Why a Shared Control Changes the Calibration
When several treatments are compared with the same control, their estimated differences are not independent: each difference contains the same estimated control mean. Treating the comparisons as independent ignores this dependence; applying a separate marginal cutoff also fails to use the joint distribution. A Dunnett-type procedure calibrates the largest treatment-versus-control statistic using the dependence induced by the shared control.
Here the comparisons are multivariate. Each group has a \(p\)-dimensional mean vector, and each treatment-control contrast is assessed as a whole with a Hotelling statistic. The resulting regions provide simultaneous coverage for the entire collection of treatment-control mean differences.
Model and Joint Calibration
Suppose there are \(m\) treatment groups and one control group, indexed by \(0,1,\ldots,m\). In group \(i\), observe \(n_i\) independent \(p\)-dimensional vectors from \(N_p(\mu_i,\Sigma)\). Assume all groups are independent, share the same positive-definite covariance matrix \(\Sigma\), and have normally distributed observations. Let \(\bar X_i\) be the sample mean in group \(i\), let \(N=\sum_{i=0}^m n_i\), and let \(\nu=N-(m+1)\). The pooled within-group covariance estimator is denoted by \(S\).
The calibration is not a collection of independent Hotelling statistics. To describe the dependence, define the \(m\times m\) matrix \(R\) by \(R_{ii}=1\) and \[ R_{ij}=\frac{1/n_0}{\sqrt{c_ic_j}},\qquad i\ne j. \] Let \(Z_1,\ldots,Z_m\) be jointly Gaussian \(p\)-vectors with \(E[Z_i]=0\) and \(\operatorname{Cov}(Z_i,Z_j)=R_{ij}\Sigma\). Independently, let \(S\) have the pooled covariance distribution, equivalently \(\nu S\sim W_p(\Sigma,\nu)\), where \(W_p\) denotes the Wishart distribution. The calibration statistics are \(Z_i^{\mathsf T}S^{-1}Z_i\); their maximum has the required all-null distribution.
Deriving the Shared-Control Dependence
For distinct treatments, the only random quantity shared by \(D_i=\bar X_i-\bar X_0\) and \(D_j=\bar X_j-\bar X_0\) is \(\bar X_0\). Independence across groups therefore gives \[ \operatorname{Cov}(D_i,D_j) =\operatorname{Cov}(\bar X_i-\bar X_0,\bar X_j-\bar X_0) =\operatorname{Var}(\bar X_0) =\Sigma/n_0. \] For a single treatment, the two group means are independent, so \[ \operatorname{Var}(D_i) =\operatorname{Var}(\bar X_i)+\operatorname{Var}(\bar X_0) =(1/n_i+1/n_0)\Sigma=c_i\Sigma. \] Dividing these covariances by \(\sqrt{c_ic_j}\) yields the stated correlation factor \(R_{ij}\). Normality makes the full vector of contrasts jointly Gaussian. Finally, within each normal group, the sample mean is independent of the within-group scatter; independence of the groups then gives independence of the pooled within-group covariance \(S\) from all group means. This establishes the joint calibration representation. \(\square\)
For a single contrast, the exact marginal distribution is the familiar Hotelling distribution: \[ \frac{\nu-p+1}{p\nu}T_i^2\sim F_{p,\nu-p+1} \quad\text{under }H_{0i}, \] when \(\nu\ge p\). This marginal fact does not determine the distribution of the maximum: the dependence between contrasts matters. Let \(q_{1-\alpha}\) be the \(1-\alpha\) quantile of the maximum under the joint centered calibration above. In general, it is computed from that joint distribution, often by multivariate numerical integration or simulation.
Worked Example: Computing the Shared-Control Correlation
There are two treatments and one control, with sample sizes \(n_0=8\), \(n_1=6\), and \(n_2=10\). Find the correlation factor between the standardized treatment-control contrasts.
The variance multipliers are \(c_1=1/6+1/8=7/24\) and \(c_2=1/10+1/8=9/40\). Thus \[ R_{12}=\frac{1/8}{\sqrt{(7/24)(9/40)}} =\frac{1/8}{\sqrt{21/320}} =\sqrt{\frac{5}{21}} \approx 0.488. \] The standardized contrasts therefore have covariance \(R_{12}\Sigma\), not zero. The positive dependence comes from the shared control mean; the treatment sample means themselves are independent.
Simultaneous Regions and Strong Error Control
Use the same joint cutoff for every treatment contrast. The resulting regions are ellipsoids centered at the estimated contrasts, each with its own scale \(c_iS\).
Under the model, the errors \(D_i-\delta_i\) have the centered joint distribution used for calibration, regardless of the actual values of the mean contrasts. Therefore the probability that all \(m\) true contrasts lie in their respective regions is \(1-\alpha\), provided the cutoff is chosen so that \(P(M\le q_{1-\alpha})=1-\alpha\). Testing \(H_{0i}:\delta_i=0\) at the simultaneous cutoff amounts to rejecting it when \(T_i^2>q_{1-\alpha}\).
To prove the claim, let \(I_0\subseteq\{1,\ldots,m\}\) be the set of indices for which \(H_{0i}\) is true. For every \(i\in I_0\), the contrast \(D_i\) is centered at zero. The subvector of centered contrasts indexed by \(I_0\), together with \(S\), has exactly the same distribution as the corresponding subvector in the joint calibration: its covariance factors are the corresponding entries of \(R\), and \(S\) is independent of the means. Thus the probability of any false rejection among true nulls is \[ P\left(\max_{i\in I_0}T_i^2>q_{1-\alpha}\right). \] Couple this centered subvector with a full centered calibration vector \((Z_1,\ldots,Z_m)\) and the same \(S\). In every realization, \[ \max_{i\in I_0}Z_i^{\mathsf T}S^{-1}Z_i \le \max_{1\le i\le m}Z_i^{\mathsf T}S^{-1}Z_i. \] Consequently, \[ P\left(\max_{i\in I_0}T_i^2>q_{1-\alpha}\right) \le P(M>q_{1-\alpha})=\alpha. \] If \(I_0\) is empty, the probability of rejecting a true null is zero, so the conclusion also holds. This proves strong familywise error control. \(\square\)
The subset argument is important: when some null hypotheses are false, their test statistics are generally shifted and the maximum over all \(m\) observed statistics does not have the centered calibration distribution. Strong control follows by considering only the statistics for true nulls and bounding their maximum by the full centered calibration maximum.
Worked Example: Calculating Two Multivariate Statistics
Suppose there are two treatments, each with five observations, and a control group with five observations. The response dimension is \(p=2\). Suppose the pooled covariance estimate and observed contrasts are \[ S=\begin{pmatrix}4&1\\1&2\end{pmatrix}, \qquad D_1=\begin{pmatrix}1\\0.5\end{pmatrix}, \qquad D_2=\begin{pmatrix}0.2\\1\end{pmatrix}. \] For each comparison \(c_i=1/5+1/5=0.4\), and \[ S^{-1}=\frac{1}{7}\begin{pmatrix}2&-1\\-1&4\end{pmatrix}. \]
For treatment 1, \[ D_1^{\mathsf T}S^{-1}D_1 =(1,0.5)\frac{1}{7}\begin{pmatrix}1.5\\1\end{pmatrix} =\frac{2}{7}, \qquad T_1^2=\frac{2/7}{0.4}=\frac{5}{7}\approx0.714. \] For treatment 2, \[ D_2^{\mathsf T}S^{-1}D_2 =(0.2,1)\frac{1}{7}\begin{pmatrix}-0.6\\3.8\end{pmatrix} =\frac{3.68}{7}, \qquad T_2^2=\frac{3.68/7}{0.4}\approx1.314. \] Here \(N=15\), there are three groups, and \(\nu=12\). The simultaneous decision compares both values to the joint cutoff \(q_{1-\alpha}\) for this design, not to two independent marginal cutoffs. These calculations identify the statistics; the joint calibration supplies the decision threshold.
How to Obtain the Cutoff
Because \(R\) depends on the sample sizes, the critical value is design-specific. An exact model-based simulation can approximate it as follows:
Generate jointly Gaussian \(Z_1,\ldots,Z_m\) with cross-covariance \(R_{ij}\Sigma\), or use a convenient common-covariance transformation of a Gaussian vector with correlation matrix \(R\).
Independently generate \(\nu S\sim W_p(\Sigma,\nu)\).
For each simulated draw, calculate \(Z_i^{\mathsf T}S^{-1}Z_i\) for every treatment and retain the largest value.
Estimate \(q_{1-\alpha}\) as the empirical \(1-\alpha\) quantile of the simulated maxima. A numerical quantile has simulation error, which can be reduced by increasing the number of draws.
The cutoff does not depend on the unknown common covariance matrix in its final value: replacing all vectors by \(\Sigma^{-1/2}\)-transformed vectors leaves the statistics and their joint law determined by \(p,\nu\), and \(R\). Thus the calibration can be performed with \(\Sigma=I_p\), the identity matrix.
Worked Example: A Two-Treatment Known-Covariance Calibration
Consider one-dimensional outcomes with known variance \(\sigma^2\), a control sample of size four, and two treatment samples also of size four. Define \(Z_i=(\bar X_i-\bar X_0)/(\sigma\sqrt{1/4+1/4})\) under the null contrasts. Each \(Z_i\) has variance one, and \[ \operatorname{Cov}(Z_1,Z_2) =\frac{1/4}{\sqrt{(1/2)(1/2)}}=\frac12. \] The exact two-sided simultaneous cutoff \(c\) at familywise level \(\alpha=0.05\) is defined by \[ P(-c\le Z_1\le c,\,-c\le Z_2\le c)=0.95, \qquad (Z_1,Z_2)\sim N_2\left( \begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}1&1/2\\1/2&1\end{pmatrix} \right). \] For observed differences \(0.8\sigma\) and \(1.2\sigma\), the absolute standardized values are \(0.8/\sqrt{1/2}\approx1.131\) and \(1.2/\sqrt{1/2}\approx1.697\). The simultaneous intervals are \(D_i\pm c\sigma\sqrt{1/2}\); whether either excludes zero is decided using the same \(c\). This calculation uses the exact correlated-normal calibration rather than treating the two comparisons as independent.
Common Pitfall and Scope
A common mistake is to use the all-null distribution of the maximum over all \(m\) observed statistics as though it remained the distribution of that maximum when some treatment effects are nonzero. False-null statistics are generally noncentral, so that reasoning is incorrect. The strong-control proof instead uses the centered joint distribution of the subset of true null statistics and bounds their maximum by the full centered calibration maximum.
Another tempting shortcut is to calibrate each contrast separately and ignore the shared-control dependence. Marginal Hotelling cutoffs provide individual, not automatically simultaneous, coverage. Bonferroni adjustment can provide a conservative familywise procedure, but it does not use the actual dependence among these comparisons. The Dunnett-type maximum calibration does. Its exact justification here relies on multivariate normality and a common covariance matrix; with heterogeneous group covariances, this pooled-covariance calibration is not exact.
Check Your Understanding
Answer each question using the common-covariance model and definitions in this tutorial.
- Why are two treatment-control contrasts positively dependent even when the treatment groups are independent of one another?
- For \(n_0=10\), \(n_1=5\), and \(n_2=20\), calculate \(c_1\), \(c_2\), and \(R_{12}\).
- What information, in addition to the response dimension and residual degrees of freedom, determines the joint cutoff?
- Explain why the maximum over all observed statistics need not have its centered calibration distribution when some null hypotheses are false.
- In the strong-control proof, why does the maximum over the true-null subset not exceed the maximum over all indices in the centered calibration vector?