# A sufficiency certificate from every fibre conditional law

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derivation-mko-finite-sufficient-statistic-01 · ocme-new142-reader-adoption-20261021



## 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

S01 · impliesBefore

P_0,P_1

After

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

Conditions: none

Rule: work · Uses: G1, work

Leaf basis: work

Both complete laws are normalized and nonnegative.



S02 · impliesBefore

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

After

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

Conditions: none

Rule: work · Uses: G1, S01, work

Leaf basis: work

S is total and its three nonempty fibres form a complete partition.



S03 · impliesBefore

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

After

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

Conditions: none

Rule: work · Uses: G1, S02, work

Leaf basis: work

Compute every fibre mass before deciding where conditioning is defined.



S04 · impliesBefore

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}

After

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

Conditions: none

Rule: work · Uses: G1, S03, work

Leaf basis: work

The conditional masses of a,b agree across theta; the other three outcomes have mass zero.



S05 · impliesBefore

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

After

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

Conditions: none

Rule: work · Uses: G1, S04, work

Leaf basis: work

The second positive-mass fibre also has a fully invariant conditional law.



S06 · impliesBefore

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

After

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

Conditions: none

Rule: work · Uses: G1, S05, work

Leaf basis: work

Do not invent a0/0 distribution; this fibre has no valid theta conditions to compare.



S07 · impliesBefore

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

After

S\text{ is sufficient for this finite model}

Conditions: none

Rule: work · Uses: G1, S06, work

Leaf basis: work

Both parameters and all three fibres are exhausted, giving sufficiency by the finite conditional-law definition.



S08 · impliesBefore

R(x)\equiv0

After

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

Conditions: none

Rule: work · Uses: G1, S07, work

Leaf basis: work

The constant statistic R has the single fibre Omega, with probability 1 under each theta. The conditional distribution of the sample outcome given R=0 is P_theta itself and varies with theta, so R is not sufficient for this finite model.







## 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 firstCompute each(theta,t) fibre mass before division.

Complete conditionalsList 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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