{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-discrete-probability-distribution-01",
  "mko_id": "mko-finite-discrete-probability-distribution",
  "title": {
    "zh-Hant": "有限離散機率分布：核 PMF 資格、事件機率、期望與推前分布",
    "en": "Finite discrete probability distribution: validating a PMF, event probability, expectation and pushforward"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-discrete-probability-distribution-01",
    "source": {
      "mko_id": "mko-finite-discrete-probability-distribution",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "a77aa248d456c6fff168a9df6f8b4ad85be410912a036130257ff1d3e0aa1b2a",
      "english_sha256": "2d1802954848cf7c228069df4e5277244c06e16e2e2bd85ef8da7a4334958a87"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "frac_rules",
      "finite_sum_rules",
      "prob_axioms",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "隨機變數 X 取值 −1、0、1、2，質量 p(−1)=1/6、p(0)=1/3、p(1)=1/4、p(2)=1/4。（一）先核 p 是否為合格 PMF；（二）求 P(X≥1)；（三）求 E[X]；（四）求 Y=X² 的推前分布、證明它仍是 PMF，求 E[Y] 並以 Σx²p(x) 交叉核對；（五）判定兩個候選 q、r 不是 PMF：q 每項介於 0 與 1 但總和不為 1，r 總和為 1 但有負項。",
      "en": "A random variable X takes the values −1, 0, 1, 2 with masses p(−1)=1/6, p(0)=1/3, p(1)=1/4, p(2)=1/4. (1) First check that p is a valid PMF; (2) find P(X≥1); (3) find E[X]; (4) find the pushforward distribution of Y=X², prove it is again a PMF, find E[Y] and cross-check it against Σx²p(x); (5) decide that two candidates q and r are not PMFs: every mass of q lies between 0 and 1 but the total is not 1, while r sums to 1 but has a negative mass."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X 的值域 R_X={−1,0,1,2}，四個值互異；p(−1)=1/6，p(0)=1/3，p(1)=1/4，p(2)=1/4，都是精確有理數。",
          "en": "The range of X is R_X={−1,0,1,2}, four distinct values; p(−1)=1/6, p(0)=1/3, p(1)=1/4, p(2)=1/4, all exact rationals."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "g(x)=x²，Y=g(X)。",
          "en": "g(x)=x² and Y=g(X)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "同一值域上的兩個候選：q(−1)=1/2、q(0)=1/3、q(1)=1/4、q(2)=0；r(−1)=1/2、r(0)=2/3、r(1)=−1/6、r(2)=0。",
          "en": "Two candidates on the same range: q(−1)=1/2, q(0)=1/3, q(1)=1/4, q(2)=0; r(−1)=1/2, r(0)=2/3, r(1)=−1/6, r(2)=0."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "這些分布都是給定的數學分布，不是由觀測資料估計或擬合而來（MKO 假設 a2）。",
          "en": "These distributions are given mathematical distributions, not estimated or fitted from observed data (MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_pmf_def",
          "G1"
        ],
        "before_tex": "p:R_X\\to\\mathbb{Q}",
        "after_tex": "p\\ \\text{is a PMF}\\iff \\forall x\\in R_X:\\ p(x)\\ge 0\\ \\land\\ \\sum_{x\\in R_X}p(x)=1",
        "relation": "definition",
        "rule": "mko_pmf_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，PMF 的資格有兩項：每個質量非負，且全部質量和恰為 1。MKO 第一段要求先核資格；不合格時不能繼續把它當機率分布求事件機率或期望。",
          "en": "By the MKO a PMF must meet two conditions: every mass is nonnegative and all masses sum to exactly 1. The MKO's first paragraph requires checking this first; if it fails, the function may not be used as a distribution for event probabilities or expectation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_pmf_def",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出資格條件，不推出新事實；兩項條件對應機率公理的非負與總機率為 1。",
            "en": "This step only states the PMF conditions and derives no new fact; the two conditions match the axioms of nonnegativity and total probability 1."
          }
        },
        "leaf_basis": "mko_pmf_def + prob_axioms"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "R_X=\\{-1,0,1,2\\}",
        "after_tex": "|R_X|=4,\\ \\text{values distinct}",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個值 −1、0、1、2 兩兩相異，值域有限，每個值恰有一個質量。這是 MKO 假設 a1 的要求：若同一值出現兩次，總和會重複計算。",
          "en": "The four values −1, 0, 1, 2 are pairwise distinct, the range is finite, and each value has exactly one mass. This is what MKO assumption a1 requires: if a value appeared twice, its mass could be counted twice in the total."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "只核值域的形狀；不涉及質量的大小。",
            "en": "Only the shape of the range is checked; nothing is said about the sizes of the masses."
          }
        },
        "leaf_basis": "finite_counting_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G1"
        ],
        "before_tex": "p(-1)=\\tfrac16,\\ p(0)=\\tfrac13,\\ p(1)=\\tfrac14,\\ p(2)=\\tfrac14",
        "after_tex": "\\forall x\\in R_X:\\ p(x)\\ge 0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個質量都是正分子除以正分母，所以都大於 0，當然非負。四個值逐一核完。",
          "en": "Each of the four masses is a positive numerator over a positive denominator, hence greater than 0 and in particular nonnegative. All four values are checked."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "非負只是資格的一半；它不推出總和為 1。",
            "en": "Nonnegativity is only one of the PMF conditions; it does not give total 1."
          }
        },
        "leaf_basis": "order_basics + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "\\sum_{x\\in R_X}p(x)=\\tfrac16+\\tfrac13+\\tfrac14+\\tfrac14",
        "after_tex": "1",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "因值互異（S02），總和恰是四個質量各加一次。通分到 12 再相加，見子步 S04.1、S04.2。",
          "en": "Since the values are distinct (S02), the total adds each of the four masses exactly once. Bring them to the denominator 12 and add, as in sub-steps S04.1 and S04.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數求值，兩邊是同一個數（N1），雙向成立。",
            "en": "Exact rational evaluation; both sides are the same number (N1), so the step is two-way."
          }
        },
        "leaf_basis": "frac_rules + finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S04.1",
        "uses": [
          "frac_rules"
        ],
        "before_tex": "\\tfrac16+\\tfrac13+\\tfrac14+\\tfrac14",
        "after_tex": "\\tfrac{2}{12}+\\tfrac{4}{12}+\\tfrac{3}{12}+\\tfrac{3}{12}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "2,4,3\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "分子分母同乘非零整數不改變分數：1/6=2/12（乘 2），1/3=4/12（乘 4），1/4=3/12（乘 3）。",
          "en": "Multiplying numerator and denominator by the same nonzero integer does not change a fraction: 1/6=2/12 (by 2), 1/3=4/12 (by 4), 1/4=3/12 (by 3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "約分即反向，前提是乘數不為零。",
            "en": "Reducing the fractions is the backward move, given the multipliers are nonzero."
          }
        },
        "leaf_basis": "frac_rules"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "frac_rules"
        ],
        "before_tex": "\\tfrac{2}{12}+\\tfrac{4}{12}+\\tfrac{3}{12}+\\tfrac{3}{12}",
        "after_tex": "\\tfrac{12}{12}=1",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "同分母相加只加分子：2+4+3+3=12，得 12/12=1。總和恰為 1，不是約等於 1。",
          "en": "Fractions with a common denominator add by their numerators: 2+4+3+3=12, giving 12/12=1. The total is exactly 1, not approximately 1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數加法與分數規則都是等式，可獨立重算（N2）。",
            "en": "Integer addition and the fraction rule are equations, independently recomputable (N2)."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S03",
          "S04"
        ],
        "before_tex": "p\\ge 0,\\ \\sum p=1",
        "after_tex": "p\\ \\text{is a PMF on } R_X",
        "relation": "implies",
        "rule": "mko_pmf_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S03 給非負，S04 給總和恰為 1，依 S01 的資格條件，p 是合格的 PMF，之後才可以求事件機率與期望。",
          "en": "S03 gives nonnegativity and S04 gives total exactly 1, so by the PMF conditions in S01, p is a PMF, and only now may event probabilities and expectations be computed."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "mko_pmf_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：p 是 PMF。足夠條件 H：非負且總和為 1；H⇒G 由定義。H 由 S03、S04 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: p is a PMF. Sufficient condition H: nonnegative with total 1; H⇒G by the definition. H is proved forwards in S03 and S04; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_pmf_def + and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "G1"
        ],
        "before_tex": "A=\\{x\\in R_X\\mid x\\ge 1\\}",
        "after_tex": "A=\\{1,2\\}",
        "relation": "equivalent",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一判定值域中的值是否滿足 x≥1：−1 否、0 否、1 是（等號成立）、2 是。所以事件 X≥1 對應 A={1,2}。",
          "en": "Test each value of the range for x≥1: −1 no, 0 no, 1 yes (equality holds), 2 yes. So the event X≥1 corresponds to A={1,2}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "集合建構記號的成員條件是雙向的；四個值都逐一判定過。",
            "en": "Membership in set-builder notation is two-way, and all four values were tested."
          }
        },
        "leaf_basis": "set_builder + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "event_sum"
        ],
        "before_tex": "P(X\\ge 1)=\\sum_{x\\in A}p(x)",
        "after_tex": "\\tfrac14+\\tfrac14=\\tfrac12",
        "relation": "equivalent",
        "rule": "event_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "事件機率是事件所含值的質量和（MKO 第二段）：p(1)+p(2)=1/4+1/4=1/2；事件外的 −1 與 0 不加入。",
          "en": "An event's probability is the sum of the masses of the values it contains (MKO paragraph 2): p(1)+p(2)=1/4+1/4=1/2; the values −1 and 0 outside the event are not added."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "event_sum",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，兩邊同一個數（N3）。前提是 S05 已確認 p 是 PMF。",
            "en": "Exact evaluation; both sides are the same number (N3). The prerequisite is S05, which confirmed that p is a PMF."
          }
        },
        "leaf_basis": "event_sum + frac_rules + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S05",
          "S06",
          "S07",
          "event_sum"
        ],
        "before_tex": "P(X\\in R_X\\setminus A)=\\tfrac16+\\tfrac13",
        "after_tex": "\\tfrac12,\\quad P(A)+P(R_X\\setminus A)=1",
        "relation": "implies",
        "rule": "event_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "核對：補事件 {−1,0} 的機率是 1/6+1/3=1/2；兩者相加為 1，符合機率公理中 P(Aᶜ)=1−P(A)。",
          "en": "Check: the complementary event {−1,0} has probability 1/6+1/3=1/2; the two add up to 1, in line with P(Aᶜ)=1−P(A) from the probability axioms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "event_sum",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "這是一致性核對，不是新的推論；它確認 S07 沒有漏算或多算。",
            "en": "This is a consistency check, not a new inference; it confirms that S07 neither missed nor double-counted a mass."
          }
        },
        "leaf_basis": "event_sum + prob_axioms + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "expectation_def"
        ],
        "before_tex": "E[X]=\\sum_{x\\in R_X}x\\,p(x)",
        "after_tex": "\\tfrac{7}{12}",
        "relation": "equivalent",
        "rule": "expectation_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "期望是值與質量的加權和（MKO）。逐項相乘見 S09.1，相加見 S09.2。",
          "en": "The expectation is the mass-weighted sum of the values (MKO). The terms are formed in S09.1 and added in S09.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "expectation_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值（N5），兩邊同一個數。",
            "en": "Exact evaluation (N5); both sides are the same number."
          }
        },
        "leaf_basis": "expectation_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S09.1",
        "uses": [
          "G1",
          "expectation_def"
        ],
        "before_tex": "\\sum_{x\\in R_X}x\\,p(x)",
        "after_tex": "(-1)\\cdot\\tfrac16+0\\cdot\\tfrac13+1\\cdot\\tfrac14+2\\cdot\\tfrac14",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依值域的四個值展開有限和，每個值乘上它自己的質量。值 0 的項為 0，但仍屬於和的一項。",
          "en": "Expand the finite sum over the four values of the range, each value multiplied by its own mass. The term for the value 0 is 0 but still belongs to the sum."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "有限和的展開與收回互為反向。",
            "en": "Expanding a finite sum and folding it back are inverse moves."
          }
        },
        "leaf_basis": "finite_sum_rules"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "frac_rules"
        ],
        "before_tex": "(-1)\\cdot\\tfrac16+0\\cdot\\tfrac13+1\\cdot\\tfrac14+2\\cdot\\tfrac14",
        "after_tex": "-\\tfrac{2}{12}+0+\\tfrac{3}{12}+\\tfrac{6}{12}=\\tfrac{7}{12}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "各項為 −1/6=−2/12、0、1/4=3/12、2/4=6/12；分子相加 −2+0+3+6=7，得 7/12。期望不必是值域中的值。",
          "en": "The terms are −1/6=−2/12, 0, 1/4=3/12 and 2/4=6/12; adding numerators, −2+0+3+6=7, gives 7/12. An expectation need not be a value in the range."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數運算，可獨立重算。",
            "en": "Exact rational arithmetic, independently recomputable."
          }
        },
        "leaf_basis": "frac_rules + neg_mul + mul_zero + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "G2",
          "pushforward_def"
        ],
        "before_tex": "g(-1)=1,\\ g(0)=0,\\ g(1)=1,\\ g(2)=4",
        "after_tex": "R_Y=\\{0,1,4\\};\\ g^{-1}(0)=\\{0\\},\\ g^{-1}(1)=\\{-1,1\\},\\ g^{-1}(4)=\\{2\\}",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一計算 g(x)=x²：(−1)²=1，0²=0，1²=1，2²=4。新值域是像集 {0,1,4}；每個新值的原像是映到它的原值集合。−1 與 1 都映到 1。",
          "en": "Compute g(x)=x² value by value: (−1)²=1, 0²=0, 1²=1, 2²=4. The new range is the image {0,1,4}; the preimage of each new value is the set of old values mapped to it. Both −1 and 1 map to 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "image_preimage_def",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "三個原像兩兩不交且聯集為 R_X，這是 S12.2 會用到的分割性質；g 不是單射，所以不能把 Y 的值一對一對回 X。",
            "en": "The three preimages are pairwise disjoint with union R_X, the partition property used in S12.2; g is not injective, so values of Y cannot be matched one-to-one with values of X."
          }
        },
        "leaf_basis": "image_preimage_def + function_def + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G1",
          "pushforward_def"
        ],
        "before_tex": "p_Y(y)=\\sum_{x\\in g^{-1}(y)}p(x)",
        "after_tex": "p_Y(0)=\\tfrac13,\\ p_Y(1)=\\tfrac16+\\tfrac14=\\tfrac{5}{12},\\ p_Y(4)=\\tfrac14",
        "relation": "equivalent",
        "rule": "pushforward_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依推前定義把映到同一新值的原質量相加（MKO 第三段）：p_Y(1)=p(−1)+p(1)=2/12+3/12=5/12；其餘兩個原像各只有一個元素。",
          "en": "By the pushforward definition, add the original masses mapped to the same new value (MKO paragraph 3): p_Y(1)=p(−1)+p(1)=2/12+3/12=5/12; the other two preimages have one element each."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "pushforward_def",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個 p_Y(y) 都是定義式的精確求值（N6），雙向成立。",
            "en": "Each p_Y(y) is an exact evaluation of its defining sum (N6), so the step is two-way."
          }
        },
        "leaf_basis": "pushforward_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S11"
        ],
        "before_tex": "p_Y\\ \\text{on } R_Y=\\{0,1,4\\}",
        "after_tex": "p_Y\\ \\text{is a PMF};\\ \\tfrac13+\\tfrac{5}{12}+\\tfrac14=1",
        "relation": "theorem_application",
        "rule": "L_pushforward_total",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_pushforward_total（證明在 S12.1、S12.2，對任意有限值域成立），由 p 是 PMF 推出 p_Y 是 PMF。直接核對：4/12+5/12+3/12=12/12=1，與引理一致。",
          "en": "By Lemma L_pushforward_total (proved in S12.1 and S12.2 for any finite range), p being a PMF makes p_Y a PMF. Direct check: 4/12+5/12+3/12=12/12=1, consistent with the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "L_pushforward_total"
          ],
          "explanation": {
            "zh-Hant": "引理的前提是 p 為 PMF（S05）與 g 為函數（G2）；數值核對 N7 是一致性檢查，不取代引理。",
            "en": "The lemma needs p to be a PMF (S05) and g to be a function (G2); the numeric check N7 is a consistency check and does not replace the lemma."
          }
        },
        "leaf_basis": "L_pushforward_total + numeric_evaluation"
      },
      {
        "id": "S12.1",
        "uses": [
          "pushforward_def",
          "order_basics"
        ],
        "before_tex": "p(x)\\ge 0\\ \\forall x",
        "after_tex": "p_Y(y)=\\sum_{x\\in g^{-1}(y)}p(x)\\ge 0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意 y∈R_Y，p_Y(y) 是若干非負數之和，非負數之和仍非負（依序與加法相容，逐項累加）。這證明了推前質量的非負性。",
          "en": "For any y∈R_Y, p_Y(y) is a sum of nonnegative numbers, and a sum of nonnegative numbers is nonnegative (order is compatible with addition, applied term by term). This proves nonnegativity of the pushforward masses."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "pushforward_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只用 p 的非負性；不宣稱 p_Y(y)>0，也不宣稱反向。",
            "en": "Only nonnegativity of p is used; neither p_Y(y)>0 nor any converse is claimed."
          }
        },
        "leaf_basis": "order_basics + finite_sum_rules"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "pushforward_def",
          "sum_over_partition"
        ],
        "before_tex": "\\sum_{y\\in R_Y}p_Y(y)=\\sum_{y\\in R_Y}\\sum_{x\\in g^{-1}(y)}p(x)",
        "after_tex": "\\sum_{x\\in R_X}p(x)=1",
        "relation": "equivalent",
        "rule": "sum_over_partition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "因 g 是函數，每個 x∈R_X 恰屬於一個原像 g⁻¹(g(x))：原像兩兩不交，聯集為 R_X。依分割求和，雙重和等於對 R_X 的和，而 p 是 PMF，所以等於 1。S12.1 與 S12.2 合起來證得引理 L_pushforward_total。",
          "en": "Since g is a function, each x∈R_X lies in exactly one preimage g⁻¹(g(x)): the preimages are pairwise disjoint with union R_X. By summing over a partition, the double sum equals the sum over R_X, which is 1 because p is a PMF. S12.1 and S12.2 together prove Lemma L_pushforward_total."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "sum_over_partition",
            "function_def"
          ],
          "explanation": {
            "zh-Hant": "分割求和是等式，兩個方向都成立；關鍵條件是每個 x 恰在一個區塊中，這由 g 是函數保證。",
            "en": "Summing over a partition is an equation and holds both ways; the key condition, each x in exactly one block, holds because g is a function."
          }
        },
        "leaf_basis": "sum_over_partition + function_def + prob_axioms"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12",
          "expectation_def"
        ],
        "before_tex": "E[Y]=\\sum_{y\\in R_Y}y\\,p_Y(y)",
        "after_tex": "0\\cdot\\tfrac13+1\\cdot\\tfrac{5}{12}+4\\cdot\\tfrac14=\\tfrac{17}{12}",
        "relation": "equivalent",
        "rule": "expectation_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p_Y 已是 PMF（S12），可求期望：0+5/12+1=5/12+12/12=17/12。",
          "en": "p_Y is a PMF (S12), so its expectation may be computed: 0+5/12+1=5/12+12/12=17/12."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "expectation_def",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值（N8），兩邊同一個數。",
            "en": "Exact evaluation (N8); both sides are the same number."
          }
        },
        "leaf_basis": "expectation_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S12",
          "S13"
        ],
        "before_tex": "\\sum_{y\\in R_Y}y\\,p_Y(y)",
        "after_tex": "\\sum_{x\\in R_X}x^2p(x)=\\tfrac16+0+\\tfrac14+1=\\tfrac{17}{12}",
        "relation": "equivalent",
        "rule": "L_regroup",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交叉核對：依引理 L_regroup（證明在 S14.1），Y 的期望也可以直接在 X 的值域上算 Σx²p(x)=1·(1/6)+0·(1/3)+1·(1/4)+4·(1/4)=2/12+0+3/12+12/12=17/12，與 S13 相同。",
          "en": "Cross-check: by Lemma L_regroup (proved in S14.1), the expectation of Y can also be computed directly on the range of X: Σx²p(x)=1·(1/6)+0·(1/3)+1·(1/4)+4·(1/4)=2/12+0+3/12+12/12=17/12, the same as S13."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_regroup",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "引理是等式，雙向成立；兩條獨立算法得到同一個 17/12（N8、N9）。",
            "en": "The lemma is an equation and holds both ways; the two independent computations give the same 17/12 (N8, N9)."
          }
        },
        "leaf_basis": "L_regroup + frac_rules + numeric_evaluation"
      },
      {
        "id": "S14.1",
        "uses": [
          "pushforward_def",
          "sum_over_partition",
          "distributive"
        ],
        "before_tex": "\\sum_{y\\in R_Y}y\\sum_{x\\in g^{-1}(y)}p(x)",
        "after_tex": "\\sum_{y\\in R_Y}\\sum_{x\\in g^{-1}(y)}g(x)p(x)=\\sum_{x\\in R_X}g(x)p(x)",
        "relation": "equivalent",
        "rule": "sum_over_partition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 p_Y(y) 的定義代入後，用分配律把 y 乘進內層和；內層的每個 x 都滿足 g(x)=y，所以 y 可換成 g(x)；最後依分割求和把雙重和合併成對 R_X 的和。S14.1 證得引理 L_regroup。",
          "en": "After substituting the definition of p_Y(y), distribute y into the inner sum; every x in the inner sum satisfies g(x)=y, so y may be replaced by g(x); finally summing over the partition merges the double sum into a sum over R_X. S14.1 proves Lemma L_regroup."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_replace",
            "sum_over_partition"
          ],
          "explanation": {
            "zh-Hant": "分配律、等值代換與分割求和都是等式，反向依序還原。",
            "en": "Distributivity, replacement of equals and summing over a partition are all equations; undoing them in order restores the original."
          }
        },
        "leaf_basis": "distributive + eq_replace + sum_over_partition + finite_sum_rules"
      },
      {
        "id": "S15",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "\\sum_{x}q(x)=\\tfrac12+\\tfrac13+\\tfrac14+0",
        "after_tex": "\\tfrac{13}{12}\\neq 1;\\ q\\ \\text{is not a PMF}",
        "relation": "implies",
        "rule": "mko_pmf_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "q 的每一項都介於 0 與 1 之間，但總和為 6/12+4/12+3/12+0=13/12≠1，不符合資格，所以 q 不是 PMF，不能拿它求事件機率或期望。",
          "en": "Every mass of q lies between 0 and 1, but the total is 6/12+4/12+3/12+0=13/12≠1, which fails the PMF conditions, so q is not a PMF and may not be used for event probabilities or an expectation."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "G3",
            "mko_pmf_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "目標 G：q 不是 PMF。足夠條件 H：總和不為 1；H⇒G 由定義。H 由精確求和得到（N10、N11），沒有把 G 當作前提。",
            "en": "Goal G: q is not a PMF. Sufficient condition H: the total is not 1; H⇒G by the definition. H comes from exact summation (N10, N11); G is never used as a premise."
          }
        },
        "leaf_basis": "mko_pmf_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "r(1)=-\\tfrac16,\\ \\sum_x r(x)=1",
        "after_tex": "r\\ \\text{is not a PMF}",
        "relation": "implies",
        "rule": "mko_pmf_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "r 的總和 3/6+4/6−1/6+0=6/6=1，但 r(1)=−1/6<0，違反非負，所以 r 不是 PMF。資格的兩項條件互相獨立，缺一不可。",
          "en": "r sums to 3/6+4/6−1/6+0=6/6=1, but r(1)=−1/6<0 violates nonnegativity, so r is not a PMF. The two PMF conditions are independent and both are required."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "G3",
            "mko_pmf_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "目標 G：r 不是 PMF。足夠條件 H：有一項為負；H⇒G 由定義。H 直接由 G3 讀出（N13），沒有把 G 當作前提。",
            "en": "Goal G: r is not a PMF. Sufficient condition H: some mass is negative; H⇒G by the definition. H is read off G3 (N13); G is never used as a premise."
          }
        },
        "leaf_basis": "mko_pmf_def + order_basics + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "p 非負且總和恰為 1，是合格的 PMF。P(X≥1)=p(1)+p(2)=1/2；E[X]=7/12。Y=X² 的推前分布為 p_Y(0)=1/3、p_Y(1)=5/12、p_Y(4)=1/4，仍是 PMF（對任意有限值域證明），E[Y]=17/12，與 Σx²p(x) 一致。q 每項介於 0 與 1 但總和 13/12，r 總和為 1 但 r(1)=−1/6，兩者都不是 PMF。",
      "en": "p is nonnegative with total exactly 1, so it is a PMF. P(X≥1)=p(1)+p(2)=1/2 and E[X]=7/12. The pushforward of Y=X² is p_Y(0)=1/3, p_Y(1)=5/12, p_Y(4)=1/4, again a PMF (proved for any finite range), and E[Y]=17/12, agreeing with Σx²p(x). q has every mass between 0 and 1 but total 13/12, and r sums to 1 but r(1)=−1/6, so neither is a PMF."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "q 的每一項都介於 0 與 1，所以 q 是 PMF",
          "en": "Every mass of q lies between 0 and 1, so q is a PMF"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "q 的總和是 13/12≠1。",
          "en": "q sums to 13/12≠1."
        }
      },
      {
        "claim": {
          "zh-Hant": "r 的總和為 1，所以 r 是 PMF",
          "en": "r sums to 1, so r is a PMF"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "r(1)=−1/6<0，違反非負。",
          "en": "r(1)=−1/6<0 violates nonnegativity."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(X≥1)=p(2)=1/4，因為只有 2 大於 1",
          "en": "P(X≥1)=p(2)=1/4, since only 2 is greater than 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "事件是 x≥1，包含 1；P(X≥1)=1/4+1/4=1/2。",
          "en": "The event is x≥1, which includes 1; P(X≥1)=1/4+1/4=1/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "E[X²]=(E[X])²=49/144",
          "en": "E[X²]=(E[X])²=49/144"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "E[X²]=17/12，與 49/144 不同。",
          "en": "E[X²]=17/12, which differs from 49/144."
        }
      },
      {
        "claim": {
          "zh-Hant": "p_Y(1)=p(1)=1/4，因為 g(1)=1",
          "en": "p_Y(1)=p(1)=1/4, because g(1)=1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "g(−1)=1 也映到 1，必須相加：p_Y(1)=5/12。",
          "en": "g(−1)=1 maps to 1 as well and must be added: p_Y(1)=5/12."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "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": "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": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "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": "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": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "mko_pmf_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-discrete-probability-distribution",
          "sha256": "a77aa248d456c6fff168a9df6f8b4ad85be410912a036130257ff1d3e0aa1b2a",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "有限離散隨機變數的機率質量函數 p 對每個有限值域元素指定非負機率，且所有質量和為 1。事件機率由相應質量相加，期望為值與質量的加權和。",
          "en": "The probability mass function p of a finite discrete random variable assigns a nonnegative probability to each element of its finite value range, and all masses sum to 1. Event probabilities add the corresponding masses; expectation is the mass-weighted sum of values."
        }
      },
      {
        "id": "event_sum",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對值域的子集 A，P(X∈A)=Σ_{x∈A} p(x)；A 以外的值不加入。",
          "en": "For a subset A of the range, P(X∈A)=Σ_{x∈A} p(x); values outside A are not added."
        }
      },
      {
        "id": "expectation_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限值域的期望：E[X]=Σ_{x∈R_X} x·p(x)。",
          "en": "Expectation on a finite range: E[X]=Σ_{x∈R_X} x·p(x)."
        }
      },
      {
        "id": "pushforward_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對函數 g，Y=g(X) 的值域是 R_Y=g(R_X)，推前質量 p_Y(y)=Σ_{x∈R_X, g(x)=y} p(x)，即把映到同一新值的原質量相加。",
          "en": "For a function g, Y=g(X) has range R_Y=g(R_X) and pushforward mass p_Y(y)=Σ_{x∈R_X, g(x)=y} p(x): the original masses mapped to the same new value are added."
        }
      },
      {
        "id": "sum_over_partition",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "若有限集合 S 是兩兩不交的子集 S_y（y∈Y）的聯集，則 Σ_{x∈S} a_x=Σ_{y∈Y} Σ_{x∈S_y} a_x（有限和的結合與交換律）。",
          "en": "If a finite set S is the union of pairwise disjoint subsets S_y (y∈Y), then Σ_{x∈S} a_x=Σ_{y∈Y} Σ_{x∈S_y} a_x (associativity and commutativity of finite sums)."
        }
      },
      {
        "id": "L_pushforward_total",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2"
        ],
        "statement": {
          "zh-Hant": "若 p 是有限值域 R_X 上的 PMF、g 是定義在 R_X 上的函數，則推前質量 p_Y 是 R_Y=g(R_X) 上的 PMF。",
          "en": "If p is a PMF on the finite range R_X and g is a function on R_X, then the pushforward mass p_Y is a PMF on R_Y=g(R_X)."
        }
      },
      {
        "id": "L_regroup",
        "kind": "lemma",
        "proved_in": [
          "S14.1"
        ],
        "statement": {
          "zh-Hant": "在 L_pushforward_total 的設定下，Σ_{y∈R_Y} y·p_Y(y)=Σ_{x∈R_X} g(x)·p(x)。",
          "en": "In the setting of L_pushforward_total, Σ_{y∈R_Y} y·p_Y(y)=Σ_{x∈R_X} g(x)·p(x)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "1/6+1/3+1/4+1/4",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S04.2",
        "lhs": "2+4+3+3",
        "rhs": "12"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "1/4+1/4",
        "rhs": "1/2"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "1/6+1/3",
        "rhs": "1/2"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "(-1)*(1/6)+0*(1/3)+1*(1/4)+2*(1/4)",
        "rhs": "7/12"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "1/6+1/4",
        "rhs": "5/12"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "1/3+5/12+1/4",
        "rhs": "1"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "0*(1/3)+1*(5/12)+4*(1/4)",
        "rhs": "17/12"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "(-1)**2*(1/6)+0**2*(1/3)+1**2*(1/4)+2**2*(1/4)",
        "rhs": "17/12"
      },
      {
        "id": "N10",
        "step": "S15",
        "lhs": "1/2+1/3+1/4+0",
        "rhs": "13/12"
      },
      {
        "id": "N11",
        "step": "S15",
        "lhs": "13/12",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N12",
        "step": "S16",
        "lhs": "1/2+2/3+(-1/6)+0",
        "rhs": "1"
      },
      {
        "id": "N13",
        "step": "S16",
        "lhs": "-1/6",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N14",
        "lhs": "(7/12)**2",
        "relation": "!=",
        "rhs": "17/12"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理有限值域、值互異、質量為精確有理數的數學分布（MKO 假設 a1）；不宣稱分布來自樣本估計、資料擬合或真實族群（MKO 假設 a2）。推前分布與重組求和的兩個引理對任意有限值域證明；其餘計算只對給定數字。不處理聯合分布或 Markov 鏈；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only mathematical distributions with a finite range of distinct values and exact rational masses (MKO assumption a1); no claim is made that the distribution comes from a sample estimate, a data fit or a real population (MKO assumption a2). The two lemmas on pushforwards and regrouped sums are proved for any finite range; the other computations concern the given numbers only. Joint distributions and Markov chains are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "R_X=\\{-1,0,1,2\\}",
        "X 的有限值域",
        "the finite range of X"
      ],
      [
        "p(x)",
        "X 的機率質量函數",
        "the probability mass function of X"
      ],
      [
        "P(X\\in A)",
        "事件 A 的機率",
        "the probability of the event A"
      ],
      [
        "E[X]",
        "X 的期望",
        "the expectation of X"
      ],
      [
        "g(x)=x^2,\\ Y=g(X)",
        "推前所用的函數與新隨機變數",
        "the map used for the pushforward and the new random variable"
      ],
      [
        "p_Y(y)",
        "Y 的推前質量",
        "the pushforward mass of Y"
      ]
    ],
    "md_error_controls": [
      {
        "title": "每項介於 0 與 1 不代表總和為 1 / Masses between 0 and 1 do not make the total 1",
        "zh": "MKO 指出的誤解是只因每項介於 0 與 1 就忽略總和。q 的每一項都在 [0,1] 內，總和卻是 13/12；r 總和為 1 卻有負項。資格必須同時核兩項，並且用精確分數相加，不用小數近似。",
        "en": "The MKO names the mistake of ignoring the total because every mass lies between 0 and 1. Every mass of q lies in [0,1], yet the total is 13/12; r sums to 1 yet has a negative mass. Both conditions must be checked, adding exact fractions rather than decimal approximations."
      },
      {
        "title": "理論分布不是資料擬合 / A theoretical PMF is not a data fit",
        "zh": "MKO 的另一個誤解是把理論 PMF 冒充資料擬合結果。本稿的 1/6、1/3、1/4、1/4 是給定的數學分布（G4），沒有任何觀測資料；說「12 次試驗中恰有 2 次得到 −1」或「這是由樣本估計的分布」都超出本稿能支持的範圍。推前分布也只是精確的數學轉換。",
        "en": "Another mistake identified in the MKO is passing off a theoretical PMF as the result of a data fit. The masses 1/6, 1/3, 1/4, 1/4 here are a given mathematical distribution (G4) with no observed data; saying 'exactly 2 of 12 trials gave −1' or 'this distribution was estimated from a sample' goes beyond anything this guide supports. The pushforward is likewise an exact mathematical transformation."
      },
      {
        "title": "推前要合併同值的質量 / A pushforward merges masses with the same image",
        "zh": "−1 與 1 都映到 1，所以 p_Y(1)=1/6+1/4=5/12。若只取 p(1)=1/4，或把值 1 列兩次，總和就不是 1 或值不再互異。另外 E[X²]=17/12 與 (E[X])²=49/144 不同：平方的期望一般不等於期望的平方。",
        "en": "Both −1 and 1 map to 1, so p_Y(1)=1/6+1/4=5/12. Taking only p(1)=1/4, or listing the value 1 twice, breaks the total of 1 or the distinctness of values. Also E[X²]=17/12 differs from (E[X])²=49/144: the expectation of a square is in general not the square of the expectation."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "引理 L_pushforward_total 與 L_regroup 只對有限值域證明；可數或連續情形需要級數或積分，本稿不處理。",
        "en": "Lemmas L_pushforward_total and L_regroup are proved for finite ranges only; countable or continuous cases need series or integrals and are not treated here."
      },
      {
        "zh-Hant": "本稿不處理聯合分布、離散隨機過程或有限 Markov 鏈（MKO 的 formal next），也不做任何統計推論。",
        "en": "Joint distributions, discrete random processes and finite Markov chains (the MKO's formal next) are not treated, and no statistical inference is made."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c76a1162145f6b40efb60946aa3c825a45dc36a902c036a45e633fa39245b697",
    "original_source_sha256": "c76a1162145f6b40efb60946aa3c825a45dc36a902c036a45e633fa39245b697",
    "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."
  }
}