A monotone adversary observes an i.i.d. labeled sample and appends correctly labeled examples. Recent work determined the worst-budget expected error but left the prescribed-budget and high-confidence laws open. We prove two sharp advances. First, for VC dimension $d\geq2$, $n$ clean examples, and exact known budget $m$, the worst-class minimax expected error is at least $c\min \left\{1,\frac dn\left[1+\log \left(1+\frac{\min\{m,n\}}d\right)\right]\right\}.$ Thus exactly $m=n$ honest insertions already realize the full $\Theta(1\wedge(d/n)\log(e+n/d))$ worst-budget penalty. The previous all-learner construction used a much larger budget and gave no interpolation in $m$. The proof introduces reciprocal completion: after a rare disagreement is selected from the clean sample, the adversary draws the missing side from its conditional clean law. The two target orientations then induce exactly the same final multiset while every inserted label remains correct. Second, we determine the worst-budget PAC sample complexity. At $d=1$ it is $\Theta(\varepsilon^{-1}\log(1/\delta))$, with no insertion penalty at any confidence. At every $d\geq2$ it is $\Theta \left(\frac{d\log(1/\varepsilon)+\log(1/\delta)}{\varepsilon}\right).$ The dimension-one upper bound follows from a new nested-disagreement argument that gives the exact tail $\mathbb P\{\operatorname{err}>\varepsilon\}\leq(1-\varepsilon)^n$. These results close the high-confidence question and locate a quantitative finite-budget obstruction.
Keywords: data insertion, monotone adversary, confidence bounds, sample budgets, robust learning, sample complexity