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For $P\ll Q$, the chi-squared divergence is ${\chi }^2(P\Vert Q)={\mathbb{E}}_Q[(dP/dQ-1{)}^2]$. Write $L_j=d{Q}_j/d{P}_0$ for the likelihood ratios of the complete $n$-ball sample. Since $d{P}_1/d{P}_0=d^{-1}{\sum }_j{L}_j$ and ${\mathbb{E}}_0{L}_j=1$, the second moment of a mixture likelihood ratio givesIf instead $L_j$ denotes a one-ball likelihood ratio, its cross moment must be raised to the $n$th power for the full sample, by independence.

Let $a={\pi }^2({\epsilon }^{-1}-1)$. From the supplied second moment of a mixture likelihood ratio, dropping a nonpositive term givesThe last step uses the assumed separation. It has a real square root only when $16{\nu }^{2}d\ge 1$; otherwise no parameters satisfy it. The chi-squared testing lower bound follows from the Cauchy-Schwarz inequality: $\Vert P_1-P_0{\Vert }_{\mathrm{TV}}\le \frac{1}{2}\sqrt{{\chi }^2(P_1\Vert P_0)}\le 2\nu$. Every statistical hypothesis testing rule has sum of its Type I error and Type II error at least $1-\Vert P_1-P_0{\Vert }_{\mathrm{TV}}$. Its larger error is at least half the sum, hence

Define $P_{\theta ,n}=d^{-1}{\sum }_{S\in \mathcal{S}}{P}_{\theta ,S}^{\otimes n}$. Expanding its chi-squared divergence by the second moment of a mixture likelihood ratio, and using the supplied one-observation cross moment and independence, givesHere $S,T$ are independent uniform members of the specified family of cyclic intervals. Their overlap $R=|S\cap T|$ does not have the hypergeometric distribution of arbitrary uniform $k$-subsets. The family's geometry must be retained.

Let $x={\theta }^2{R}^2/k^2<1/4$. Since $-\log (1-x)\le x/(1-x)\le 2x$ and $R^2/k^2\le R/k$, we obtain $(1-x{)}^{-n/2}\le e^{nx}\le e^{n{\theta }^{2}R/k}$. Taking the expected value proves the upper bound. The factor $n$ in the last exponent is present in the PDF and missing in the TeX aid.