Canonical Joint-Normal Statistics Across Arms and Stages

Why a Joint Model Is Needed

In a multi-arm, multi-stage trial, each planned look produces a statistic for each active treatment-control contrast. These statistics are not generally independent: repeated looks use accumulating data, and contrasts may share data. A useful representation puts all of them into one joint probability model. It describes their means and dependencies together, rather than treating each arm and look as if it were an isolated test.

This matters even before a boundary is chosen. If an interim rule selects an arm because its observed statistic is large, the statistic for the selected arm is no longer distributed like an unselected standard normal statistic. The joint model describes the statistics before selection and therefore provides the probability structure needed to reason about what selection has done.

Definition: Let $\theta_j$ be the true treatment-control contrast for arm $j$, with positive values defined as favorable to treatment. At planned look $k$, let $\widehat{\theta}_{jk}$ estimate that contrast and let $I_{jk}$ be its information, conventionally the reciprocal of its variance. The standardized statistic is $$ Z_{jk}=\frac{\widehat{\theta}_{jk}}{1/\sqrt{I_{jk}}} =\widehat{\theta}_{jk}\sqrt{I_{jk}}. $$ In the canonical model, the vector of all planned $Z_{jk}$ values is multivariate normal, has unit marginal variances, and has means and covariances determined by the treatment effects, information, and dependence between score increments.

Assumptions and the Canonical Form

The model is exact when the relevant contrast estimates are jointly normal with known variances and the assumed score-increment covariance. It is often used as a large-sample approximation when the estimates are not exactly normal or variances are estimated. The planned looks, contrast definitions, and information values must be specified in a way that matches the trial. For the canonical information-time representation, assume that information accumulates and that score increments from disjoint information intervals are independent. Assume also that the treatment effects remain fixed across looks.

Write the standardized score as $Z_{jk}=U_j(I_{jk})/\sqrt{I_{jk}}$, where $U_j(I_{jk})$ is the cumulative score for arm $j$ at its information level. Its mean is $\theta_j I_{jk}$. Under the canonical model, the cumulative scores have covariance $\operatorname{Cov}\{U_j(s),U_l(t)\}=\rho_{jl}\min(s,t)$. Here $\rho_{jl}$ is the correlation between the score increments for arms $j$ and $l$ over a common information interval; $\rho_{jj}=1$. The matrix of these increment correlations must be positive semidefinite. In an actual design, the appropriate cross-arm correlations come from the joint contrast construction and cannot be chosen arbitrarily.

\operatorname{E}(Z_{jk})=\theta_j\sqrt{I_{jk}}, \qquad \operatorname{Var}(Z_{jk})=1, \qquad \operatorname{Cov}(Z_{jk},Z_{lm}) = \rho_{jl} \frac{\min(I_{jk},I_{lm})}{\sqrt{I_{jk}I_{lm}}}.

The covariance expression applies when the score processes share the stated information-time structure. When all arms have the same information $I_k$ at look $k$, it simplifies to $\rho_{jl}\sqrt{I_{\min(k,m)}/I_{\max(k,m)}}$. Setting $j=l$ gives the within-arm correlation between repeated looks. Setting $k=m$ gives the cross-arm correlation at the same information level. Thus the joint covariance matrix has unit diagonal entries but need not have zero off-diagonal entries.

Worked Example: One Arm at Three Looks

Suppose one hypothetical treatment-control contrast has true effect $\theta_1=0.20$ outcome units, with planned information values $I_{11}=40$, $I_{12}=90$, and $I_{13}=160$ in inverse outcome-units squared. Assume the canonical model and information accumulation assumptions hold.

The mean at each look is $\theta_1\sqrt{I_{1k}}$. Therefore the mean vector is $(0.20\sqrt{40},0.20\sqrt{90},0.20\sqrt{160})=(1.265,1.897,2.530)$, to three decimal places. Each statistic has variance 1.

For looks 1 and 2, the correlation is $\min(40,90)/\sqrt{40\cdot90}=40/60=0.6667$. For looks 1 and 3 it is $40/\sqrt{40\cdot160}=40/80=0.5000$; for looks 2 and 3 it is $90/\sqrt{90\cdot160}=90/120=0.7500$. Hence the covariance matrix is

\begin{pmatrix} 1 & 0.6667 & 0.5000\\ 0.6667 & 1 & 0.7500\\ 0.5000 & 0.7500 & 1 \end{pmatrix}.

The later statistic is more precise, but it is not an independent replication of the earlier statistic: it incorporates the earlier information. The increasing means here reflect the hypothetical positive effect and the increasing information, not a change in the underlying effect.

Why the Covariance Has This Form

The covariance formula follows by standardizing the cumulative scores. Under the assumed information-time model, scores accumulated up to two information levels share the increments up to the smaller level. Their covariance is therefore $\rho_{jl}\min(I_{jk},I_{lm})$. Dividing by the standard deviations $\sqrt{I_{jk}}$ and $\sqrt{I_{lm}}$ gives the covariance of the standardized statistics:

\operatorname{Cov}(Z_{jk},Z_{lm}) = \operatorname{Cov}\left( \frac{U_j(I_{jk})}{\sqrt{I_{jk}}}, \frac{U_l(I_{lm})}{\sqrt{I_{lm}}} \right) = \frac{\rho_{jl}\min(I_{jk},I_{lm})} {\sqrt{I_{jk}}\sqrt{I_{lm}}}.

For one arm, take $j=l$, so $\rho_{jj}=1$. If $I_{jk}\le I_{jm}$, the covariance reduces to $I_{jk}/\sqrt{I_{jk}I_{jm}}=\sqrt{I_{jk}/I_{jm}}$. Since both variances equal 1, this covariance is also their correlation. The result is the familiar square-root information-fraction correlation for repeated looks.

Worked Example: Two Arms at Two Looks

Consider two hypothetical arms with common information $I_1=50$ and $I_2=100$. Let their true contrasts be $\theta_1=0.15$ and $\theta_2=0.10$ outcome units. Suppose the score-increment correlation between arms is $\rho_{12}=0.30$, and assume the canonical joint model holds. Order the statistics as $(Z_{11},Z_{12},Z_{21},Z_{22})$.

The mean vector is $(0.15\sqrt{50},0.15\sqrt{100},0.10\sqrt{50},0.10\sqrt{100})=(1.061,1.500,0.7071,1.000)$. Each variance is 1. The within-arm correlation across looks is $\sqrt{50/100}=0.7071$. At the same look, the cross-arm correlation is $0.30$. Across arms and looks, the correlation between $Z_{11}$ and $Z_{22}$, for example, is $0.30\sqrt{50/100}=0.2121$.

Applying the same formula to each pair gives

\operatorname{Cov} \begin{pmatrix} Z_{11}\\Z_{12}\\Z_{21}\\Z_{22} \end{pmatrix} = \begin{pmatrix} 1 & 0.7071 & 0.3000 & 0.2121\\ 0.7071 & 1 & 0.2121 & 0.3000\\ 0.3000 & 0.2121 & 1 & 0.7071\\ 0.2121 & 0.3000 & 0.7071 & 1 \end{pmatrix}.

For example, $Z_{12}$ and $Z_{21}$ have covariance $0.30\min(100,50)/\sqrt{100\cdot50}=0.2121$. The correlation $0.30$ is an assumption supplied by the joint score structure in this example, not a universal value for multi-arm trials.

Selection Changes the Distribution You Observe

The joint model describes the full vector before applying an interim selection rule. A selected statistic is a function of that vector, so its distribution generally differs from its marginal distribution. For instance, selecting the arm with the largest interim statistic selects partly on random variation as well as on treatment effect. The marginal fact $Z_{jk}\sim N(\theta_j\sqrt{I_{jk}},1)$ does not imply that the statistic for the selected arm has that same distribution conditional on selection.

Worked Example: Selecting the Larger of Two Null Statistics

Suppose two hypothetical arm statistics at one look are independent standard normal variables under the global null. A rule selects the arm with the larger statistic. For either arm considered separately, the probability of exceeding 1.645 is approximately 0.0500. But the selected statistic is the maximum of the two.

The selected maximum is at most 1.645 exactly when both statistics are at most 1.645. Independence gives $P(\max(Z_{11},Z_{21})\le1.645)=0.95^2=0.9025$. Therefore $P(\max(Z_{11},Z_{21})>1.645)=1-0.9025=0.0975$.

Thus a threshold with a 5% marginal exceedance probability for each statistic has a 9.75% probability of being exceeded by at least one in this example. This calculation illustrates why post-selection reasoning must account for the joint distribution. It is not a general boundary-calibration rule: if the statistics are dependent, the joint probability changes.

Checking the Model and Knowing Its Limits

A covariance matrix proposed for the full vector must be symmetric, have ones on the diagonal for standardized statistics, and be positive semidefinite. These are necessary consistency checks: a matrix that fails them cannot be a covariance matrix. The information values should also be increasing across planned looks for a given accumulating contrast, and the implied means should use the same effect scale and information units throughout.

The canonical structure can be inadequate if information does not accumulate as assumed, if score increments are dependent across disjoint intervals, if the treatment effect changes over time, or if the model for the outcomes produces materially non-normal statistics. Unplanned changes to sample size or analysis timing can also make a fixed planned-information representation inappropriate. A good diagnostic checks whether the information and covariance assumptions reflect the actual design and whether simulated or observed joint behavior is compatible with the assumed model. Covariance estimates from limited data can be unstable; a plausible-looking individual correlation is not enough to establish that the entire matrix is valid.

An alternative is to work directly with the vector of contrast estimates and its covariance matrix, rather than standardizing every component. This can be useful when the variances differ or the contrast structure is complicated, but the units and scaling remain visible and calculations may be less compact. Treating armwise statistics as independent is simpler, but it is justified only when the design truly gives zero cross-arm covariance; it does not remove the dependence across repeated looks. The canonical representation is concise when its assumptions hold, while a direct covariance model is preferable when they do not.

The model specifies probability distributions; by itself, it does not choose a stopping rule or establish an error rate. Nor does it guarantee that a normal approximation is accurate in a small sample. Its value is that it makes dependence explicit, so subsequent analyses can use a coherent joint distribution instead of silently ignoring shared information or selection.

Key takeaway: Under the canonical joint-normal model, every planned standardized contrast has variance 1, its mean is the true contrast multiplied by the square root of its information, and every off-diagonal covariance must reflect the shared score increments and the specified cross-arm dependence.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

For one arm with information 25 at the first look and 100 at the second, what is the canonical correlation between the two standardized statistics?
If $\theta_j=0.12$ outcome units and $I_{jk}=64$ inverse outcome-units squared, what is the mean of $Z_{jk}$?
With common information 40 and 90 for two arms and score-increment correlation $0.25$, what is the cross-arm, cross-look covariance?
Why is a statistic selected as the largest of several interim statistics generally not distributed as a single marginal standard normal statistic?
Name two checks on a proposed covariance matrix and one circumstance that can undermine the canonical information-time model.

Solutions. (1) $\sqrt{25/100}=0.5000$. (2) $0.12\sqrt{64}=0.12(8)=0.9600$. (3) $0.25\min(40,90)/\sqrt{40\cdot90}=0.25(40/60)=0.1667$. (4) Selection depends on the realized values; conditioning on being the largest changes the distribution. (5) It must be symmetric and positive semidefinite, with unit diagonal for standardized statistics; for example, dependent increments across disjoint information intervals can violate the canonical model.