What it answers
This is the record’s reply of 2026-10-10 to CLAIM-218 (granted, and right), which showed that A86’s D_Q(U, V) = sup_q |R_q(U) − R_τ(q)(V)| is not Blackwell’s order: one number per task needs a prior or a minimax summary; subtracting the risks of two tasks needs their losses on one scale; the absolute value counts improvement as loss and makes D_Q symmetric; and τ on tasks is unanchored. The reply restates the term so that the first three points are met. The fourth is met for report fidelity, where τ is the identity (CLAIM-212), and stays open for re-performance (QUESTION-005). Steps (a) to (d) below are elementary and checkable. That receivers’ judged fidelity behaves like this term is a proposal, and the defeat condition tests it.
The claim
Definition. Let U be the source and V the rendering, both experiments about one situation Z, which is finite in any design. A decision problem q has a finite set of actions and a loss L_q(z, a) in [0, 1]. Let π be a stated prior on Z, taken from the design’s distribution of situations. The Bayes risk r_q,π(U) is the least expected loss, under π, over decision rules that read U. Then
D⁺_Q(U → V; π) = sup over q ∈ Q of [r_q,π(V) − r_q,π(U)]⁺,
where [x]⁺ = max(x, 0). It is the largest increase in risk, over the problems in Q, from using the rendering in place of the source. The same problem is used on both sides, so no scale between tasks is needed.
(a) A garbling never gains. Suppose V is a garbling of U: there is a Markov kernel G with P(V | Z) = G P(U | Z). Any rule δ that reads V gives a rule on U, namely apply G to U and then δ, and under each z it has the same distribution of actions, so the same risk. So the best rule on U does at least as well as the best rule on V, and r_q,π(U) ≤ r_q,π(V) for every q and π. Hence D⁺_Q(V → U; π) = sup_q [r_q,π(U) − r_q,π(V)]⁺ = 0 always. This is the easy direction, the one the manuscript uses.
(b) The converse. Suppose D⁺(U → V; π) = 0 for every finite problem with losses in [0, 1] and every prior π. Then r_q,π(V) ≤ r_q,π(U) for all of them. Rescaling a bounded loss to [0, 1], by L ↦ (L − m)/(M − m) with m and M its least and greatest values, rescales every Bayes risk by the same increasing affine map, so the inequality holds for every bounded loss. By the Bayes form of Blackwell’s theorem for finitely many states (THEORY-156), V is then at least as informative as U. The form the record holds needs the source’s outcome space to be standard Borel (CLAIM-326). So D⁺(U → V; ·) vanishes for every problem and prior exactly when the rendering loses nothing the source offered. A rendering that is a strict garbling has D⁺(U → V) > 0 for some problem and prior, and D⁺(V → U) = 0 for all.
(c) It is directed. Let Z be uniform on {0, 1}, U = Z, and V a constant. Take q = “guess Z” with 0–1 loss. From U the guess is always right: r(U) = 0. From V nothing is learnt, and any guess is wrong with probability 1/2: r(V) = 1/2. So D⁺(U → V) = [1/2 − 0]⁺ = 1/2, and D⁺(V → U) = [0 − 1/2]⁺ = 0. The term charges the rendering for its loss and does not charge the source for being better.
(d) A86’s D_Q is its symmetrization. With τ the identity on tasks and the same prior, D_Q(U, V) = sup_q |r_q(U) − r_q(V)|. For any real x, |x| = max([x]⁺, [−x]⁺), and a supremum of a maximum is the maximum of the suprema. So
D_Q(U, V) = max(sup_q [r_q(V) − r_q(U)]⁺, sup_q [r_q(U) − r_q(V)]⁺) = max(D⁺(U → V), D⁺(V → U)).
This is CLAIM-218’s third point, that D_Q counts improvement as distortion, stated exactly. In CLAIM-258’s worked case, the rendering that answers “Is Z = 3?” loses 1/3 on “Is Z = 1?” and gains 1/3 on “Is Z = 3?”. So D⁺(U → V) = 1/3 and D⁺(V → U) = 1/3, and D_Q = 1/3. A rendering that told its reader nothing at all would also have D_Q = 1/3 on these two problems (it loses 1/3 on “Is Z = 1?”, since the risk with no information is 1/3, and gains nothing), but its directed pair is (1/3, 0). The directed pair tells the two renderings apart, and D_Q does not.
(e) A prior-free form. Taking the supremum over priors, or using minimax risks, gives a task-restricted analogue of Le Cam’s deficiency. The exact relation, with its constants, depends on how losses and deficiency are normalised and is in Torgersen (LIT-778), which the record has not read, so it is not stated here.
What it does not say
- It does not say Q is fixed. Which tasks are named is a choice (CLAIM-252, QUESTION-075), and D⁺ is relative to it.
- It does not say improvements do not matter. They are reported in the other direction, D⁺(V → U), as CLAIM-106’s asymmetry asks.
- It does not supply a correspondence of tasks for re-performance fidelity. That stays QUESTION-005.
Note of 2026-10-15: split under ADR-043
ADR-043’s full classification found this claim compound: the properties of D⁺_Q, steps (a) to (d), are demonstrated, and that it should be the decision term is a proposal whose defeat condition is empirical and untested. The properties are now CLAIM-326 (demonstrated), on which this claim rests. This claim keeps the proposal, and its defeat condition, and is marked interpretive. Its title is kept: it leads with “should be”, and its second half names the properties the proposal relies on, which CLAIM-326 now carries.