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Type I Error Control in Re-Estimation · Tutorial 466 of 699

Why Naive Pooled Tests Fail After Re-Estimation

Learn why pooling stage data with fixed-design calibration can be invalid after an interim-driven sample-size change, and how to diagnose the resulting error-control risk.

Advanced 11 min read

What You'll Learn

  • Write the pooled z-statistic as a weighted sum of independent stage statistics.
  • Explain why its usual standard normal calibration depends on a fixed sample size.
  • Derive the conditional distribution of the pooled statistic after re-estimation.
  • Work through a two-stage normal design with an inflated one-sided Type I error rate.
  • Distinguish conditional rejection probabilities from the overall Type I error rate.
  • Recognize why combination or conditional-error methods need a calibration that accounts for adaptation.

Why pooling is not automatically safe

Pooling observations is usually an efficient way to estimate a common mean. In a fixed-design test with a known variance, the pooled mean has a familiar standard error, and the corresponding z-statistic has a standard normal distribution under the null hypothesis. But after an interim analysis, the total sample size may depend on the first-stage data. The final sample size is then part of the selection rule, not a fixed feature of the design. Applying a fixed-design calibration to the pooled observations can therefore fail to control the Type I error rate.

The issue is not that the arithmetic of pooling becomes wrong. The pooled statistic can still be calculated correctly. The issue is that its null distribution may no longer be the distribution used to choose the critical value. This distinction matters in a re-estimation design: correct arithmetic does not by itself provide a valid test.

Model and setup: Suppose observations in two stages are independent, with \(X_{ij}\sim N(\mu,\sigma^2)\), where the variance \(\sigma^2\) is known and common across stages. Stage one contains \(n_1\) observations. After seeing stage one, the design chooses a stage-two size \(n_2\), which may depend on the stage-one data. Stage-two observations are collected independently of the interim data. For testing \(H_0:\mu=\mu_0\), define \(Z_i=\sqrt{n_i}(\bar{x}_i-\mu_0)/\sigma\) for a stage with \(n_i>0\).

The pooled statistic and its calibration

If the total sample size \(N=n_1+n_2\) is fixed in advance, the pooled mean is \((n_1\bar{x}_1+n_2\bar{x}_2)/N\). Its standardized statistic can be written either in terms of the stage means or in terms of stagewise z-statistics:

$$ Z_{\mathrm{pool}} =\frac{n_1(\bar{x}_1-\mu_0)+n_2(\bar{x}_2-\mu_0)} {\sigma\sqrt{n_1+n_2}} =\sqrt{\frac{n_1}{N}}Z_1+\sqrt{\frac{n_2}{N}}Z_2. $$

When the sample sizes are fixed, \(Z_1\) and \(Z_2\) are independent standard normal variables under \(H_0\). The squared weights sum to one, so \(Z_{\mathrm{pool}}\sim N(0,1)\). A one-sided test that rejects for \(Z_{\mathrm{pool}}>c\), with \(c\) chosen so that \(P(Z>c)=\alpha\) for \(Z\sim N(0,1)\), consequently has Type I error rate \(\alpha\).

Worked Example: Checking a Fixed-Size Pooled Calculation

Suppose a hypothetical normal-mean experiment has \(n_1=n_2=10\), known \(\sigma=1\), and tests \(H_0:\mu=0\). The observed stage means are \(\bar{x}_1=0.40\) and \(\bar{x}_2=-0.10\). Calculate the pooled statistic.

The stagewise statistics are \(Z_1=\sqrt{10}(0.40)=1.2649\) and \(Z_2=\sqrt{10}(-0.10)=-0.3162\). Directly from the pooled-mean formula,

$$ Z_{\mathrm{pool}} =\frac{10(0.40)+10(-0.10)}{\sqrt{20}} =\frac{3}{4.4721} \approx 0.6708. $$

The weighted stagewise calculation agrees:

$$ \frac{\sqrt{10}(1.2649)+\sqrt{10}(-0.3162)}{\sqrt{20}} \approx 0.6708. $$

These are the same pooled statistic, not competing calculations. With the total size fixed at 20, its null distribution is standard normal under the stated model. A different conclusion about error control would require a different design feature, such as choosing the second-stage size using the first-stage result.

The key change under re-estimation is visible by conditioning on the interim data. Write \(N=n_1+n_2(Z_1)\) when the second-stage size is a function of the first-stage result. Conditional on \(Z_1\), the second-stage statistic is still standard normal under \(H_0\), provided the model assumptions hold. The pooled statistic then has conditional mean and variance

$$ E_0(Z_{\mathrm{pool}}\mid Z_1) =\sqrt{\frac{n_1}{N}}Z_1, \qquad \operatorname{Var}_0(Z_{\mathrm{pool}}\mid Z_1) =\frac{n_2}{N}. $$

Both quantities depend on the interim result through \(n_2\), and the conditional mean also depends directly on \(Z_1\). Thus the final statistic is not generally a standard normal variable conditional on the information that determined the sample size. Its unconditional null distribution is a mixture over the interim outcomes and the corresponding sample sizes. That mixture is not guaranteed to be standard normal.

A two-stage example with inflated Type I error

The following deliberately simple hypothetical design gives a direct demonstration. It uses a one-sided test at nominal level \(\alpha=0.025\), with known \(\sigma=1\), and sets \(\mu_0=0\). There is one observation in stage one, \(X_1\sim N(\mu,1)\), so \(Z_1=X_1\) under the null. Let \(c=1.96\), the usual approximate upper-tail standard normal critical value for a one-sided 0.025 test.

Use this interim rule: if \(Z_1>c\), stop and reject \(H_0\) using the single-stage test. Otherwise, collect one independent stage-two observation \(X_2\), and reject if the ordinary pooled statistic for two observations exceeds the same fixed-design critical value \(c\). This is a sample-size decision based on interim data: the total sample size is one or two. On the two-observation branch, “pooling” means using \((X_1+X_2)/\sqrt{2}\) as the z-statistic under \(H_0\).

Worked Example: Calculating the Error Inflation

Under \(H_0\), let \(Z_1=X_1\) and \(Z_2=X_2\). They are independent standard normal variables. Rejection occurs either on the early-stop branch or on the continuation branch. These events are disjoint, so the actual Type I error rate is

$$ \begin{aligned} P_0(\text{reject}) &=P(Z_1>c) +P\left(Z_1\leq c,\frac{Z_1+Z_2}{\sqrt{2}}>c\right)\\ &=P(Z_1>c)+\int_{-\infty}^{c} \phi(z)\left[1-\Phi(\sqrt{2}c-z)\right]\,dz, \end{aligned} $$

where \(\phi\) and \(\Phi\) are the standard normal density and distribution functions. The integral is strictly positive: for example, the event \(0\leq Z_1\leq 0.5\) and \(Z_2>\sqrt{2}c\) is contained in the continuation-branch rejection event. Independence gives the following positive lower bound for the extra rejection probability:

$$ P(0\leq Z_1\leq 0.5)P(Z_2>\sqrt{2}(1.96)) \approx 0.0005329. $$

Therefore the actual Type I error rate is strictly greater than 0.025; this lower bound alone shows it is at least about 0.02553. The second term in the exact expression includes all continuation-branch rejections, not just the event used for the lower bound. No numerical value for that full integral is needed to establish inflation. The reason is selection: the data that determine whether there is a second observation also enter the pooled test statistic.

This example uses early stopping to make the calculation transparent. The same structural concern arises whenever \(n_2\) is selected from interim results: the weights and the distribution of the final pooled statistic are then linked to those results. The example does not establish that every adaptation inflates Type I error by the same amount, or that every sample-size rule must inflate it. It establishes that nominal fixed-design calibration alone does not guarantee the desired error rate after adaptation.

What the interim conditioning reveals

Worked Example: Conditional Rejection Chance After Re-Estimation

Consider a hypothetical design with \(n_1=10\), known \(\sigma=1\), and a one-sided critical value \(c=1.96\). Suppose the interim rule chooses \(n_2=10\) when \(Z_1>1.5\) and \(n_2=90\) otherwise. At an interim result \(Z_1=2.00\), the rule selects \(N=20\). Under \(H_0\), the conditional distribution of the pooled statistic is

$$ Z_{\mathrm{pool}}\mid Z_1=2.00 \sim N\left( \sqrt{\frac{10}{20}}(2.00), \frac{10}{20} \right) =N(1.414,0.500). $$

The second argument here is the variance, so the conditional standard deviation is \(\sqrt{0.500}=0.7071\). The conditional probability of crossing 1.96 is

$$ P_0(Z_{\mathrm{pool}}>1.96\mid Z_1=2.00) =1-\Phi\left(\frac{1.96-1.414}{0.7071}\right) \approx 1-\Phi(0.772) \approx 0.220. $$

This is a conditional rejection chance for a specified interim result, not the overall Type I error rate and not a p-value for the observed study. The overall rate averages rejection probabilities across all possible first-stage outcomes, with the sample-size rule applied to each. In a fixed-size design that average is calibrated by the standard normal result; after re-estimation, it must account for the rule.

How to analyze the design responsibly

The pooled estimate can remain useful as an estimate of the common mean. What cannot be assumed is that the usual fixed-design p-value or cutoff remains valid just because the observations have been combined correctly. A valid inferential procedure must preserve the Type I error rate under the actual adaptation rule, including its thresholds, possible stopping decisions, and second-stage sample sizes.

The earlier material on conditional error and combination tests provides approaches designed for adaptive testing. Their central practical distinction from naive pooling is that the rejection rule is calibrated to the adaptive design, rather than borrowed unchanged from a fixed-total-size experiment. For a re-estimation plan, the analysis should be specified alongside the adaptation rule, not chosen afterward based on which calculation is most favorable.

Useful checks include verifying that the stage-two size is determined only by the allowed interim information, confirming independence of future stage-two data from that information under the model, and checking that the final rejection rule is calibrated over the full range of interim outcomes. If the variance is estimated rather than known, or if adaptation changes the statistic or its distribution, the known-variance normal argument above is not sufficient. Those departures require a procedure justified for the relevant model and design.

Common mistake and limitations

  • Mistake: “The pooled z-statistic has a standard normal distribution for any sample size, so the usual fixed-design critical value remains valid.” Correction: The standard normal result follows for fixed stage sizes under the stated model. When the second-stage size depends on stage-one data, the statistic’s weights and conditional distribution depend on the interim result; its null calibration must be checked under that adaptive rule.
  • Mistake: “If the pooled estimate is calculated correctly, its p-value must be correct.” Correction: Estimation arithmetic and test calibration are separate. The estimate can be correct while the reference distribution used for the p-value is not.
  • Mistake: “A large conditional rejection chance at one interim value is the overall Type I error.” Correction: Overall Type I error is the probability of rejection averaged over all possible interim outcomes under \(H_0\), including how each outcome determines the sample size.

The example assumes independent normal observations and known common variance. Actual applications may involve estimated variances, discrete outcomes, covariate adjustment, unequal variances, missing data, or adaptation based on nuisance estimates. Such settings can change the distribution of stagewise statistics, so the simple normal calculation should not be transferred without justification. Nor should the example be read as a universal numerical inflation factor: the amount, and sometimes the direction, of a calibration error depends on the adaptation and rejection rules.

Key takeaway: A fixed-design pooled z-test is calibrated when the stage sizes are fixed under its model. If sample size depends on interim data, pooling remains an arithmetic operation, but fixed-design Type I error control is no longer automatic.

Check Your Understanding

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

  1. For fixed \(n_1\) and \(n_2\), why does the weighted pooled statistic have variance one under \(H_0\)?
  2. In the two-stage inflation example, identify a positive-probability event that contributes to rejection beyond the early-stop probability.
  3. For the conditional example with \(Z_1=2.00\) and \(n_2=10\), what are the conditional mean and variance of \(Z_{\mathrm{pool}}\)?
  4. Does a conditional rejection chance of 0.220 mean that the overall Type I error rate is 0.220? Explain.
  5. Name one feature of the design that a valid adaptive analysis must incorporate.

Solutions. (1) The independent stage statistics each have variance one, and the squared weights sum to \(n_1/N+n_2/N=1\). (2) For example, \(0\leq Z_1\leq0.5\) and \(Z_2>\sqrt{2}(1.96)\) leads to continuation and pooled rejection, and has positive probability. (3) The conditional mean is \(\sqrt{10/20}(2.00)=1.414\), and the conditional variance is \(10/20=0.500\). (4) No. It is a conditional probability at a particular interim result; the overall error rate averages over all interim outcomes and sample-size decisions under \(H_0\). (5) It must incorporate the interim adaptation rule, including how stage-two size or stopping depends on interim data.

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