Asymptotic normality

An estimator is asymptotically normal when a rescaled estimation error converges in distribution to a normal law.

Slutsky's theorem combines convergence in distribution with convergence in probability through continuous algebraic operations.

For estimating equations, asymptotic covariance often has the form $A^{-1}B{A}^{-1}$, called a sandwich covariance.

Regression heteroskedasticity-consistent standard errors use a specific sandwich covariance estimate; general estimating equations can produce sandwich covariances outside regression.

If $\sqrt{n}(T_n-\theta )$ is asymptotically normal, differentiability gives $\sqrt{n}(g(T_n)-g(\theta ))$ the variance multiplied by $g^{\prime }(\theta {)}^2$.

For independent Bernoulli random variables of fixed success probability $0<p<1$, put $\hat{p}=n^{-1}{\sum }_i{X}_i$ and $o=p/(1-p)$. The central limit theorem gives $\sqrt{n}(\hat{p}-p)\stackrel{d}{\to }N(0,p(1-p))$. Applying the delta method to the odds map $g(p)=p/(1-p)$, whose derivative is $(1-p{)}^{-2}$, givesThe ratio is undefined on the all-success event, whose probability is $p^n$. Any finite extension there has the same limiting probability distribution, since that probability tends to zero. This is an asymptotic statement and does not assert an exact finite-sample variance of the unextended ratio. It is useful for uncertainty in estimated odds when the success probability stays away from one.

If a response has mean $\mu$ and variance $V(\mu )$, the delta method gives $\mathrm{Var}(g(Y))\approx g^{\prime }(\mu {)}^{2}V(\mu )$. Choosing $g^{\prime }(\mu )$ proportional to $V(\mu {)}^{-1/2}$ makes this first-order variance independent of $\mu$. For $V(\mu )=k{\mu }^2$ on positive means, the resulting transformation is the natural logarithm. This is a local approximation and requires the transformation to be defined on the observations; it does not make arbitrary normal responses positive or preserve an exact normal distribution.

For $Y\sim N(\mu ,k{\mu }^2)$ with $\mu \ne 0$ and fixed $k>0$, represent $Y=\mu (1+\sqrt{k}Z)$ with $Z$ having the standard normal distribution. Taking the natural logarithm of the absolute value gives the displayed identity. Its random second term has a common distribution, independent of $\mu$, so this variance-stabilizing transformation gives an exactly constant variance. That variance is finite: the normal density is bounded near zero and ${\int }_0^1|\log t{|}^{2}dt<\infty$, while normal tails control logarithmic growth at infinity. The transformation is defined almost surely, loses the sign of the mean, and does not produce an exact normal distribution. A zero mean makes the original observation identically zero and is outside its domain.

If $r_n(X_n-s)$ converges in distribution to $X$ and $\Phi$ is Hadamard differentiable at $s$ tangentially to a space containing the limit, the displayed transformation rule holds under the usual measurability and tight separable-support conditions. The variables $X_n$ must belong to the domain of $\Phi$.

For a centered bivariate normal sample,This is the multivariate delta method for $g(a,c,b)=c/\sqrt{ab}$.

For an independent sample from $\mathrm{Uniform}[-\theta ,\theta ]$, the maximum likelihood estimator is $M_n={\max }_i|X_i|$ andThe parameter-dependent support makes the convergence rate $n$ rather than $\sqrt{n}$, so the regular maximum-likelihood central limit theorem does not apply.