How tight can a simulator-assisted robot certificate be? A 1/m floor, even for a perfect simulator

A short proof that no valid 95% lower bound on a robot policy's mean return from m real trials plus any simulation can beat a gap of about 2.3/m.

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Three earlier steps of our sim2real-inference programme found that certified bounds on a robot policy's performance did not get tighter when a simulator was added. Was that our constructions, or the problem?

Result. Take any 95% lower confidence bound on mean return in [0, 1] that is valid for every joint law of real and simulated outcomes, and give it m paired real and simulated runs plus unlimited simulation. Then there is a law, with a perfect simulator, under which the bound sits at least 1 - (0.10)^(1/m) (about 2.3/m) below the truth at least half the time.

Why. Compare a policy that always succeeds with one that fails with probability eps. The simulator is identical in both. All m real trials look identical with probability (1 - eps)^m, so a bound valid for the second policy must stay below 1 - eps on the first with that probability.

Checked. A Monte Carlo of the indistinguishability probability matches 0.10 for m from 10 to 1000. On the all-successes sample the exact bound is within about 1.3 times of the floor.

What it says about our earlier steps. At m = 240 the floor is 0.0095. Our bounded-discrepancy bound had a gap of 0.193 and plain empirical Bernstein 0.060, so there is room above the floor, but no simulator can remove the 1/m term. It can only shrink the part that scales with the square root of the variance. What is still open is a bound that depends on the real-sim correlation.

Code, log and tests: research/sim2real-inference. AI-assisted; the proof is two lines but has not been peer reviewed.

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