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Repeated Significance Testing · Tutorial 8 of 699

How Dependence Changes Error Inflation

See how overlapping information changes the probability of at least one false rejection, and why dependence usually reduces—but does not remove—error inflation.

Beginner 10 min read

What You'll Learn

  • Distinguish per-look significance from the probability of any rejection.
  • Derive the correlation between two standardized statistics using nested data.
  • Compare independent and positively dependent rejection events numerically.
  • Explain when positive dependence reduces the chance of at least one rejection.
  • Identify why correlation alone does not determine familywise error in every setting.
  • Recognize the limits of using independence as a benchmark.

Why Dependence Matters

When a trial is tested repeatedly, the probability of a false rejection at any look depends not only on each test’s significance level but also on how the rejection events overlap. If two tests use much of the same data, their results tend to move together. A false rejection at one look is then more likely to coincide with a false rejection at another, so the tests create fewer distinct opportunities for a false positive than independent tests with the same per-look error rate.

That reduction is not the same as eliminating inflation. If either look can trigger a rejection, the relevant probability is the chance of at least one rejection under the trial-wise null hypothesis \(H_0\). Even with positive dependence, that probability can remain larger than the per-look significance level \(\alpha\).

Definition: Let \(A_j\) be the event that test \(j\) rejects \(H_0\) when \(H_0\) is true. The any-look error probability is \(P(A_1\cup\cdots\cup A_K\mid H_0)\), the probability of at least one false rejection across the \(K\) looks. The events \(A_j\) are positively dependent when their joint occurrence is more likely than it would be under independence; for two events, this means \(P(A_1\cap A_2\mid H_0)>P(A_1\mid H_0)P(A_2\mid H_0)\).

Dependence in Cumulative Tests

A standard model makes the source of dependence explicit. Suppose \(X_1,\ldots,X_{n_2}\) are independent observations from \(N(\mu,\sigma^2)\), with known \(\sigma\), and consider testing \(H_0:\mu=\mu_0\). At two planned looks, \(n_1<n_2\), define the standardized statistics

$$ Z_j=\frac{\sqrt{n_j}(\bar{x}_j-\mu_0)}{\sigma},\qquad j=1,2, $$

where \(\bar{x}_j\) is the mean of the first \(n_j\) observations. Under \(H_0\), each statistic has a standard normal distribution. The second statistic reuses the observations from the first look; the two statistics are therefore not independent.

To derive their correlation, write \(Z_j=S_j/\sqrt{n_j}\), where \(S_j=\sum_{i=1}^{n_j}(X_i-\mu_0)/\sigma\). Under \(H_0\), each summand has mean zero and variance one. The first \(n_1\) summands appear in both sums, while the remaining summands in \(S_2\) are independent of \(S_1\). Thus \(\operatorname{Cov}(S_1,S_2)=n_1\), and

$$ \operatorname{Corr}(Z_1,Z_2) =\frac{\operatorname{Cov}(S_1,S_2)}{\sqrt{n_1n_2}} =\sqrt{\frac{n_1}{n_2}}. $$

This correlation is positive, and it is larger when the looks are closer in sample size. The formula relies on independent observations, the normal model with known variance, the null hypothesis, and a cumulative design in which the earlier observations are retained. In other models, the exact correlation formula may differ.

Worked Example: Correlation From Nested Sample Sizes

Suppose a hypothetical trial has planned looks after 100 and 400 observations. Assume the observations satisfy the model above under \(H_0\), and each look uses all observations collected so far. Find the correlation between the two test statistics.

Substitute \(n_1=100\) and \(n_2=400\) into the cumulative-test formula:

$$ \operatorname{Corr}(Z_1,Z_2) =\sqrt{\frac{100}{400}} =\sqrt{0.25} =0.5. $$

The statistics have correlation \(0.5\), not zero. This is evidence that the two statistics move together under this model. It does not, by itself, give the probability of any rejection: that probability also depends on the rejection thresholds and the joint distribution of the test statistics.

How Overlap Changes the Any-Look Probability

For two rejection events, the probability of at least one rejection follows directly from inclusion–exclusion:

$$ P(A_1\cup A_2\mid H_0) =P(A_1\mid H_0)+P(A_2\mid H_0)-P(A_1\cap A_2\mid H_0). $$

The intersection is subtracted because outcomes in which both tests reject would otherwise be counted twice. If both marginal error probabilities are \(\alpha\), then the any-look probability is \(2\alpha-P(A_1\cap A_2\mid H_0)\). Under independence, the intersection equals \(\alpha^2\). Under positive dependence, it is larger than \(\alpha^2\), so the union probability is smaller than the independent benchmark. Unless the events overlap completely, however, the union can still exceed \(\alpha\).

Worked Example: Two Correlated Two-Sided Tests

Consider two hypothetical, standard normal test statistics under \(H_0\), each with correlation \(0.5\). Each test rejects when \(|Z_j|>1.96\), giving a marginal two-sided significance level of approximately \(0.05\). For this bivariate normal model, numerical integration gives a joint rejection probability of approximately \(P(A_1\cap A_2\mid H_0)=0.00925\). Calculate the chance of at least one rejection and compare it with the independent case.

Using inclusion–exclusion, the dependent any-look probability is

$$ P(A_1\cup A_2\mid H_0) \approx 0.05+0.05-0.00925 =0.09075 \approx 0.0908. $$

Thus the probability is about \(9.08\%\), still above the \(5\%\) per-test level. For two independent tests with the same marginal error probabilities, it would be

$$ 1-(1-0.05)^2 =1-(0.95)^2 =0.0975. $$

In this example, positive dependence reduces the any-look probability from \(9.75\%\) to about \(9.08\%\). It does not reduce it to \(5\%\). The numerical joint probability depends on the bivariate normal model and the stated cutoff; it should not be inferred from the correlation alone.

Worked Example: Rejection Events With Specified Overlap

For a separate hypothetical illustration, suppose two tests each have a \(0.05\) false-rejection probability, and their joint false-rejection probability is \(0.03\). Compare their any-look probability with that of independent tests.

The events are positively dependent because \(0.03>0.05\times0.05=0.0025\). Their union probability is

$$ P(A_1\cup A_2\mid H_0) =0.05+0.05-0.03 =0.07. $$

The any-look error is \(7\%\), which is larger than either marginal error of \(5\%\), but smaller than the independent benchmark of \(9.75\%\). The example illustrates the role of overlap directly: greater joint rejection probability means fewer outcomes in which at least one rejection occurs.

Positive Dependence: What It Does and Does Not Guarantee

For two tests with equal marginal error probabilities, the comparison with independence is exact once the joint rejection probability is known. If positive dependence means \(P(A_1\cap A_2\mid H_0)>\alpha^2\), then

$$ P(A_1\cup A_2\mid H_0) =2\alpha-P(A_1\cap A_2\mid H_0) <2\alpha-\alpha^2 =1-(1-\alpha)^2. $$

The conclusion is that the dependent tests have a lower any-look error probability than independent tests with the same marginal error probabilities. It does not say the error probability equals \(\alpha\), or that it is necessarily close to \(\alpha\).

For more than two looks, the same principle is that overlapping rejection events leave fewer distinct outcomes in their union. But knowing only that every pair of tests has positive correlation does not, in an arbitrary model, determine the probability of the union. The joint distribution of all the rejection events matters. In the cumulative normal setting, the statistics are jointly normal with nonnegative correlations, and their rejection patterns reflect the shared data. The exact familywise probability still depends on the schedule, the boundaries, and the joint distribution.

It is also important to distinguish correlation between test statistics from positive dependence between rejection events. A positive correlation describes how the numerical statistics co-vary. Rejection events are threshold-based, and their overlap depends on the thresholds and on the distribution’s shape. The bivariate normal example supports the expected comparison for its specific assumptions; it is not a universal conversion from a correlation value to an error probability.

Comparison, Diagnostics, and Limitations

ApproachWhat it tells youMain limitation
Independent-test benchmarkComputes the any-look error when tests are independent and their marginal error rates are known.Can overstate the error for positively dependent cumulative tests.
Exact joint-probability calculationUses the joint distribution and rejection regions to calculate the actual union probability.Requires a suitable model and can be difficult for many looks or complex boundaries.
Union boundGives \(P(A_1\cup\cdots\cup A_K\mid H_0)\leq\sum_{j=1}^K P(A_j\mid H_0)\), regardless of dependence.Does not use the overlap between events and may be loose.

A practical diagnostic question is whether later tests reuse earlier observations. If they do, treating their statistics as independent is usually inconsistent with the data structure. Also check that the event being discussed is the union of all rejection events, rather than a rejection at one particular look. The introductory material on per-look versus overall significance distinguishes these error rates; dependence explains why the overall rate cannot be obtained by simply treating each look as an independent new trial.

The independence calculation remains useful as a reference point, not as a default answer for cumulative tests. Conversely, positive dependence should not be used as a justification for ignoring repeated testing. The exact amount of inflation must be assessed for the design and analysis being used. These comparisons do not specify adjusted stopping boundaries or account for departures from the assumed model.

A Common Mistake

Wrong reasoning: “The interim statistics are positively correlated, so repeated testing does not inflate the false-positive rate.” Positive dependence generally increases overlap, but it does not make all rejection events identical. In the two-look example, the any-look probability remains about \(9.08\%\), not \(5\%\).

Correction: Calculate or otherwise justify the joint rejection probability under the null, then use the union probability. Correlation gives information about dependence, but it is not itself the familywise error rate.

Key takeaway: With the same per-look error rates, positive dependence generally makes the chance of at least one rejection lower than under independence because false rejections overlap more. The overall error can nevertheless remain above the per-look level, and its exact value depends on the joint distribution and rejection rules.

Check Your Understanding

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

  1. Two tests each have marginal false-rejection probability \(0.04\) and are independent. What is the probability of at least one false rejection?
  2. For two tests with marginal error probabilities \(0.05\) and joint rejection probability \(0.02\), calculate the any-look error. Is the dependence positive relative to independence?
  3. Under the cumulative normal model, find the correlation of statistics at sample sizes \(n_1=80\) and \(n_2=320\).
  4. Why does positive correlation between test statistics not, by itself, determine the any-look error probability?
  5. Does the union bound require the tests to be independent? State the bound for three tests with marginal error probabilities of \(0.05\) each.

Solutions. (1) Independence gives \(1-(1-0.04)^2=1-0.9216=0.0784\), or \(7.84\%\). (2) Inclusion–exclusion gives \(0.05+0.05-0.02=0.08\), or \(8\%\). Since \(0.02>0.05^2=0.0025\), the events have positive dependence by the two-event definition; the independent benchmark is \(0.0975\). (3) \(\sqrt{80/320}=\sqrt{0.25}=0.5\). (4) Rejection depends on both the joint distribution and the thresholds; correlation between numerical statistics alone does not specify the joint probability of their threshold-crossing events. (5) No. The bound is \(P(A_1\cup A_2\cup A_3\mid H_0)\leq0.05+0.05+0.05=0.15\), regardless of dependence.

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