Example. A paired trace-selection calculation [ftip-00PA]

The selective-refinement comparison motivates contrasting repeatability and coverage under a fixed retained budget [dang2026selecting, Sections 6-7; Tables 4-5, 10]. Consider two arms from Definition [ftip-00P8] with the same P0 boundary, model, task split, verifier, budget, and paired assignment law. They differ only in the selector: \(a\) retains the highest teacher-forced scores and \(a'\) retains a coverage-stratified subset. Both have \(Q_{\rm hold}=\{q_1,q_2\}\) and \(\mathfrak S=\{s_1,s_2\}\). The paired assignment law gives the same held-out task at indices 1 and 2; arm \(a'\) has one additional admitted coverage run at index 3. Take \(g_a=g_{a'}=\operatorname {id}_{\{0,1\}}\). The receipts are

\[ \begin {array}{c|cc|ccc} &q_{a,i}&s_{a,i}&A_{a,i}&I^{\rm prod}_{a,i}&Y_{a,i}\\\hline i=1&q_1&s_1&1&1&1\\ i=2&q_1&s_1&1&1&0 \end {array} \qquad \begin {array}{c|cc|ccc} &q_{a',i}&s_{a',i}&A_{a',i}&I^{\rm prod}_{a',i}&Y_{a',i}\\\hline i=1&q_1&s_1&1&1&1\\ i=2&q_1&s_1&1&1&1\\ i=3&q_2&s_2&1&1&1 \end {array} \]

The first two rows are the common paired indices, so \(N_a=2\), \(N_{a'}=3\), and \(N_{\rm pair}=2\). Both arms have \(\widehat p^{\rm prod}=1\); the third row is included in arm \(a'\)'s coverage and run-rate denominators but not in the paired contrast. Thus \(\widehat C^{\rm task}_a=\widehat C^{\rm strat}_a=\tfrac 12\) and \(\widehat C^{\rm task}_{a'}=\widehat C^{\rm strat}_{a'}=1\), while \(\widehat \Delta _{a,a'}=-\tfrac 12\). The calculation shows why a repeatability rate alone cannot identify coverage: the two selectors have the same productive rate but different admitted task and strategy support. This is a finite FTIP construction using the typed receipt and pairing definitions, not a source result.