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There is a genuine qualification: the first printed inequality is valid as a useful general assertion for $0<\eta <1$, but is false for arbitrary $\eta >0$. Also the displayed likelihood ratio presupposes $P_{f_1}\ll P_{f_0}$. We first prove the intended bound and then give a counterexample to the unrestricted quantifier.

For any estimator $T$, set $A=\{d(T,f_0)<r_n\}$. By the triangle inequality, $d(T,f_1)\ge r_n$ on $A$, while $d(T,f_0)\ge r_n$ on $A^c$. HenceLet $C=\{|Z-1|\le \eta \}$ with $0<\eta <1$. On $C$, $Z\ge 1-\eta$. ThereforeThe last step is Markov inequality. Taking the supremum over the model and then the infimum over estimators proves the intended metric two-point risk bound. At $\eta =1$ the right side is zero. For $\eta >1$ a negative right side is also harmless, but a positive right side need not be a lower bound.

Here is an explicit counterexample with positive probability density functions and $n=3$. Let $A_0=[0,1/2)$, $\delta =1/100$, and take the model consisting ofUse the L2 norm metric, for which $D=d(f_0,f_1)=2(1-2\delta )$, and take $r_n=D/2$ for every $n$. The majority-half decision selects $f_0$ if at least two observations lie in $A_0$. Each error probability is $p=3{\delta }^2-2{\delta }^3=0.000298$, so its maximum normalized risk function is $2p=0.000596$. This is also the likelihood-ratio test, giving ${\mathbb{E}}_{f_0}|Z-1|=2(1-2p)$. At $\eta =3/2$ the proposed right side is $1/12-2p/3>0.083$, contradicting the risk of this explicit estimator. Thus the intended range restriction cannot be omitted.

For the second request use the intended L2 norm metric $d(f,g)=({\int }_0^1(f-g{)}^2{)}^{1/2}$. The PDF defines its square, not the square itself as a metric. PutBoth probability density functions belong to the model, and $d(f_0,f_{1,n})=a_n=2r_n$. The two-point choice may depend on $n$, because the lower bound is applied separately at each sample size. Under $f_0$, the product likelihood ratio $Z_n={\prod }_i(1+a_{n}g(X_i))$ satisfiesThis is the product chi-squared divergence. By Cauchy-Schwarz inequality, ${\mathbb{E}}_0|Z_n-1|<1/4$. With $\eta =1/2$, the proved metric two-point risk bound has normalized right side greater than $1/8$. ConsequentlyThe proof works for every $n\ge 1$ despite the introductory $n>2$. Measurability of the functions defining the probability density functions is understood. This establishes the requested lower bound; it does not assert a matching upper bound over this unrestricted class.

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.

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.

Two independent cyclic intervals intersect with probability at most $\min \{1,(2k-1)/d\}\le 2k/d$. For $k\le d/2$, precisely the $2k-1$ starting-point offsets $-(k-1),\dots ,k-1$ can overlap; for larger $k$, the upper bound is automatic. The cyclic interval overlap bound and the preceding chi-squared divergence estimate therefore giveAs in 1(e), the square-root hypothesis is meaningful when $\alpha d\ge 1$. The chi-squared testing lower bound bounds the total variation distance by $\sqrt{2{\alpha }^{\epsilon }}/2$. The worst individual Type II error dominates the error under the uniform mixture model. ThusTaking $\alpha$ small makes this lower bound informative. This argument uses the actual cyclic-interval alternative, without substituting a different family of subsets.

The chain is a random walk on the finite abelian group $(\mathbb{Z}/n\mathbb{Z}{)}^d$, so its stationary distribution is uniform and its characters diagonalize the transition operator. For one coordinate and $\theta =2\pi k/n$, the eigenvalue isUniformly in $p\in [0,1]$,for an absolute $c>0$. Hence the one-coordinate chi-squared distance after $t\ge n^2$ is at most $C{e}^{-ct/n^2}$. The coordinates evolve independently, so the product formula for chi-squared distance givesThe chi-squared divergence bound on total variation distance now yieldsuniformly in $p_1,\dots ,p_d$.

For $d=1$, the first nonconstant character has eigenvalue modulus $1-O(n^{-2})$, uniformly in $p$. Testing against its real or imaginary part gives a fixed positive total-variation distance until time $c{n}^2$. Thus

The pushforward measure of $\mu$ under $g$ isThe Lebesgue decomposition theorem says that for sigma-finite measures $P,Q$ there are unique measures $P_{\mathrm{ac}}$ and $P_{\mathrm{s}}$ such that

For a convex lower-semicontinuous function $f$ with $f(1)=0$, choose any measure $\lambda$ dominating probability measures $P,Q$, put $p=dP/d\lambda$ and $q=dQ/d\lambda$, and define the f-divergenceusing the lower-semicontinuous perspective value when $q=0$. This definition includes the singular part and is independent of $\lambda$.

To prove the data processing inequality for f-divergences, let $\mathcal{G}=\sigma (g)$ and use $\lambda =P+Q$. The densities of $g_{\ast }P$ and $g_{\ast }Q$, pulled back to $\mathcal{G}$, are ${\mathbb{E}}_{\lambda }(p\mid \mathcal{G})$ and ${\mathbb{E}}_{\lambda }(q\mid \mathcal{G})$. Since the perspectiveis jointly convex, conditional Jensen inequality givesIntegration proves

The chi-squared divergence iswhen $P\ll Q$, and is infinite otherwise.

Fix a probability measure $Q$. Let $J$ be uniform on $\{1,\dots ,M\}$ and, conditionally on $J=j$, draw $X$ from $P_j$. Compare this joint law with the reference law under which $J$ is uniform and independent of $X\sim Q$. Forthe target expectation is $M^{-1}{\sum }_j{P}_j(A_j)$, while its reference expectation isbecause the $A_j$ form a partition. Its reference variance is $M^{-1}(1-M^{-1})$. The Cauchy-Schwarz inequality applied to the likelihood ratio givesThe last divergence separates over $J$:Taking the infimum over all probability measures $Q$ proves the required inequality.

The elementary logarithm inequality ${\log }_{2}u\le ({\log }_{2}e)(u-1)$ givesSince ${\sum }_x(P(x)-Q(x))=0$,This proves $D(P\Vert Q)\le (\log e){\chi }^2(P\Vert Q)$, relating relative entropy to chi-squared divergence.

If $L_j=d{P}_j/dQ$ and the nonnegative mixture weights sum to one, the likelihood ratio of the mixture model is $L={\sum }_j{w}_j{L}_j$. Expanding ${\mathbb{E}}_Q{L}^2-1$ proves the displayed chi-squared divergence identity. For independent random variables sampled from a fixed component, one-observation cross moments multiply across observations. The component is drawn once for the whole sample; redrawing it per observation gives a different mixture model.