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Write $\ell (x,\theta )=\log f(x,\theta )$ and let ${\dot{\ell }}_{{\theta }_0}(x)$ be its score function. A second-order Taylor expansion around ${\theta }_0$ gives, for fixed $h$,where every ${\theta }_{i,n}$ lies between ${\theta }_0$ and ${\theta }_0+h/\sqrt{n}$. The mean-zero score identity, the Fisher information matrix, and the multivariate central limit theorem giveThe assumed regularity and a uniform law of large numbers giveConsequently, by the Slutsky theorem,

For the stated Gaussian shift model, direct expansion of the two normal distribution densities givesSince $I({\theta }_0)X\sim N(0,I({\theta }_0))$, this has exactly the limiting distribution above. Equivalently, both limits are normal with mean $-\frac{1}{2}{h}^{T}I({\theta }_0)h$ and variance $h^{T}I({\theta }_0)h$. This is the local asymptotic normality of the model at ${\theta }_0$.

By the strong law of large numbers,almost surely. Brownian scaling and a maximal inequality show that changing Brownian time by $o(n)$ changes its value by $o_{\mathbb{P}}(\sqrt{n})$; explicitly, first restrict to $|T_n-n{\sigma }^2|\le \delta n$, bound the Brownian maximum over a time interval of length $2\delta n$, and then let $\delta \downarrow 0$. Consequentlyin probability.

Butfor every $n$. Since $B_{T_n}$ has the law of $S_n$, Slutsky theorem proves the Central limit theorem from the Skorokhod embedding:

Let $z_{1-\gamma /2}$ be the corresponding standard normal distribution quantile. Slutsky theorem permits replacing $\beta$ by the consistent estimator $\hat{\beta }$, giving the confidence intervalAfter standardization, the preceding convergence in distribution and continuity of the normal distribution show that its coverage tends to $1-\gamma$.

Under $H_0$, conditional independence givesMultiplication by the measurable sign $s(z_1)$ and the law of total expectation prove the first identity.

WriteThenConditional orthogonality under $H_0$ and the variance bounds in (ii) giveand similarly for the term containing ${\xi }_i{a}_i$. The Cauchy-Schwarz inequality bounds the last term bywhich is $o_p(n^{-1/2})$ by $n{\mathrm{MSPE}}_f{\mathrm{MSPE}}_g\to 0$.

The leading summands ${ϵ}_i{\xi }_{i}s(z_i)$ are independent, centered, and have variance $\mathrm{Var}({ϵ}_1{\xi }_1)$ because $s^2=1$. The central limit theorem and consistency of ${\tau }_D$ therefore give, by the Slutsky theorem,This is the generalized covariance measure statistic.

Without the null, condition first on $(x_1,z_1)$. Sincethe term involving $\mathbb{E}(x_1\mid z_1)$ vanishes, givingFor the specified alternative, independence and $\mathbb{E}{x}_1=0$ make the right sideThe constant choice $s=1$ gives zero because $\mathbb{E}h(z_1)=0$, so no first-order power is expected. Takinginstead gives $\mathbb{E}(x_1^2)\mathbb{E}|h(z_1)|>0$, producing asymptotic power.

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.

Letand define the functional principal component scoresUse empirical centered scores ${\hat{\xi }}_{ik},{\hat{\eta }}_{il}$ from the sample covariance eigenfunctions and formThe finite-dimensional statistic is

Under $H_0$, $Y=\epsilon$ is independent of $X$. The population cross-covariances vanish, andThe multivariate central limit theorem, consistency of the empirical eigenpairs, and Slutsky theorem therefore giveRejecting above the $(1-\alpha )$ quantile gives an asymptotic level-$\alpha$ test.

Under a fixed alternative, $\mathbb{E}[{\eta }_l{\xi }_k]$ is the $(l,k)$ coordinate of the cross-covariance operator $C_{YX}=B{C}_X$, where $B$ is the Hilbert-Schmidt operator with kernel $\beta$. ThusThe test is consistent whenever this retained block contains a nonzero cross-covariance; fixed truncation can miss alternatives outside the selected principal-component subspaces.

Continuous differentiability gives the first-order expansionThe assumed convergence in distribution implies $\sqrt{n}({\hat{\theta }}_n-{\theta }_0)=O_p(1)$, so after multiplication by $\sqrt{n}$ the remainder is $o_p(1)$. The Slutsky theorem gives the multivariate delta method

Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta >0$ there is $M$ such that $\mathbb{P}(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\stackrel{P}{\to }0$,Letting $\eta \downarrow 0$ gives $X_n(Y_n-c)\stackrel{P}{\to }0$. Also, the continuous-mapping theorem gives $c{X}_n\stackrel{d}{\to }cX$. Sincethe Slutsky theorem yields

Now write the requested statistic asThe central limit theorem givesSince $\mathbb{E}[Z_i^2]=1<\infty$, the strong law of large numbers givesalmost surely, and therefore its reciprocal converges in probability to one. Applying the product result provesThis is a self-normalized central limit theorem with a second-moment denominator.

The given identities implyThus, writingwe have the exact identityThe sample variance satisfies $s_n^2\stackrel{P}{\to }1$, soThe Slutsky theorem and part (c) now give

WriteExpanding $R_i^2$ gives the leading term ${\epsilon }_i^2{\xi }_i^2$ and terms of the forms treated in parts b and c, together with their versions obtained by interchanging $(\epsilon ,F)$ and $(\xi ,G)$. For example, the pure error terms are ${\epsilon }_i^2{G}_i^2$, ${\xi }_i^2{F}_i^2$, and $F_i^2{G}_i^2$, and each cross term is controlled by Cauchy-Schwarz from these. HenceAssuming this limit is positive, the continuous mapping theorem yields ${\tau }_D\stackrel{p}{\to }\{\mathbb{E}({\epsilon }_1^2{\xi }_1^2){\}}^{1/2}$. Combining this with the assumed convergence in distribution of $\sqrt{n}{\tau }_N$ and applying the Slutsky theorem givesThis is the studentization of the generalized covariance measure statistic.

Let $({\lambda }_j,{\varphi }_j)$ be the ordered eigenpairs of the common covariance operator $C_X$, and write $\delta =\mu -{\mu }^{\ast }$. Estimate $C_X$ by the pooled covariance operatorand denote its first $K$ eigenpairs by $({\hat{\lambda }}_j,{\hat{\varphi }}_j)$. The sign ambiguity of each eigenfunction disappears after squaring. Consider the Two-sample FPCA mean statistic

Under $H_0$, the Hilbert-space central limit theorem giveswhere $G$ is a centered Gaussian random element with covariance $C_X$. The assumed eigenvalue gaps give consistency of the estimated eigenvalues and eigenfunctions, so Slutsky's theorem yieldsAn asymptotic level-$\alpha$ test therefore rejects when $T_{n,K}$ exceeds the $(1-\alpha )$-quantile of the chi-squared distribution with $K$ degrees of freedom.

Under a fixed alternative,The test is consequently consistent whenever the mean difference has a nonzero projection onto one of the retained principal components. A difference orthogonal to their span is invisible to this fixed-$K$ test, so $\mu \ne {\mu }^{\ast }$ alone does not guarantee consistency. Increasing $K$ with $n$ can recover such alternatives, but requires additional eigenvalue and approximation control.

The multivariate Wald statistic for a candidate parameter $\theta \in {\mathbb{R}}^p$ isReplacing $I({\hat{\theta }}_n)$ by any consistent estimator of $I({\theta }_0)$ gives the same limit.

Under $P_{{\theta }_0}$, part (a) and consistency of the information matrix imply, by Slutsky theorem, thatConsequentlyusing the characterization of the chi-squared distribution as the squared norm of a standard normal vector.

If $q_{p,1-\alpha }$ is the $(1-\alpha )$ quantile of ${\chi }_p^2$, an asymptotic $(1-\alpha )$ confidence region is

LetThen the null hypothesis is $R{\theta }_0=r$. Define the Wald statistic for linear restrictions

Under $H_0$, part (a) and the continuous mapping theorem givewhereThe matrix $V_0$ is positive definite because $I({\theta }_0)$ is positive definite and $R$ has full row rank. Consistency givesTherefore Slutsky theorem yieldsand applying the continuous squared-norm map proves rigorously that

Let $q_{k,1-\alpha }$ be the $(1-\alpha )$ quantile of ${\chi }_k^2$. Rejecting $H_0$ whenhas rejection probability tending to $\alpha$ under every fixed parameter satisfying $H_0$, and is therefore an asymptotically valid level-$\alpha$ test.

Each completed visit to radius $r_2$ begins a new radial excursion. By part (a), the conditional probability that the following excursion reaches radius $R$ before radius $r_1$ isThe Strong Markov property at the successive stopping times ${\tau }_{2k}$ makes these trials independent with the same success probability. Consequently $N(R)$ has a geometric distribution on $\{1,2,\dots \}$ with parameter $q_R$:As $R\to \infty$, $q_R\to 0$ andSlutsky theorem now gives

Let ${\hat{p}}_k=X_k/n_k$, where $X_k\sim \mathrm{Binomial}(n_k,p_k)$ independently, and putThe Wald statistic is $Z=({\hat{p}}_1-{\hat{p}}_0)/\sqrt{\hat{V}}$. The central limit theorem and Slutsky theorem give $Z\dot{\sim }N(0,1)$ under $H_0$. When $p_1-p_0={\delta }^{\ast }>0$,

Condition on the independent external sample, and abbreviate ${\delta }_{\pi }={\hat{\pi }}_m-\pi$ and ${\delta }_{\mu }={\hat{\mu }}_m-\mu$. Expanding the plug-in estimator around the true nuisance functions giveswhere $\sqrt{n}{R}_{n,m}=o_p(1)$ because the two nuisance mean squared errors vanish, the evaluation sample is independent of the nuisance fits, $A$ is bounded, and $Y$ has bounded support. The linear term is the empirical average of the influence function, so the central limit theorem givesThe remaining bias obeys the Cauchy-Schwarz inequalityThus the claimed conclusion follows under the standard product-rate conditionequivalently $\sqrt{n}\sqrt{\mathrm{MSE}({\hat{\pi }}_m)operatornameMSE({\hat{\mu }}_m)}\to 0$. Slutsky theorem then yields

As printed, the paper instead assumes only $\sqrt{n}\mathrm{MSE}({\hat{\pi }}_m)\mathrm{MSE}({\hat{\mu }}_m)\to 0$, which is insufficient with its stated definition of MSE. For example, deterministic nuisance errors of size $n^{-3/16}$ have both MSEs equal to $n^{-3/8}$ and satisfy the printed condition, while the scaled product bias is $\sqrt{n}{n}^{-3/8}=n^{1/8}$. The result therefore requires the stronger condition above, or “MSE” in the printed rate must be read as root mean squared error.

Factorizing and using part (c),by the Slutsky theorem, since $\hat{\theta }\to {\theta }_0$ in probability. Thus the limiting variable is a negative scaled exponential, with $Z\sim \mathrm{Exp}(1/{\theta }_0)$.

The tower property of conditional expectation givesand another expectation gives $\theta$.

Use the augmented inverse-probability-weighted estimatorCondition on the independently trained nuisance estimators. Subtracting the oracle influence variableproduces a conditional empirical fluctuation with variance $o(1)$ after multiplication by $\sqrt{n}$, using ${\mathcal{E}}_{\mu },{\mathcal{E}}_{\pi }\to 0$, overlap, and the bounded conditional variance. Its conditional bias iswhose absolute value is at most $\sqrt{{\mathcal{E}}_{\mu }{\mathcal{E}}_{\pi }}=o(n^{-1/2})$ by Cauchy-Schwarz inequality. ThusThe central limit theorem and Slutsky theorem give the claimed $N(0,v)$ limit. Without auxiliary data, use cross-fitting: split the sample into folds, train both nuisance estimators away from each observation's fold, and average the same score over held-out observations.

The conditional multivariate normal distribution givesFor a jointly Gaussian vector, the residual is independent of the regressor, so $y=z^T\beta +\sigma e$ with $e\sim N(0,1)$ independent of $z$.

The Square-root Lasso minimizesAt a nonzero residual $R$, its KKT condition iswhich proves the required inequality.

Writing the response vector as $Y=f+\sigma e$, the reverse triangle inequality givesThe strong law of large numbers gives $\Vert e{\Vert }_2/\sqrt{n}{\to }_{P}1$, hence $\hat{\sigma }{\to }_P\sigma$.

Under $X\perp \perp Y\mid Z$, Gaussianity makes $e$ independent of $(X,Z)$ and hence of $R$. Conditionally on $R$, $R^{T}e/\Vert R{\Vert }_2\sim N(0,1)$. The remaining numerator term obeysCombining this with $\hat{\sigma }{\to }_P\sigma$ and Slutsky theorem proves the standard-normal limit.

If the numerator of the generalized covariance measure statistic obeys a central limit theorem with variance $\mathbb{E}({\epsilon }^2{\xi }^2)$ and its empirical residual-product second moment converges in probability to the same positive quantity, the Slutsky theorem makes the studentized statistic converge in distribution to $N(0,1)$.