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

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Uniform payoff convergence does not imply gradient convergence

Statement refuted

For a parameter-dependent random payoff, pointwise weak convergence of its output law, even together with uniform convergence of its expected payoff, implies convergence of derivatives of the expected payoffs.

Facts & Assumptions

Given: Weak convergence of probability laws means convergence of integrals against every bounded continuous real function.

[F1]

The derivatives of sine and cosine are cosine and minus sine: $(\sin x)^{'} = \cos x$ and $\cos 0 = 1$ .

[F2]

Probability measures and probability spaces: the one-point space with unit mass is a probability space; Random elements and real random variables, Law or distribution of a random element and The Dirac set function at a point make each deterministic payoff's law the Dirac measure at its value.

[F3]

Expectation of a nonnegative or integrable random variable and Change of variables for expectation: on the one-point space an integrable deterministic variable has expectation equal to its value, and a bounded continuous test function integrates against an output law as the expectation of that test function of the output.

Counterexample

Given: Use a one-point probability space, so each payoff below is deterministic and its law is a Dirac probability measure.

1.1givenF2F3algebra

On the one-point probability space, for $n \geq 1$ put $Y_{n} (θ) = n^{- 1} \sin (n^{2} θ)$ and let $Y (θ) = 0$ . By [F2], their laws are $δ_{Y_{n} (θ)}$ and $δ_{0}$ . For every $θ$ , $∣ Y_{n} (θ) - Y (θ) ∣ \leq 1 / n$ , so their expectations converge uniformly to zero. For every bounded continuous $ψ$ , [F3] gives $\int ψ d δ_{Y_{n} (θ)} = ψ (Y_{n} (θ)) \to ψ (0) = \int ψ d δ_{0}$ ; hence the output laws converge weakly at every fixed $θ$ .

2.1F1step 1.1algebra∎

The expected payoff functions are $J_{n} (θ) = Y_{n} (θ)$ and $J (θ) = 0$ . By [F1], $J_{n}^{'} (0) = n^{- 1} n^{2} \cos 0 = n$ , whereas $J^{'} (0) = 0$ . Thus the derivatives do not converge, despite the stronger uniform convergence of the expected payoffs.

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