Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 42 / 4

A scalar M-estimator is a selected root of $n^{-1}{\sum }_i\psi (Y_i,\hat{\theta })=0$, or equivalently a minimizer of a differentiable empirical objective whose derivative is this estimating equation. Its population statistical functional $T(F)={\theta }_F$ solves $\int \psi (x,{\theta }_F)dF(x)=0$. Assume local uniqueness, consistency of the sample root, a valid differentiable local expansion, finite ${\mathbb{E}}_F{\psi }^2$, andThese conditions include domination sufficient to differentiate under the integral and the required local uniform control of the sample derivative.

For $F_{\epsilon }=(1-\epsilon )F+\epsilon {\delta }_z$, differentiate the population estimating equation at $\epsilon =0$. Since its uncontaminated integral is zero,Thus the influence function is $\mathrm{IF}(z;T,F)=-\psi (z,{\theta }_F)/A$, with a proportionality constant independent of $z$.

Expanding the sample equation at its population root givesThe summands have mean zero. The central limit theorem and Slutsky's theorem therefore yield the scalar sandwich variance of an M-estimatorThis $V$ is the limiting variance of the $\sqrt{n}$-scaled error; the estimator's variance is approximately $V/n$. In a correctly centred location family with $\psi (x,\theta )=\psi (x-\theta )$, let $m={\mathbb{E}}_F{\psi }^{\prime }(X-\theta )$. Then $A=-m$, so the influence function is $\psi (z-\theta )/m$ and $V=\mathbb{E}{\psi }^2/m^2$. The minus sign from differentiating $x-\theta$ is essential.

The clipped Huber score in the subsequent subparts requires $0<b<\infty$. The PDF only displays an upper bound: at $b=0$ the estimating function vanishes identically, and at $b<0$ it is a constant with no root. A positive clipping threshold is the regular interpretation.

At the standard normal distribution, symmetry and the odd Huber score give population root zero. For $0<b<\infty$, the derivative of the clipped score is $1_{\{|x|<b\}}$ except at its two corners, which have zero normal probability. Put $m_b=2\Phi (b)-1>0$ and write $\varphi$ for the standard normal density. The sandwich variance of an M-estimator has numeratorUsing ${\varphi }^{\prime }(z)=-z\varphi (z)$ and integration by parts,ThereforeAs checks, it tends to $1$ as $b\to \infty$, the sample mean limit, and to $\pi /2$ as $b\downarrow 0$, the sample median limiting variance. The estimator variance for a sample of size $n$ is approximately this quantity divided by $n$.

The root calculation for the Huber location estimator givesSince $|{\psi }_b(z)|\le b$ and equality is attained for $|z|\ge b$, its gross-error sensitivity isThus it is a B-robust estimator at the normal reference distribution. At a general centred distribution the same conclusion requires $m={\mathbb{E}}_F{\psi }_b^{\prime }(X)>0$, in which case the bound is $b/m$. Boundedness of the score alone does not rescue a zero sensitivity denominator.