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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.

Write $p=p^{\ast }$, ${\mu }_z=\mathbb{E}(Y\mid Z=z)$, and ${\tau }^{\ast }={\mu }_1-{\mu }_0$. The summand has finite variance, so the central limit theorem giveswhere

For $k\ge n$, $p_k(n)=\left(\genfrac{}{}{0ex}{}{k-1}{n-1}\right)p^n(1-p{)}^{k-n}$. It is the sum of $n$ independent geometric variables, so the central limit theorem applies with mean $n/p$ and standard deviation $\sqrt{n(1-p)}/p$. Take
$k_a(n)=\lceil n/p+a\sqrt{n(1-p)}/p\rceil$ and $k_b(n)=\lfloor n/p+b\sqrt{n(1-p)}/p\rfloor$.

The evaluation map $f\mapsto f(1)$ is continuous on $C[0,1]$ in the uniform norm. The continuous mapping theorem applied to part (c) therefore givesThis recovers the central limit theorem from the functional version.

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.

Write $k=2r+1$ and $m=n/k$. The central limit theorem gives jointlywhere the $Z_j$ are independent standard normal variables. HenceIts limiting cumulative distribution function is

Assuming $g$ is measurable, applying it separately to independent and identically distributed random variables preserves those properties. Thus $W_i=g(X_i)$ have common mean ${\mu }_g$ and variance ${\sigma }_g^2$. For ${\sigma }_g^2>0$, the central limit theorem givesHence, for large $n$, the approximate probability distribution isThe second argument is the variance. If ${\sigma }_g^2=0$, this sample mean equals ${\mu }_g$ almost surely, so the appropriate description is a point mass at ${\mu }_g$.

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed random variables with finite mean $\mu$ and finite positive variance ${\sigma }^2$, thenHere the arrow means convergence in distribution: for every real $z$, the probability that the expression on the left is at most $z$ tends to $\Phi (z)$, the standard normal distribution function. If ${\sigma }^2=0$, the variables equal $\mu$ almost surely and their sample mean is exactly $\mu$.

Apply the central limit theorem to the independent and identically distributed random variables $X_i^2$, whose mean and variance are $\theta$ and ${\theta }^2$. It gives the approximate standard normal distributionThe supplied standard normal quantile $c=1.96$ leaves probability $0.025$ in each tail. HenceSince $\theta >0$, this event is equivalent toFor $n>1.96^2$, equivalently integer $n\ge 4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ confidence intervalThe approximation is justified for large $n$; positivity at $n\ge 4$ by itself does not guarantee accurate coverage. This construction is a confidence interval from a relative normal pivot: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed statistical parameter $\theta$ with probability approximately $0.95$ under repeated sampling.

No, not from the stated assumptions alone. The centered slope is a weighted sum,and the weighted errors need not be identically distributed. The ordinary central limit theorem therefore does not apply directly.

For a concrete counterexample, for every $n\ge 3$ chooseThese satisfy all three centring and orthogonality assumptions, and both explanatory columns are nonzero. Let the errors be independent with Rademacher distribution, taking values $\pm 1$ with equal probability. Thenwhich takes values $-1,0,1$ with probabilities $1/4,1/2,1/4$ for every $n$. Increasing the sample size adds no information about this slope, and the standardized estimator does not approach a normal distribution.

A useful additional assumption for asymptotic normality of a weighted regression slope is that no single design value dominates:Under this condition and the given finite error variance, the Lindeberg-Feller central limit theorem yieldsjustifying the approximation $N({\beta }_1,{\sigma }^2/S_{xx})$ for the slope as well.

The algebraic identity ${\hat{\beta }}_0-{\beta }_0=n^{-1}{\sum }_i{ϵ}_i$ does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large $n$,Its mean ${\beta }_0$ and variance ${\sigma }^2/n$ remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\tau }^2<\infty$, thenHere $\stackrel{d}{\to }$ denotes convergence in distribution.

The central limit theorem givesWrite $z=z_{1-\alpha /2}$ for the indicated standard normal quantile. Inverting the approximate pivot inequality and substituting $Y/n=1/\hat{\theta }$ gives the approximate confidence intervalIts coverage is approximately $1-\alpha$ for large $n$. The intersection respects the positive rate parameter; asymptotically the lower endpoint is positive. This is an asymptotic interval, not an exact finite-sample confidence interval.

At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.

Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

By the addition rule for independent Poisson random variables, $X_n$ has the same distribution as a sum of $n$ independent variables $Z_j\sim \mathrm{Poisson}(\lambda )$. Each has expectation and variance $\lambda$. The central limit theorem therefore givesFor $\lambda =0$, every variable is zero, and the conclusion is the degenerate normal law with zero variance.

For the uniform distribution on $[0,1]$, the mean is $1/2$ and the variance is $1/12$. The central limit theorem for independent identically distributed variables with finite nonzero variance givesChoose $\boxed{\alpha =1/2,\beta =1/2}$. ThenThe standard normal distribution function is continuous and strictly increasing, so its value at the finite positive number $\sqrt{12}$ is strictly between $1/2$ and one, as required.