Independent Increments and Stagewise Independence

Why Independent Increments Matter

In a group sequential design, statistics at successive looks usually use overlapping data. Their cumulative test statistics are therefore correlated. This dependence is not a defect: it is the expected consequence of repeatedly examining accumulating information. But when evidence is combined across stages, it matters whether the inputs are cumulative statistics or evidence calculated from new information at each stage. The latter can be independent under the canonical model, giving a direct basis for calibrating some combination tests.

The distinction is especially important in adaptive sample size re-estimation. A second-stage sample size may depend on first-stage results, while the second-stage observations are still new. Under suitable sampling and test-calibration conditions, the new stage's p-value can remain independent of the first-stage p-value. Independent increments explain the statistical basis for that conclusion; they do not make every adaptive combination rule valid automatically.

The Canonical Model and Its Increments

Index information by $I$, with $I>0$. In the canonical normal model, let $S(I)$ denote the cumulative score-like statistic at information $I$. Assume that for $0\leq I_a<I_b$, the increment $S(I_b)-S(I_a)$ is independent of information accumulated through $I_a$, and

S(I_b)-S(I_a)\sim N\!\left(\theta(I_b-I_a),\,I_b-I_a\right).

Here $\theta$ is the standardized effect per square root of information, and the variance of an increment equals its information length. Under the null hypothesis $H_0:\theta=0$, the increment has mean zero. This model is exact in settings with independent normal increments and is also used as an approximation for many large-sample test statistics. Its use requires the information scale and the statistic's null distribution to be correctly specified.

Definition: An independent increment is the change in the cumulative score-like statistic between two information times. Under the canonical model, increments over nonoverlapping information intervals are independent. A stagewise statistic standardizes an increment using only the information added during that stage: $$ Z_i^{\mathrm{inc}}=\frac{S(I_i)-S(I_{i-1})}{\sqrt{I_i-I_{i-1}}}. $$

For two stages, write $I_1$ for the first-stage information and $I_2>I_1$ for total information at the second look. The two stagewise statistics are $Z_1^{\mathrm{inc}}=S(I_1)/\sqrt{I_1}$ and $Z_2^{\mathrm{inc}}=[S(I_2)-S(I_1)]/\sqrt{I_2-I_1}$. Under $H_0$, each has a standard normal distribution, and the independent-increment assumption makes them independent. The same independence holds under a fixed alternative in this model, although their means are then nonzero.

Worked Example: Cumulative Dependence and Independent Stage Evidence

Suppose the first look occurs at information $I_1=100$ and the second at total information $I_2=225$. The cumulative statistics are $Z(I)=S(I)/\sqrt I$. Find their correlation under the canonical model, and identify the information in the second-stage increment.

Because $S(I_1)$ is part of $S(I_2)$, the covariance of the cumulative scores under $H_0$ is $\operatorname{Cov}\{S(I_1),S(I_2)\}=I_1=100$. Thus $$ \operatorname{Corr}\{Z(I_1),Z(I_2)\} =\frac{100}{\sqrt{100\cdot225}} =\frac{100}{150} =0.6667. $$ The cumulative statistics are positively correlated; treating them as independent would be incorrect.

The second-stage information is $I_2-I_1=125$. Its increment statistic is $$ Z_2^{\mathrm{inc}}=\frac{S(225)-S(100)}{\sqrt{125}}. $$ Under $H_0$, this statistic is standard normal and independent of $Z(I_1)$, because it uses only the increment over the nonoverlapping interval from information 100 to 225.

Why the Stagewise Statistics Are Independent

The cumulative statistics can be written in terms of the first-stage and second-stage increments. Let $Z_1=S(I_1)/\sqrt{I_1}$ and $U_2=[S(I_2)-S(I_1)]/\sqrt{I_2-I_1}$. Under $H_0$, $Z_1$ and $U_2$ are independent standard normal variables. The cumulative statistic at the second look is

Z(I_2)= \sqrt{\frac{I_1}{I_2}}\,Z_1+ \sqrt{\frac{I_2-I_1}{I_2}}\,U_2.

The squared coefficients sum to one, so $Z(I_2)$ also has variance one. Its correlation with $Z_1$ is the coefficient $\sqrt{I_1/I_2}$. The following calculation makes clear why the corresponding increment, rather than the cumulative statistic, is independent of the first look.

Express the cumulative second-look statistic.
Since $S(I_2)=S(I_1)+[S(I_2)-S(I_1)]$, the standardized statistic is the weighted sum shown above.

Remove the first-stage contribution.
Rearranging gives $U_2=[\sqrt{I_2}Z(I_2)-\sqrt{I_1}Z_1]/\sqrt{I_2-I_1}$. This is the standardized second-stage increment.

Apply the model's independence assumption.
The increment $S(I_2)-S(I_1)$ is independent of $S(I_1)$. Standardizing each by a fixed positive constant preserves independence, so $U_2$ and $Z_1$ are independent.

This is a derivation from the canonical model, not a claim that any two statistics calculated at different calendar times are independent. The inputs must correspond to nonoverlapping information increments, or another justified source of independent stagewise data.

From Independent Increments to Independent P-Values

Consider a one-sided test whose direction is specified in advance, with larger values of the stagewise statistic supporting the alternative. Define the stagewise p-value by $p_i=1-\Phi(Z_i^{\mathrm{inc}})$, where $\Phi$ is the standard normal cumulative distribution function. Under $H_0$, if $Z_i^{\mathrm{inc}}\sim N(0,1)$, then $p_i$ is uniform on $(0,1)$. If the stagewise statistics are independent, applying this function separately to each statistic preserves independence. Thus the stagewise p-values are independent uniform random variables under the null.

This conclusion extends beyond normally distributed test statistics if each stagewise p-value is valid under the null and the p-values are independent. In continuous settings, exact null p-values are uniform. For a discrete test they are generally not exactly uniform; validity instead means $P(p_i\leq u\mid H_0)\leq u$ for every $u$ between zero and one. Independence of discrete p-values does not turn them into exact uniforms.

Key condition: Independent increments support combination-test calibration when each stagewise p-value is computed from its own increment, is correctly calibrated under the null, and the p-values are independent under the null. The test direction, stagewise procedures, and any combination weights must also satisfy the requirements of the chosen combination rule.

Worked Example: Calibrating an Inverse-Normal Combination

Suppose two one-sided stagewise p-values are $p_1=0.20$ and $p_2=0.40$. Assume they are independent and uniform under $H_0$, and use fixed weights $w_1=0.60$ and $w_2=0.80$. These weights satisfy $w_1^2+w_2^2=0.36+0.64=1$. Calculate the inverse-normal statistic and its one-sided p-value.

The statistic is $T=w_1\Phi^{-1}(1-p_1)+w_2\Phi^{-1}(1-p_2)$. Substitution gives $$ T=(0.60)\Phi^{-1}(0.80)+(0.80)\Phi^{-1}(0.60) =(0.60)(0.8416212)+(0.80)(0.2533471) =0.7076504\approx0.7077. $$ Under $H_0$, each inverse-normal term before weighting is standard normal. Independence and the squared-weight condition imply $T\sim N(0,1)$. Therefore the one-sided combined p-value is $1-\Phi(0.7076504)\approx0.2396$. At significance level $\alpha=0.05$, the combined result is not statistically significant. This calculation uses the unrounded normal quantiles before rounding the final statistic.

The calibration follows from three ingredients: the null distribution of each stagewise p-value, their independence, and the fixed normalized weights. Independence alone would not justify the same reference distribution if the weights were chosen using the observed evidence.

Adaptive Stage-Two Sampling

Independence can remain relevant when the second-stage sample size is selected after examining first-stage data. The key is conditional calibration: given all first-stage information and the resulting sample-size choice, the second-stage test must have its stated null distribution using only new second-stage data. If its p-value is conditionally uniform under $H_0$, then it is independent of the first-stage information in the following precise sense.

P(p_1\leq a,\ p_2\leq b\mid H_0) =E\!\left[\mathbf{1}\{p_1\leq a\} P(p_2\leq b\mid\mathcal F_1,H_0)\mid H_0\right] =ab,

for $0\leq a,b\leq1$, where $\mathcal F_1$ contains the first-stage data and the sample-size choice, and where $p_1$ is exactly uniform under $H_0$. The second equality uses $P(p_2\leq b\mid\mathcal F_1,H_0)=b$. This establishes independence in the exact continuous case. If stage two uses a discrete test, the conditional probability is generally at most $b$, not equal to it, so the exact independence-and-uniformity conclusion needs qualification.

Worked Example: Checking Conditional Calibration After Re-Estimation

In a hypothetical trial, an interim rule chooses either 80 or 120 new participants for stage two based on the first-stage result. Assume that, under $H_0$, the new participants' outcomes are independent of all first-stage outcomes, and that for either chosen sample size the prespecified stage-two test produces an exact one-sided uniform p-value conditional on the first-stage data and the chosen size. What does this imply about the stagewise p-values?

For every $b$ in $[0,1]$, the stated assumption gives $P(p_2\leq b\mid\mathcal F_1,H_0)=b$, whether the selected size is 80 or 120. If $p_1$ is also exactly uniform under $H_0$, then for any $a,b$ in $[0,1]$, $$ P(p_1\leq a,\ p_2\leq b\mid H_0) =P(p_1\leq a\mid H_0)b =ab. $$ Thus $p_1$ and $p_2$ are independent uniform p-values under the null. The data-dependent sample size does not itself destroy this independence under these assumptions. This conclusion would not follow merely from saying that stage two uses “new data”: the stage-two test must remain correctly calibrated conditional on the first-stage information and the size selected.

Diagnostics, Alternatives, and Limitations

A practical diagnostic is to trace exactly which observations, scores, and estimated quantities enter each stagewise p-value. If stage two reuses first-stage observations, estimates a variance from pooled data, or selects a test after inspecting first-stage evidence, its p-value may not be conditionally calibrated as assumed. A model-based claim of independent increments should also be checked against the design: information must be defined consistently, increments must be correctly centered and scaled under $H_0$, and the relevant observations or score increments must satisfy the independence or conditional-independence assumptions. In large samples, canonical independence may be an approximation rather than an exact finite-sample property.

Approach	What is combined	Dependence or calibration issue
Cumulative group sequential statistic	Statistics based on all data available at each look	Looks overlap, so cumulative statistics are correlated and require a reference distribution that accounts for that correlation.
Stagewise combination test	P-values calculated from distinct stage increments	Can use an independence-based reference distribution if stagewise p-values are valid and independent under $H_0$.
Single pooled final test	All observations in one final analysis	Requires a valid final-test distribution under the actual sampling and adaptation rule; it is not automatically equivalent to a stagewise combination test.

An important limitation is that independence of the p-values does not validate every way of combining them. For example, the inverse-normal calculation above uses fixed weights with squared sum one. If weights depend on the observed first-stage evidence, the resulting statistic need not be standard normal under $H_0$, even when the two p-values themselves are independent. Likewise, a p-value computed from a misspecified test, an incorrectly handled nuisance parameter, or a two-sided procedure treated as a one-sided normal transform may not have the distribution the combination calculation assumes.

A common mistake is to reason, “The second look happens later, so its statistic is independent of the first.” That is false for cumulative statistics: the second-look statistic contains the first-stage data. The correction is to separate the cumulative statistic into its first-stage contribution and the new stage increment, and then establish independence for the increment under the stated model. Another mistake is to infer exact validity from a sample-size rule alone. The relevant question is whether the stage-two p-value has the required null distribution conditional on the interim information and the rule's chosen design.

Key takeaway: In the canonical model, nonoverlapping information increments are independent even though cumulative statistics are correlated. This supports independent stagewise evidence when each stage's p-value uses its own increment and is correctly calibrated. The combination method's weights and reference distribution still require their own justification.

Check Your Understanding

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

At information times $I_1=64$ and $I_2=144$, what is the correlation of the cumulative canonical statistics under $H_0$?
Why does transforming independent continuous standard normal stagewise statistics into one-sided p-values preserve independence under $H_0$?
A stage-two sample size is chosen from interim data. State the key condition on the stage-two p-value needed for the conditional-independence argument.
Two stagewise p-values are independent, but the inverse-normal weights are selected using the first p-value. Does independence alone establish a standard normal combination statistic? Explain.

Solutions. (1) The correlation is $\sqrt{I_1/I_2}=\sqrt{64/144}=2/3\approx0.6667$; the cumulative statistics share the first-stage information. (2) A measurable transformation applied separately to each of two independent variables preserves independence; under the null, the normal upper-tail transformations are uniform. (3) The stage-two p-value must have its stated null distribution conditional on the interim information and the selected sample size, using a correctly calibrated stage-two procedure. (4) No. Data-dependent weights can be associated with the first-stage evidence, so the weighted sum need not have the standard normal reference distribution; independence of the unweighted p-values is not enough.