Question 18180: A Total Statistic and a Qualified Model Are Separate Checks

Original question and givens

Use one common finite model with parameter indices=Fin(2)={0,…,1}, outcomes=Fin(3)={0,…,2}, complete candidate laws=[[1/8, 1/8, 6/8], [1/2, 1/3, 1/6]], and the single total statistic T=[0, 5, 10] in outcome order. Each law must be nonnegative and sum to 1 before model qualification. T depends on the observation, not the unknown parameter. An observed value is an estimate, while T is the estimator function; no prior or posterior is supplied. Give (parameter_count, outcome_count, statistic, model_qualified, statistic_total, certificate).

• G001 — Use one common finite model with parameter indices=Fin(2)={0,…,1}, outcomes=Fin(3)={0,…,2}, complete candidate laws=[[1/8, 1/8, 6/8], [1/2, 1/3, 1/6]], and the single total statistic T=[0, 5, 10] in outcome order. Each law must be nonnegative and sum to 1 before model qualification. T depends on the observation, not the unknown parameter. An observed value is an estimate, while T is the estimator function; no prior or posterior is supplied. Give (parameter_count, outcome_count, statistic, model_qualified, statistic_total, certificate).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-statistical-estimator

Step-by-step derivation

source operation=estimator-qualification;original snapshot and mathematics retained

explicit ordered source output fields=[parameter_count, outcome_count, statistic, model_qualified, statistic_total, certificate]

estimator parameter0 complete law sum := sum(1/8, 1/8, 3/4)

estimator parameter0 complete law sum = 1

original parameter=0;SAME outcome carrier Fin(3);aligned reduced candidate law=[1/8, 1/8, 3/4]

sum=1;nonnegative=Yes;law qualified=Yes

estimator parameter1 complete law sum := sum(1/2, 1/3, 1/6)

estimator parameter1 complete law sum = 1

original parameter=1;SAME outcome carrier Fin(3);aligned reduced candidate law=[1/2, 1/3, 1/6]

sum=1;nonnegative=Yes;law qualified=Yes

original parameter carrier Fin(2);common outcome carrier Fin(3);ONE original statistic array=[0, 5, 10]

all model laws qualified=Yes;one value per outcome=Yes;SAME statistic map T(o) at EVERY parameter

Reverse checks

• S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S004 · premise_and_result_audit — Recompute estimator parameter1 complete law sum with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S002 · premise_and_result_audit — Recompute estimator parameter0 complete law sum with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

(2, 3, [0, 5, 10], Yes, Yes, single_total_statistic_independent_of_parameter)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.