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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 $1-2\alpha$.

For $0\le \alpha <1/2$, an $\alpha$-trimmed mean sorts $n$ observations, removes the smallest and largest $k=\lfloor \alpha n\rfloor$, and averages the remaining $n-2k$ values. It limits the effect of a small number of extreme observations while still averaging a central portion of the data.

For $(1,2,3,4,100)$ and $\alpha =1/5$, removing $1$ and $100$ gives trimmed mean $(2+3+4)/3=3$, compared with sample mean $22$. Replacing $100$ by a larger value leaves this trimmed mean unchanged. Replacing enough observations can still move the retained central values arbitrarily far.

For a probability distribution with quantile function $F^{-1}$, the population statistical functional isThe normalizing denominator is the retained fraction of probability. At a regular symmetric reference probability distribution, the influence function of a trimmed mean is a clipped version of the observation, scaled by this fraction. Centering and symmetry assumptions matter when using that simpler expression.