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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 functional statistic has the form $T(F_n)$, where $F_n=n^{-1}{\sum }_i{\delta }_{X_i}$ is the empirical distribution and $T$ is a statistical functional on a class of distributions. It separates the population quantity $T(F)$ from its sample estimate. Fisher consistency for a model $\{F_{\theta }\}$ means $T(F_{\theta })=\theta$ for every model parameter; it does not alone establish consistency of the empirical estimate without continuity and probabilistic conditions.

The influence function measures infinitesimal contamination at a point:when this derivative exists. A bounded influence function limits first-order sensitivity to an arbitrarily remote observation.

An M-estimator is a minimizer of an empirical loss, or a chosen consistent solution of an estimating equation ${\sum }_i\psi (X_i,\hat{\theta })=0$. Its population functional solves ${\mathbb{E}}_F\psi (X,T(F))=0$. Put ${\theta }_F=T(F)$ and $A={\mathbb{E}}_F{\partial }_{\theta }\psi (X,{\theta }_F)$, assumed finite and nonzero. Differentiating the contaminated population equation givessince the unperturbed expected score is zero. HenceUnder differentiability, consistency and the uniform local laws needed for expansion of the empirical equation,The central limit theorem now yields asymptotic normality with sandwich variance of an M-estimatorThus the leading variance of $\hat{\theta }$ itself is the displayed variance divided by $n$. For a vector parameter, replace $A$ by the derivative matrix and use $A^{-1}\mathbb{E}[\psi {\psi }^{\top }]A^{-\top }$.

In a location model with known scale $s$, choose $\psi (x,\theta )={\psi }_0((x-\theta )/s)$. For a symmetric reference distribution, an odd score is Fisher consistent. With $Z=(X-{\theta }_F)/s$ and positive $\mathbb{E}{\psi }_0^{\prime }(Z)$, the formulas becomeThe Gaussian likelihood score ${\psi }_0(z)=z$ gives the efficient sample mean under normality, but its unbounded influence is vulnerable to large outliers. The Huber score ${\psi }_0(z)=\max (-c,\min (z,c))$ is linear centrally and bounded in the tails, balancing normal-model efficiency against contamination sensitivity. Its denominator is $\mathbb{P}(|Z|<c)$ for a continuous law. The sign score gives the median and needs a nonsmooth derivation using the density at the median, rather than incorrectly setting its ordinary derivative to zero. Redescending scores can reject very extreme observations, but may have multiple roots and require a consistent root-selection rule.

Thus appropriate choices seek Fisher consistency, a nonzero stable derivative, finite variance, reasonable reference-model efficiency and bounded influence. Robust scale estimation may be needed for the tuning constant; joint location-scale estimation uses the matrix influence calculation rather than treating an estimated scale as known. Bounded influence describes local robustness and is not by itself a global breakdown point theorem.

Let ${\Delta }_n={\sup }_{\theta \in \Theta }|Q_n(\theta )-Q(\theta )|$. For any $\epsilon >0$, the setis compact. If it is empty there is nothing to prove. Otherwise continuity and the unique minimum give a strictly positive separation gapTake any measurable attained minimizer of $Q_n$. Its defining inequality givesConsequentlyThis proves argmin consistency under uniform convergence in probability. No continuity of $Q_n$ is needed once the minimizer exists; compactness and continuity concern the deterministic separation gap. The uniform convergence in probability assumption controls all candidate parameters, including the random minimizer.

For the estimating equation, fix $\epsilon >0$ and put $a={\theta }_0-\epsilon$, $b={\theta }_0+\epsilon$. The prescribed signs make $\eta =\frac{1}{2}\min \{-S(a),S(b)\}>0$. Pointwise convergence in probability at just these two points impliesand similarly $\mathrm{Pr}(S_n(b)\le 0)\to 0$. With probability tending to one, $S_n(a)<0<S_n(b)$. The intermediate value theorem then gives a zero inside $(a,b)$, and uniqueness identifies it with ${\hat{\theta }}_n$. HenceThis is consistency of a uniquely bracketed zero. It needs no monotonicity, no continuity of the limit $S$, and no uniform convergence of the $S_n$. The bracket interval must lie in the domain of $S_n$: read literally, the printed sign condition for every positive $\epsilon$ puts every real point into $\Theta$, so this requirement is satisfied. More generally it is enough to have an interval about ${\theta }_0$ and sign brackets arbitrarily close to it.

The homogeneous treatment effect and consistency implyInstrument validity gives $Z\perp R\mid X$.

If Assumption 1 holds, put $V=R-{\alpha }_{0}X$. Then $\mathbb{E}[V\mid X]=0$, so conditional instrument independence givesConsequently both population estimating equations vanish at $({\beta }_0,{\alpha }_0)$ for any probability limit of $\hat{\gamma }$.

If Assumption 2 holds, choose the linear-projection coefficientThen $\mathbb{E}[(R-{\alpha }_{\ast }X)X]=0$, whileThe first term is zero by conditional instrument independence and the second by the definition of ${\alpha }_{\ast }$. Thus $({\beta }_0,{\alpha }_{\ast })$ solves the population equations. Under the nonsingularity condition from part b, the root is unique, so standard estimating equation consistency proveswhenever either Assumption 1 or Assumption 2 holds. This is double robustness.

LetAssumption 2 implies ${\gamma }_0=q/d$. Although the printed $\hat{\gamma }$ need not converge to ${\gamma }_0$, its first-order effect vanishes because $\mathbb{E}[VX]=0$. Inverting the $2\times 2$ Jacobian of the remaining estimating equations gives the influence functionThe first-stage condition yieldsHence the asymptotic variance of $\sqrt{n}(\hat{\beta }-{\beta }_0)$ is the sandwich expression

There is a defect in the printed assumptions: $\mathrm{Var}(V\mid A,X)={\sigma }^2$ alone does not determine $\mathbb{E}[V^2{R}_0^2]$, because $R_0$ depends on $Z$ and $V$ can have a nonzero conditional mean given $(A,X)$ under unmeasured confounding. Under the standard intended strengthening $\mathbb{E}[V^2\mid X,Z]={\sigma }^2$, the formula simplifies toThe asymptotic variance of $\hat{\beta }$ itself is ${\mathcal{V}}_{\beta }/n$.

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

For $\vartheta =(q,t)$ use the estimating equationIts empirical mean vanishes exactly at $(\hat{p},\hat{\tau })$. The population equation has the unique root $(p,{\tau }^{\ast })$, so the Z-estimator is consistent. Linearizing the equation, or simplifying the corresponding sandwich covariance, gives the influence functionHence