With densities $p,q$,
$$
\operatorname{TV}(P,Q)=\frac12\int|p-q|d\mu,
\qquad
H(P,Q)^2=\int(\sqrt p-\sqrt q)^2d\mu.
$$
By <Cauchy-Schwarz inequality>,
$$
\operatorname{TV}(P,Q)
\leq\frac12H(P,Q)
\left(\int(\sqrt p+\sqrt q)^2d\mu\right)^{1/2}
\leq H(P,Q).
$$
Also $(\sqrt p-\sqrt q)^2\leq|p-q|$, so $H^2\leq2\operatorname{TV}$.

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
$$
H^2(P^n,Q^n)=2-2\left(1-\frac12H^2(P,Q)\right)^n.
$$

<Le Cam two-point lemma> states, for squared-error estimation at parameter points $\theta_0,\theta_1$, that
$$
\inf_{\widehat\theta}\max_{j=0,1}
\mathbb E_j(\widehat\theta-\theta_j)^2
\geq\frac{(\theta_1-\theta_0)^2}{8}
\left(1-\operatorname{TV}(P_0,P_1)\right).
$$
Take $\theta_0=0$, $\theta_1=\delta=1/(4n)$. The one-observation uniform densities overlap on length $1-\delta$, so $H^2(P_0,P_1)=2\delta$. Therefore
$$
H^2(P_0^n,P_1^n)
=2-2(1-\delta)^n\leq2n\delta=\frac12.
$$
The first distance inequality gives $\operatorname{TV}(P_0^n,P_1^n)\leq1/\sqrt2$. Le Cam's lemma now yields
$$
\inf_{\widehat\theta}\sup_{\theta\in\mathbb R}
\mathbb E_\theta(\widehat\theta-\theta)^2
\geq\frac{1-1/\sqrt2}{128n^2}.
$$
This proves the claim with the displayed universal positive constant.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-210/4/solution