Split conformal marginal coverage

Define $g(\omega ,z)=(s(H(\omega ),z_i){)}_{i=1}^{m+1}$. For each $i$, the map $(\omega ,z)\mapsto (H(\omega ),z_i)$ is measurable into $\mathcal{H}\times \mathcal{Z}$: the preimage of a rectangle $A\times B$ is $H^{-1}(A)\times \{z:z_i\in B\}$, which belongs to $\mathcal{T}\otimes {\mathcal{Z}}^{\otimes (m+1)}$, and such rectangles generate the target product sigma-algebra. Joint measurability of $s$ therefore makes every coordinate of $g$ measurable. The finite tuple $g$ is measurable into the product Borel sigma-algebra because preimages of Borel rectangles are finite intersections of coordinate preimages. For each fixed $\omega$, $g(\omega ,\cdot )$ is permutation-equivariant because the same fixed rule $s(H(\omega ),\cdot )$ is applied coordinatewise. By [F2] its pushforward is a regular conditional distribution of the score vector; by the sectionwise exchangeability in the hypothesis and [F3], that score-vector distribution is exchangeable.

Put $n=m+1$ and $k=\lceil n(1-\alpha )\rceil$. If $k=n$, [F5] sets $q=+\mathrm{\infty }$, so every true response is covered. If $k\le m$ and the true test score $R_n$ exceeds the $k$th calibration score $q$, then at least $k$ calibration scores are strictly below $R_n$; consequently at most $n-k$ of all $n$ scores are at least $R_n$, and $p_n\le (n-k)\mathrm{/}n\le \alpha$. Thus $p_n>\alpha$ implies coverage in every case.

Apply [F4] to the conditional exchangeable score law from step 1.1. The event $p_n>\alpha$ has conditional probability at least $1-\alpha$, and step 1.2 includes it in the nonrandomized coverage event. For the randomized version, the second assertion of [F2] makes $Q_{\omega }\otimes \lambda$ a regular conditional distribution of $(Z,U)$ given $\mathcal{T}$. The map that sends $(\omega ,z,u)$ to the tie-randomized rank computed from $g(\omega ,z)$ and $u$ is jointly measurable: the score coordinates are measurable by step 1.1, comparisons of real-valued measurable functions are measurable, and the rank formula uses finite sums, products, and scalar division. Applying the first assertion of [F2] to this map gives the conditional law of $p_{U,n}$. For almost every $\omega$, $Q_{\omega }$ is exchangeable, $H(\omega )$ is fixed, and $\lambda$ is independent uniform; [F4] therefore gives conditional probability exactly $1-\alpha$ for $p_{U,n}>\alpha$, which is precisely randomized-set coverage.

The conditional-kernel identity in [F1], applied with the conditioning event $H=\mathrm{\Omega }$, integrates the conditional coverage probabilities from step 2.1 to their unconditional probabilities. The lower bound and the exact equality therefore also hold marginally.