Full conformal marginal coverage

Let $(\mathcal{X},\mathcal{X})$ and $(\mathcal{Y},\mathcal{Y})$ be measurable feature and response spaces, let $m\ge 0$, and let $Z_i=(X_i,Y_i)$, $1\le i\le m+1$, be random elements of the product measurable space $\mathcal{Z}=(\mathcal{X}\times \mathcal{Y},\mathcal{X}\otimes \mathcal{Y})$. Suppose their joint law is invariant under every coordinate permutation, as in Exchangeable random elements. Suppose a measurable, permutation-equivariant full score assignment $A:{\mathcal{Z}}^{m+1}\to {\mathbb{R}}^{m+1}$ is used for every augmented candidate sample, as in Full conformal prediction set. Then, for $0<\alpha <1$, its nonrandomized full conformal set satisfies $\mathbb{P}\{Y_{m+1}\in C(X_{m+1})\}\ge 1-\alpha \mathrm{.}$ If the tie-randomized set uses one $U$ with Lebesgue probability law on $[0,1]$, independent of $\sigma (Z_1,\dots ,Z_{m+1})$ and shared across the candidate scan, then $\mathbb{P}\{Y_{m+1}\in C_U(X_{m+1})\}=1-\alpha \mathrm{.}$ These are marginal guarantees over the exchangeable examples and any stated tie randomization.

The theorem gives marginal coverage, not coverage conditional on the test features or on a subgroup. It does not provide a distribution-shift guarantee.