannot be inverted in the ordinary full-dimensional statistic; this exact test therefore requires enough residual degrees of freedom.

Key takeaway: For one fixed contrast, estimate its covariance by $qS_p$, with $q=\sum_i c_i^2/n_i$. Under independent normal groups with common covariance, the resulting Hotelling statistic has an exact $F_{p,\nu-p+1}$ calibration after multiplication by $(\nu-p+1)/(\nu p)$.

Check Your Understanding

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

For coefficients $(1/2,1/2,-1)$, what group means does the contrast compare, and why do the coefficients qualify as a contrast?
Explain why the covariance estimate for $\hat\psi$ is $qS_p$, and write $q$ for a general contrast.
With $N=36$, $g=4$, and outcome dimension $p=3$, give $\nu$ and the null F degrees of freedom, assuming the model conditions hold.
Why must the observations be centered by population means, rather than sample means, in the orthogonal-transform argument for independence?
What changes in the estimated contrast and in $q$ if every contrast coefficient is multiplied by the same nonzero constant? What happens to $T^2$?