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For $X_1\sim \mathrm{Poisson}(\theta )$, its probability generating function isThus $\mathbb{E}[X_1]=G^{\prime }(1)=\theta$ andThe log-likelihood is, up to an additive constant,so the maximum-likelihood estimator is ${\hat{\theta }}_{\mathrm{MLE}}=\overline{X}$. The central limit theorem gives

Under the null model, $\pi =1$, which is a boundary point of the parameter space. Standard interior-point asymptotic normality of the maximum-likelihood estimator therefore need not apply.

Indeed, put $a=1-e^{-\lambda }$. Under $\pi =1$, $N_{+}\sim \mathrm{Binomial}(n,a)$. For the unconstrained estimator $\tilde{\pi }=N_{+}/(na)$, the central limit theorem givesSince $\hat{\pi }=\min \{1,\tilde{\pi }\}$, the continuous mapping theorem yieldsThis distribution has probability $1/2$ at zero and a continuous normal half-density on the negative half-line, so it is not normal.

Let $S_n={\sum }_{i=1}^n{X}_i$. The log-likelihood iswith score function $n/\theta -S_n$ and second derivative $-n/{\theta }^2<0$. The unique maximum-likelihood estimator is thereforeFor an exponential distribution of rate $\theta$,The central limit theorem givesApplying the delta method to $h(x)=1/x$, for which $h^{\prime }(1/\theta )=-{\theta }^2$, yields the exponential-rate maximum-likelihood estimator limit

PutThe central limit theorem gives $\sqrt{n}({\hat{\mu }}_n-\mu ({\theta }_0))\Rightarrow N(0,{\sigma }_h^2)$. Provided ${\mu }^{\prime }({\theta }_0)\ne 0$, the delta method applied to ${\mu }^{-1}$ yieldsStrict monotonicity alone does not exclude ${\mu }^{\prime }({\theta }_0)=0$; in that exceptional case this root-$n$ limit need not hold and higher-order analysis is required.

One does not necessarily have $\mathrm{Var}({\hat{\theta }}_n)\ge [nI({\theta }_0){]}^{-1}$ for every $n$, because ${\hat{\theta }}_n$ need not be unbiased. The Cramer-Rao bound for an estimator with bias $B_n(\theta )$ is insteadunder the corresponding regularity assumptions.

The assumed maximum-likelihood estimator is the sample mean of $h(X_i)$. Its central limit theorem isFor a regular model, the regular maximum-likelihood limit isThe two asymptotic variances must agree, soConsequentlyto first asymptotic order. The estimator therefore attains the Cramer-Rao bound asymptotically: its precision, the reciprocal variance, exhausts the information in the sample.

PutThe log-likelihood, apart from an additive constant, isOver the full parameter space its maximum-likelihood estimates are $(\hat{\mu },{\hat{\sigma }}^2)=(\overline{X},S/n)$. Under $H_0$, maximizing over the nuisance parameter $\mu$ gives $({\hat{\mu }}_0,{\sigma }_0^2)=(\overline{X},1)$. Thus the generalized likelihood-ratio test statistic, in the convention $2({\sup }_{\Theta }{\ell }_n-{\sup }_{{\Theta }_0}{\ell }_n)$, is

Under $H_0$, the normal-sample decomposition givesWriting $Z_n=(S-(n-1))/\sqrt{2(n-1)}$, the central limit theorem for a sum of squared independent standard normal variables yields $Z_n\stackrel{d}{\to }N(0,1)$. Hence Slutsky theorem givesand in particular ${\hat{\sigma }}^2\stackrel{P}{\to }1$.

For $h(x)=x-1-\log x$, Taylor theorem at $x=1$ givesConsequentlyby the continuous mapping theorem and Slutsky's theorem. Since the square of a standard normal variable has the chi-squared distribution with one degree of freedom,

The log-likelihood is, up to constants, $n\alpha \log \beta -\beta {\sum }_i{X}_i$, so the maximum-likelihood estimator is $\hat{\beta }=\alpha /\overline{X}$. By the central limit theorem,Applying the delta method to $g(x)=\alpha /x$ gives

The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta ]$. HencesoThus $\tilde{\theta }$ is unbiased.

Also,The central limit theorem givesReplacing $\theta$ in the asymptotic standard error by the consistent estimator $\tilde{\theta }$ gives the Wald confidence intervalIt may be intersected with the parameter space $(0,1)$.

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.

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.

For observations $X_1,\dots ,X_n$, the log-likelihood is, up to an additive constant,Its unique maximum isNow $\mathbb{E}[X_i^2]={\sigma }^2$ and, using $\mathbb{E}[Z^4]=3$, $\mathrm{Var}(X_i^2)=2{\sigma }^4$. The central limit theorem followed by the delta method for $g(x)=\sqrt{x}$ gives

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 central limit theorem givesApplying the delta method to $g(x)=x^2$ yieldsFor $\mu =0$ this denotes the degenerate distribution at zero, which is the correct limit at the $\sqrt{n}$ scale.

The Poisson central limit theorem, or the ordinary central limit theorem applied to the $X_i$, gives $Y_n\stackrel{d}{\to }Y$ for a standard normal random variable $Y$. The negative-part map $x\mapsto x^{-}:=\max (-x,0)$ is continuous, so the continuous mapping theorem gives

The preceding representation and the central limit theorem giveSinceone may takeFor $0<p<1$ this gives the required convergence to $\Phi (x)$ at every real $x$.

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.

Let $W=Y-\theta \sim G_0$ and suppose $a={\mathbb{E}}_{G_0}[{\psi }^{\prime }(W)]\ne 0$. A first-order expansion of the estimating equation $n^{-1}{\sum }_i\psi (Y_i-T_n)=0$ gives the asymptotic linear representationThe central limit theorem therefore yields

Let ${\sigma }^2=\mathrm{Var}(2^{X_1})\le \ell 2^b$. If $\sigma =0$ the result is immediate. Otherwise the central limit theorem applies to $L$, while the displayed threshold is at least $\alpha \sigma /\sqrt{n}$. Hence

Under $\delta ={\delta }_A$,andThe central limit theorem therefore gives an asymptotic normal distribution with this mean and variance.