Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 36 / 3 / a / Solution

Fix any estimator $\hat{\eta }$ and put $A(x)=d(\hat{\eta }(x),f)$, $B(x)=d(\hat{\eta }(x),g)$ and $D=d(f,g)$. The triangle inequality for the metric gives $A+B\ge D$. 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.