Noninteger Information Ratios

Why Noninteger Ratios Need No Special Case

Group sequential calculations are often illustrated with information times that are simple multiples of a base unit. Actual analysis times need not have that form. An interim analysis might occur at information $t_1=1.3$, followed by another at $t_2=2.8$; the ratio $t_2/t_1$ is not an integer. The canonical joint distribution still applies directly. Its formulas use the information times themselves and their ratios, not a count of equal-sized blocks.

Here $t_i$ denotes the cumulative information at look $i$, expressed on a common scale. Assume $0<t_1<\cdots<t_K$, and let $Z_i$ be the standardized statistic at look $i$. Under the canonical model with drift parameter $\theta$, the mean of $Z_i$ is $\theta\sqrt{t_i}$; under the null, $\theta=0$. The model treats the vector of statistics as jointly normal, with covariance determined by the ordering and spacing of the information times.

Canonical joint model: For positive, strictly increasing information times $t_1,\ldots,t_K$, the statistics are jointly normal with $$ E_\theta(Z_i)=\theta\sqrt{t_i}, \qquad \operatorname{Cov}_\theta(Z_i,Z_j)=\sqrt{\frac{t_i}{t_j}}\quad (i\le j). $$ In particular, $\operatorname{Var}(Z_i)=1$, and for $i<j$, the correlation between $Z_i$ and $Z_j$ is $\sqrt{t_i/t_j}$. The times may be arbitrary positive real numbers; they do not have to be integer multiples of a reference amount.

The convention in this formula is that $t_i$ is the earlier time when $i<j$. Equivalently, the covariance is $\sqrt{\min(t_i,t_j)/\max(t_i,t_j)}$. A larger gap in information generally means a smaller correlation between the cumulative standardized statistics, but it does not make the statistics independent.

Transition Formula at Arbitrary Information Times

For two looks with $t_i<t_j$, the conditional distribution at the later look is also available in closed form. It is useful for calculating conditional probabilities or describing how a current statistic relates to a future one under a specified drift.

Conditional transition: Under the canonical joint model, $$ Z_j\mid Z_i=z_i \sim N\left( \sqrt{\frac{t_i}{t_j}}z_i+ \theta\frac{t_j-t_i}{\sqrt{t_j}}, \;1-\frac{t_i}{t_j} \right). $$ The second argument of $N$ is the variance. The conditional standard deviation is therefore $\sqrt{1-t_i/t_j}$.

This follows from the conditional distribution formula for a jointly normal pair. Since both standardized statistics have variance one and their covariance is $r=\sqrt{t_i/t_j}$, the conditional variance of $Z_j$ given $Z_i$ is $1-r^2=1-t_i/t_j$. The conditional mean is the later mean plus $r$ times the deviation of the observed earlier statistic from its mean:

\begin{aligned} E_\theta(Z_j\mid Z_i=z_i) &=\theta\sqrt{t_j} +\sqrt{\frac{t_i}{t_j}}\left(z_i-\theta\sqrt{t_i}\right)\\ &=\sqrt{\frac{t_i}{t_j}}z_i +\theta\frac{t_j-t_i}{\sqrt{t_j}}. \end{aligned}

The final expression makes clear why a noninteger ratio causes no difficulty: it involves only positive real-valued times and their square roots. The conditional variance is positive whenever $t_j>t_i$. If the times are equal, the later statistic is not a new look under this model and the transition variance is zero.

Worked Example: A Conditional Probability With an Irregular Ratio

Suppose a hypothetical trial has looks at $t_1=1.3$ and $t_2=2.8$, and the canonical model is being evaluated under drift $\theta=0.20$. The observed first-look statistic is $Z_1=1.10$. Find the conditional probability that $Z_2$ exceeds 1.645.

The information ratio is $2.8/1.3$, which is not an integer. Apply the transition formula directly. The conditional mean is $$ \sqrt{\frac{1.3}{2.8}}(1.10) +0.20\left(\sqrt{2.8}-\frac{1.3}{\sqrt{2.8}}\right) \approx 0.92881. $$ Using unrounded values, the conditional standard deviation is $$ \sqrt{1-\frac{1.3}{2.8}}\approx 0.73192505. $$

Standardize the threshold using those unrounded model values: $$ \frac{1.645-0.92880795}{0.73192505}\approx 0.97850. $$ Therefore, $$ P_\theta(Z_2>1.645\mid Z_1=1.10) =1-\Phi(0.97850)\approx 0.1640. $$

Under the stated canonical model and drift, the conditional probability that the second statistic exceeds 1.645, given the observed first statistic, is about 16.4%. This is a conditional probability for a future statistic under a specified model, not a probability that a hypothesis is true.

Independent Increments Behind the Formula

The canonical representation also explains the transition variance and why the size of an information increment can be any positive real number. Define the unstandardized canonical score at time $t_i$ as $S_i=\sqrt{t_i}Z_i$. Under the model, its mean is $\theta t_i$, its variance is $t_i$, and its covariance with $S_j$ is $\min(t_i,t_j)$. Thus, for successive times, the increment $S_j-S_i$ has mean $\theta(t_j-t_i)$ and variance $t_j-t_i$. In the jointly normal model, increments over disjoint information intervals are independent.

S_j-S_i\sim N\left(\theta(t_j-t_i),\,t_j-t_i\right), \qquad Z_j= \frac{S_i+(S_j-S_i)}{\sqrt{t_j}}.

Dividing the earlier score by $\sqrt{t_j}$ contributes $\sqrt{t_i/t_j}Z_i$ to the later statistic. The new increment contributes its mean $\theta(t_j-t_i)/\sqrt{t_j}$, and its variance after scaling is $(t_j-t_i)/t_j$. Since the increment is independent of the earlier score, this gives the transition formula and conditional variance $1-t_i/t_j$. No step requires splitting the interval into an integer number of equal increments.

Worked Example: Checking a Covariance Matrix at Irregular Times

Consider three hypothetical looks at $t_1=0.75$, $t_2=1.4$, and $t_3=2.1$. Relative to the first time, the later ratios are $1.4/0.75\approx1.867$ and $2.1/0.75=2.8$, so these are not all integer multiples of the first information amount.

Use the covariance formula for each pair: $$ \operatorname{Cov}(Z_1,Z_2)=\sqrt{\frac{0.75}{1.4}}\approx0.7319,\quad \operatorname{Cov}(Z_1,Z_3)=\sqrt{\frac{0.75}{2.1}}\approx0.5976,\quad \operatorname{Cov}(Z_2,Z_3)=\sqrt{\frac{1.4}{2.1}}\approx0.8165. $$ The resulting canonical covariance matrix is $$ \begin{pmatrix} 1 & 0.7319 & 0.5976\\ 0.7319 & 1 & 0.8165\\ 0.5976 & 0.8165 & 1 \end{pmatrix}. $$

The order $0.75<1.4<2.1$ gives positive information increments $0.75$, $0.65$, and $0.70$. The independent-increment representation therefore provides a direct validity check: the three cumulative scores are formed from independent normal increments with positive variances $0.75$, $0.65$, and $0.70$. The noninteger ratios do not obstruct a valid joint covariance structure.

Using a Transition Over an Unequal Increment

A transition calculation can be read as the distribution of the next standardized statistic given the current one. The gap $t_j-t_i$, rather than whether $t_j/t_i$ is an integer, determines the amount of new information in the interval. The formula is the same for a short increment, a long increment, or an irregularly placed look.

Worked Example: Probability of Crossing a Later Threshold

In a hypothetical analysis, suppose the current information is $t_i=1.6$, the next look is at $t_j=2.5$, the observed statistic is $Z_i=1.25$, and the assumed drift is $\theta=0.30$. Find the conditional probability that the next statistic exceeds 1.96.

The ratio $2.5/1.6=1.5625$ is not an integer. The conditional mean is $$ \sqrt{\frac{1.6}{2.5}}(1.25) +0.30\frac{2.5-1.6}{\sqrt{2.5}} =0.8(1.25)+0.30\frac{0.9}{\sqrt{2.5}} \approx1.17076. $$ The conditional standard deviation is $$ \sqrt{1-\frac{1.6}{2.5}}=\sqrt{0.36}=0.60. $$

Thus the standardized threshold is $$ \frac{1.96-1.17076}{0.60}\approx1.3154, $$ and the conditional crossing probability is $1-\Phi(1.3154)\approx0.0942$. Under the specified model and drift, the probability that the next statistic exceeds 1.96, conditional on the current value 1.25, is about 9.42%.

Interpretation, Checks, and Limits

The key practical check is to use cumulative information times consistently. They must be positive and ordered, and the covariance between two looks must use the earlier time in the numerator. For every later look, $0<t_i/t_j<1$, so the conditional variance $1-t_i/t_j$ lies strictly between zero and one. If a calculation produces a negative conditional variance, an off-diagonal covariance greater than one, or a covariance that increases when the later time is moved farther away, check the time ordering and the ratio.

An irregular information schedule can be handled in two equivalent ways: apply the covariance formula to the cumulative $Z$ statistics, or work with independent increments of the unstandardized score and then standardize. The first is convenient for joint distributions of reported statistics. The second makes the contribution from each interval explicit. The cumulative $Z$ statistics themselves are correlated; it is the score increments over disjoint information intervals that are independent under the canonical model.

A common mistake is to interpret a noninteger ratio as requiring a fractional number of analyses or to round the ratio to the nearest integer before calculating a correlation. Neither operation is justified. For example, with $t_i=1.3$ and $t_j=2.8$, the canonical correlation is $\sqrt{1.3/2.8}$, not a value obtained by rounding $2.8/1.3$. The information times locate the looks on a scale; their ratio is not a count of discrete blocks.

The formulas do not establish that a particular analysis really follows the canonical model. They rely on the joint-normal canonical approximation, the chosen information scale, and the assumed drift. If those assumptions are poor, a numerically correct transition calculation may still be a poor description of the trial. Noninteger times are allowed by the formulas; that fact alone does not validate the model. In applications, the information values should represent the analysis information on a coherent scale, and each look should occur later than the previous one.

Key takeaway: In the canonical model, arbitrary positive, increasing real-valued information times are permitted. Covariances, conditional means, and conditional variances follow from the same formulas as in equally spaced examples; only the actual information values change.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

At $t_1=0.8$ and $t_2=2.0$, what is the canonical correlation between $Z_1$ and $Z_2$?
For the same two times, what is the conditional variance of $Z_2$ given $Z_1$?
Under drift $\theta=0$, write the conditional mean of $Z_2$ given $Z_1=z_1$ for $t_1=0.8$, $t_2=2.0$.
Why does a noninteger ratio $t_2/t_1$ not invalidate the canonical covariance formula?

Solutions. (1) The correlation is $\sqrt{0.8/2.0}=\sqrt{0.4}\approx0.6325$. (2) The conditional variance is $1-0.8/2.0=0.60$. (3) With $\theta=0$, the conditional mean is $\sqrt{0.8/2.0}\,z_1\approx0.6325z_1$. (4) The canonical formulas require positive ordered information times, not integer multiples; they use the real-valued ratio $t_1/t_2$ directly.

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