 Solution

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.