= Metric squared-loss two-point bound
{title2=$R^*\geq\tfrac14d(f,g)^2(1-\operatorname{TV}(P_f,P_g))$}
For any <estimator> $T$ in a <metric> parameter space, $d(T,f)^2+d(T,g)^2\geq d(f,g)^2/2$ by the <triangle inequality>. Average the two <risk functions> and replace their <probability density functions> by their minimum. This proves $\max(R_f,R_g)\geq d(f,g)^2\int\min(p_f,p_g)/4$. The overlap equals one minus <total variation distance>. The same proof applies to squared error for a real-valued <statistical functional> of the parameter, without requiring that functional to be injective.
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