Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 / 2

Put $q_s=F^{-1}(s)$ and $k=-F^{-1}(\alpha )>0$. Symmetry gives $F^{-1}(1-\alpha )=k$. Differentiating the trimmed mean functional along $F_t=(1-t)F+t{\Delta }_x$ givesUnder the substitution $y=q_s$, symmetry implieswhileEvaluating the last integral in the three regions $x<-k$, $|x|\le k$, and $x>k$ yieldsThis is the influence function of a trimmed mean, namely the Huber score with clipping parameter $k$, divided by $1-2\alpha$.

Let $m=\lfloor \alpha n\rfloor$. If at most $m$ observations are replaced arbitrarily, every order statistic retained between ranks $m+1$ and $n-m$ remains between the minimum and maximum of the unreplaced observations. The trimmed mean is therefore bounded as the replacement values diverge.

If more than $m$ observations are replaced by a common value tending to $+\infty$, at least one replacement remains after the largest $m$ observations are trimmed, and the trimmed mean tends to $+\infty$. The analogous construction tends to $-\infty$. Thus the largest fraction of arbitrary replacements for which boundedness is guaranteed isThis is the convention for the finite-sample replacement breakdown point used in the question; the alternative convention based on the smallest breaking fraction reports $(m+1)/n$.