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Type I Error Control in Re-Estimation · Tutorial 444 of 699

Fisher Combination Tests With Adaptive Stage Two

Learn when Fisher’s combination test controls Type I error after adaptive sample-size re-estimation, how to calculate its combined p-value, and how to diagnose invalid applications.

Advanced 12 min read

What You'll Learn

  • Derive the chi-square reference distribution for Fisher’s statistic from independent uniform p-values.
  • State the conditional-validity requirement for a stage-two p-value after adaptive sample-size selection.
  • Calculate Fisher’s statistic and combined p-value for two stages.
  • Distinguish exact chi-square calibration from situations where it is not justified.
  • Identify design and analysis failures that can undermine Type I error control.
  • Compare Fisher combination with inverse-normal combination and an unadjusted pooled analysis.

Why Combine Stagewise P-Values?

In a two-stage design, the second-stage sample size may be selected using information observed at the first stage. That flexibility creates a calibration question: does the final test still reject the global null hypothesis \(H_0\) with probability at most the chosen significance level \(\alpha\)? Fisher’s combination test offers a way to combine evidence from both stages without treating the adaptive sample-size rule as though it had been fixed in advance. Its validity, however, depends on how the stagewise p-values behave under \(H_0\).

The essential condition is not simply that the stages use different participants. The stage-two p-value must remain valid conditional on the stage-one information used to choose the second-stage sample size. When that condition holds, adaptive sample size does not by itself invalidate Fisher’s reference distribution.

Definition: Given two stagewise p-values \(p_1\) and \(p_2\), Fisher’s combination statistic is \(T=-2\log(p_1p_2)=-2\log(p_1)-2\log(p_2)\). Small p-values produce large values of \(T\), which count against \(H_0\).

Reference Distribution and Its Derivation

Suppose that under \(H_0\), \(p_1\) and \(p_2\) are independent and each has a Uniform(0,1) distribution. For a single Uniform(0,1) random variable \(U\), define \(Y=-2\log U\). For \(y\geq 0\),

$$ P(Y\leq y) =P(-2\log U\leq y) =P(U\geq e^{-y/2}) =1-e^{-y/2}. $$

The last expression is the cumulative distribution function of a chi-square random variable with two degrees of freedom. Thus \(-2\log p_1\) and \(-2\log p_2\) are independent chi-square variables with two degrees of freedom. Their sum has a chi-square distribution with four degrees of freedom.

$$ T=-2\log(p_1p_2)\sim\chi^2_4 \qquad \text{under }H_0. $$

For this reference distribution, the upper-tail probability at an observed value \(t\) is

$$ P(\chi^2_4\geq t)=e^{-t/2}\left(1+\frac{t}{2}\right), \qquad t\geq 0. $$

This formula gives the combined p-value. Equivalently, since \(t=-2\log(p_1p_2)\), it can be evaluated directly as

$$ p_{\mathrm{Fisher}}=p_1p_2\{1-\log(p_1p_2)\}. $$

The combined p-value is compared with \(\alpha\); reject \(H_0\) when \(p_{\mathrm{Fisher}}\leq\alpha\). At \(\alpha=0.05\), the upper-tail critical value for \(\chi^2_4\) is approximately 9.4877.

When Stage Two Is Adaptive

Let \(D_1\) denote all stage-one information available when the sample-size decision is made, and let the rule choose a stage-two sample size \(n_2=g(D_1)\). The relevant condition is that, under \(H_0\), the stage-two p-value is conditionally Uniform(0,1) given \(D_1\), including the selected value of \(n_2\). In symbols, for every relevant stage-one outcome,

$$ P(p_2\leq u\mid D_1)=u,\qquad 0\leq u\leq 1. $$

If \(p_1\) is also Uniform(0,1) under \(H_0\), then \(p_1\) and \(p_2\) are independent. To see this, for any \(a,b\in[0,1]\), use that \(p_1\) is determined by \(D_1\):

$$ P(p_1\leq a,\ p_2\leq b) =E\left[I(p_1\leq a)P(p_2\leq b\mid D_1)\right] =E\left[I(p_1\leq a)b\right] =ab. $$

This is the joint distribution of two independent Uniform(0,1) variables, so the chi-square reference distribution follows. The argument shows why adaptive sample size can be compatible with Fisher combination: conditional calibration, rather than a fixed value of \(n_2\), is what supports the reference law.

A standard setting satisfying the condition is one in which the stage-two observations are fresh and independent of stage one, the chosen \(n_2\) is fixed once \(D_1\) is observed, and the stage-two test produces an exact null p-value for that sample size. For example, with a known standard deviation and a normally distributed test statistic under \(H_0\), the properly calibrated one-sided normal-tail p-value is uniform at each fixed \(n_2\). That exact conditional property must be justified for the actual analysis; it should not be assumed merely because the samples are separate.

Worked Example: Adaptive Sample Size With a Valid Stage-Two P-Value

Hypothetical design. A one-sided test has independent, normally distributed stagewise observations with known standard deviation. A pre-specified rule chooses \(n_2=80\) when the stage-one p-value is between 0.10 and 0.20, and \(n_2=120\) otherwise. In one trial, \(p_1=0.12\), so the rule selects 80 stage-two observations. The stage-two normal statistic is \(Z_2=1.80\), with larger values supporting the alternative. Calculate the Fisher combined p-value.

Under the null, conditional on the stage-one data and the selected sample size, the new stage-two observations have the null distribution. With the known standard deviation and a correctly standardized normal statistic, the one-sided stage-two p-value is therefore conditionally uniform. The assumed independence of the stage samples and the fixed-after-selection sample size support the required calibration here.

The stage-two p-value is the upper tail of a standard normal distribution:

$$ p_2=P(Z\geq 1.80)=1-\Phi(1.80)\approx 0.03593. $$

The product of the stagewise p-values is \(0.12(0.0359303)=0.00431164\). Hence

$$ T=-2\log(0.00431164)\approx 10.8929. $$

The combined p-value is

$$ P(\chi^2_4\geq 10.8929) =e^{-10.8929/2}\left(1+\frac{10.8929}{2}\right) \approx 0.02779. $$

At the 0.05 significance level, this result rejects \(H_0\). The decision is justified by the conditional validity argument, not by ignoring the adaptive rule. The p-value of about 0.02779 measures how often a \(\chi^2_4\) statistic would be at least this large under the calibrated null model; it is not the probability that \(H_0\) is true.

Worked Calculations and Interpretation

Worked Example: Combining Two Stagewise P-Values

Hypothetical data. Suppose two independent stages yield valid one-sided p-values \(p_1=0.04\) and \(p_2=0.20\), and both p-values support the same pre-specified alternative. Find the Fisher statistic and combined p-value.

The product is \(0.04(0.20)=0.008\), so

$$ T=-2\log(0.008)=9.65663. $$

Using the chi-square tail formula,

$$ p_{\mathrm{Fisher}} =e^{-9.65663/2}\left(1+\frac{9.65663}{2}\right) =0.008\{1-\log(0.008)\} \approx 0.04663. $$

At \(\alpha=0.05\), reject \(H_0\). The second-stage p-value alone is not small, but Fisher’s statistic combines the evidence from both stages. The interpretation depends on the two p-values satisfying the stated null calibration and independence conditions.

Worked Example: A Combined Result That Does Not Reject

Hypothetical data. Let \(p_1=0.08\) and \(p_2=0.20\) be independent Uniform(0,1) p-values under \(H_0\). Compute the test result at \(\alpha=0.05\).

Their product is \(0.08(0.20)=0.0160\), giving

$$ T=-2\log(0.0160)=8.27033. $$

The upper-tail probability is

$$ p_{\mathrm{Fisher}} =e^{-8.27033/2}\left(1+\frac{8.27033}{2}\right) =0.0160\{1-\log(0.0160)\} \approx 0.08216. $$

Because \(0.08216>0.05\), do not reject \(H_0\) at the 0.05 level. This is not evidence that the null hypothesis has been established; it means the combined test did not reach the specified rejection threshold.

Assumptions, Diagnostics, and Alternatives

Before using the \(\chi^2_4\) reference distribution, check the conditions that create it. For the \(\chi^2_4\) reference distribution, the p-values must be independent and Uniform(0,1) under the global null; compatible, pre-specified directions may aid interpretation of the combined evidence but are not required for this null calibration. Under \(H_0\), each stagewise p-value must have the required uniform distribution, and \(p_2\) must remain uniform conditional on the stage-one information used for adaptation. If the design uses conservative p-values rather than exact uniform p-values, the exact chi-square claim does not automatically follow; the null calibration of the combined test needs separate justification.

Practical diagnostics focus on the design and analysis process, not on testing independence from the observed p-values in a single trial. Check that the sample-size rule was specified and implemented as intended, that stage-two data were not used in the rule before the second-stage test was formed, and that the calculation of \(p_2\) matches the selected sample size and test. Investigate whether exclusions, outcome definitions, or analysis choices were changed after looking at stage-two outcomes. Such changes can undermine conditional calibration even if the sample-size rule itself was permissible.

Fisher combination is one option among several. The inverse-normal combination test transforms stagewise p-values to normal scores and combines them using chosen weights; it can offer a transparent way to express the relative contribution of stages. Fisher’s method instead uses a fixed transformation of the p-values and, under the stated assumptions, a \(\chi^2_4\) reference distribution. An unadjusted pooled analysis may be efficient in a fixed design, but when the sample size or analysis is adapted, its ordinary fixed-design calibration cannot simply be presumed valid. The methods have different structures; none removes the need to justify the null distribution used for the final test.

Fisher’s test also has limits. It is a test of the global null, not an estimator of the treatment effect, and the combined p-value does not quantify the effect’s size or precision. The method can be sensitive to a very small p-value from one stage, so the combined result need not mean that both stages separately provide strong evidence. If the stage-two p-value is not conditionally valid, the standard calculation may not control Type I error as claimed. Incompatible alternatives can make the combined result difficult to interpret, but do not affect that null calibration when the p-values are independent and Uniform(0,1) under the null being tested.

Common Mistake

Common mistake: “The stage-two sample size was chosen using stage-one data, so the two p-values cannot be independent; Fisher’s chi-square test is therefore invalid.” This reasoning is too broad. Dependence is not ruled out merely by describing the stages, but conditional uniformity of \(p_2\) given the information used for adaptation implies independence from \(p_1\) when \(p_1\) is determined by that information. The correction is to establish the conditional null distribution of \(p_2\) for every relevant stage-one outcome and selected sample size. If that condition fails, the usual \(\chi^2_4\) calibration is not justified.

Check Your Understanding

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

  1. Under what conditions does Fisher’s two-stage statistic have a \(\chi^2_4\) null distribution?
  2. Why can a stage-two sample size depend on stage-one data without necessarily invalidating Fisher combination?
  3. For independent valid stagewise p-values \(p_1=0.10\) and \(p_2=0.10\), calculate \(T\) and the Fisher combined p-value. Does the result reject at \(\alpha=0.05\)?
  4. A researcher selects \(n_2\) using stage-one data, then uses a stage-two p-value whose null distribution is not calibrated conditional on the selected \(n_2\). Can the usual \(\chi^2_4\) reference law be assumed? Explain.

Solutions. (1) Under \(H_0\), the two stagewise p-values must be independent Uniform(0,1) variables; in an adaptive design, a sufficient route is for \(p_1\) to be uniform and for \(p_2\) to be conditionally uniform given the stage-one information. (2) If the stage-two p-value is conditionally uniform for each relevant stage-one outcome and selected sample size, it is independent of \(p_1\), so the Fisher reference distribution remains valid. (3) The product is \(0.01\), and \(T=-2\log(0.01)=9.21034\). The combined p-value is \(0.01\{1-\log(0.01)\}\approx 0.05605\), so do not reject at 0.05. (4) No. Without conditional calibration, the stated assumptions for the \(\chi^2_4\) distribution have not been established, so a different justification or calibrated method is needed.

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