Chi-squared divergence of product measures 2026-10-07
If , independence gives . Since , this proves the displayed identity. The Cauchy-Schwarz inequality also gives . Thus perturbations of size have a uniformly bounded joint total variation distance, a useful input to a metric squared-loss two-point bound.
For estimating the expected value over all continuous probability density functions on , compare and . Their means differ by , and . The chi-squared divergence of product measures bounds joint total variation distance by . The metric squared-loss two-point bound then gives the displayed uniform minimax risk lower bound.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 36 3 a Solution Created 2026-10-03 Updated 2026-10-07
Fix any estimator and put , and . The triangle inequality for the metric gives . HenceThe larger of the two risk functions dominates their average. Using nonnegativity to replace both probability density functions by their minimum,The right side is independent of the estimator, so taking the infimum provesThis metric squared-loss two-point bound uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 36 3 b Solution Created 2026-10-03 Updated 2026-10-07
Apply the preceding argument to the two product observation laws and the real-valued estimator . Alternatively, directly useThe average of the two mean squared errors is at least times the overlap of the two joint probability density functions. Their overlap equalsTaking the infimum over estimators therefore yieldsHere is the minimax risk for this functional. The factor is the total variation distance, so the same metric squared-loss two-point bound applies to a functional even when distinct density parameters have the same functional value.