Result · DS-EFFICIENT-SCORE-GLOBAL-UPPER

An efficient-score problem bounds profiled optimization from above

Proved herepublication: internal

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Meaning

Optimizing efficient-score information over the same bin budget bounds the global profiled optimum.

Statement

Taking determinants/suprema in efficient-score domination yields a lower-dimensional D-optimal quantization problem whose optimum is an upper bound on the globally optimal in-bin Ds value.

Relationships

Rests on

The full chain beneath it

Enables

Open next door

Local graph
Local map of D-optimal efficient-score quantization upper-bounds global Ds1 prerequisites to the left, 3 results it enables or questions it raises to the right, 0 boundary cases below.Full-data efficient-score dominationFull-data efficient-s…Scalar efficient-score upper problem is exactly solvable by dynamic programmingScalar efficient-scor…Best practical Ds solver with theorem-backed certificateBest practical Ds sol…Exact off-class global basin transfers to empirical exchange-stable optimaExact off-class globa…D-optimal efficient-score quantization upper-bounds global DsD-optimal efficient-s…
Prerequisites to the left, what it enables or raises to the right, boundary cases below. Open in the map.

Proof

DS7. Full-data efficient-score domination

Let

S^=Sψ−B∗Sλ,B∗=Iψλfull(Iλλfull)−1.\widehat S=S_\psi-B^*S_\lambda, \qquad B^*=I^{\rm full}_{\psi\lambda}(I^{\rm full}_{\lambda\lambda})^{-1}.

For every quantizer qq,

Sψ(Iq)⪯Var⁡(E[S^∣q(S)]).\boxed{ S_\psi(I_q) \preceq \operatorname{Var}(E[\widehat S\mid q(S)]). }

Interpretation: profiling nuisance information from the bins cannot beat first projecting to the full-data efficient score and then measuring the retained between-cell information of that projection.

Taking log-determinants and suprema gives

sup⁡qFs(q)≤best D-optimal K-bin value for the dψ-dimensional efficient score.\boxed{ \sup_q F_s(q) \le \text{best D-optimal K-bin value for the }d_\psi\text{-dimensional efficient score}. }

For the last equality to deterministic quantization of S^\widehat S, atomlessness is required for the efficient-score law itself.

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