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A functional statistic evaluates a statistical functional $T$ at the empirical distribution $F_n=n^{-1}{\sum }_i{\delta }_{X_i}$. The map $T$ acts on distributions, whereas $T(F_n)$ is a statistic. If $T$ has an asymptotic linear representation, thenA square-integrable influence function with mean zero consequently gives asymptotic variance $n^{-1}\int \mathrm{IF}(x;T,F{)}^{2}dF(x)$ by the central limit theorem.

Let $q$ be the unique $p$th quantile, with $0<p<1$, and suppose the density $f$ is continuous and positive at $q$. Define quantiles of contaminated distributions using the generalized inverse. For $z\ne q$, differentiating $(1-\epsilon )F(q_{\epsilon })+\epsilon 1_{z\le q_{\epsilon }}=p$ givesAt $z=q$, the generalized-inverse quantile stays exactly $q$ under this contamination, so its derivative is zero. The step formula thus holds $F$-almost everywhere and determines the asymptotic variance $p(1-p)/(nf(q{)}^2)$. Its gross-error sensitivity is $\max (p,1-p)/f(q)$, but its local-shift sensitivity is infinite because of the jump.

For a symmetric degree-$m$ U-statistic kernel, the population functional is $T(F)=\int hd{F}^m$. Contaminating each of the $m$ identical distribution arguments and differentiating gives the displayed influence function. It equals $m$ times the first Hoeffding projection and gives leading asymptotic variance $m^2\mathrm{Var}(h_1(X))/n$. If that projection vanishes, the usual root-$n$ normal limit degenerates and higher-order projections determine the limiting law.

An M-estimator based on an estimating equation chooses a consistent root ofIts population target $T(F)={\theta }_F$ solves ${\mathbb{E}}_F\psi (X,{\theta }_F)=0$. For a scalar parameter, suppose the root is locally unique andFor the contaminated distribution $F_{\epsilon }=(1-\epsilon )F+\epsilon {\delta }_x$, differentiate its population equation at zero contamination. The influence function satisfiesConsequentlyThe constant of proportionality is independent of $x$. A local expansion of the sample estimating equation similarly givesThe central limit theorem yields asymptotic normality with the sandwich variance of an M-estimatorThus the first-order variance of the unscaled estimator is $V/n$. For a vector parameter the same differentiation gives $\mathrm{IF}=A^{-1}\psi$ and $V=A^{-1}B{A}^{-\mathsf{T}}$, with $A=-\mathbb{E}{\partial }_{\theta }\psi$ and $B=\mathbb{E}\psi {\psi }^{\mathsf{T}}$.

For the location problem, take the target location to be zero and let $Z$ have the standard normal distribution, with density $\varphi$ and distribution function $\Phi$. An odd score has $\mathbb{E}\psi (Z)=0$. For a differentiable score, integration by parts givesThe last expression also gives the appropriate derivative of the population equation for bounded monotone scores with jumps, by differentiating the shifted normal density. Assume $A>0$ and put $h=\psi /A$. ThenMultiplying a score by a positive constant changes neither its estimator nor this normalized influence function.

First establish feasibility. The bound impliesso necessarily $C\ge \sqrt{\pi /2}$. For $C>\sqrt{\pi /2}$, let $K>0$ solveThis solution is unique: the ratio tends to $\sqrt{\pi /2}$ as $K\downarrow 0$, tends to infinity as $K\to \infty$, and has positive derivative becauseTake the Huber score ${\psi }_K(x)=\max (-K,\min (x,K))$. Its derivative is one on $(-K,K)$ and zero outside, so $A=p_K$. HenceIt is odd, nondecreasing, meets the bound, and satisfies $\mathbb{E}[Z{h}_K(Z)]=1$.

To prove optimality, put $a=1/p_K$. For each fixed $x$, the minimum of $u^2-2axu$ over $|u|\le C$ is attained by projecting $ax$ onto $[-C,C]$, which is exactly $h_K(x)$. Therefore every competing normalized influence function $h$ satisfiesTaking expectations, the right-hand side vanishes because both functions satisfy $\mathbb{E}[Zh(Z)]=1$. Thus $\mathbb{E}{h}^2\ge \mathbb{E}{h}_K^2$. This proves the optimal bounded influence function for normal location, even over the larger class of all bounded normalized functions, and hence over the required odd nondecreasing class. Its minimum asymptotic variance isThe numerator is $\mathbb{E}[Z^2{1}_{\{|Z|\le K\}}]+K^2\mathrm{Pr}(|Z|>K)$.

At the included boundary $C=\sqrt{\pi /2}$, equality in the feasibility bound forces $h(x)=C\mathrm{sgn}(x)$ almost everywhere. This is the influence function of the sample median, with $V=\pi /2$. It is obtained from the rescaled scores ${\psi }_K/K$ as $K\downarrow 0$. The boundary solution is the sign score, or equivalently the median estimator. Literal substitution of $K=0$ in the clipped score gives the identically zero function and no identifiable estimating equation; the printed clipped formula therefore requires this limiting interpretation at equality.

A statistical functional is a map $T$ defined on a class of probability distributions. Its associated functional statistic is $T(F_n)$, where $F_n=n^{-1}{\sum }_i{\delta }_{X_i}$ is the empirical distribution and each ${\delta }_{X_i}$ is a Dirac measure. For example, $T(F)=\int xdF(x)$ yields the sample mean. The influence function records the derivative in the direction of point contamination:It is a local, infinitesimal robustness diagnostic, not a description of arbitrary large contamination.

Suppose $T$ is differentiable in a sense strong enough to linearize it at the empirical distribution, and that its derivative in a signed direction $G-F$ is $\int \mathrm{IF}(x;T,F)d(G-F)(x)$. Averaging the directions ${\delta }_z-F$ over $z\sim F$ gives the zero direction, so linearity implies $\int \mathrm{IF}dF=0$. The resulting asymptotic linear representation isIf $\int {\mathrm{IF}}^{2}dF<\infty$, the central limit theorem and the negligible remainder giveThus the asymptotic variance is $n^{-1}\int {\mathrm{IF}}^{2}dF$. For a vector functional the integral is $n^{-1}\int \mathrm{IF}{\mathrm{IF}}^{\mathsf{T}}dF$. Convergence of the actual finite-sample variance to this leading expression additionally requires control of second moments of the remainder, or an appropriate uniform integrability condition; convergence in distribution alone does not provide that. Existence of a contamination derivative by itself also does not guarantee the required empirical linearization.

Several complementary robustness measures follow. Gross-error sensitivity is ${\gamma }^{\ast }={\sup }_z|\mathrm{IF}(z)|$. Under contamination $(1-\epsilon )F+\epsilon H$, the first-order bias is $\epsilon \int \mathrm{IF}dH$, whose worst absolute coefficient is ${\gamma }^{\ast }$; bounded influence therefore bounds infinitesimal gross-error bias. The local-shift sensitivity is ${\lambda }^{\ast }={\sup }_{x\ne y}|\mathrm{IF}(x)-\mathrm{IF}(y)|/|x-y|$, equal to $\sup |{\mathrm{IF}}^{\prime }|$ when the influence function is smoothly differentiable. It measures the effect of moving a contaminant a small distance. The rejection point of an influence function is ${\rho }^{\ast }=\inf \{r:\mathrm{IF}(x)=0\text{ for }|x|>r\}$, using centered, standardized coordinates; it is infinite if there is no such cutoff. A finite cutoff completely rejects sufficiently extreme observations to first order. These measures address different aspects of robustness. The integral of squared influence controls asymptotic precision and efficiency, while breakdown point concerns finite amounts of contamination and is not determined by bounded influence alone.

For the quantile, assume $0<p<1$, a unique solution $q=q_p(F)$, and a density continuous and positive at $q$. Define quantiles of the contaminated distribution by generalized inversion, since a distribution with a point mass need not attain $p$ as an equality. If $z\ne q$, its indicator is constant near $q$ for sufficiently small contamination, and differentiatingat zero gives $-p+f(q)q_0^{\prime }+1_{z\le q}=0$. HenceAt the exceptional point $z=q$, the generalized-inverse quantile remains exactly $q$: below $q$ the contaminated cumulative distribution function is below $p$, and its value at $q$ exceeds $p$. Its actual derivative there is zero. The usual step-function expression is an almost-everywhere version, which is all that the asymptotic variance needs. Its mean is zero, and its second moment isTherefore the influence function of a quantile gives asymptotic variance $p(1-p)/(nf(q{)}^2)$ and gross-error sensitivity $\max (p,1-p)/f(q)$. Its jump makes local-shift sensitivity infinite, and its nonzero constant tails make its rejection point infinite. If $f(q)=0$ or the quantile is not unique, this regular derivation does not apply.

This is a symmetric beta distribution with parameters $(2,2)$, so its mean and median are both $1/2$. At the median its probability density function is $f(1/2)=3/2$, giving asymptotic variance $1/9$ for the sample median. Direct integration givesConsequently $\sqrt{n}({\bar{X}}_n-1/2)\stackrel{d}{\to }N(0,1/20)$. The sample median has $20/9$ times the asymptotic variance of the sample mean, with asymptotic relative efficiency

For the standard normal distribution, $m=0$ and $f(0)=1/\sqrt{2\pi }$, so the sample median has asymptotic variance $\pi /2$ under scaling by $\sqrt{n}$. The central limit theorem gives $\sqrt{n}{\bar{X}}_n\stackrel{d}{\to }N(0,1)$ for the sample mean. Thus the sample median has $\pi /2$ times the asymptotic variance of the sample mean, and its asymptotic relative efficiency against the sample mean is

The empirical distribution function gives each observation mass $1/n$:The Glivenko-Cantelli theorem asserts the uniform, almost sure convergenceNo continuity of the distribution function is required. Here is a proof that also handles its jumps. Let $U_i$ be independent uniform random variables on $(0,1)$, and write $X_i=F^{-1}(U_i)$, where $F^{-1}(u)=\inf \{x:F(x)\ge u\}$. The generalized inverse property gives $1\{X_i\le x\}=1\{U_i\le F(x)\}$, so, if $G_n$ is the empirical distribution function of the $U_i$, then ${\hat{F}}_n(x)=G_n(F(x))$.

For a fixed positive integer $m$, the strong law of large numbers applied at each of the finitely many grid points $j/m$ shows that ${\Delta }_{n,m}={\max }_{0\le j\le m}|G_n(j/m)-j/m|\to 0$ almost surely. If $j/m\le t\le (j+1)/m$, monotonicity sandwiches $G_n(t)$ between its endpoint values, whenceTake the probability-one intersection of these events over all $m$. On that event the limit superior is at most $1/m$ for every $m$, hence is zero. Therefore ${\sup }_x|G_n(F(x))-F(x)|\to 0$. The coupled sequence has the same joint law as any independent sample from $F$, which proves the asserted almost sure conclusion for the original sample.

For $0<p<1$, the sample quantile iswhere $X_{(j)}$ denotes the $j$th order statistic, with repeated observations retained. Suppose the population median $m=F^{-1}(1/2)$ is unique and $F$ is continuously differentiable in a neighbourhood of $m$, with probability density function $f(m)>0$. ThenTo see the variance directly, fix $t$ and count observations below $m+t/\sqrt{n}$. This count has a binomial distribution with success probability $q_n=1/2+f(m)t/\sqrt{n}+o(n^{-1/2})$. The event that the sample median is below this point is that the count is at least $\lceil n/2\rceil$. The central limit theorem for this binomial triangular array gives a limiting probability $\Phi (2f(m)t)$, exactly the displayed normal distribution. Centering is necessary for stating the limiting distribution; subtracting the deterministic population median has no effect on the asymptotic variance requested in the comparisons.

First, the absolute values in the printed limit are an error: the correct long-run variance of a stationary process is the signed sum of autocovariances. Directly,soEach coefficient tends to $1$ and has absolute value at most $1$. Absolute summability and the dominated convergence theorem therefore giveThe sum of absolute values is an upper bound, not the general limit. For a concrete counterexample to the printed assertion, take $X_t={\eta }_t-{\eta }_{t-1}$ with unit-variance iid noise. Then $\gamma (0)=2$, $\gamma (\pm 1)=-1$, and other covariances vanish. The absolute sum is $4$, but ${\overline{X}}_T=({\eta }_T-{\eta }_0)/T$, so $T\mathrm{Var}({\overline{X}}_T)=2/T\to 0$.

For correlated data the central limit theorem can still hold under suitable strong mixing of a stationary process and moment conditions, but the limiting variance is the long-run variance of a stationary process, rather than the one-observation variance. When it is positive,Positive serial dependence usually increases the standard error, while negative dependence can reduce it. The approximate effective sample size of a stationary sample is $T{\gamma }_X(0)/{\sum }_h{\gamma }_X(h)$ when that denominator is positive. A long-memory time series can require a different normalization or a different limit law. Absolute covariance summability alone does not prove a CLT: if $X_t=V{\eta }_t$ with a common independent random scale $V$ taking values $1$ and $2$ with equal probabilities and iid ${\eta }_t$ with the standard normal distribution, the off-diagonal covariances are zero, yet $\sqrt{T}{\overline{X}}_T$ has the nonnormal scale-mixture law $VN(0,1)$. The example is not an ergodic stationary process. If the long-run variance of a stationary process is zero, the usual nondegenerate square-root-$T$ CLT is unavailable.

For the causal AR(1), $|\varphi |<1$ andThe conditional distribution follows because the current innovation is independent of the past.

For a fixed known $\varphi$, condition on the observed $X_1$ and use the $n=T-1$ transitions $t=2,\dots ,T$. Their conditional maximum likelihood criterion is, up to constants,Differentiating in $\mu$ yieldsSince $X_t-\varphi X_{t-1}=(1-\varphi )\mu +{\epsilon }_t$,Thus it is an unbiased estimator, even conditionally on $X_1$, andIt has statistical consistency with mean-square convergence, and the iid-noise strong law of large numbers also gives almost-sure statistical consistency. Its conditional distribution is exactly normal with the displayed mean and variance. If $X_0$ is also observed and all $T$ transitions are used, replace $T-1$ by $T$.

The fixed-$\varphi$ qualification is necessary for the exact finite-sample claims. If $\varphi$ is jointly estimated, conditional likelihood function is linear regression with an intercept: writing ${\bar{X}}_{-}=n^{-1}{\sum }_{t=2}^T{X}_{t-1}$ and ${\bar{X}}_{+}=n^{-1}{\sum }_{t=2}^T{X}_t$, the unconstrained estimators areThis ratio is not generally an unbiased estimator and does not have the preceding finite-sample variance. Under the usual stationary regression conditions it has statistical consistency; its asymptotic variance is ${\sigma }_{\epsilon }^2/(1-\varphi {)}^2$. Indeed, it differs from ${\bar{X}}_{+}$ by $\hat{\varphi }(X_T-X_1)/(n(1-\hat{\varphi }))$, an asymptotically negligible endpoint term. Profiling an unknown innovation variance does not change the fixed-$\varphi$ estimate of $\mu$.

Let $p=\Phi (1)-\Phi (0)$, and write $\phi$ for the standard normal distribution density. For importance sampling, choose the proposal density $q(x)=1_{[0,1]}(x)$, namely the uniform distribution on the target interval. Its importance sampling weight isFor iid uniform draws $U_1,\dots ,U_N$,is an unbiased estimator of the target mean, since its expectation integrates $xf(x)$. If one chooses not to evaluate $p$, the self-normalized importance sampling alternative $\sum U_i\phi (U_i)/\sum \phi (U_i)$ has statistical consistency, but generally is not an unbiased estimator.

Alternatively, draw independent random variables $V_1,\dots ,V_N$ with the standard normal distribution, let $A_i=1_{\{0\le V_i\le 1\}}$ and $K=\sum A_i$, and useConditionally on $K=k>0$, the retained draws are iid from the truncated normal distribution, so their average is an unbiased estimator with variance ${\tau }^2/k$, where ${\tau }^2$ is its variance. If $K=0$ the ratio is undefined; one can continue drawing until a prescribed number of retained observations is obtained.

Here importance sampling uses every draw inside the interval, whereas only $p\approx 0.3413$ of the normal proposals are retained. The advantage can also be quantified. Integration using ${\phi }^{\prime }(x)=-x\phi (x)$ givesFor uniform importance sampling,For the retained-draw ratio with a fixed total proposal budget, its asymptotic variance coefficient is ${\tau }^2/p\approx 0.233347$. Thus this importance choice has substantially smaller error for the same proposal count; it also avoids wasting about two thirds of the standard normal distribution draws. This comparison is specific to this well-matched importance sampling proposal; arbitrary proposal choices do not share this guarantee.

Under Neyman allocation, sample sizes are proportional to the arm standard deviations. HereFor total size $n_{\max }$, the minimized asymptotic variance isEqual allocation givesThe Neyman allocation therefore reduces the large-sample variance by $0.04/n_{\max }$, about $5.9\%$ of the equal-allocation variance.