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Repeated unadjusted tests inflate global-null error

Example

Let K≥2 be an integer, let α∈(0,1), and under a global-null law P0 let P1,…,PK be mutually independent random variables with P0(Pi≤u)=u(0≤u≤1). At each of the K looks, reject the global null if the current value is at most α, without adjusting the threshold. The probability of at least one rejection is P0 ⁣(⋃i=1K{Pi≤α})=1−(1−α)K>α. The first rejection time is a stopping time for the filtration Fk=σ(P1,…,Pmin⁡(k,K)) for k≥0, with F0 trivial; thus the filtration stays constant after look K.

Facts & Assumptions

Given: Assume AC for the comparison with the A-page theorem. Under a global null law P0, K≥2 independent uniform [0,1] random variables P1,…,PK are observed successively. Here uniform means exactly P0(Pi≤u)=u for every u∈[0,1].

[F1]

Independent random elements: random variables are independent when the sigma-algebras they generate are independent.

[F2]

Independent random elements have product joint law: the joint law of finitely many independent random elements is the product of their marginal laws.

[F3]

Discrete stopping time: σ is a stopping time when {σ≤k}∈Fk for every k.

[F4]

Discrete Ville inequality and anytime threshold validity: under each null law, a nonnegative test supermartingale with initial expectation at most one has all-time threshold-crossing probability at most the threshold level.

[A1]

AC is required only for applying [F4]; the exact product probability calculation below uses no choice principle (The Axiom of Choice).

Verification

1.1given

For each i, the uniform-law assumption gives P0(Pi>α)=1−α.

1.2F3givenconstruct

Define σ=inf⁡{1≤i≤K:Pi≤α}, with inf⁡∅=∞. For every k≥0, {σ≤k}=⋃i=1min⁡(k,K){Pi≤α}∈Fk, where the union is empty for k=0. Thus [F3] makes σ a stopping time, and the rejection event by look K is {σ≤K}.

2.1F1F2step 1.1

By [F1]–[F2], the probability that every look misses is the product of the marginal probabilities, so P0(⋂i=1K{Pi>α})=(1−α)K.

3.1step 2.1algebra

The event of at least one rejection is the complement of the event in step 2.1. Therefore its probability is 1−(1−α)K.

4.1givenstep 3.1algebra

Since 0<1−α<1 and K≥2, (1−α)K<1−α, so step 3.1 is strictly greater than α.

5.1F4A1step 1.2step 4.1∎

[F4] controls arbitrary threshold stopping only when its nonnegative test-supermartingale hypotheses hold. The calculation above concerns repeated fixed-look uniform p-values and does not assume such a process, so the excess rejection probability is not a violation of Ville's inequality.

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