Garbling Increases Bayes Risk

Abstract

Blackwell dominance is reflexive and transitive, includes measurable deterministic post-processing, and its garblings cannot improve optimal Bayes risk.

Lemma 1.1 (Blackwell dominance is reflexive).

$$\forall P:\mathrm{Kernel}\left(\theta ,X\right),\mathrm{BlackwellDominates}\left(P,P\right).$$

Proof. Machine-checked in Lean as D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_refl (✓ std3). ∎

Source. Repository-derived.

Commentary.

Every experiment dominates itself because the identity kernel is a Markov kernel and garbling by that kernel leaves the experiment unchanged.

Thus the experiment itself is recovered by an admissible garbling, which supplies the witness required by Blackwell dominance.

Lemma 1.2 (Blackwell dominance is transitive).

$$\begin{array}{c}\forall P:\mathrm{Kernel}\left(\theta ,X\right),Q:\mathrm{Kernel}\left(\theta ,X_1\right),R:\mathrm{Kernel}\left(\theta ,X_2\right),\\ \mathrm{BlackwellDominates}\left(P,Q\right)\wedge \mathrm{BlackwellDominates}\left(Q,R\right)\Rightarrow \\ \mathrm{BlackwellDominates}\left(P,R\right).\end{array}$$

Proof. Machine-checked in Lean as D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_trans (✓ std3). ∎

Source. Repository-derived.

Commentary.

Suppose Q is obtained from P through one Markov garbling and R is obtained from Q through another. Composing the two garbling kernels gives a Markov kernel directly from the output of P to the output of R.

Associativity of kernel composition identifies this composite garbling with R, so P Blackwell-dominates R.

Lemma 1.3 (Measurable maps are Blackwell garblings).

$$\begin{array}{c}\forall P:\mathrm{Kernel}\left(\theta ,X\right),f:X\to X_1,\\ \mathrm{Measurable}\left(f\right)\Rightarrow \\ \mathrm{BlackwellDominates}\left(P,\mathrm{map}\left(P,f\right)\right).\end{array}$$

Proof. Machine-checked in Lean as D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_map (✓ std3). ∎

Source. Repository-derived.

Commentary.

A measurable transformation of the observation space determines a deterministic Markov kernel. Applying that kernel after an experiment is exactly the mapped experiment.

Consequently every measurable deterministic post-processing of an experiment is a Blackwell garbling of the original experiment.

Theorem 1.4 (Garbling cannot decrease optimal Bayes risk).

$$\begin{array}{c}\forall P:\mathrm{Kernel}\left(\theta ,X\right),Q:\mathrm{Kernel}\left(\theta ,X_1\right),\\ \mathrm{BlackwellDominates}\left(P,Q\right)\Rightarrow \\ \forall \ell :\theta \to Y\to ENNReal,\pi :\mathrm{Measure}\left(\theta \right),\\ \mathrm{bayesRisk}\left(\ell ,P,\pi \right)\le \mathrm{bayesRisk}\left(\ell ,Q,\pi \right).\end{array}$$

Proof. Machine-checked in Lean as D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.bayesRisk_le_of_blackwellDominates (✓ std3). ∎

Source. Repository-derived.

Commentary.

If Q is obtained by applying a Markov garbling to P, then every decision procedure based on Q can also be run after observing P: first garble the observation and then apply that procedure.

Taking the infimum over all Markov decision rules therefore gives no larger Bayes risk for P than for Q. The comparison holds for every ENNReal-valued loss and every measure used as the prior.

References

• Truth anchor: D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.bayesRisk_le_of_blackwellDominates
• Truth anchor: D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_map
• Truth anchor: D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_refl
• Truth anchor: D5/S3/Estimation/DecisionRisk/GarblingIncreasesBayesRisk.blackwellDominates_trans