Worked MaxT Boundary Calculation with Shared Control

Why Shared Control Changes the Boundary

In a multi-arm trial, each experimental treatment may be compared with the same control group. The resulting test statistics are not independent: a control-group mean that happens to be low makes all treatment-minus-control contrasts look more favorable, while a high control mean moves them in the opposite direction. A maximum-statistic, or MaxT, boundary accounts for this dependence when setting a common rejection threshold.

This tutorial considers one-sided treatment-superiority tests at a single analysis stage. Its target is the probability, under the global null hypothesis, that at least one treatment statistic crosses its stagewise boundary. That probability is a stagewise familywise crossing probability. It is not, by itself, the probability of at least one false rejection over several stages.

Definition: Suppose $K$ treatment arms are compared with a shared control at one analysis stage. Let $Z_i$ be the standardized statistic for treatment $i$ versus control, with larger values favoring treatment $i$. A one-sided MaxT rule rejects $H_{0i}$ if $Z_i>c$, where the common critical value $c$ is calibrated so that $$ P\!\left(\max_{1\leq i\leq K} Z_i>c\mid H_{01},\ldots,H_{0K}\right)=\alpha. $$ The probability on the left is the global-null stagewise familywise crossing probability.

Model and Shared-Control Covariance

Assume independent observations across participants, independent treatment groups, and a common known outcome variance $\sigma^2$. At this stage, the sample means are normally distributed. Under the global null, each treatment has the same mean as control. Write $\bar{Y}_i$ for the mean in treatment arm $i$, $\bar{Y}_0$ for the control mean, and $n_i$ and $n_0$ for their sample sizes. The standardized treatment-control statistic is $$ Z_i=\frac{\bar{Y}_i-\bar{Y}_0}{\sigma\sqrt{1/n_i+1/n_0}}. $$ Under the global null, each $Z_i$ has a standard normal distribution. For two distinct treatment arms $i$ and $j$, their contrast covariance is $$ \operatorname{Cov}(\bar{Y}_i-\bar{Y}_0,\bar{Y}_j-\bar{Y}_0) =\operatorname{Var}(\bar{Y}_0) =\frac{\sigma^2}{n_0}. $$ The treatment-arm means are independent of each other and of the control mean. The shared control term is therefore the only source of covariance between the two contrasts. Dividing by their standard deviations gives $$ \operatorname{Corr}(Z_i,Z_j) = \frac{1/n_0}{\sqrt{(1/n_i+1/n_0)(1/n_j+1/n_0)}}. $$ This correlation is positive. With equal group sizes, it is $1/2$, regardless of the common sample size.

These expressions give a multivariate normal model for the vector of test statistics under the global null: each component has mean zero and variance one, and the off-diagonal correlations are determined by the allocation. A MaxT calculation uses this joint distribution, not just the individual standard normal distributions. In practice, the variance is often estimated rather than known; then the joint statistic may not follow this exact normal model. The calculation below should not be applied unchanged to such a design.

Worked Example: Deriving the Correlation for Equal Allocation

Suppose two experimental arms and one control each have 100 participants, with a common known outcome standard deviation of 10 units. Under the global null, calculate the correlation between the two standardized treatment-control statistics.

For either contrast, the variance of the difference in sample means is $$ \sigma^2\left(\frac{1}{100}+\frac{1}{100}\right) =100(0.02)=2. $$ The covariance of the two contrasts comes from their shared control mean: $$ \operatorname{Cov}(\bar{Y}_1-\bar{Y}_0,\bar{Y}_2-\bar{Y}_0) =\frac{100}{100}=1. $$ Thus the correlation of the standardized statistics is $$ \operatorname{Corr}(Z_1,Z_2)=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}=0.500. $$ The standard deviation of 10 units and the sample sizes determine the contrast variances, but cancel out in the correlation. The two statistics are positively correlated, not independent.

Calibrating a Common MaxT Critical Value

For a specified covariance matrix, choose $c$ so that the probability all $K$ statistics remain at or below $c$ is $1-\alpha$. Equivalently, the probability their maximum exceeds $c$ is $\alpha$. For most nontrivial correlation matrices, this probability has no simple closed form and is evaluated by multivariate-normal integration or a sufficiently accurate numerical algorithm.

The calibration has a simple justification. As $c$ increases, the event that all statistics are at most $c$ can only become more likely. Thus its probability is nondecreasing in $c$, and the upper-tail crossing probability is nonincreasing. For a nonsingular multivariate normal distribution, the crossing probability varies continuously from 1 toward 0 as $c$ moves from very small to very large values. A numerical root-finder can therefore locate a value achieving the target $\alpha$, to the chosen numerical tolerance.

For two equally allocated treatment arms, the correlation is $1/2$. A useful representation is $$ Z_1=\frac{U+E_1}{\sqrt{2}}, \qquad Z_2=\frac{U+E_2}{\sqrt{2}}, $$ where $U,E_1,E_2$ are independent standard normal variables. The common component $U$ represents the shared-control contribution. Conditioning on $U=u$ makes the two statistics independent, so $$ P(Z_1\leq c,Z_2\leq c) =\int_{-\infty}^{\infty}\phi(u)\Phi(\sqrt{2}c-u)^2\,du, $$ where $\phi$ and $\Phi$ are the standard normal density and distribution functions. The critical value solves this integral equal to $1-\alpha$. This conditioning argument also illustrates why treating the two statistics as independent would use the wrong crossing probability.

Worked Example: Calibrating the Two-Arm Boundary

Continue with two equally allocated experimental arms sharing one control. Find the one-sided MaxT critical value for a global-null stagewise familywise crossing probability of $\alpha=0.05$.

The correlation matrix is $$ \begin{pmatrix} 1 & 0.5\\ 0.5 & 1 \end{pmatrix}. $$ Using the conditional-normal integral, solve $$ \int_{-\infty}^{\infty}\phi(u)\Phi(\sqrt{2}c-u)^2\,du=0.95. $$ Numerical integration and root-finding give $c\approx1.913$ (reported to three significant figures). The resulting stagewise rule is to reject either treatment’s one-sided null when its statistic exceeds 1.913. By construction, under the stated global-null model, the probability that at least one of the two statistics exceeds this threshold is approximately 0.0500.

For comparison, using the ordinary one-sided 5% standard-normal cutoff for each arm would give $c\approx1.645$. That cutoff controls each individual test at 5%, but it does not make the probability of at least one crossing equal to 5%. The MaxT cutoff is larger because it is calibrated to the joint event across both treatment comparisons.

Checking a Candidate Boundary and Interpreting Results

The same joint distribution can evaluate a proposed threshold rather than solve for one. For two arms with correlation $\rho$, the global-null stagewise crossing probability is $$ P(Z_1>c\ \text{or}\ Z_2>c) =1-P(Z_1\leq c,Z_2\leq c). $$ Equivalently, it is the sum of the two marginal upper-tail probabilities minus the probability that both statistics exceed $c$. Subtracting the joint exceedance avoids counting outcomes in which both treatments cross twice.

A boundary check should distinguish the design-level probability from the observed analysis. The design-level calculation asks how often the rule would produce one or more crossings under repeated trials generated by the null model. In an actual trial, comparing the observed maximum statistic with $c$ determines whether a crossing occurred; it does not estimate the design’s error rate.

Worked Example: Evaluating a Proposed Boundary

For the two-arm shared-control model with correlation $0.5$, suppose a designer proposes $c=1.96$. Calculate the stagewise global-null crossing probability and assess whether this boundary achieves 0.05.

The probability of no crossing is the bivariate-normal probability $$ P(Z_1\leq1.96,Z_2\leq1.96) =\int_{-\infty}^{\infty}\phi(u)\Phi(\sqrt{2}(1.96)-u)^2\,du. $$ Numerical evaluation gives a no-crossing probability of approximately $0.9547$, so $$ P(\max(Z_1,Z_2)>1.96)=1-0.9547=0.0453. $$ The proposed boundary is more conservative than a 0.05 stagewise target: its crossing probability is about 0.0453, not 0.0500. The difference is a consequence of the boundary being higher than the calibrated value, not evidence that the observed trial has a 4.53% probability of a false conclusion.

Now suppose the observed statistics are $Z_1=2.05$ and $Z_2=1.84$. At the calibrated boundary $c\approx1.913$, the first statistic crosses and the second does not. This identifies a stagewise rejection under the specified rule. The global-null crossing probability remains a repeated-sampling property of the rule and covariance model; it is not recalculated from these two observed values.

Unequal Allocation, Diagnostics, and Scope

Unequal allocations generally produce unequal correlations. For example, with $n_0=100$, $n_1=80$, and $n_2=50$, the contrast variances apart from the common factor $\sigma^2$ are $1/80+1/100=0.0225$ and $1/50+1/100=0.0300$. The covariance apart from $\sigma^2$ is $1/100=0.0100$, giving $$ \operatorname{Corr}(Z_1,Z_2) =\frac{0.0100}{\sqrt{0.0225(0.0300)}}\approx0.385. $$ The correct MaxT calculation uses this correlation rather than assuming the equal-allocation value of 0.5. With more arms, compute every off-diagonal correlation using the allocation formula, assemble the full matrix, and evaluate the multivariate-normal probability for the maximum.

A practical calibration check is to recompute the crossing probability at the reported boundary using an independent probability routine or a simulation under the same covariance matrix. Simulation is a diagnostic, not a substitute for checking the inputs: a finite simulation can fluctuate around the target, especially when the target probability is small. Also check that the covariance matrix is symmetric, has ones on the diagonal, and is positive semidefinite. A matrix that fails these checks may reflect an incorrect allocation input, inconsistent covariance formula, or rounding error.

Several common failures have recognizable symptoms:

Ignoring the shared control. Treating the statistics as independent understates their positive dependence and changes the maximum’s distribution. Use the covariance induced by the shared control.
Using the wrong tail. For a superiority test in which larger statistics favor treatment, the event is $Z_i>c$. A lower-tail or two-sided calculation answers a different question.
Using a marginal cutoff for a familywise target. A 5% cutoff for each comparison does not generally give a 5% probability of any crossing. Calibrate the joint maximum.
Confusing a stage with a sequence of looks. This calculation controls the crossing probability at the specified stage under its model. Combining crossings over multiple stages requires their joint dependence over time and an appropriate sequential calibration.

MaxT calibration uses the specified dependence structure and can be less conservative than assigning each comparison a threshold that ignores dependence. Its validity is limited by the model: incorrect covariance inputs, non-normal small-sample statistics, variance estimation, or unaccounted design changes can invalidate the numerical calibration. In addition, calibration under the global null directly establishes weak familywise control. Strong control under configurations where only some treatment nulls are true requires an additional justification, such as the subset pivotality property: the joint distribution of the statistics for true null hypotheses must not depend on which other hypotheses are false. Under that condition, the probability that any true-null statistic crosses is bounded by the global-null maximum probability, since the true-null statistics form a subset of the full set.

Key takeaway: A shared control induces positive correlation among treatment-control statistics. A stagewise MaxT threshold is obtained by calibrating the maximum under the corresponding joint null distribution; the covariance matrix, tail direction, analysis stage, and target error probability must all match the design.

Check Your Understanding

Answer each question, then compare with the concise solutions.

Two treatment arms and a control each have equal sample sizes. What is the correlation between the standardized treatment-control statistics under the common-variance normal model, and why is it positive?
For two arms with correlation $0.5$, describe the equation that determines a one-sided MaxT critical value for stagewise familywise error 0.05.
Why does using the one-sided 5% standard-normal cutoff separately for every arm not generally give a 5% stagewise familywise crossing probability?
With $n_0=100$, $n_1=80$, and $n_2=50$, calculate the correlation between the two treatment-control statistics.
Does calibrating a boundary under the global null alone automatically establish strong familywise error control? State a condition that can justify the extension.

Solutions. (1) The correlation is $1/2$; both contrasts contain the same control mean with a negative sign, so their covariance is the positive variance of that shared mean. (2) Solve $\int_{-\infty}^{\infty}\phi(u)\Phi(\sqrt{2}c-u)^2\,du=0.95$, giving $c\approx1.913$. (3) The familywise event is that at least one statistic crosses, a joint event whose probability is not the individual 0.05 tail probability. (4) The correlation is $0.0100/\sqrt{0.0225(0.0300)}\approx0.385$. (5) No. Strong control can follow when subset pivotality holds, so the joint distribution of the true-null statistics is unchanged by the status of other hypotheses.