Metric squared-loss two-point bound ($R^{\ast }\ge \frac{1}{4}d(f,g{)}^2(1-\mathrm{TV}(P_f,P_g))$)

For any estimator $T$ in a metric parameter space, $d(T,f{)}^2+d(T,g{)}^2\ge d(f,g{)}^2/2$ by the triangle inequality. Average the two risk functions and replace their probability density functions by their minimum. This proves $\max (R_f,R_g)\ge d(f,g{)}^2\int \min (p_f,p_g)/4$. 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.

For estimating the expected value over all continuous probability density functions on $[0,1]$, compare $f=1$ and $g_n(u)=1+(u-1/2)/\sqrt{n}$. Their means differ by $1/(12\sqrt{n})$, and ${\chi }^2(P_{g_n}\Vert P_f)=1/(12n)$. The chi-squared divergence of product measures bounds joint total variation distance by $1/2$. The metric squared-loss two-point bound then gives the displayed uniform minimax risk lower bound.