CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Correct level marginals do not justify a dependent product

Statement refuted

Correct marginal success probabilities at every level suffice for the product of level sample means to be unbiased when the level samples are dependent.

Facts & Assumptions

Given: Let $Ω = {1, 2, 3, 4}$ with uniform $P$ , $A_{1} = {1, 2}$ , $A_{2} = {1}$ , and use one sample at each level.

[F1]

Nested rare-event levels and conditional laws: The level probabilities are $p_{1} = P (A_{1})$ , $p_{2} = P (A_{2}) / P (A_{1})$ , and the target is $P (A_{2}) = p_{1} p_{2}$ .

[F2]

Independent fixed-level splitting: The two-level product estimator is the product of the two level indicator means; the prescribed algorithm requires independent level arrays.

[F3]

Independent splitting estimator moments and work: Under level independence, the product estimator has expectation $p_{1} p_{2}$ .

Counterexample

1.1F1givenconstruct

Let $U, V$ be independent Bernoulli $(1 / 2)$ . Construct the level-one sample by $X_{1} = 1 + V$ if $U = 1$ , and $X_{1} = 3 + V$ if $U = 0$ ; then $X_{1} \sim P$ . Construct $X_{2} = 1$ if $U = 1$ , and $X_{2} = 2$ if $U = 0$ ; then $X_{2} \sim P (\cdot ∣ A_{1})$ .

2.1F1F2step 1.1algebra

The correct marginal success probabilities are $P (X_{1} \in A_{1}) = 1 / 2$ and $P (X_{2} \in A_{2}) = 1 / 2$ . But the two indicators both equal $U$ , so their product has expectation $1 / 2$ , while $P (A_{2}) = 1 / 4$ . Thus dependence biases the product despite both correct marginals, and it violates the independence condition in [F2].

3.1F2F3step 2.1∎

The unbiasedness conclusion in [F3] has the stated independence hypothesis; it cannot be applied to this coupling. The target counterexample refutes the proposed implication.

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