Why Independent Increments Are Not Independent Z Statistics

Why the distinction matters

In a group sequential design, information accumulates across planned analyses. The score collected by a later look includes the score already collected at an earlier look. The new score increments can be independent under the canonical model, but the cumulative standardized statistics reuse earlier information. They are therefore correlated. Confusing these two statements leads to incorrect probabilities for repeated testing and incorrect interpretations of what a later statistic says given an earlier one.

The key is to keep track of both the cumulative score and the cumulative information. We use $U_j$ for the score through look $j$, $I_j$ for its information, and $Z_j=U_j/\sqrt{I_j}$ for the standardized cumulative statistic. Set $U_0=I_0=0$, and write $\Delta U_j=U_j-U_{j-1}$ and $\Delta I_j=I_j-I_{j-1}$.

Definition: In the canonical Gaussian model, information levels are fixed and increasing, so $0=I_0<I_1<\cdots<I_K$. For a fixed canonical parameter $\theta$, the score increments are independent and satisfy $\Delta U_j\sim N(\theta\Delta I_j,\Delta I_j)$. The null hypothesis is $\theta=0$. The cumulative score is $U_j=\sum_{\ell=1}^j\Delta U_\ell$, and the cumulative Z statistic is $Z_j=U_j/\sqrt{I_j}$.

This model is exact in settings such as normal observations with known variance, and it is a standard asymptotic approximation for regular score statistics. It assumes the stated information levels correspond to the score increments, the increments have the specified means and variances, and the increments are independent (or adequately approximated as independent). Under a fixed alternative, the same $\theta$ applies across looks. The distinction in this tutorial follows from the cumulative construction; it does not require a new testing rule.

How independent increments create correlated Z statistics

For each look, the cumulative score is a sum of all increments up to that look. Thus if $j<k$, the scores $U_j$ and $U_k$ share the first $j$ increments. Independence of the later increments from the earlier ones does not remove that shared part.

Result: Under the canonical model, $E(Z_j)=\theta\sqrt{I_j}$, $\operatorname{Var}(Z_j)=1$, and, for $j\leq k$, $$ \operatorname{Cov}(Z_j,Z_k)=\sqrt{\frac{I_j}{I_k}}. $$ Therefore the correlation between the cumulative Z statistics at looks $j$ and $k$ is $\sqrt{I_j/I_k}$, which is positive when $j<k$.

To justify the result, $U_j$ has mean $\theta I_j$ and variance $I_j$, because its independent increments have means summing to $\theta I_j$ and variances summing to $I_j$. For $j\leq k$, write $U_k=U_j+(U_k-U_j)$. The second term is a sum of increments after look $j$, so it is independent of $U_j$. Consequently, $\operatorname{Cov}(U_j,U_k)=\operatorname{Var}(U_j)=I_j$. Dividing by $\sqrt{I_j I_k}$ gives the covariance formula. Both Z statistics have variance one, so their covariance is also their correlation.

The same result can be seen directly by standardizing each increment. Let $A_\ell=\Delta U_\ell/\sqrt{\Delta I_\ell}$. The $A_\ell$ are independent, each with variance one, and $$ Z_j=\sum_{\ell=1}^j\sqrt{\frac{\Delta I_\ell}{I_j}}\,A_\ell. $$ For two looks, the terms up to the earlier look appear in both sums. The later statistic has additional independent terms, but it does not discard the earlier ones.

Worked Example: Correlation at Two Information Levels

Suppose a hypothetical trial has planned information levels $I_1=40$ and $I_2=100$. The canonical model has independent increments. Find the correlation between the two cumulative Z statistics.

Since the first look precedes the second, the covariance result gives $$ \operatorname{Corr}(Z_1,Z_2)=\sqrt{\frac{I_1}{I_2}} =\sqrt{\frac{40}{100}}=\sqrt{0.4}=0.6325. $$

Thus $Z_1$ and $Z_2$ are not independent. This conclusion holds under the null and under a fixed alternative in the canonical model: the alternative changes the means, not this covariance. Independent increments describe the newly accumulated score, not the two cumulative Z statistics.

The conditional distribution at a later look

Because the canonical score vector is jointly normal, the conditional distribution at a later look can be obtained by separating the score already observed from the future increments. This gives a direct description of what the later cumulative statistic can be, conditional on the earlier statistic.

Result: For $j<k$, conditional on $Z_j=z$, the later statistic has distribution $$ Z_k\mid Z_j=z\sim N\left(\sqrt{\frac{I_j}{I_k}}\,z+ \frac{\theta(I_k-I_j)}{\sqrt{I_k}},\;1-\frac{I_j}{I_k}\right). $$ Under the null, set $\theta=0$. The second argument of $N$ is the variance.

For a justification, the future score $U_k-U_j$ is independent of $U_j$, with mean $\theta(I_k-I_j)$ and variance $I_k-I_j$. Given $Z_j=z$, the earlier score is $U_j=\sqrt{I_j}z$. Therefore $$ Z_k=\frac{U_j+(U_k-U_j)}{\sqrt{I_k}} =\sqrt{\frac{I_j}{I_k}}\,z+\frac{U_k-U_j}{\sqrt{I_k}}. $$ The conditional mean is the fixed first term plus the mean of the future increment divided by $\sqrt{I_k}$; the conditional variance is $(I_k-I_j)/I_k$. This proves the stated distribution.

The conditional mean under an alternative includes a contribution from future information. It is not generally just the correlation times $z$. Under $H_0$, the drift term is zero, so the conditional mean is $\sqrt{I_j/I_k}\,z$. In either case, the conditional variance is less than one: given the earlier statistic, part of the later statistic is already determined by the shared score.

Worked Example: A Later Statistic Given an Earlier Result

Consider the hypothetical information levels $I_1=40$ and $I_2=100$. Under $H_0$, suppose the observed first-look statistic is $Z_1=1.5$. Find the conditional mean and standard deviation of $Z_2$, and calculate the conditional probability that $Z_2$ exceeds 1.96.

Under the null, the conditional mean and variance are $$ E(Z_2\mid Z_1=1.5)=\sqrt{\frac{40}{100}}(1.5) =0.6325(1.5)=0.9487, $$ $$ \operatorname{Var}(Z_2\mid Z_1=1.5)=1-\frac{40}{100}=0.6. $$ The conditional standard deviation is $\sqrt{0.6}=0.7746$. Standardizing the threshold gives $$ P(Z_2>1.96\mid Z_1=1.5) =1-\Phi\left(\frac{1.96-0.9487}{0.7746}\right) =1-\Phi(1.306)\approx 0.0958. $$

Conditional on this first-look result and the null model, the probability of exceeding 1.96 at the second look is about 0.096. This is a conditional probability for the future statistic, not the probability that $H_0$ is true.

Seeing the dependence in an increment representation

The increment representation makes the source of dependence especially clear. Standardized score increments are independent, but a cumulative Z statistic is a weighted sum of them. The weights depend on the information accumulated by that look. A later statistic contains the earlier increment components as well as new ones.

Worked Example: Independent Increments, Dependent Cumulative Statistics

Suppose a hypothetical two-look study has $I_1=25$, $I_2=100$, and fixed alternative $\theta=0.12$. The increment information is 25 at the first look and 75 at the second. State the distributions of the score increments, standardize them, and show how they form the cumulative statistics.

By the canonical model, $$ \Delta U_1\sim N(0.12(25),25)=N(3,25),\qquad \Delta U_2\sim N(0.12(75),75)=N(9,75), $$ and the two increments are independent. Define $A_1=\Delta U_1/5$ and $A_2=\Delta U_2/\sqrt{75}$. Then $A_1$ and $A_2$ are independent, each with variance one, and their means are $$ E(A_1)=\frac{3}{5}=0.6,\qquad E(A_2)=\frac{9}{\sqrt{75}}=1.039. $$ At the first look, $Z_1=A_1$. At the second look, $$ Z_2=\frac{\Delta U_1+\Delta U_2}{10} =0.5A_1+\frac{\sqrt{75}}{10}A_2 =0.5A_1+0.8660A_2. $$

The representation gives $E(Z_1)=0.6$ and $E(Z_2)=0.5(0.6)+0.8660(1.039)\approx1.20$, as required by $E(Z_j)=\theta\sqrt{I_j}$. Since $Z_1=A_1$ and $A_2$ is independent of $A_1$, $\operatorname{Cov}(Z_1,Z_2)=0.5\operatorname{Var}(A_1)=0.5$. This equals $\sqrt{25/100}=0.5$. The independent new increment $A_2$ adds uncertainty at the second look, but the earlier increment $A_1$ remains part of $Z_2$.

Consequences for sequential calculations

A tempting shortcut is to treat the Z statistics at the planned looks as independent because the score increments are independent. That shortcut is wrong. The joint probability of crossing one or more boundaries depends on the joint distribution of the cumulative statistics, including their covariance. For example, under $H_0$, a two-look calculation must account for $\operatorname{Corr}(Z_1,Z_2)=\sqrt{I_1/I_2}$; multiplying two marginal probabilities as though the looks were independent does not generally give the probability of a sequential event.

There are two equivalent ways to do correct calculations under the canonical model. One is to work with the jointly normal vector of cumulative Z statistics and its covariance matrix. The other is to work with independent score increments and reconstruct each cumulative statistic from the increments. Conditional calculations use the distribution derived above. These are different representations of the same joint model, not competing assumptions.

The independent-increment representation is often convenient for simulation and for understanding how information is added. The covariance or conditional-distribution representation is often convenient for evaluating probabilities involving cumulative statistics. Neither representation licenses treating cumulative Z statistics as independent. In particular, the marginal fact $Z_j\sim N(\theta\sqrt{I_j},1)$ at each look is not enough to calculate a probability involving multiple looks; the dependence must also be included.

Assumptions, diagnostics, and a common pitfall

The formulas rely on the canonical model: increasing fixed information levels, independent normal score increments with the specified means and variances, and a common fixed $\theta$ under the alternative. In regular models, independence and normality may be asymptotic rather than exact. If increments are not adequately independent, if their variances do not match the information increments, or if the information levels used do not correspond to the score process, the canonical covariance and conditional formulas may not describe the data well.

A useful check is to calculate $\sqrt{I_j/I_k}$ for every pair of planned looks. The resulting correlations should be between zero and one, with correlation one on the diagonal. For three or more looks, the implied matrix should also be a valid covariance matrix. When a model or simulation is available, compare the observed correlations of cumulative statistics with these canonical values. Substantial discrepancies call for examining the score construction, information values, or independence approximation rather than silently assuming independent Z statistics.

Wrong reasoning: “The score added after each look is independent of the earlier score, so the Z statistics from the looks are independent.”
Correction: The later cumulative score includes the earlier score. Independent future increments are added to a shared earlier component, producing correlated cumulative Z statistics.
Another frequent error: Using the null conditional mean formula under an alternative. Under a fixed $\theta$, include the drift term $\theta(I_k-I_j)/\sqrt{I_k}$.

The canonical model is a working description of the joint score process, not a guarantee that every real analysis has exact normal increments. The correlation formula also depends on the information ordering and the specified increment variances. It should not be transferred unchanged to a setting with a different joint score structure.

Key takeaway: Independent increments do not imply independent cumulative Z statistics. At looks $j\leq k$, the canonical correlation is $\sqrt{I_j/I_k}$, and the later statistic conditional on an earlier one has variance $1-I_j/I_k$, with a drift term under a fixed alternative.

Check Your Understanding

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

Under the canonical null model, $I_1=36$ and $I_2=144$. What is $\operatorname{Corr}(Z_1,Z_2)$?
For $I_1=25$, $I_2=100$, and $\theta=0$, give the conditional distribution of $Z_2$ given $Z_1=0.8$.
Why does independence of $\Delta U_1$ and $\Delta U_2$ not imply independence of $Z_1$ and $Z_2$?
For $I_1=20$, $I_2=80$, and $\theta=0.1$, calculate the conditional mean of $Z_2$ given $Z_1=1$. State the conditional variance as well.

Solutions. (1) $\sqrt{36/144}=0.5$. (2) The conditional mean is $\sqrt{25/100}(0.8)=0.4$, and the variance is $1-25/100=0.75$. Thus $Z_2\mid Z_1=0.8\sim N(0.4,0.75)$. (3) $Z_2$ contains the earlier score increment as part of its cumulative score, so the two cumulative statistics share a random component. (4) The conditional mean is $\sqrt{20/80}(1)+0.1(80-20)/\sqrt{80}=0.5+0.6708=1.171$. The conditional variance is $1-20/80=0.75$.