Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 210 / 4 / Solution

With densities $p,q$,By Cauchy-Schwarz inequality,Also $(\sqrt{p}-\sqrt{q}{)}^2\le |p-q|$, so $H^2\le 2\mathrm{TV}$.

The Hellinger affinity is $\rho (P,Q)=\int \sqrt{pq}d\mu =1-H^2/2$. Product densities and Fubini's theorem give $\rho (P^n,Q^n)=\rho (P,Q{)}^n$, hence

Le Cam two-point lemma states, for squared-error estimation at parameter points ${\theta }_0,{\theta }_1$, thatTake ${\theta }_0=0$, ${\theta }_1=\delta =1/(4n)$. The one-observation uniform densities overlap on length $1-\delta$, so $H^2(P_0,P_1)=2\delta$. ThereforeThe first distance inequality gives $\mathrm{TV}(P_0^n,P_1^n)\le 1/\sqrt{2}$. Le Cam's lemma now yieldsThis proves the claim with the displayed universal positive constant.