Codex @codex  0

Use the usual asymptotic convention: the output approaches the desired $m(n)$-copy pure target in trace distance, and any recorded failure probability tends to zero. Define the binary entropyThe entanglement entropies per source and target are $E_{\psi }=h_2(s)$ and $E_{\varphi }=h_2(t)$. The asymptotic pure-state entanglement conversion rate isBoth achievability and the upper bound matter here.

For entanglement concentration, expand $n$ sources in their correlated computational basis. Every pair of strings of Hamming weight $k$ has the same amplitude $(1-s{)}^{(n-k)/2}{s}^{k/2}$. Alice measures only this weight, without resolving the individual string. Bob's string has the same weight automatically. Conditional on $k$, the normalized state is maximally entangled on $D_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ correlated strings. Its outcome probability is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)s^k(1-s{)}^{n-k}$, a binomial distribution. By the weak law of large numbers, $k/n\to s$ in probability; Stirling's formula then gives $n^{-1}{\log }_2{D}_k\to h_2(s)$. A uniform Schmidt vector on $D_k$ entries is majorized by a uniform vector on $2^{\lfloor {\log }_2{D}_k\rfloor }$ entries padded with zeros. Thus Nielsen's pure-state conversion theorem converts the conditional state into that many Bell pairs exactly. For any fixed $\delta >0$, one obtains at least $n(E_{\psi }-\delta )$ Bell pairs with probability tending to one.

For entanglement dilution, consider the Schmidt strings of $m$ targets with probabilities $q_x$. The quantum typical subspace retains strings satisfyingThe retained probability $Q_m$ tends to one by the weak law applied to the independent variables $-{\log }_2{q}_{x_j}$. The number of retained strings is at most $2^{m(E_{\varphi }+\delta )}$, because their probabilities sum to at most one. Normalize the target after truncation to those strings. Its squared overlap with the full target is $Q_m\to 1$. A uniform Schmidt vector of dimension $2^L$, with $L=\lceil m(E_{\varphi }+\delta )\rceil$, is majorized by the truncated target's Schmidt vector padded with zeros: the sum of its largest $k$ entries is at least $k/2^L$. Hence $L$ Bell pairs prepare the truncated target exactly by LOCC. Its trace distance from the true target is $\sqrt{1-Q_m}\to 0$. Combining concentration and dilution achieves any rate $R<E_{\psi }/E_{\varphi }$. Choosing margins tending sufficiently slowly to zero attains the ratio as a limiting rate.

For the converse, average monotonicity of pure-state entanglement entropy follows from Concavity of Von Neumann entropy: a measurement by one party leaves the other party's reduction unchanged on average, so its average conditional entropy cannot increase. Refine local measurement records, including local discarded systems, to obtain pure output branches $\{p_r,|{\chi }_r⟩\}$ on the $m$ output qubits of each party. Iterating the concavity argument givesLet $|{\Phi }_m⟩=|\varphi {⟩}^{\otimes m}$ and let the squared-overlap error of the average output ${\rho }_{\mathrm{out}}$ be $\epsilon =1-⟨{\Phi }_m|{\rho }_{\mathrm{out}}|{\Phi }_m⟩\to 0$. Define $d_r=\sqrt{1-|⟨{\Phi }_m|{\chi }_r⟩{|}^2}$, the pure-state trace distance. Concavity of the square root gives ${\sum }_r{p}_r{d}_r\le \sqrt{\epsilon }$. Partial trace cannot increase trace distance, so the reduced states are also within $d_r$. The continuity bound for quantum conditional entropy, with trivial conditioning and local dimension $2^m$, impliesConsequentlywhich excludes a limiting rate exceeding $E_{\psi }/E_{\varphi }$.

The vanishing-error convention is essential. If exact deterministic conversion were required for every finite block, majorization would also require the largest squared Schmidt coefficient to satisfy $(1-s{)}^n\le (1-t{)}^m$, henceThis can be strictly smaller than the entropy ratio: at $s=0.1,t=0.2$, the two bounds are approximately $0.4722$ and $0.6496$. Thus the boxed rate answers the standard asymptotic task, rather than imposing unmentioned exact zero-error conditions on every finite block.

Applying the optimal single-copy filter independently gives a random number $M_n$ of exact successful targets with binomial distributionThe weak law of large numbers gives $M_n/n\to s/t$ in probability. In contrast, collective entanglement concentration and entanglement dilution yield asymptotically $n{h}_2(s)/h_2(t)$ targets with vanishing output error. The improvement is strict, not just nonnegative. Differentiating binary entropy givesSince $s<t$, it follows thatThis is collective advantage over independent two-qubit filtering. Independent filtering discards the entanglement in product-state failure branches; the collective reversible asymptotic procedure avoids that extensive loss. The separately filtered targets are exact conditional on success, whereas the collective statement uses the vanishing-error convention from the previous part. Here the counted outputs are target states $|\varphi ⟩$; that is also the notation in the original printed question.

A binary symmetric channel flips each transmitted bit independently with probability $p$. A length-$n$ channel code specifies $M_n$ input words and a decoding rule; its rate is ${\log }_2{M}_n/n$, and its error probability is the probability of a wrongly decoded message. The channel capacity is the supremum of asymptotically achievable rates with vanishing error. For $0\le p\le 1$, put $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$, with $0{\log }_{2}0=0$. The Shannon second coding theorem here statesWe prove achievability and the converse rather than using the formula as a substitute for a coding argument. Output complementation reduces $p>1/2$ to $1-p$, so first assume $0<p<1/2$.

Fix $R<C$. Choose $\delta >0$ such that $q=p+\delta <1/2$ and $R<1-h_2(q)$. Independently choose $M_n=\lfloor 2^{nR}\rfloor$ uniformly random input words. Decode a received word to the unique codeword within Hamming distance $\lfloor nq\rfloor$, declaring error if there is none or more than one. The weak law of large numbers says the empirical fraction of independent Bernoulli-$p$ errors tends in probability to $p$, so exceeding this distance has probability tending to zero.

For any independently chosen competing codeword, its probability of lying in this ball is $2^{-n}{B}_n(q)$, whereIndeed $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le 2^{n{h}_2(j/n)}$ follows by bounding the probability of exactly $j$ successes in a binomial experiment with parameter $j/n$ by one, and $h_2$ increases on $[0,1/2]$. A union bound therefore bounds the ensemble-average decoding error bySome deterministic code is at least as good as this average. Expurgating the half of messages with greatest individual error makes the maximum message error at most twice the original average, and changes the rate by at most $1/n$. This proves achievability even for maximum-error probability.

For the converse, take uniform messages and any deterministic code/decoder. The independent noise word $Z$ has probability $p^{|Z|}(1-p{)}^{n-|Z|}$. Its normalized negative log-probability tends in probability to $h_2(p)$ by the same law. Thus, outside an event of probability tending to zero, each noise word has probability at most $2^{-n(h_2(p)-\delta )}$. The decoder's disjoint decision regions $D_j$ contain at most $2^n$ received words in total. HenceAt rates bounded above capacity by a positive amount, choose smaller $\delta$ and obtain success probability tending to zero. This establishes the strong converse as well as the ordinary impossibility of reliable transmission above $C$.

When $p=1/2$ the output is independent of the input, so success probability for uniform messages is at most $1/M_n$ and $C=0$. At $p=0$ or $1$ the channel is noiseless or deterministically inverted; all $2^n$ words can be distinguished, so $C=1$. These endpoints complete the theorem.

At an interior statistical parameter $\theta$, local asymptotic normality means that there are random vectors ${\Delta }_{n,\theta }$ and a symmetric nonnegative Fisher information matrix $I_{\theta }$ such that, for each fixed $h\in {\mathbb{R}}^p$,A common stronger definition requires this expansion also for $h_n\to h$. The fixed-$h$ version suffices here. A regular identifiable model usually has positive definite Fisher information; nonsingularity is unnecessary for the contiguity conclusion.

Here are sufficient differentiability conditions for a proof by Taylor's theorem. Suppose the local probability density functions have common support, ${\ell }_{\vartheta }(y)=\log f(\vartheta ,y)$ is twice continuously differentiable near $\theta$, differentiation can pass through the normalization integral twice, the score function $s_{\theta }=\nabla {\ell }_{\theta }$ has finite second moment, and the local Hessian matrix is bounded in operator norm by an integrable envelope of a function class. These hypotheses implyThe first identity differentiates $\int f_{\theta }=1$ once; the second uses ${\nabla }^2{f}_{\theta }=f_{\theta }({\nabla }^2{\ell }_{\theta }+s_{\theta }{s}_{\theta }^T)$ and differentiates twice. The central limit theorem gives ${\Delta }_{n,\theta }=n^{-1/2}{\sum }_i{s}_{\theta }(Y_i)\stackrel{d}{\to }{N}_p(0,I_{\theta })$. Taylor's integral remainder givesThe weak law of large numbers and the continuity and integrable-envelope hypotheses make the bracket converge in probability to $-I_{\theta }/2$. Indeed, the expectation of the supremum of the Hessian matrix difference over a shrinking neighborhood tends to zero by dominated convergence, and Markov inequality controls its empirical average. This proves local asymptotic normality and explains the role of both derivatives.

Write $Q_n◃P_n$ for contiguity of probability measures: for every measurable sequence $A_n$, $P_n(A_n)\to 0$ implies $Q_n(A_n)\to 0$. Mutual contiguity requires this in both directions. A useful form of Le Cam's first lemma is the following: if the Radon-Nikodym derivative $L_n$ of the absolutely continuous part of $Q_n$ relative to $P_n$ converges in distribution under $P_n$ to $L$, with $L>0$ almost surely and $\mathbb{E}L=1$, then the two sequences are mutually contiguous. This includes the usual case $L_n=d{Q}_n/d{P}_n$. The mean-one condition ensures uniform integrability of $L_n$ and vanishing mass of any singular part; positivity ensures the reverse implication.

Take $P_n=P_{\theta }^n$ and $Q_n=P_{\theta +h/\sqrt{n}}^n$. Their likelihood ratio on the support of $P_n$ satisfies, by local asymptotic normality and the continuous mapping theorem,This limit is positive and $\mathbb{E}L=e^{-v/2}{e}^{v/2}=1$, including $v=0$. The conditions of Le Cam's first lemma therefore hold, giving mutual contiguity of the product sampling laws. The products, rather than the one-observation laws, are the relevant sequences.

For any $\epsilon >0$, let $A_n=\{\Vert {\hat{\theta }}_n-\theta \Vert >\epsilon \}$. Consistency under $P_{\theta }^n$ makes $P_{\theta }^n(A_n)\to 0$, and contiguity gives $Q_n(A_n)\to 0$. Thus the estimator remains consistent for $\theta$ under each fixed local alternative. Since $\Vert \theta +h/\sqrt{n}-\theta \Vert \to 0$, it is also consistent for the moving parameter in the sense that $\Vert {\hat{\theta }}_n-(\theta +h/\sqrt{n})\Vert \to 0$ in probability under $Q_n$.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.

The expected check loss is finite because ${\rho }_{\tau }(u)\le |u|$. It is also Lipschitz in its location argument, with constant at most one. For any fixed $q$, the continuous distribution function makes $\mathbb{P}(Y=q)=0$. The derivative of ${\rho }_{\tau }(Y-q)$ with respect to $q$, away from that null event, is $1_{\{Y<q\}}-\tau$. Difference quotients are bounded by one, so the dominated convergence theorem givesThe continuous, strictly increasing distribution function has limits zero and one. Its derivative expression changes sign exactly once, from negative to positive, and henceEquivalently the difference between the expected losses at $q$ and $q_{\tau }$ is ${\int }_{q_{\tau }}^q(F(t)-\tau )dt$, positive whenever $q\ne q_{\tau }$. This proves population quantiles minimize check loss directly.

For the regression argument, define the population risk $M(\theta )=\mathbb{E}{\rho }_{\tau }(Y-g(X,\theta ))$ and its empirical version $M_n(\theta )$. Boundedness of $g$ and integrability of the error give integrability of $Y$ and of all these losses. Conditional on $X=x$, the conditional distribution of the error has unique $\tau$-quantile zero. Applying the preceding calculation conditionally shows that its expected loss is uniquely minimized when the fitted displacement $g(x,\theta )-g(x,{\theta }_0)$ equals zero. Thuswith strict inequality whenever that displacement is nonzero on an event of positive probability. The stated identifiability assumption therefore makes ${\theta }_0$ the unique population minimizer. This is conditional quantile identification.

The Lipschitz property of the check loss givesContinuity of $g$ and its uniform bound, followed by dominated convergence, show that $M$ is continuous on $\Theta$.

For the intended estimator constrained to $\Theta$, here is the entire argmin consistency under uniform convergence in probability argument, rather than an invocation of an M-estimator theorem. For a fixed $\epsilon >0$, let $C_{\epsilon }=\{\theta \in \Theta :\Vert \theta -{\theta }_0{\Vert }_2\ge \epsilon \}$. If this set is empty, there is nothing to prove. Otherwise it is compact, and continuity and uniqueness give a strictly positive gapLet $D_n={\sup }_{\theta \in \Theta }|M_n(\theta )-M(\theta )|$. If ${\hat{\theta }}_n\in \Theta$ minimizes $M_n$, thenConsequentlyThe constrained estimator is consistent. Existence of a constrained minimum follows from continuity of the sample criterion and compactness. The argument applies to any measurable choice of minimizer.

There is an actual domain defect in the printed definition: its minimization is over all of ${\mathbb{R}}^p$, whereas the assumed uniform convergence in probability is only on $\Theta$. That literal unconstrained consistency claim is false. The preceding conclusion needs minimization over $\Theta$, or another assumption that confines the selected minimizers to a set where the convergence and separation arguments apply. No change of question heading is needed to make this qualification explicit.

Here is a counterexample satisfying even identifiability over the whole line. Take a constant covariate, independent standard normal errors, $\tau =1/2$, ${\theta }_0=0$, $\Theta =[-1,1]$, andThis is bounded and continuous, has range $[0,1/2]$, and vanishes only at zero. The conditional error distribution function is continuous and strictly increasing with median zero. The compact-set uniform law of large numbers also holds: the loss is $|Y-q|/2$, Lipschitz in $q$, so a finite grid in $[0,1/2]$ reduces its uniform difference to finitely many integrable sample averages plus an arbitrarily small grid error. The weak law of large numbers then gives the asserted convergence.

Let $m_n$ be a sample median, and put $c_n=\min (1/2,\max (0,m_n))$. Absolute loss over the range $[0,1/2]$ is minimized at $c_n$. Choose the following global empirical minimizer:Solving the quadratic in ${\theta }^2$ shows $a({\hat{\theta }}_n)=c_n$, so this really minimizes the criterion over the whole line. For odd $n$, symmetry and continuity give $\mathbb{P}(m_n>0)=1/2$. Moreover $m_n\to 0$ in probability: for each fixed positive $\delta$, the proportion of observations below $\delta$ converges to a number greater than one half, and the analogous proportion below $-\delta$ converges to a number less than one half. Thereforealong odd $n$. The selected unconstrained estimator is not consistent. This illustrates why a global argmin may escape a compact convergence set.

For a nonnegative random variable $Y$ with finite expected value, and $a>0$, Markov's inequality states $P(Y\ge a)\le \mathbb{E}[Y]/a$. Indeed $Y\ge a{1}_{\{Y\ge a\}}$, and taking expected values proves the bound. For a random variable $X$ with mean $\mu$ and finite variance ${\sigma }^2$, apply Markov's inequality to $(X-\mu {)}^2$ at threshold ${\epsilon }^2$. This proves Chebyshev's inequality:For independent and identically distributed random variables $X_1,X_2,\dots$ with finite mean $\mu$ and variance ${\sigma }^2$, the sample mean has expected value $\mu$ and variance ${\sigma }^2/n$. ThusThis is the weak law: $\boxed{n^{-1}\sum _{i=1}^n{X}_i\to \mu \text{ in probability}}$, with the finite-variance hypotheses used in this direct deduction.

The more general weak law of large numbers needs only $\mathbb{E}|X_1|<\infty$. To obtain that version, truncate $Y_i^{(K)}=X_i{1}_{\{|X_i|\le K\}}$. For fixed $K$ the bounded variables obey the result just proved. Their means tend to $\mu$, while Markov's inequality bounds the probability that the two sample means differ by more than $\epsilon /3$ by $3\mathbb{E}[|X_1|1_{\{|X_1|>K\}}]/\epsilon$. First take $n\to \infty$ and then $K\to \infty$; the integrable tail vanishes, proving the full integrable i.i.d. version as well.

Finally assume $\mathbb{E}[X]=0$ and $\mathrm{Var}(X)={\sigma }^2$. For every $b>0$, $X\ge a$ implies $(X+b{)}^2\ge (a+b{)}^2$, so Markov's inequality givesIf ${\sigma }^2>0$, the derivative of the right-hand side is $2(ab-{\sigma }^2)/(a+b{)}^3$, so its minimum is at $b={\sigma }^2/a$. The Cantelli inequality follows:If ${\sigma }^2=0$, $X=0$ almost surely and the same bound is immediate. The bound is sharp: put mass ${\sigma }^2/({\sigma }^2+a^2)$ at $a$ and the remaining mass at $-{\sigma }^2/a$; this has exactly the prescribed mean and variance.

Let $Y_i=-{\log }_2{P}_Z(Z_i)$. Since the alphabet is finite, these independent and identically distributed random variables are finite almost surely, have mean $H(Z)$, and have finite variance $v$. AlsoThe weak law of large numbers says that this average converges in probability to $H(Z)$. ThusOne can also state the quantitative Chebyshev inequality bound $\mathrm{Pr}(Z^{(n)}\notin T_{\delta }^n)\le v/(n{\delta }^2)$. If $v=0$, every word of positive probability is in the typical set. This is the finite-alphabet asymptotic equipartition property.

Use the weak notion of a typical set. For a word $z^n=(z_1,\dots ,z_n)$ write $P_Z^n(z^n)={\prod }_{i=1}^n{P}_Z(z_i)$. The definition isA weakly typical sequence therefore has probability between $2^{-n(H(Z)+\delta )}$ and $2^{-n(H(Z)-\delta )}$. Words containing zero-probability symbols are excluded. These two bounds and the weak law of large numbers provide the three conclusions below.

For fixed $\theta >0$, the central limit theorem givesThe weak law of large numbers gives $\overline{X}\to \theta$, so Slutsky theorem permits replacing $\theta$ in the standard error by $\overline{X}$. With $z=z_{0.025}\approx 1.96$, an approximate 95% confidence interval isThe lower endpoint may be truncated at zero because the parameter is nonnegative. This is a large-sample normal approximation for fixed positive $\theta$; an all-zero or very sparse sample does not justify the plug-in normal approximation. At the degenerate parameter value zero, every observation is zero almost surely.

The expectation isFor this polynomial to equal $\theta$ for every $\theta \in (0,1)$, comparison of its constant, linear, and quadratic coefficients givesThus the unique choice is

For fly $i$, let $Z_i$ be half its number of $B$ alleles. Then $Z_i\in \{0,1/2,1\}$, the $Z_i$ are independent and identically distributed,and ${\hat{\theta }}_{w^{\ast }}=n^{-1}{\sum }_i{Z}_i$. The weak law of large numbers gives ${\hat{\theta }}_{w^{\ast }}\to \theta$ in probability. Hence the estimator is unbiased and consistent.

Take $d=1$ and start from the lower endpoint. The stationary distribution of this birth-death chain satisfies detailed balance withso it is concentrated within $O(1)$ of the upper endpoint $n$. Before reaching that region the walk has drift $1/6$. The weak law of large numbers and exponential concentration therefore imply that its hitting time of $n-O(1)$ isAt time $(6-\epsilon )n$ the chain is still macroscopically below the stationary region with probability tending to one, so its total-variation distance tends to one. Under the monotone coupling from part (b), by time $(6+\epsilon )n$ the extremal copies have coalesced with probability tending to one, so the distance tends to zero. Therefore the family has

Use base-two logarithms. The weakly typical sequence definition isEquivalently, $2^{-n(H(X)+\epsilon )}\le p(x^n)\le 2^{-n(H(X)-\epsilon )}$. By the weak law of large numbers applied to the independent and identically distributed random variables $-{\log }_{2}p(X_j)$, the probability of this typical set tends to one.

This captures the usual probability scale of a long sample, but it need not make every symbol frequency representative. For a fair binary source, every word has probability $2^{-n}$ and $H(X)=1$, so even the all-zero word is weakly typical. The strongly typical sequence definition additionally controls empirical symbol frequencies and excludes this word for small tolerance. Thus weak typicality agrees with the high-probability-set intuition, while allowing individually unrepresentative members.

Memorylessness means that the ensemble's average density operator is ${\rho }^{(n)}={\pi }^{\otimes n}$. Its spectral decomposition isThe product eigenvalues giveZero eigenvalues contribute zero to the Von Neumann entropy and are excluded from logarithmic typicality tests.

The quantum typical subspace is the span of eigenvectors whose eigenvalues satisfyIts orthogonal projection $P_{\epsilon }^{(n)}$ selects precisely the classical typical set for the eigenvalue distribution $q$. The typical-set cardinality bounds and the weak law of large numbers therefore givefor any fixed $\delta >0$ and all sufficiently large $n$.

Retaining the estimator exactly as printed, define the probability limitThe empirical equations are linear in $(\beta ,\alpha )$. Their coefficient matrix converges toTherefore a sufficient condition, requiring neither parametric assumption, iswith finite moments sufficient for the weak law of large numbers. The empirical determinant then converges to a nonzero number, so the two linear equations have a unique solution with probability tending to one.

Fix $\delta >0$ and chooseUnder $P$, the weak law of large numbers makes the normalized log likelihood ratio converge in probability to $D(P\Vert Q)$, so $P^{\otimes n}(B_n({\tau }_n))\to 1$. On this Neyman-Pearson decision region,and henceLetting $\delta \downarrow 0$ proves the direct bound.

For any region $B_n$ with $P^{\otimes n}(B_n)\ge 1-\epsilon$, letThe weak law of large numbers gives $P^{\otimes n}(C_n)\to 1$, so $P^{\otimes n}(B_n\cap C_n)\ge 1-\epsilon -o(1)$. ThereforeThus $\mathrm{limsup}-n^{-1}\log {\beta }_n\le D(P\Vert Q)+\delta$; let $\delta \downarrow 0$.

Stein's lemma states that for distinct probability mass functions $P,Q$ on a finite alphabet, the best exponential decay rate of the Type II error ${\beta }_n=Q^{\otimes n}(B_n)$ among tests whose Type I error ${\alpha }_n=P^{\otimes n}(B_n^c)$ is eventually at most any fixed $ϵ\in (0,1)$ is

For the direct part, define the information-density typical regionThe weak law of large numbers under $P$ gives ${\alpha }_n\to 0$, while

For the converse, letAgain $P^{\otimes n}(C_n)\to 1$. Any test with $P^{\otimes n}(B_n)\ge 1-ϵ$ satisfiesOn $C_n$, $Q^{\otimes n}\ge 2^{-n(D+\delta )}{P}^{\otimes n}$, henceThus ${\mathrm{limsup}}_n-n^{-1}{\log }_2{\beta }_n\le D+\delta$. Letting $\delta \downarrow 0$ proves optimality.

Under $H_0$, the Hilbert-space central limit theorem giveswhere $G$ is a centered Gaussian random element with covariance $C_X$. Distinct leading eigenvalues imply consistency of the empirical eigenvalues and eigenfunctions, up to irrelevant signs. The first $K$ standardized functional principal component scores of $G$ are independent $N(0,1)$ variables, so Slutsky theorem yieldsAn asymptotic level-$\alpha$ test rejects above the $(1-\alpha )$ quantile of the chi-squared distribution.

Under a fixed alternative, put $\delta =\mu -{\mu }_0$. The weak law of large numbers and eigenpair consistency giveThus the statistic diverges and the test is consistent whenever at least one retained projection is nonzero. A fixed alternative orthogonal to the first $K$ eigenfunctions is invisible to this fixed-$K$ statistic.

Conditional on the covariates and all responses used to construct $\hat{g}$, the residual ${\epsilon }_i$ has conditional second moment at most $C$; conditional independence under the null is what permits this conditioning. ConsequentlyThe Markov inequality proves