The excess-tail functional cannot upper-bound the trimmed mean

This law is absolutely continuous with density $1\mathrm{/}(4w)$ on each interval $[-1-w,-1+w]$ and $[1-w,1+w]$, has mean zero and variance ${\sigma }^2=1+w^2\mathrm{/}3$. The replacement rule is a measurable finite-coordinate rule and makes at most $20000=2\eta N$ changes. Put $\epsilon =.08+12\log 40\mathrm{/}{10}^6$, so $.08<\epsilon <.081<1\mathrm{/}4$, and $\delta \ge e^{-N\mathrm{/}4}$. At lower tail mass $2\epsilon$ the quantile is $-1-w+8w\epsilon$. Integration over the lower interval gives $E[(Q_{2\epsilon }-X){\mathbf{1}}_{\{X\le Q_{2\epsilon }\}}]=(8w\epsilon {)}^2\mathrm{/}(8w)=8w{\epsilon }^2$; symmetry gives the same upper-tail excess. Hence $3E_{\mathrm{e}\mathrm{x}\mathrm{c}}(4\epsilon )+2\sigma \sqrt{\log 40\mathrm{/}{10}^6}<.0042$, using $\log 40<4$, $\epsilon <.081$ and ${\sigma }^2<1.000001$.

Let $A$ be the event that the clean first-half sign average has absolute value at most $.01$, and let $B$ be the event that the second-half negative-sign count lies between $N\mathrm{/}4$ and $3N\mathrm{/}4$. The sign average has variance $1\mathrm{/}N$, so [F3] gives $P(A^c)\le .01$. The second-half count has variance $N\mathrm{/}4$, so $P(B^c)\le 4\mathrm{/}N$. On $A$ there are at least $.495N$ negative first-half signs, so the replacement changes exactly $20000$ of them. The resulting sign average is at least $-.01+.04=.03$. On $B$, the ranks in [F1] put $\alpha$ in the negative interval and $\beta$ in the positive interval, since $\epsilon <.081<1\mathrm{/}4$. Every clipped value is at least its altered sign minus $w$: a negative value is at least $-1-w$, and a positive value, including a replacement $+1$, clips to at least $1-w$. Therefore $\widehat{\mu }\ge .03-w=.029$ on $A\cap B$, whose probability is at least $.989996$. This exceeds the proposed threshold $.0042$ with probability greater than $\delta =.1$, disproving the asserted coverage.