The Canonical Joint Normal Model

Why a Joint Model Is Needed

At each scheduled interim look, an information-standardized statistic has a marginal distribution. But a sequence of such statistics is not a collection of unrelated results: earlier observations contribute to later cumulative statistics. To describe the probability of a whole monitoring path, we need a joint distribution that accounts for both the mean at every look and the dependence between looks.

The canonical joint normal model supplies this description. It represents the statistics at the scheduled looks by a multivariate normal vector whose mean changes with information and whose covariance is determined by the information times. The model gives a common probability framework for studying paths under the null or a fixed alternative; it does not, by itself, specify a stopping boundary or a testing procedure.

Definition: Suppose there are $K$ scheduled looks, with strictly increasing cumulative information $0<I_1<\cdots<I_K$. Let $Z_k$ be the signed, information-standardized statistic at look $k$, and let $\delta$ be the fixed effect on the scale for which the statistic has unit information-based variance. The canonical joint normal model is $$ (Z_1,\ldots,Z_K)^\mathsf{T}\sim N_K(\boldsymbol{\mu},\boldsymbol{\Sigma}), \qquad \mu_k=\delta\sqrt{I_k}, \qquad \Sigma_{ij}=\sqrt{\frac{I_i}{I_j}}\ \text{if }i\leq j, $$ with $\Sigma_{ij}=\Sigma_{ji}$ if $i>j$. Equivalently, using information fractions $t_k=I_k/I_K$ and $\theta=\delta\sqrt{I_K}$, the mean is $\mu_k=\theta\sqrt{t_k}$ and the covariance entries are $\Sigma_{ij}=\sqrt{t_i/t_j}$ for $i\leq j$.

Here $N_K$ denotes a $K$-dimensional normal distribution, $\boldsymbol{\mu}$ is the vector of means, and $\boldsymbol{\Sigma}$ is the covariance matrix. Each $Z_k$ has variance one, so its marginal distribution is $N(\delta\sqrt{I_k},1)$. The covariance matrix has diagonal entries one. Because every $Z_k$ is standardized, the off-diagonal covariance entries are also correlations.

The information fractions satisfy $0<t_1<\cdots<t_K=1$. The parameter $\theta$ is the signed effect measured in standard-error units at the planned final information. Under the null hypothesis of no effect, $\delta=0$ and $\theta=0$; under a fixed alternative, the means generally increase in magnitude with information. The sign convention matters: a positive $\delta$ shifts the statistics upward, while a negative $\delta$ shifts them downward.

Assumptions and Interpretation

The model is used when the scheduled looks have known or prespecified cumulative information and the standardized statistics are jointly normal, exactly or as a suitable approximation. Its canonical form assumes the same underlying effect parameter applies across looks and that the covariance structure is represented by the information ordering above. The statistic must use a consistent direction and standardization at each look. In applications where the model is only approximate, normality and the canonical covariance structure need to be reasonable for the study design and analysis.

A multivariate normal model is more than a list of marginal distributions. If each $Z_k$ is individually standard normal under the null, that alone does not specify the probability of a particular sequence of values. The covariance matrix is essential for joint probabilities, such as the chance of crossing a boundary at one or more looks. In particular, treating the looks as independent would replace the off-diagonal entries with zero and would generally describe a different monitoring process.

The canonical covariance entries are positive and increase as two looks become closer in information. The model makes no claim that information is equally spaced in calendar time. Two looks a month apart can have very different information increments from two other looks a month apart; the joint model is indexed by $I_k$ or $t_k$, not by elapsed time alone.

Worked Example: Constructing a Three-Look Model

Consider a hypothetical design with information fractions $t_1=0.25$, $t_2=0.60$, and $t_3=1$, and suppose the fixed standardized effect at final information is $\theta=0.80$. Construct the mean vector and covariance matrix.

The mean at look $k$ is $\mu_k=\theta\sqrt{t_k}$. Thus, $$ \boldsymbol{\mu} = \begin{pmatrix} 0.80\sqrt{0.25}\\ 0.80\sqrt{0.60}\\ 0.80\sqrt{1} \end{pmatrix} = \begin{pmatrix} 0.4000\\ 0.6197\\ 0.8000 \end{pmatrix}. $$

The diagonal entries of $\boldsymbol{\Sigma}$ are one. The off-diagonal entries are $$ \Sigma_{12}=\sqrt{\frac{0.25}{0.60}}=0.6455,\qquad \Sigma_{13}=\sqrt{\frac{0.25}{1}}=0.5000,\qquad \Sigma_{23}=\sqrt{\frac{0.60}{1}}=0.7746. $$ Therefore, $$ \boldsymbol{\Sigma} = \begin{pmatrix} 1 & 0.6455 & 0.5000\\ 0.6455 & 1 & 0.7746\\ 0.5000 & 0.7746 & 1 \end{pmatrix}. $$

Under this model, $Z_2$ has marginal distribution $N(0.6197,1)$, and $Z_1$ and $Z_3$ have correlation $0.5000$. The reported means and correlations are model parameters, not observed results from a real study. The joint model also asserts that all three statistics together have a trivariate normal distribution.

Why the Mean Has This Form

The mean vector follows from the way an information-standardized statistic scales. On the canonical effect scale, the expected cumulative signal at information $I_k$ is $\delta I_k$, while the standard deviation used to standardize it is $\sqrt{I_k}$. Dividing the expected signal by that standard deviation gives $$ \frac{\delta I_k}{\sqrt{I_k}}=\delta\sqrt{I_k}. $$ Thus, standardization makes the variance one, but the mean under a fixed effect still grows with the square root of information. At final information $I_K$, define $\theta=\delta\sqrt{I_K}$. Since $t_k=I_k/I_K$, it follows that $$ \delta\sqrt{I_k} =\delta\sqrt{I_Kt_k} =\theta\sqrt{t_k}. $$ This establishes the stated mean-vector parameterization. The canonical covariance formula specifies how the standardized cumulative statistics co-vary across looks; its detailed connection to shared data is a separate derivation.

This scaling also gives a useful check: at the final look, $t_K=1$, so the final mean is $\theta$. At an earlier look, the mean is $\theta\sqrt{t_k}$, not $\theta t_k$. Information accrues linearly, whereas the standardized mean accrues with its square root.

Worked Example: Null and Alternative Mean Vectors

Suppose a hypothetical two-look design has $t_1=0.36$ and $t_2=1$. Compare the mean vectors under the null and under a fixed alternative with $\theta=1.20$.

Under $H_0$, $\theta=0$, so $$ \boldsymbol{\mu}_0 = \begin{pmatrix} 0\sqrt{0.36}\\ 0\sqrt{1} \end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}. $$

Under the fixed alternative, $$ \boldsymbol{\mu}_a = \begin{pmatrix} 1.20\sqrt{0.36}\\ 1.20\sqrt{1} \end{pmatrix} = \begin{pmatrix} 0.7200\\ 1.200 \end{pmatrix}. $$

The covariance matrix is the same under these two parameter values in the canonical model: $$ \boldsymbol{\Sigma} = \begin{pmatrix} 1 & \sqrt{0.36/1}\\ \sqrt{0.36/1} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0.6000\\ 0.6000 & 1 \end{pmatrix}. $$

For example, the marginal probability under the alternative that the first statistic exceeds 1.96 is $$ P(Z_1>1.96)=1-\Phi(1.96-0.7200) =1-\Phi(1.240)\approx 0.1075, $$ where $\Phi$ is the standard normal cumulative distribution function. This is a one-look probability under the specified alternative. It is not the probability of crossing at either look, which depends on the joint distribution and both looks.

Using the Joint Model for a Path Quantity

Once the mean and covariance are specified, linear combinations of the statistics can be evaluated using the normal model. If $\boldsymbol{a}$ is a fixed vector, then $\boldsymbol{a}^\mathsf{T}\boldsymbol{Z}$ is normally distributed with mean $\boldsymbol{a}^\mathsf{T}\boldsymbol{\mu}$ and variance $\boldsymbol{a}^\mathsf{T}\boldsymbol{\Sigma}\boldsymbol{a}$. This follows from the defining linear-transformation property of a multivariate normal vector.

Worked Example: A Difference Between Two Looks

Return to the hypothetical three-look model with $t_1=0.25$, $t_3=1$, and $\theta=0.80$. Find the distribution of $Z_3-Z_1$, then calculate the probability that this difference exceeds 1.

Its mean is the difference of the marginal means: $$ E(Z_3-Z_1)=0.8000-0.4000=0.4000. $$

Using $\operatorname{Var}(Z_1)=\operatorname{Var}(Z_3)=1$ and $\operatorname{Cov}(Z_1,Z_3)=0.5000$, $$ \operatorname{Var}(Z_3-Z_1) =1+1-2(0.5000)=1.000. $$ Hence $Z_3-Z_1\sim N(0.4000,1.000)$, where the second argument is the variance. The requested probability is $$ P(Z_3-Z_1>1) =1-\Phi\left(\frac{1-0.4000}{\sqrt{1.000}}\right) =1-\Phi(0.6000) \approx 0.2743. $$

In this hypothetical model, there is about a 27.4% probability that the final statistic exceeds the first-look statistic by more than one standardized unit. The calculation uses the covariance between looks; treating the two statistics as independent would give a different variance and answer.

Checks, Alternatives, and Limitations

Before using the canonical model, check that the information times are strictly increasing, that the final fraction equals one, and that all statistics use the same sign convention and standardization. The covariance matrix should be symmetric, have ones on its diagonal, and have entries between zero and one for distinct ordered looks. For the two-look case, its determinant is $1-t_1$, which is positive when $0<t_1<1$. More generally, the prescribed information fractions generate a valid covariance structure under the canonical model.

Diagnostics focus on whether the assumptions are credible rather than on a single observed path. Review the analysis plan for fixed scheduled looks and a consistent statistic; check that information fractions correspond to the intended cumulative information; and investigate departures from the assumed normal approximation, effect stability, or information calculation. If the statistic, estimand, or analysis changes across looks, the canonical parameterization may not adequately represent the sequence. A numerical covariance matrix with invalid entries or inconsistent ordering is a direct implementation warning.

The main alternatives are useful for different purposes, but are not interchangeable with the joint model:

Separate marginal normal models can describe each look individually, but cannot determine probabilities involving more than one look without a dependence specification.
An independence assumption is simpler to compute with, but contradicts the canonical covariance entries for nested looks and can misstate path probabilities.
Simulation can estimate probabilities for complicated monitoring rules. It must still generate statistics from a dependence model appropriate to the design; simulation does not repair an incorrect joint model.

The canonical model is a statistical representation, not a guarantee that data follow it exactly. It is most directly suited to a fixed effect and a prespecified information schedule. It does not explain how to estimate covariance from a particular data structure, how to derive transition distributions, or how to choose monitoring boundaries. Those require additional design and inferential considerations. Under substantial model departures, calculated joint probabilities may be inaccurate even when each marginal approximation appears reasonable.

Common pitfall: “Each look has a standard normal statistic under $H_0$, so the statistics are independent.” This reasoning confuses marginal distributions with a joint distribution. Under the canonical null model, each marginal mean is zero and variance is one, but $\operatorname{Cov}(Z_i,Z_j)=\sqrt{I_i/I_j}$ for $i<j$, which is positive. Use the full covariance matrix whenever the question concerns multiple looks or a monitoring path.

Key takeaway: The canonical model specifies a multivariate normal vector with mean $\theta\sqrt{t_k}$ at look $k$ and covariance $\sqrt{t_i/t_j}$ for $i\leq j$. The means describe the fixed-effect drift; the covariance describes dependence across information times.

Check Your Understanding

Use the canonical model to answer each question. Information fractions are relative to final information.

For $t=(0.49,1)$ and $\theta=1.00$, find the mean vector and covariance matrix.
Under $H_0$, what are the marginal mean and variance of each $Z_k$? Does that make the statistics independent?
For $t_1=0.25$ and $t_2=1$, calculate $\operatorname{Var}(Z_2-Z_1)$.
With $t_1=0.25$, $t_2=1$, and $\theta=0.80$, calculate $P(Z_1>1.96)$ approximately.

Solutions. (1) The mean vector is $(1.00\sqrt{0.49},1.00\sqrt{1})^\mathsf{T}=(0.7000,1.000)^\mathsf{T}$. The covariance matrix is $\begin{pmatrix}1&0.7000\\0.7000&1\end{pmatrix}$. (2) Each marginal mean is zero and each variance is one. They are not independent under this joint normal model when there are distinct looks, because their covariance is positive. (3) The covariance is $\sqrt{0.25/1}=0.5000$, so the variance is $1+1-2(0.5000)=1.000$. (4) The first-look mean is $0.80\sqrt{0.25}=0.4000$, so $P(Z_1>1.96)=1-\Phi(1.56)\approx 0.0594$.