A sufficiency certificate from every fibre conditional law

Goal and scope

Audit invariance across theta on every positive-mass fibre, retaining undefined null conditions.

No factorization theorem or estimator-quality inference.

Givens and symbols

• work

• G1 — Theta={0,1}, Omega={a,b,c,d,z}; P0=(1/8,3/8,1/4,1/4,0), P1=(1/16,3/16,3/8,3/8,0). S(a)=S(b)=0, S(c)=S(d)=1, S(z)=2.

• S — total statistic
• t — fibre label

Rules and sources

• mko_definition · cited — A statistic is a total map on the full sample carrier and its fibers form a partition. For every (theta,t) first compute P_theta(T=t). Define and calculate the exact fiber conditional law only on a positive-mass cell, then compare all valid theta laws at each t. Leave zero-mass conditions undefined, encoded as null in these outputs. Separate explicit nonsufficiency witnesses, the identity statistic and the constant-statistic boundary. No factorization theorem or estimator-quality conclusion is introduced. · mko-finite-sufficient-statistic
• work · prerequisite — Finite sufficiency: on each fibre compare full conditional laws only where conditioning has positive mass, requiring theta invariance; null quotients remain undefined.

Main derivation

P_0,P_1

1/8+3/8+1/4+1/4=1,\quad1/16+3/16+3/8+3/8=1

S^{-1}(0),S^{-1}(1),S^{-1}(2)

\{a,b\},\{c,d\},\{z\}

P_\theta(S=t),\quad t=(0,1,2)

P_0^S=(1/2,1/2,0),\quad P_1^S=(1/4,3/4,0)

t=0:\quad P_0(a\mid S=0)=\frac{1/8}{1/2},\quad P_1(a\mid S=0)=\frac{1/16}{1/4}

P_0(\cdot\mid S=0)=P_1(\cdot\mid S=0)=(1/4,3/4,0,0,0)

t=1:\quad\frac{1/4}{1/2}=\frac{3/8}{3/4}=1/2

P_0(\cdot\mid S=1)=P_1(\cdot\mid S=1)=(0,0,1/2,1/2,0)

P_0(S=2)=P_1(S=2)=0

P_\theta(\cdot\mid S=2)\text{ undefined for both }\theta

\text{all positive fibre conditionals invariant across valid }\theta

S\text{ is sufficient for this finite model}

R(x)\equiv0

P_0(a\mid R=0)=1/8\ne1/16=P_1(a\mid R=0)

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · G1, S07, work — The identity statistic instead has point-mass conditionals on valid fibres and is sufficient; the boundaries differ.
• S07 · premise_audit · G1, S06, work — No factorization theorem replaced the table audit.
• S06 · premise_audit · G1, S05, work — Undefined is not a zero-vector probability law.
• S05 · premise_audit · G1, S04, work — Both theta denominators on this cell are positive.
• S04 · premise_audit · G1, S03, work — List the full five-point conditional law, not only a.
• S03 · premise_audit · G1, S02, work — Sufficiency does not require the marginal law of S to be invariant.
• S02 · premise_audit · G1, S01, work — Do not omit null fibre{z}.
• S01 · premise_audit · G1, work — z has zero mass in both laws but remains in the carrier.

Error controls and negative controls

Mass first

Compute each(theta,t) fibre mass before division.

Complete conditionals

List all five sample probabilities on each valid cell.

• Sufficiency requires an invariant law of S. → reject — S is sufficient here although P0(S=0)=1/2 and P1(S=0)=1/4.
• Use(0,0,0,0,0) as the law at S=2. → reject — It sums to 0 and conditioning has zero denominator, so it is undefined.
• Constant R is sufficient here. → reject — Its conditional probability of a varies as1/8 or1/16.

Result and limitations

S is sufficient and constant R is not; null-fibre conditional laws remain undefined.

Verification and source binding

3 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 7b86518f187ccbc75abb75a795f774026b2bfa13c74cea0f969179bcaa062304
• original source SHA-256: 99f4f978f27d4111a01876f24ceb88ab0e95d184de6346874988ea73556fac8f
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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