{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-joint-distribution-01",
  "mko_id": "mko-finite-joint-distribution",
  "title": {
    "zh-Hant": "有限聯合分布：驗證聯合質量表並由同一 joint law 聚合邊際與事件",
    "en": "Finite joint distribution: validating a joint mass table and aggregating marginals and events from one joint law"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-joint-distribution-01",
    "source": {
      "mko_id": "mko-finite-joint-distribution",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "d04aa2266aeda810e9e7e400220d366a61da23cdc33192e7286eac1d0fed0340",
      "english_sha256": "56edeacfb8cda68555a194d3d171ecf5208fd8129a3c2a5884aacd132751bb06"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "cartesian_product_def",
      "frac_rules",
      "finite_sum_rules",
      "order_basics",
      "numeric_evaluation",
      "finite_sum_regroup"
    ],
    "goal": {
      "zh-Hant": "給定支撐 S={(0,0),(0,1),(1,0),(1,1),(2,1)} 與質量 p(0,0)=1/6、p(0,1)=1/3、p(1,0)=1/4、p(1,1)=1/12、p(2,1)=1/6。驗證 (S,p) 是二維有限聯合分布；由同一個 joint law 算出 X 與 Y 的邊際、矩形事件 P(X∈{1,2}, Y=1) 與單點質量 P(X=2, Y=0)；最後構造一個邊際相同但不同的 joint law，說明兩個邊際一般不能反推聯合分布。",
      "en": "Given the support S={(0,0),(0,1),(1,0),(1,1),(2,1)} with masses p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, verify that (S,p) is a two-dimensional finite joint distribution; from this one joint law compute the marginals of X and Y, the rectangle event P(X∈{1,2}, Y=1) and the single-pair mass P(X=2, Y=0); finally construct a different joint law with the same marginals, showing that two marginals do not in general determine the joint distribution."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "表：S={(0,0),(0,1),(1,0),(1,1),(2,1)}，p(0,0)=1/6、p(0,1)=1/3、p(1,0)=1/4、p(1,1)=1/12、p(2,1)=1/6，皆為精確有理數。",
          "en": "Table: S={(0,0),(0,1),(1,0),(1,1),(2,1)} with p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, all exact rationals."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "X 與 Y 是同一個值對的兩個座標：在 (x,y) 上 X=x、Y=y。所有事件機率都由這同一張表聚合。",
          "en": "X and Y are the two coordinates of the same pair: at (x,y), X=x and Y=y. Every event probability is aggregated from this one table."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "A={1,2}，B={1}。",
          "en": "A={1,2} and B={1}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{table }(S,p)",
        "after_tex": "\\text{(i) distinct pairs, (ii) } p(s)\\ge 0 \\text{ exact, (iii) } \\sum_{s\\in S}p(s)=1",
        "relation": "definition",
        "rule": "mko_joint_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，一張表要成為二維有限聯合分布，必須同時滿足三項資格：(i) 列出的值對互異；(ii) 每個質量是非負的精確數；(iii) 質量總和恰好等於 1，不是近似等於 1。之後每一項都逐一核對，三項都通過之前不計算任何邊際或事件。",
          "en": "By the MKO, a table is a two-dimensional finite joint distribution only if it meets three requirements at once: (i) the listed pairs are distinct; (ii) every mass is a nonnegative exact number; (iii) the masses sum to exactly 1, not approximately 1. Each requirement is checked below, and no marginal or event is computed before all three pass."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_joint_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義、列出要核的三項，不推出新事實。",
            "en": "This step only unfolds the definition and lists the three obligations; it derives no new fact."
          }
        },
        "leaf_basis": "mko_joint_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1",
          "cartesian_product_def"
        ],
        "before_tex": "(0,0),\\ (0,1),\\ (1,0),\\ (1,1),\\ (2,1)",
        "after_tex": "\\text{five pairwise distinct pairs}",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個有序對相等必須兩個座標都相等。依第一座標分組：第一座標不同的兩對必然不同；第一座標為 0 的 (0,0)、(0,1) 第二座標不同；第一座標為 1 的 (1,0)、(1,1) 第二座標不同；第一座標為 2 的只有 (2,1)。所以 C(5,2)=10 組比較全部得到「不相等」，資格 (i) 成立。",
          "en": "Two ordered pairs are equal only if both coordinates agree. Group by first coordinate: two pairs with different first coordinates differ; with first coordinate 0, (0,0) and (0,1) differ in the second coordinate; with first coordinate 1, (1,0) and (1,1) differ in the second coordinate; first coordinate 2 has only (2,1). So all C(5,2)=10 comparisons give 'not equal', and requirement (i) holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "分組涵蓋全部 10 組比較，沒有遺漏；一個座標不同就足以推出兩對不同，不需要反向。",
            "en": "The grouping covers all 10 comparisons with none missed; one differing coordinate suffices for two pairs to differ, and no converse is needed."
          }
        },
        "leaf_basis": "cartesian_product_def + case_analysis"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G1"
        ],
        "before_tex": "p=\\tfrac16,\\ \\tfrac13,\\ \\tfrac14,\\ \\tfrac1{12},\\ \\tfrac16",
        "after_tex": "p(s)>0\\ \\text{for every } s\\in S",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個質量都是兩個正整數的商，所以是正的精確有理數，特別是非負。資格 (ii) 成立；精確有理數也符合 MKO 假設 a1。",
          "en": "Each mass is a quotient of two positive integers, so it is a positive exact rational and in particular nonnegative. Requirement (ii) holds, and exact rationals also meet MKO assumption a1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只需逐一看五個質量的符號；正推出非負，反向不宣稱。",
            "en": "It needs only the sign of each of the five masses; positive implies nonnegative, and no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "G1"
        ],
        "before_tex": "\\sum_{s\\in S}p(s)=\\tfrac16+\\tfrac13+\\tfrac14+\\tfrac1{12}+\\tfrac16",
        "after_tex": "\\sum_{s\\in S}p(s)=1",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用精確分數運算求總和（細節見 S04.1、S04.2）：結果恰為 1，資格 (iii) 成立。不能用四捨五入的小數代替。",
          "en": "Compute the total with exact fraction arithmetic (details in S04.1 and S04.2): the result is exactly 1, so requirement (iii) holds. Rounded decimals may not be used instead."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "左右兩式表示同一個有理數，可由 S04.2 倒推回 S04.1 的形式，雙向成立。",
            "en": "Both sides name the same rational number, and S04.2 can be read backwards to the form in S04.1, so the step holds both ways."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S04.1",
        "uses": [
          "G1"
        ],
        "before_tex": "\\tfrac16+\\tfrac13+\\tfrac14+\\tfrac1{12}+\\tfrac16",
        "after_tex": "\\tfrac{2}{12}+\\tfrac{4}{12}+\\tfrac{3}{12}+\\tfrac{1}{12}+\\tfrac{2}{12}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "k\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 a/b=(ak)/(bk)（k≠0）把每個分數改寫成分母 12：1/6 取 k=2，1/3 取 k=4，1/4 取 k=3，1/12 取 k=1。每一項的值不變。",
          "en": "Use a/b=(ak)/(bk) for k≠0 to give every fraction the denominator 12: k=2 for 1/6, k=4 for 1/3, k=3 for 1/4 and k=1 for 1/12. No term changes its value."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "反向是約去同一個非零因數 k，回到原分數；兩邊逐項相等。",
            "en": "Backward, cancel the same nonzero factor k to return to the original fractions; the two sides agree term by term."
          }
        },
        "leaf_basis": "frac_rules"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1"
        ],
        "before_tex": "\\tfrac{2}{12}+\\tfrac{4}{12}+\\tfrac{3}{12}+\\tfrac{1}{12}+\\tfrac{2}{12}",
        "after_tex": "\\tfrac{12}{12}=1",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "同分母相加：a/c+b/c=(a+b)/c，分子 2+4+3+1+2=12，所以總和是 12/12=1。",
          "en": "Add over the common denominator, a/c+b/c=(a+b)/c: the numerators give 2+4+3+1+2=12, so the total is 12/12=1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數加法與同分母合併都是等值改寫，可逐步倒推。",
            "en": "Integer addition and combining over a common denominator are rewrites of equal values and can be undone step by step."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S04"
        ],
        "before_tex": "\\text{(i), (ii), (iii) checked}",
        "after_tex": "(S,p)\\ \\text{is a finite joint law}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三項資格分別在 S02、S03、S04 成立，合取後依 S01 的定義，(S,p) 是二維有限聯合分布。從此可以由這一張表聚合任何事件的機率。",
          "en": "The three requirements hold by S02, S03 and S04; combining them, by the definition in S01, (S,p) is a two-dimensional finite joint distribution. From now on any event probability may be aggregated from this one table."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S04",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(S,p) 是有限 joint law。足夠條件 H：資格 (i)(ii)(iii) 同時成立；H⇒G 由定義。H 由 S02 到 S04 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: (S,p) is a finite joint law. Sufficient condition H: requirements (i), (ii) and (iii) all hold; H⇒G by the definition. H is proved forwards in S02 to S04; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_joint_def + and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G2",
          "joint_event_sum"
        ],
        "before_tex": "C\\subseteq\\mathbb{Z}^2",
        "after_tex": "P((X,Y)\\in C)=\\sum_{s\\in S\\cap C}p(s)",
        "relation": "definition",
        "rule": "joint_event_sum",
        "conditions": [
          "(S,p)\\ \\text{valid}"
        ],
        "explanation": {
          "zh-Hant": "有效的 joint law 決定每個事件的機率：把事件中屬於支撐的值對質量相加；不在 S 中的值對貢獻 0。這是互斥單點事件的可加性。例如 C=ℤ² 時得到 S04 的總和 1，與 P(Ω)=1 一致。",
          "en": "A valid joint law determines every event probability: add the masses of the support pairs inside the event; pairs outside S contribute 0. This is additivity over disjoint single-pair events. For example C=ℤ² gives the total 1 of S04, in line with P(Ω)=1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "joint_event_sum",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "需要 S05：若質量和不是 1 或有負質量，所得的數就不是機率。此步是定義的使用，不推出其他事實。",
            "en": "It needs S05: if the masses did not sum to 1 or some mass were negative, the numbers obtained would not be probabilities. The step applies a definition and derives nothing else."
          }
        },
        "leaf_basis": "joint_event_sum + prob_axioms"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "marginal_def",
          "G1"
        ],
        "before_tex": "p_X(x)=\\sum_{y:(x,y)\\in S}p(x,y)",
        "after_tex": "p_X(0)=\\tfrac12,\\ p_X(1)=\\tfrac13,\\ p_X(2)=\\tfrac16",
        "relation": "implies",
        "rule": "marginal_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "X 的邊際是事件 {X=x}={x}×ℤ 的機率，依 S06 把該事件中支撐值對的質量相加。先列出每個 x 的值對（S07.1），再相加（S07.2）。",
          "en": "The marginal of X is the probability of the event {X=x}={x}×ℤ; by S06 add the masses of the support pairs in that event. First list the pairs for each x (S07.1), then add (S07.2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "marginal_def"
          ],
          "explanation": {
            "zh-Hant": "要求列出的值對是 S 與 {x}×ℤ 的完整交集；漏掉一對就會得到錯的邊際。不宣稱反向：由邊際不能還原這些值對的個別質量。",
            "en": "It requires the listed pairs to be the complete intersection of S with {x}×ℤ; missing one pair gives a wrong marginal. No converse is claimed: the individual masses cannot be recovered from the marginal."
          }
        },
        "leaf_basis": "marginal_def + joint_event_sum + frac_rules"
      },
      {
        "id": "S07.1",
        "uses": [
          "G1",
          "set_operations_def"
        ],
        "before_tex": "S\\cap(\\{x\\}\\times\\mathbb{Z}),\\ x=0,1,2",
        "after_tex": "\\{(0,0),(0,1)\\},\\ \\{(1,0),(1,1)\\},\\ \\{(2,1)\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一檢查 S 的五個值對的第一座標，把它們放進對應的集合。每個值對恰屬一個集合，三個集合兩兩不交且聯集是 S。",
          "en": "Check the first coordinate of each of the five pairs of S and put it into the matching set. Each pair lies in exactly one set; the three sets are pairwise disjoint and their union is S."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "五個值對都已逐一判定，兩邊是同一個集合（逐元素相同），雙向成立。",
            "en": "All five pairs were decided one by one, so both sides are the same sets element by element, and the step holds both ways."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "G1"
        ],
        "before_tex": "p(0,0)+p(0,1),\\ p(1,0)+p(1,1),\\ p(2,1)",
        "after_tex": "\\tfrac12,\\ \\tfrac13,\\ \\tfrac16",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1/6+1/3=1/6+2/6=3/6=1/2；1/4+1/12=3/12+1/12=4/12=1/3；p(2,1)=1/6。",
          "en": "1/6+1/3=1/6+2/6=3/6=1/2; 1/4+1/12=3/12+1/12=4/12=1/3; p(2,1)=1/6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確分數運算是等值改寫，左右兩邊是同樣的三個有理數。",
            "en": "Exact fraction arithmetic rewrites equal values, so the two sides are the same three rationals."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07.1",
          "S07.2",
          "finite_sum_regroup"
        ],
        "before_tex": "\\sum_{x}p_X(x)=\\sum_{x}\\sum_{y:(x,y)\\in S}p(x,y)",
        "after_tex": "\\sum_{x}p_X(x)=\\sum_{s\\in S}p(s)=1",
        "relation": "equivalent",
        "rule": "finite_sum_regroup",
        "conditions": [
          "\\text{the fibres }S\\cap(\\{x\\}\\times\\mathbb{Z})\\text{ partition }S"
        ],
        "explanation": {
          "zh-Hant": "S07.1 的三個集合分割 S，依有限和重組，三個邊際值相加等於整個支撐的質量和，即 1。這對任何有效的有限 joint law 都成立，不只這張表；本例的數值核對是 1/2+1/3+1/6=1。",
          "en": "The three sets of S07.1 partition S, so by regrouping the finite sum, adding the three marginal values gives the total mass of the support, which is 1. This holds for every valid finite joint law, not only this table; the numeric check here is 1/2+1/3+1/6=1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_regroup",
            "S07.1"
          ],
          "explanation": {
            "zh-Hant": "重組只改變有限和的分組，不改變其值；兩邊是同一個數，雙向成立。條件是分割：每個值對恰被算一次。",
            "en": "Regrouping changes only how the finite sum is grouped, not its value; both sides are the same number, so the step is two-way. The condition is the partition: each pair is counted exactly once."
          }
        },
        "leaf_basis": "finite_sum_regroup + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S06",
          "marginal_def",
          "G1"
        ],
        "before_tex": "p_Y(y)=\\sum_{x:(x,y)\\in S}p(x,y)",
        "after_tex": "p_Y(0)=\\tfrac{5}{12},\\ p_Y(1)=\\tfrac{7}{12}",
        "relation": "implies",
        "rule": "marginal_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依第二座標分組：Y=0 的值對是 (0,0)、(1,0)，質量和 2/12+3/12=5/12；Y=1 的值對是 (0,1)、(1,1)、(2,1)，質量和 4/12+1/12+2/12=7/12。兩者相加為 1，與 S08 同理的重組一致。",
          "en": "Group by second coordinate: the pairs with Y=0 are (0,0) and (1,0), with total 2/12+3/12=5/12; the pairs with Y=1 are (0,1), (1,1) and (2,1), with total 4/12+1/12+2/12=7/12. They add to 1, consistent with regrouping as in S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "marginal_def"
          ],
          "explanation": {
            "zh-Hant": "要求每個 y 的值對列表完整：S 中第二座標為 0 的恰有兩對，為 1 的恰有三對，共五對，與 S 的大小相符。",
            "en": "It requires the list for each y to be complete: S has exactly two pairs with second coordinate 0 and exactly three with 1, five in all, matching the size of S."
          }
        },
        "leaf_basis": "marginal_def + joint_event_sum + frac_rules"
      },
      {
        "id": "S10",
        "uses": [
          "S06",
          "G3"
        ],
        "before_tex": "P(X\\in A,\\ Y\\in B)=P((X,Y)\\in A\\times B)",
        "after_tex": "P(X\\in A,\\ Y\\in B)=\\tfrac14",
        "relation": "implies",
        "rule": "joint_event_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "矩形事件要求兩個座標同時落入指定集合，所以它是 A×B 這個值對集合的事件。依 S06，先求 S∩(A×B)（S10.1），再把質量相加（S10.2）。",
          "en": "A rectangle event requires both coordinates to fall in their sets at the same time, so it is the event that (X,Y) belongs to A×B. By S06, first find S∩(A×B) (S10.1), then add the masses (S10.2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "G3",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "需要「同時」：(x,y)∈A×B 當且僅當 x∈A 且 y∈B。若改成「或」就是另一個事件，結果也不同。",
            "en": "It needs 'at the same time': (x,y)∈A×B exactly when x∈A and y∈B. Using 'or' instead gives a different event with a different value."
          }
        },
        "leaf_basis": "joint_event_sum + cartesian_product_def"
      },
      {
        "id": "S10.1",
        "uses": [
          "G1",
          "G3",
          "cartesian_product_def"
        ],
        "before_tex": "S\\cap(A\\times B)",
        "after_tex": "\\{(1,1),\\ (2,1)\\}",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一檢查：(0,0) 與 (0,1) 的 x=0∉A；(1,0) 的 y=0∉B；(1,1) 的 1∈A 且 1∈B；(2,1) 的 2∈A 且 1∈B。所以交集恰為 {(1,1),(2,1)}。",
          "en": "Check each pair: (0,0) and (0,1) have x=0∉A; (1,0) has y=0∉B; (1,1) has 1∈A and 1∈B; (2,1) has 2∈A and 1∈B. So the intersection is exactly {(1,1),(2,1)}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "S 的五個值對都判定過，兩邊是同一個集合，雙向成立。",
            "en": "All five pairs of S were decided, so both sides are the same set and the step holds both ways."
          }
        },
        "leaf_basis": "cartesian_product_def + set_operations_def + set_extensionality"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "G1"
        ],
        "before_tex": "p(1,1)+p(2,1)",
        "after_tex": "\\tfrac{1}{12}+\\tfrac{2}{12}=\\tfrac14",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1/12+1/6=1/12+2/12=3/12=1/4。",
          "en": "1/12+1/6=1/12+2/12=3/12=1/4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確分數運算的等值改寫，雙向成立。",
            "en": "An exact rewrite of equal fractions, two-way."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "S09",
          "S10"
        ],
        "before_tex": "P(X\\in A)\\,P(Y\\in B)=\\left(\\tfrac13+\\tfrac16\\right)\\tfrac{7}{12}",
        "after_tex": "\\tfrac{7}{24}\\neq\\tfrac{6}{24}=P(X\\in A,\\ Y\\in B)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "P(X∈A)=p_X(1)+p_X(2)=1/3+1/6=1/2（{X=1} 與 {X=2} 互斥），P(Y∈B)=p_Y(1)=7/12，乘積為 7/24；但 S10 算得的矩形機率是 1/4=6/24。所以對這個 joint law，矩形機率不是兩個邊際機率的乘積，必須由聯合表本身聚合。",
          "en": "P(X∈A)=p_X(1)+p_X(2)=1/3+1/6=1/2 (the events {X=1} and {X=2} are disjoint) and P(Y∈B)=p_Y(1)=7/12, with product 7/24; but S10 gives the rectangle probability 1/4=6/24. So for this joint law the rectangle probability is not the product of the marginal probabilities and must be aggregated from the joint table itself."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S09",
            "S10",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只比較兩個精確數；此步只說這張表上兩者不等，不宣稱任何 joint law 都如此（乘積律 Q 上兩者相等）。",
            "en": "It only compares two exact numbers; it says they differ for this table, not for every joint law (for the product law Q they agree)."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules + prob_axioms"
      },
      {
        "id": "S12",
        "uses": [
          "S06",
          "S07",
          "S09",
          "G1"
        ],
        "before_tex": "P(X=2,\\ Y=0)",
        "after_tex": "0\\neq\\tfrac{5}{72}=p_X(2)\\,p_Y(0)",
        "relation": "implies",
        "rule": "joint_event_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S 中第一座標為 2 的只有 (2,1)，所以 (2,0)∉S，事件 {(2,0)} 不含支撐值對，依 S06 其機率是空和 0。另一方面 p_X(2)p_Y(0)=(1/6)(5/12)=5/72>0。單點質量必須直接讀 joint law，不能由邊際相乘得到。",
          "en": "The only pair of S with first coordinate 2 is (2,1), so (2,0)∉S; the event {(2,0)} contains no support pair and by S06 its probability is the empty sum 0. On the other hand p_X(2)p_Y(0)=(1/6)(5/12)=5/72>0. A single-pair mass must be read from the joint law, not obtained by multiplying marginals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S09",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要支撐列表完整（假設 a1）：正因為 (2,0) 沒被列出，它的質量才是 0。",
            "en": "It needs the support list to be complete (assumption a1): the mass of (2,0) is 0 precisely because (2,0) is not listed."
          }
        },
        "leaf_basis": "joint_event_sum + finite_sum_rules + frac_rules"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "p_X\\ \\text{on}\\ \\{0,1,2\\},\\ p_Y\\ \\text{on}\\ \\{0,1\\}",
        "after_tex": "Q(x,y)=p_X(x)\\,p_Y(y)\\ \\text{is a finite joint law with marginals}\\ p_X,\\ p_Y",
        "relation": "theorem_application",
        "rule": "L_product_law",
        "conditions": [
          "p_X\\ge 0,\\ \\sum_x p_X(x)=1",
          "p_Y\\ge 0,\\ \\sum_y p_Y(y)=1"
        ],
        "explanation": {
          "zh-Hant": "把引理 L_product_law 用在 S_X={0,1,2}、S_Y={0,1} 上；引理的前提由 S07、S08、S09 提供。Q 的六個質量是 Q(0,0)=5/24、Q(0,1)=7/24、Q(1,0)=5/36、Q(1,1)=7/36、Q(2,0)=5/72、Q(2,1)=7/72。引理本身在 S13.1 到 S13.3 一般地證明。",
          "en": "Apply Lemma L_product_law with S_X={0,1,2} and S_Y={0,1}; its hypotheses come from S07, S08 and S09. The six masses of Q are Q(0,0)=5/24, Q(0,1)=7/24, Q(1,0)=5/36, Q(1,1)=7/36, Q(2,0)=5/72 and Q(2,1)=7/72. The lemma itself is proved in general in S13.1 to S13.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "S09",
            "L_product_law"
          ],
          "explanation": {
            "zh-Hant": "引理需要兩個邊際各自非負且總和為 1；S07 到 S09 已逐項確認。此步只建構 Q，不宣稱 Q 等於原來的 joint law。",
            "en": "The lemma needs each marginal to be nonnegative with total 1, which S07 to S09 confirm. The step only constructs Q and does not claim that Q equals the original joint law."
          }
        },
        "leaf_basis": "L_product_law + numeric_evaluation"
      },
      {
        "id": "S13.1",
        "uses": [
          "cartesian_product_def",
          "order_basics"
        ],
        "before_tex": "Q(x,y)=p_X(x)\\,p_Y(y),\\ (x,y)\\in S_X\\times S_Y",
        "after_tex": "\\text{distinct pairs},\\ Q(x,y)\\ge 0\\ \\text{exact rational}",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [
          "p_X(x)\\ge 0",
          "p_Y(y)\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "S_X×S_Y 的元素依第一、第二座標的選擇逐一列出，每個選擇給出一個不同的有序對，所以值對互異。兩個非負精確有理數的乘積仍是非負精確有理數。這給出 joint law 的資格 (i) 與 (ii)。",
          "en": "The elements of S_X×S_Y are listed by the choice of first and second coordinate, and each choice gives a different ordered pair, so the pairs are distinct. The product of two nonnegative exact rationals is again a nonnegative exact rational. This gives requirements (i) and (ii) of a joint law."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "cartesian_product_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "非負乘非負得非負；不宣稱反向。此子步不使用 L_product_law。",
            "en": "Nonnegative times nonnegative is nonnegative; no converse is claimed. This sub-step does not use L_product_law."
          }
        },
        "leaf_basis": "cartesian_product_def + order_basics + frac_rules"
      },
      {
        "id": "S13.2",
        "uses": [
          "S13.1",
          "finite_sum_rules",
          "distributive",
          "finite_sum_regroup"
        ],
        "before_tex": "\\sum_{x\\in S_X}\\sum_{y\\in S_Y}p_X(x)\\,p_Y(y)",
        "after_tex": "\\sum_{x\\in S_X}p_X(x)\\cdot\\sum_{y\\in S_Y}p_Y(y)=1\\cdot 1=1",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [
          "\\sum_x p_X(x)=1",
          "\\sum_y p_Y(y)=1"
        ],
        "explanation": {
          "zh-Hant": "依第一座標把 S_X×S_Y 上的和重組成兩層和；內層 Σ_y p_X(x)p_Y(y) 中 p_X(x) 是常數，可提出得 p_X(x)Σ_y p_Y(y)=p_X(x)·1；外層再相加得 Σ_x p_X(x)=1。資格 (iii) 成立。",
          "en": "Regroup the sum over S_X×S_Y by first coordinate into a double sum; in the inner sum Σ_y p_X(x)p_Y(y) the factor p_X(x) is constant and comes out, giving p_X(x)Σ_y p_Y(y)=p_X(x)·1; the outer sum then gives Σ_x p_X(x)=1. Requirement (iii) holds."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "finite_sum_rules",
            "finite_sum_regroup"
          ],
          "explanation": {
            "zh-Hant": "提出常數因子與重組有限和都是等值改寫，可倒推；所用條件是兩個邊際各自總和為 1。此子步不使用 L_product_law。",
            "en": "Pulling out a constant factor and regrouping a finite sum are rewrites of equal values and can be reversed; the conditions used are that each marginal totals 1. This sub-step does not use L_product_law."
          }
        },
        "leaf_basis": "distributive + finite_sum_rules + finite_sum_regroup + mul_one"
      },
      {
        "id": "S13.3",
        "uses": [
          "S13.1",
          "finite_sum_rules",
          "marginal_def"
        ],
        "before_tex": "\\sum_{y\\in S_Y}Q(x,y),\\ \\sum_{x\\in S_X}Q(x,y)",
        "after_tex": "p_X(x)\\sum_{y}p_Y(y)=p_X(x),\\ p_Y(y)\\sum_{x}p_X(x)=p_Y(y)",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [
          "\\sum_x p_X(x)=1",
          "\\sum_y p_Y(y)=1"
        ],
        "explanation": {
          "zh-Hant": "Q 的 X 邊際是固定 x 後對 y 求和：常數 p_X(x) 提出，剩下 Σ_y p_Y(y)=1，所以等於 p_X(x)；Y 邊際同理等於 p_Y(y)。S13.1 到 S13.3 合起來證得引理 L_product_law。",
          "en": "The X-marginal of Q fixes x and sums over y: the constant p_X(x) comes out and the remaining Σ_y p_Y(y) is 1, so it equals p_X(x); the Y-marginal equals p_Y(y) in the same way. S13.1 to S13.3 together prove Lemma L_product_law."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "Σ c·a_i=c·Σ a_i 與乘以 1 都是等值改寫，雙向成立。此子步不使用 L_product_law。",
            "en": "Σ c·a_i=c·Σ a_i and multiplying by 1 are rewrites of equal values, so the step is two-way. This sub-step does not use L_product_law."
          }
        },
        "leaf_basis": "finite_sum_rules + mul_one + marginal_def"
      },
      {
        "id": "S14",
        "uses": [
          "S07",
          "S09",
          "S12",
          "S13"
        ],
        "before_tex": "J(2,0)=0,\\ Q(2,0)=\\tfrac{5}{72}",
        "after_tex": "J\\neq Q\\ \\text{while}\\ J,\\ Q\\ \\text{have the same marginals}",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把兩個 joint law 看成 ℤ² 上的質量函數（支撐外為 0）。原表 J 在 (2,0) 的質量是 0（S12），Q 在 (2,0) 的質量是 5/72，兩個函數在一點不同，所以 J≠Q。但兩者的邊際都是 p_X 與 p_Y（J 由 S07、S09，Q 由 S13）。一個反例就否定「兩個邊際決定聯合分布」這個全稱命題。",
          "en": "Regard both joint laws as mass functions on ℤ² (zero off the support). The original table J has mass 0 at (2,0) (S12), while Q has mass 5/72 there; the two functions differ at a point, so J≠Q. Yet both have the marginals p_X and p_Y (J by S07 and S09, Q by S13). One counterexample refutes the universal claim that two marginals determine the joint distribution."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S12",
            "S13",
            "function_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：邊際一般不能反推 joint law。足夠條件 H：存在兩個不同 joint law 有相同邊際；H⇒G 因為否定全稱只需一個反例。H 由 S12、S13 正向構造，沒有把 G 當作前提。",
            "en": "Goal G: marginals do not in general determine the joint law. Sufficient condition H: two different joint laws with the same marginals exist; H⇒G because one counterexample refutes a universal claim. H is constructed forwards in S12 and S13; G is never used as a premise."
          }
        },
        "leaf_basis": "function_def + exists_intro + quantifier_negation"
      }
    ],
    "result": {
      "zh-Hant": "(S,p) 是有效的二維有限聯合分布（值對互異、質量為正的精確有理數、總和恰為 1）。由同一 joint law 得 p_X(0)=1/2、p_X(1)=1/3、p_X(2)=1/6，p_Y(0)=5/12、p_Y(1)=7/12，矩形事件 P(X∈{1,2}, Y=1)=1/4，單點質量 P(X=2,Y=0)=0。乘積律 Q=p_X p_Y 有相同邊際但 Q(2,0)=5/72≠0，所以兩個邊際不能反推聯合分布。",
      "en": "(S,p) is a valid two-dimensional finite joint distribution (distinct pairs, positive exact rational masses, total exactly 1). From this one joint law, p_X(0)=1/2, p_X(1)=1/3, p_X(2)=1/6, p_Y(0)=5/12, p_Y(1)=7/12, the rectangle event P(X∈{1,2}, Y=1)=1/4 and the single-pair mass P(X=2,Y=0)=0. The product law Q=p_X p_Y has the same marginals but Q(2,0)=5/72≠0, so two marginals do not determine the joint distribution."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "兩個邊際相乘就得到聯合分布：P(X=x,Y=y)=p_X(x)p_Y(y) 對本表成立",
          "en": "Multiplying the marginals gives the joint law: P(X=x,Y=y)=p_X(x)p_Y(y) holds for this table"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 (2,0)：J 的質量是 0，而 p_X(2)p_Y(0)=5/72。",
          "en": "At (2,0) the mass of J is 0 while p_X(2)p_Y(0)=5/72."
        }
      },
      {
        "claim": {
          "zh-Hant": "質量 0.166、0.333、0.25、0.083、0.166 的表是有效的有限聯合分布",
          "en": "The table with masses 0.166, 0.333, 0.25, 0.083, 0.166 is a valid finite joint distribution"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "總和是 0.998≠1；MKO 要求總和恰為 1。",
          "en": "The total is 0.998≠1; the MKO requires the total to be exactly 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(X∈{1,2}, Y=1)=P(X∈{1,2})·P(Y=1)=7/24",
          "en": "P(X∈{1,2}, Y=1)=P(X∈{1,2})·P(Y=1)=7/24"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "由 joint law 聚合得 1/12+1/6=1/4=6/24≠7/24。",
          "en": "Aggregating the joint law gives 1/12+1/6=1/4=6/24≠7/24."
        }
      },
      {
        "claim": {
          "zh-Hant": "把 (0,1) 列兩次、各給質量 1/6，仍是符合 MKO 的聯合表",
          "en": "Listing (0,1) twice with mass 1/6 each is still a joint table in the sense of the MKO"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "MKO 要求支撐列出互異值對；重複列出的值對必須合併成一筆（質量 1/3）後才是合格的表。",
          "en": "The MKO requires the support to list distinct pairs; a repeated pair must be merged into one row (mass 1/3) before the table qualifies."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "prob_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "機率公理：P(A)≥0；P(Ω)=1；兩兩互斥事件的（可數）聯集機率等於各機率之和。由此 P(∅)=0、P(Aᶜ)=1−P(A)。",
          "en": "Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A)."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_joint_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-joint-distribution",
          "sha256": "d04aa2266aeda810e9e7e400220d366a61da23cdc33192e7286eac1d0fed0340",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "二維有限聯合分布列出互異值對(x,y)及非負精確質量，總和恰為1。邊際、矩形事件及單點質量都從同一joint law聚合。",
          "en": "A two-dimensional finite joint distribution lists distinct pairs (x,y) and nonnegative exact masses summing exactly to 1. Marginals, rectangle events and single-pair masses all aggregate from that same joint law."
        }
      },
      {
        "id": "joint_event_sum",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限 joint law (S,p) 給出事件機率：對 C⊆ℤ²，P((X,Y)∈C)=Σ_{s∈S∩C} p(s)；不在 S 中的值對質量為 0。這是機率公理對互斥單點事件的可加性。",
          "en": "A finite joint law (S,p) gives event probabilities: for C⊆ℤ², P((X,Y)∈C)=Σ_{s∈S∩C} p(s); pairs not in S have mass 0. This is the additivity of the probability axioms over disjoint single-pair events."
        }
      },
      {
        "id": "marginal_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邊際：p_X(x)=P(X=x)=Σ_{y:(x,y)∈S} p(x,y)，p_Y(y)=P(Y=y)=Σ_{x:(x,y)∈S} p(x,y)，即事件 {x}×ℤ 與 ℤ×{y} 的機率。",
          "en": "Marginals: p_X(x)=P(X=x)=Σ_{y:(x,y)∈S} p(x,y) and p_Y(y)=P(Y=y)=Σ_{x:(x,y)∈S} p(x,y), the probabilities of the events {x}×ℤ and ℤ×{y}."
        }
      },
      {
        "id": "finite_sum_regroup",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和重組：若有限集合 S 分割成兩兩不交的子集 S_1、S_2 等（每個元素恰屬其中一個），則 Σ_{s∈S} f(s) 等於各子集上的和再相加。",
          "en": "Regrouping a finite sum: if a finite set S is split into pairwise disjoint pieces S_1, S_2 and so on (each element in exactly one piece), then Σ_{s∈S} f(s) equals the total of the sums over the pieces."
        }
      },
      {
        "id": "L_product_law",
        "kind": "lemma",
        "proved_in": [
          "S13.1",
          "S13.2",
          "S13.3"
        ],
        "statement": {
          "zh-Hant": "設 p_X 定義在有限整數集 S_X 上、p_Y 定義在有限整數集 S_Y 上，都是非負精確有理質量且各自總和恰為 1。則 Q(x,y)=p_X(x)p_Y(y) 在 S_X×S_Y 上是有限 joint law，且其兩個邊際恰為 p_X 與 p_Y。",
          "en": "Let p_X on a finite integer set S_X and p_Y on a finite integer set S_Y be nonnegative exact rational masses, each summing exactly to 1. Then Q(x,y)=p_X(x)p_Y(y) on S_X×S_Y is a finite joint law whose two marginals are exactly p_X and p_Y."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "1/6+1/3+1/4+1/12+1/6",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S04.2",
        "lhs": "2+4+3+1+2",
        "rhs": "12"
      },
      {
        "id": "N3",
        "step": "S07.2",
        "lhs": "1/6+1/3",
        "rhs": "1/2"
      },
      {
        "id": "N4",
        "step": "S07.2",
        "lhs": "1/4+1/12",
        "rhs": "1/3"
      },
      {
        "id": "N5",
        "step": "S08",
        "lhs": "1/2+1/3+1/6",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S09",
        "lhs": "1/6+1/4",
        "rhs": "5/12"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "1/3+1/12+1/6",
        "rhs": "7/12"
      },
      {
        "id": "N8",
        "step": "S10.2",
        "lhs": "1/12+1/6",
        "rhs": "1/4"
      },
      {
        "id": "N9",
        "step": "S11",
        "lhs": "(1/3+1/6)*(7/12)",
        "rhs": "7/24"
      },
      {
        "id": "N10",
        "step": "S11",
        "lhs": "7/24",
        "relation": "!=",
        "rhs": "1/4"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "(1/6)*(5/12)",
        "rhs": "5/72"
      },
      {
        "id": "N12",
        "step": "S13",
        "lhs": "(1/2)*(5/12)+(1/2)*(7/12)",
        "rhs": "1/2"
      },
      {
        "id": "N13",
        "step": "S13.2",
        "lhs": "(1/2+1/3+1/6)*(5/12+7/12)",
        "rhs": "1"
      },
      {
        "id": "N14",
        "lhs": "0.166+0.333+0.25+0.083+0.166",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N15",
        "lhs": "1/2+7/12-1/4",
        "rhs": "5/6"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 的範圍內：支撐為互異整數對、質量為精確有理數的有限 joint law（假設 a1），視為數學模型而非資料配適或連續密度（假設 a2）。本稿驗證一張具體的表，並一般地證明「乘積律」引理；不處理獨立性、協方差或條件期望（MKO 的後續節點），不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO: a finite joint law whose support consists of distinct integer pairs with exact rational masses (assumption a1), treated as a mathematical model, not as data fitting or a continuous density (assumption a2). The guide validates one concrete table and proves the product-law lemma in general; it does not treat independence, covariance or conditional expectation (later nodes of the MKO), and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "S",
        "支撐：列出的互異整數對",
        "support: the listed distinct integer pairs"
      ],
      [
        "p(x,y)",
        "值對 (x,y) 的精確質量；不在 S 中的值對質量為 0",
        "exact mass of the pair (x,y); pairs outside S have mass 0"
      ],
      [
        "X,\\ Y",
        "座標投影：在值對 (x,y) 上 X=x、Y=y",
        "coordinate projections: X=x and Y=y at the pair (x,y)"
      ],
      [
        "p_X,\\ p_Y",
        "X 與 Y 的邊際質量",
        "marginal masses of X and of Y"
      ],
      [
        "A\\times B",
        "矩形事件：第一座標在 A 且第二座標在 B",
        "rectangle event: first coordinate in A and second coordinate in B"
      ],
      [
        "Q",
        "兩個邊際相乘得到的另一個 joint law",
        "the other joint law formed by multiplying the two marginals"
      ]
    ],
    "md_error_controls": [
      {
        "title": "兩張邊際表不是聯合分布 / Two marginal tables are not a joint law",
        "zh": "MKO 指出的常見誤解是把兩個邊際表直接當聯合分布。本稿的 J 與 Q 有完全相同的邊際（1/2、1/3、1/6 與 5/12、7/12），但在 (2,0) 的質量一個是 0、一個是 5/72（S14）。邊際只是聯合表的聚合結果，聚合會丟失資訊，所以不能反推。",
        "en": "The common mistake named by the MKO is to take two marginal tables as the joint distribution. Here J and Q have exactly the same marginals (1/2, 1/3, 1/6 and 5/12, 7/12), yet their masses at (2,0) are 0 and 5/72 (S14). Marginals are aggregates of the joint table, and aggregation loses information, so they cannot be inverted."
      },
      {
        "title": "總和必須恰為 1 / The total must be exactly 1",
        "zh": "若把質量換成三位小數 0.166、0.333、0.25、0.083、0.166，這些有限小數本身仍是精確有理數，但總和是 0.998，不是 1，所以這張表不合 MKO 的資格（iii）；若它們只是原質量的截斷近似，也不能當作原來的精確質量。S04 用原給定分數計算得到恰好 12/12=1。",
        "en": "Replacing the masses by 0.166, 0.333, 0.25, 0.083 and 0.166 still gives exact rational numbers when the finite decimals are read literally. However, their total is 0.998 rather than 1, so the table fails requirement (iii). If intended merely as truncated approximations, they must not be treated as the original exact masses. S04 uses the original fractions and obtains exactly 12/12=1."
      },
      {
        "title": "矩形事件要兩座標同時成立 / A rectangle event needs both coordinates at once",
        "zh": "P(X∈A, Y∈B) 是兩座標同時成立的事件，本例為 1/4。把它算成 P(X∈A)+P(Y∈B)−P(X∈A, Y∈B)=1/2+7/12−1/4=5/6 是「或」事件；算成 P(X∈A)P(Y∈B)=7/24 則是把邊際當成 joint law。兩者都不是矩形機率（S10、S11）。",
        "en": "P(X∈A, Y∈B) is the event that both coordinate conditions hold at once, here 1/4. Computing P(X∈A)+P(Y∈B)−P(X∈A, Y∈B)=1/2+7/12−1/4=5/6 gives the 'or' event; computing P(X∈A)P(Y∈B)=7/24 treats the marginals as the joint law. Neither is the rectangle probability (S10, S11)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "有限和重組（finite_sum_regroup）與 joint_event_sum 是本稿的前提，不在本稿中由更基本的公理推出。",
        "en": "Regrouping of finite sums (finite_sum_regroup) and joint_event_sum are premises of this guide; they are not derived here from more basic axioms."
      },
      {
        "zh-Hant": "N14、N15 是錯誤控制中反例數值的核對，只對這些數字有效。",
        "en": "N14 and N15 check the numbers in the error controls and are valid for those numbers only."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "5b995a039b001aab1d5b78c71b0b7780f68598a4f9bf05748dd25e86da4430d9",
    "original_source_sha256": "5b995a039b001aab1d5b78c71b0b7780f68598a4f9bf05748dd25e86da4430d9",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}