{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261015-advance-20261116-r05-finite-sufficient-statistic-0026",
  "pilot_number": "P250-0126",
  "number": 126,
  "family_id": "qf-mod47-suff-conditional-table",
  "title": {
    "zh-Hant": "第 126 題：有限模型充分統計量",
    "en": "Question 126: Conditional Laws Are Defined Only on Positive-Mass Statistic Cells"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261015-advance-20261116-r05-finite-sufficient-statistic-0026",
    "batch_id": "batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-finite-sufficient-statistic"
    ],
    "title_zh": "有限模型充分統計量",
    "stem_zh": "total statistic fibers與正質量conditional laws的完整theta-invariance責任。操作=suff-conditional-table；資料={\"theta_values\":[\"a0\",\"b0\"],\"samples\":[0,1,2,3],\"probability_rows\":[[[2,5],[1,10],[2,5],[1,10]],[[1,10],[2,5],[1,10],[2,5]]],\"statistic\":[0,1,0,1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "([[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]], zero_mass_cells_remain_undefined)",
      "explanation_zh": "Wave6第十層核對=([[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]], zero_mass_cells_remain_undefined)；依完整明示資料獨立重算；zero-mass conditional保持undefined，nonsufficient需明示witness；identity/constant boundaries分開且不使用factorization theorem或宣稱估計品質。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "有限模型充分統計量"
      ]
    },
    "variant": {
      "family_id": "qf-mod47-suff-conditional-table",
      "seed": 202611165026,
      "parameters": {
        "theta_values": [
          "a0",
          "b0"
        ],
        "samples": [
          0,
          1,
          2,
          3
        ],
        "probability_rows": [
          [
            [
              2,
              5
            ],
            [
              1,
              10
            ],
            [
              2,
              5
            ],
            [
              1,
              10
            ]
          ],
          [
            [
              1,
              10
            ],
            [
              2,
              5
            ],
            [
              1,
              10
            ],
            [
              2,
              5
            ]
          ]
        ],
        "statistic": [
          0,
          1,
          0,
          1
        ],
        "domain": "Exact_finite_sufficient_statistic",
        "operation": "suff-conditional-table",
        "semantic_template": "qf-mod47-suff-conditional-table/{\"probability_rows\":[[[2,5],[1,10],[2,5],[1,10]],[[1,10],[2,5],[1,10],[2,5]]],\"samples\":[0,1,2,3],\"statistic\":[0,1,0,1],\"theta_values\":[\"a0\",\"b0\"]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、一般Banach／IVT／Taylor證明、MLE漸近性、factorization theorem、一般DAG iff或primitive-root主張、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-15",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261015-advance-20261116-r05-finite-sufficient-statistic-0026",
    "batch_id": "batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "ddbda5008ebf2a659e44b59f3cad93074629ef7c7840241418b3849478f1d1ed",
      "family_id": "qf-mod47-suff-conditional-table",
      "seed": 202611165026,
      "semantic_template": "qf-mod47-suff-conditional-table/{\"probability_rows\":[[[2,5],[1,10],[2,5],[1,10]],[[1,10],[2,5],[1,10],[2,5]]],\"samples\":[0,1,2,3],\"statistic\":[0,1,0,1],\"theta_values\":[\"a0\",\"b0\"]}"
    },
    "fields": {
      "title": "Conditional Laws Are Defined Only on Positive-Mass Statistic Cells",
      "stem": "Use complete theta labels=[a0, b0], samples=[0, 1, 2, 3], aligned exact probability rows=[[[2, 5], [1, 10], [2, 5], [1, 10]], [[1, 10], [2, 5], [1, 10], [2, 5]]] and total statistic=[0, 1, 0, 1]. Each theta law is normalized. Fibers partition the full sample carrier, retaining zero-mass cells. Give (conditional_cells, boundary).",
      "answer": {
        "kind": "expression",
        "value": "([[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]], zero_mass_cells_remain_undefined)",
        "explanation": "For every theta and fiber first sum its mass. On a positive-mass cell divide each aligned sample mass by the fiber mass. For mass0 retain null as the encoding of an undefined conditional law, not a fabricated zero distribution or empty table. The exact result, retaining the requested order, is ([[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]], zero_mass_cells_remain_undefined)."
      },
      "skills": [
        "Conditional Laws Are Defined Only on Positive-Mass Statistic Cells"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "total statistic fibers與正質量conditional laws的完整theta-invariance責任。操作=suff-conditional-table；資料={\"theta_values\":[\"a0\",\"b0\"],\"samples\":[0,1,2,3],\"probability_rows\":[[[2,5],[1,10],[2,5],[1,10]],[[1,10],[2,5],[1,10],[2,5]]],\"statistic\":[0,1,0,1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use complete theta labels=[a0, b0], samples=[0, 1, 2, 3], aligned exact probability rows=[[[2, 5], [1, 10], [2, 5], [1, 10]], [[1, 10], [2, 5], [1, 10], [2, 5]]] and total statistic=[0, 1, 0, 1]. Each theta law is normalized. Fibers partition the full sample carrier, retaining zero-mass cells. Give (conditional_cells, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "Ω=[0, 1, 2, 3]; T=[0, 1, 0, 1]",
      "after": "fibers=[[0, [0, 2]], [1, [1, 3]]]",
      "relation": "implies",
      "rule_id": "statistic_fibers",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依統計量取值收集原樣本標籤；不能把樣本位置與樣本標籤混讀。",
        "en": "Group the original sample labels by statistic value; distinguish a sample's position from its label."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "T 必須對每個樣本點都有且只有一個值。",
          "en": "T must have exactly one value at every sample point."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "[[0, [0, 2]], [1, [1, 3]]]",
      "after": "disjoint union of fibers = [0, 1, 2, 3]",
      "relation": "implies",
      "rule_id": "partition_audit",
      "conditions": [],
      "explanation": {
        "zh-Hant": "各樣本點恰落入一個 fiber，聯集覆蓋全部 Ω。",
        "en": "Each sample lies in exactly one fiber, and their union covers all of Omega."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "即使某個 fiber 在某參數下機率為零，它仍屬於這個 partition。",
          "en": "A fiber stays in the partition even when it has zero mass under a parameter."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "P_a0(Ω) := sum(2/5, 1/10, 2/5, 1/10)",
      "after": "P_a0(Ω) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐參數核對整列機率正規化。",
        "en": "Verify normalization of the probability row for this parameter."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(Ω)，不以小數近似替代。",
          "en": "Recompute P_a0(Ω) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "2/5",
          "1/10",
          "2/5",
          "1/10"
        ],
        "output": "1",
        "label": "P_a0(Ω)"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "P_a0(T=0) := sum(2/5, 2/5)",
      "after": "P_a0(T=0) = 4/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "只加總 fiber [0, 2] 的質量。",
        "en": "Sum only the masses in fiber [0, 2]."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(T=0)，不以小數近似替代。",
          "en": "Recompute P_a0(T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "2/5",
          "2/5"
        ],
        "output": "4/5",
        "label": "P_a0(T=0)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "P_a0(X=0|T=0) := div(2/5, 4/5)",
      "after": "P_a0(X=0|T=0) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 2/5，分母是剛求出的 fiber 機率 4/5。",
        "en": "The numerator is the atom mass 2/5 under the same law and the denominator is the fiber mass 4/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(X=0|T=0)，不以小數近似替代。",
          "en": "Recompute P_a0(X=0|T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2/5",
          "4/5"
        ],
        "output": "1/2",
        "label": "P_a0(X=0|T=0)"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "reconstruct P_a0(X=0) := mul(1/2, 4/5)",
      "after": "reconstruct P_a0(X=0) = 2/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_a0(X=0)，不以小數近似替代。",
          "en": "Recompute reconstruct P_a0(X=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "4/5"
        ],
        "output": "2/5",
        "label": "reconstruct P_a0(X=0)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "P_a0(X=2|T=0) := div(2/5, 4/5)",
      "after": "P_a0(X=2|T=0) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 2/5，分母是剛求出的 fiber 機率 4/5。",
        "en": "The numerator is the atom mass 2/5 under the same law and the denominator is the fiber mass 4/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(X=2|T=0)，不以小數近似替代。",
          "en": "Recompute P_a0(X=2|T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2/5",
          "4/5"
        ],
        "output": "1/2",
        "label": "P_a0(X=2|T=0)"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "reconstruct P_a0(X=2) := mul(1/2, 4/5)",
      "after": "reconstruct P_a0(X=2) = 2/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_a0(X=2)，不以小數近似替代。",
          "en": "Recompute reconstruct P_a0(X=2) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "4/5"
        ],
        "output": "2/5",
        "label": "reconstruct P_a0(X=2)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "conditional total (a0,0) := sum(1/2, 1/2)",
      "after": "conditional total (a0,0) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正質量 fiber 上的完整條件分布必須加總為 1。",
        "en": "The full conditional law on a positive-mass fiber must sum to 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 conditional total (a0,0)，不以小數近似替代。",
          "en": "Recompute conditional total (a0,0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "conditional total (a0,0)"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "P_a0(T=1) := sum(1/10, 1/10)",
      "after": "P_a0(T=1) = 1/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "只加總 fiber [1, 3] 的質量。",
        "en": "Sum only the masses in fiber [1, 3]."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(T=1)，不以小數近似替代。",
          "en": "Recompute P_a0(T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/10",
          "1/10"
        ],
        "output": "1/5",
        "label": "P_a0(T=1)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "P_a0(X=1|T=1) := div(1/10, 1/5)",
      "after": "P_a0(X=1|T=1) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 1/10，分母是剛求出的 fiber 機率 1/5。",
        "en": "The numerator is the atom mass 1/10 under the same law and the denominator is the fiber mass 1/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(X=1|T=1)，不以小數近似替代。",
          "en": "Recompute P_a0(X=1|T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "1/5"
        ],
        "output": "1/2",
        "label": "P_a0(X=1|T=1)"
      }
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "reconstruct P_a0(X=1) := mul(1/2, 1/5)",
      "after": "reconstruct P_a0(X=1) = 1/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_a0(X=1)，不以小數近似替代。",
          "en": "Recompute reconstruct P_a0(X=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/5"
        ],
        "output": "1/10",
        "label": "reconstruct P_a0(X=1)"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "P_a0(X=3|T=1) := div(1/10, 1/5)",
      "after": "P_a0(X=3|T=1) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 1/10，分母是剛求出的 fiber 機率 1/5。",
        "en": "The numerator is the atom mass 1/10 under the same law and the denominator is the fiber mass 1/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_a0(X=3|T=1)，不以小數近似替代。",
          "en": "Recompute P_a0(X=3|T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "1/5"
        ],
        "output": "1/2",
        "label": "P_a0(X=3|T=1)"
      }
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "reconstruct P_a0(X=3) := mul(1/2, 1/5)",
      "after": "reconstruct P_a0(X=3) = 1/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_a0(X=3)，不以小數近似替代。",
          "en": "Recompute reconstruct P_a0(X=3) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/5"
        ],
        "output": "1/10",
        "label": "reconstruct P_a0(X=3)"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "conditional total (a0,1) := sum(1/2, 1/2)",
      "after": "conditional total (a0,1) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正質量 fiber 上的完整條件分布必須加總為 1。",
        "en": "The full conditional law on a positive-mass fiber must sum to 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 conditional total (a0,1)，不以小數近似替代。",
          "en": "Recompute conditional total (a0,1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "conditional total (a0,1)"
      }
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "P_b0(Ω) := sum(1/10, 2/5, 1/10, 2/5)",
      "after": "P_b0(Ω) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐參數核對整列機率正規化。",
        "en": "Verify normalization of the probability row for this parameter."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(Ω)，不以小數近似替代。",
          "en": "Recompute P_b0(Ω) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/10",
          "2/5",
          "1/10",
          "2/5"
        ],
        "output": "1",
        "label": "P_b0(Ω)"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "P_b0(T=0) := sum(1/10, 1/10)",
      "after": "P_b0(T=0) = 1/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "只加總 fiber [0, 2] 的質量。",
        "en": "Sum only the masses in fiber [0, 2]."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(T=0)，不以小數近似替代。",
          "en": "Recompute P_b0(T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/10",
          "1/10"
        ],
        "output": "1/5",
        "label": "P_b0(T=0)"
      }
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "P_b0(X=0|T=0) := div(1/10, 1/5)",
      "after": "P_b0(X=0|T=0) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 1/10，分母是剛求出的 fiber 機率 1/5。",
        "en": "The numerator is the atom mass 1/10 under the same law and the denominator is the fiber mass 1/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(X=0|T=0)，不以小數近似替代。",
          "en": "Recompute P_b0(X=0|T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "1/5"
        ],
        "output": "1/2",
        "label": "P_b0(X=0|T=0)"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "reconstruct P_b0(X=0) := mul(1/2, 1/5)",
      "after": "reconstruct P_b0(X=0) = 1/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_b0(X=0)，不以小數近似替代。",
          "en": "Recompute reconstruct P_b0(X=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/5"
        ],
        "output": "1/10",
        "label": "reconstruct P_b0(X=0)"
      }
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "P_b0(X=2|T=0) := div(1/10, 1/5)",
      "after": "P_b0(X=2|T=0) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 1/10，分母是剛求出的 fiber 機率 1/5。",
        "en": "The numerator is the atom mass 1/10 under the same law and the denominator is the fiber mass 1/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(X=2|T=0)，不以小數近似替代。",
          "en": "Recompute P_b0(X=2|T=0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "1/5"
        ],
        "output": "1/2",
        "label": "P_b0(X=2|T=0)"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "reconstruct P_b0(X=2) := mul(1/2, 1/5)",
      "after": "reconstruct P_b0(X=2) = 1/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_b0(X=2)，不以小數近似替代。",
          "en": "Recompute reconstruct P_b0(X=2) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/5"
        ],
        "output": "1/10",
        "label": "reconstruct P_b0(X=2)"
      }
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "conditional total (b0,0) := sum(1/2, 1/2)",
      "after": "conditional total (b0,0) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正質量 fiber 上的完整條件分布必須加總為 1。",
        "en": "The full conditional law on a positive-mass fiber must sum to 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 conditional total (b0,0)，不以小數近似替代。",
          "en": "Recompute conditional total (b0,0) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "conditional total (b0,0)"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "P_b0(T=1) := sum(2/5, 2/5)",
      "after": "P_b0(T=1) = 4/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "只加總 fiber [1, 3] 的質量。",
        "en": "Sum only the masses in fiber [1, 3]."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(T=1)，不以小數近似替代。",
          "en": "Recompute P_b0(T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "2/5",
          "2/5"
        ],
        "output": "4/5",
        "label": "P_b0(T=1)"
      }
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "P_b0(X=1|T=1) := div(2/5, 4/5)",
      "after": "P_b0(X=1|T=1) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 2/5，分母是剛求出的 fiber 機率 4/5。",
        "en": "The numerator is the atom mass 2/5 under the same law and the denominator is the fiber mass 4/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(X=1|T=1)，不以小數近似替代。",
          "en": "Recompute P_b0(X=1|T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2/5",
          "4/5"
        ],
        "output": "1/2",
        "label": "P_b0(X=1|T=1)"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "reconstruct P_b0(X=1) := mul(1/2, 4/5)",
      "after": "reconstruct P_b0(X=1) = 2/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_b0(X=1)，不以小數近似替代。",
          "en": "Recompute reconstruct P_b0(X=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "4/5"
        ],
        "output": "2/5",
        "label": "reconstruct P_b0(X=1)"
      }
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "P_b0(X=3|T=1) := div(2/5, 4/5)",
      "after": "P_b0(X=3|T=1) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子是同一 law 的原子機率 2/5，分母是剛求出的 fiber 機率 4/5。",
        "en": "The numerator is the atom mass 2/5 under the same law and the denominator is the fiber mass 4/5."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P_b0(X=3|T=1)，不以小數近似替代。",
          "en": "Recompute P_b0(X=3|T=1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2/5",
          "4/5"
        ],
        "output": "1/2",
        "label": "P_b0(X=3|T=1)"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "reconstruct P_b0(X=3) := mul(1/2, 4/5)",
      "after": "reconstruct P_b0(X=3) = 2/5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "條件機率乘回 fiber 質量，須恢復該原子的機率。",
        "en": "Multiply by the fiber mass to recover that atom's original probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 reconstruct P_b0(X=3)，不以小數近似替代。",
          "en": "Recompute reconstruct P_b0(X=3) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "4/5"
        ],
        "output": "2/5",
        "label": "reconstruct P_b0(X=3)"
      }
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "conditional total (b0,1) := sum(1/2, 1/2)",
      "after": "conditional total (b0,1) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正質量 fiber 上的完整條件分布必須加總為 1。",
        "en": "The full conditional law on a positive-mass fiber must sum to 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 conditional total (b0,1)，不以小數近似替代。",
          "en": "Recompute conditional total (b0,1) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "conditional total (b0,1)"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "all parameter × fiber cells audited",
      "after": "[[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]]",
      "relation": "implies",
      "rule_id": "complete_conditional_table",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按原參數次序、再按統計量取值次序列回每個 cell；沒有跳過零質量分支。",
        "en": "Return every cell in parameter order and then statistic-value order; retain the zero-mass branch rule."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "本題只求條件分布表；未額外宣稱估計品質或引入 factorization theorem。",
          "en": "This task computes the conditional table; it makes no estimator-quality claim and does not invoke factorization."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "fibers": [
      [
        0,
        [
          0,
          2
        ]
      ],
      [
        1,
        [
          1,
          3
        ]
      ]
    ],
    "cells": [
      [
        "a0",
        0,
        "4/5",
        [
          [
            0,
            "1/2"
          ],
          [
            2,
            "1/2"
          ]
        ]
      ],
      [
        "a0",
        1,
        "1/5",
        [
          [
            1,
            "1/2"
          ],
          [
            3,
            "1/2"
          ]
        ]
      ],
      [
        "b0",
        0,
        "1/5",
        [
          [
            0,
            "1/2"
          ],
          [
            2,
            "1/2"
          ]
        ]
      ],
      [
        "b0",
        1,
        "4/5",
        [
          [
            1,
            "1/2"
          ],
          [
            3,
            "1/2"
          ]
        ]
      ]
    ]
  },
  "conclusion": {
    "canonical_value": "([[a0, 0, 4/5, [[0, 1/2], [2, 1/2]]], [a0, 1, 1/5, [[1, 1/2], [3, 1/2]]], [b0, 0, 1/5, [[0, 1/2], [2, 1/2]]], [b0, 1, 4/5, [[1, 1/2], [3, 1/2]]]], zero_mass_cells_remain_undefined)",
    "explanation": {
      "zh-Hant": "逐 cell 的完整條件分布如表；零質量 cell 必須保留 null。",
      "en": "The full cellwise conditional laws are listed; zero-mass cells must retain null."
    }
  },
  "definition_note": {
    "zh-Hant": "統計量在完整樣本集合上有定義；逐參數正規化，僅對正質量 fiber 定義條件分布。",
    "en": "The statistic is total on the full sample carrier. Normalize each parameter law and condition only on positive-mass fibers."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations",
      "batch_path": "public/data/questions/batches/batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations/questions.jsonl",
      "batch_sha256": "8ce239dc6b93456c36b291b3ff0d0d055f7f5a0896790e1605f94f4266893893",
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      "original_line_sha256": "9b87d73a8032ca48db51d9c64fdf85f10a90b50fa8e74642c60ca78414cbdcbd",
      "question_compact_sha256": "ddbda5008ebf2a659e44b59f3cad93074629ef7c7840241418b3849478f1d1ed",
      "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261116-r05-finite-sufficient-statistic-foundations.en.jsonl",
      "english_row_compact_sha256": "f2eb133c134ebe740aa45b61d5a56013c973576b2e24e76b3ac790fed81f5280"
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    "mko_bindings": {
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        "english_sha256": "cd5defd0a6ff5e44e92db46ae4d8f1c71413883dfb875e6539fbcd7cac62fef6",
        "original_path": "public/data/mko/finite-sufficient-statistic.json",
        "english_path": "public/data/i18n/editions/mko-finite-sufficient-statistic.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "fb325b42cfcaa25ddbae992459c2684afd767fe218028bc0b9c0823e67c83b7c",
    "original_candidate_sha256": "dc1022c91ba3fc83388e3b3691ed6cd196df77ee93dcccbc4c2c1832efed1c23",
    "original_candidate_line_sha256": "dc1022c91ba3fc83388e3b3691ed6cd196df77ee93dcccbc4c2c1832efed1c23"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "本稿是來源綁定的教學推導候選；精確重算不等於獨立數學審查或形式化證明。原題、答案、公式、AST 與歷史未被修改。",
      "en": "This is a source-bound pedagogical derivation candidate. Exact recomputation is not independent mathematical review or formal proof. Original questions, answers, formulas, ASTs and history are unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261024-002",
    "review_date": "2026-10-24",
    "owner_authorized_reader_publication": true,
    "scope": "Family prototypes and mathematical/bilingual reasoning variants reviewed in this existing context; parameter substitutions, all original/source bindings, full answer tuples and arithmetic ASTs independently reconstructed per question. Numeric binomial text slots are validated against independently regenerated n,k,C(n,k),n-k values. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "4727d1bd957a9b7f6cadb7972ddb5ab80a31f5259dc5394c1dff2b71944ed36b",
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    "bilingual_semantics": true,
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    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
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}