{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-markov-chain-01",
  "mko_id": "mko-finite-markov-chain",
  "title": {
    "zh-Hant": "有限狀態 Markov 鏈：核對行隨機矩陣，計算兩步時刻律與路徑機率",
    "en": "Finite-state Markov chain: checking a row-stochastic matrix and computing two-step time laws and path probabilities"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-markov-chain-01",
    "source": {
      "mko_id": "mko-finite-markov-chain",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "713cef8fb3c1efc7627452592ab30fb69b9ebb7992c47aad78b9ae322ee7fd55",
      "english_sha256": "531d840a993901f295df6db2e167b57a16a1d3820e22c5212389593049a3ef38"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "prob_axioms",
      "finite_sum_rules",
      "matrix_product_def",
      "distributive",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "狀態依序為 1、2、3，轉移矩陣 P 的行為 (1/2,1/2,0)、(1/4,1/2,1/4)、(0,1/3,2/3)，初始律 μ_0=(1/2,1/2,0)。(1) 核對 P 是行隨機矩陣，並說明列和不必為 1；(2) 證明一步公式 μ_{n+1}=μ_nP 且結果仍是機率向量，算出 μ_1 與 μ_2；(3) 證明路徑機率公式並計算 P(X_0=1,X_1=2,X_2=3)、P(X_0=2,X_1=3,X_2=3)；(4) 說明 P 單獨一般不能決定完整的無條件時刻律與路徑律。",
      "en": "The states are ordered 1, 2, 3; the transition matrix P has rows (1/2,1/2,0), (1/4,1/2,1/4), (0,1/3,2/3), and the initial law is μ_0=(1/2,1/2,0). (1) Check that P is row-stochastic and explain why column sums need not be 1; (2) prove the one-step formula μ_{n+1}=μ_nP and that the result is again a probability vector, and compute μ_1 and μ_2; (3) prove the path-probability formula and compute P(X_0=1,X_1=2,X_2=3) and P(X_0=2,X_1=3,X_2=3); (4) explain why P alone generally does not determine complete unconditional marginal and path laws."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "狀態次序 (1,2,3)。P 的行：第 1 行 (1/2,1/2,0)、第 2 行 (1/4,1/2,1/4)、第 3 行 (0,1/3,2/3)；行為現在狀態、列為下一步狀態（MKO 假設 a1：精確有理數）。",
          "en": "State order (1,2,3). Rows of P: row 1 (1/2,1/2,0), row 2 (1/4,1/2,1/4), row 3 (0,1/3,2/3); rows are the current state and columns the next state (MKO assumption a1: exact rationals)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "(X_n) 是以 P 為核、以 μ_0=(1/2,1/2,0) 為初始律的時間齊次 Markov 鏈（性質見前提規則 p_markov_property）。",
          "en": "(X_n) is a time-homogeneous Markov chain with kernel P and initial law μ_0=(1/2,1/2,0) (properties as in the prerequisite rule p_markov_property)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較用初始律 μ_0'=(0,0,1)，核不變。",
          "en": "Comparison initial law μ_0'=(0,0,1), with the same kernel."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_markov_def",
          "G1"
        ],
        "before_tex": "P=(P_{ij})_{i,j\\in\\{1,2,3\\}}",
        "after_tex": "P_{ij}\\ge0,\\quad \\sum_jP_{ij}=1;\\quad \\Pr(X_n=i)>0\\ \\Longrightarrow\\ P_{ij}=\\Pr(X_{n+1}=j\\mid X_n=i)",
        "relation": "definition",
        "rule": "mko_markov_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "此本地採用修訂先固定狀態次序 (1,2,3)，把 P 視為題目指定的轉移核：每行非負且精確和為 1。對正機率的現在狀態 i，第 i 行才可直接用比值條件機率讀取；若 Pr(X_n=i)=0，該比值未定義，核行仍由模型指定。本題 μ_0(3)=0 但第 3 行有完整資料。原定義的勘誤已由使用者採用於本地追加式修訂，但不改寫歷史 canonical MKO；後續 S06 與 S10 的證明已分開處理零質量情形。",
          "en": "This adopted local revision fixes the state order (1,2,3) and treats P as the prescribed transition kernel: each row is nonnegative and sums exactly to 1. For a current state i of positive probability, row i can be identified with ratio-defined conditional probabilities. If Pr(X_n=i)=0 that ratio is undefined, although the model still prescribes the kernel row. Here μ_0(3)=0 and row 3 is fully supplied. The owner has adopted the definition correction in this append-only local revision, without rewriting the historical canonical MKO; the later S06 and S10 proofs already separate zero-mass cases."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_markov_def"
          ],
          "explanation": {
            "zh-Hant": "行隨機要求作用於指定轉移核的每一行，列沒有對應要求。與比值條件機率的等同另需 Pr(X_n=i)>0；不能在零機率條件事件上作除法。",
            "en": "The row-stochastic requirements concern every prescribed kernel row, with no matching column requirement. Identification with a ratio conditional additionally needs Pr(X_n=i)>0; division at a null conditioning event is not allowed."
          }
        },
        "leaf_basis": "mko_markov_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "P_{ij}\\in\\{0,\\tfrac14,\\tfrac13,\\tfrac12,\\tfrac23\\}",
        "after_tex": "P_{ij}\\ge0\\ \\text{for all nine entries}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "九個元素的值只有 0、1/4、1/3、1/2、2/3，全部非負。",
          "en": "The nine entries take only the values 0, 1/4, 1/3, 1/2, 2/3, all nonnegative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "九格完整列舉；0 是允許的（表示不可能的轉移）。",
            "en": "All nine cells are enumerated; 0 is allowed (it marks an impossible transition)."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "\\sum_jP_{1j},\\ \\sum_jP_{2j},\\ \\sum_jP_{3j}",
        "after_tex": "1,\\ 1,\\ 1:\\ P\\ \\text{is row-stochastic}",
        "relation": "implies",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三行逐一求和（S03.1 到 S03.3），每行都精確等於 1。連同 S02，P 是行隨機矩陣，可以用來做步進計算。",
          "en": "Sum the three rows one by one (S03.1 to S03.3); each equals 1 exactly. With S02, P is row-stochastic and may be used for step computations."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_sum_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數求和；近似為 1 不算通過。",
            "en": "Exact rational sums; being approximately 1 would not pass."
          }
        },
        "leaf_basis": "finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S03.1",
        "uses": [
          "G1"
        ],
        "before_tex": "P_{11}+P_{12}+P_{13}",
        "after_tex": "\\tfrac12+\\tfrac12+0=1",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 1 行：1/2+1/2+0=1（N1）。",
          "en": "Row 1: 1/2+1/2+0=1 (N1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，可獨立重算。",
            "en": "Exact evaluation, independently recomputable."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S03.2",
        "uses": [
          "G1"
        ],
        "before_tex": "P_{21}+P_{22}+P_{23}",
        "after_tex": "\\tfrac14+\\tfrac12+\\tfrac14=1",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 2 行：1/4+1/2+1/4=1（N2）。",
          "en": "Row 2: 1/4+1/2+1/4=1 (N2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，可獨立重算。",
            "en": "Exact evaluation, independently recomputable."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S03.3",
        "uses": [
          "G1"
        ],
        "before_tex": "P_{31}+P_{32}+P_{33}",
        "after_tex": "0+\\tfrac13+\\tfrac23=1",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 3 行：0+1/3+2/3=1（N3）。",
          "en": "Row 3: 0+1/3+2/3=1 (N3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，可獨立重算。",
            "en": "Exact evaluation, independently recomputable."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "\\sum_iP_{i1},\\ \\sum_iP_{i2},\\ \\sum_iP_{i3}",
        "after_tex": "\\tfrac34,\\ \\tfrac43,\\ \\tfrac{11}{12}\\ \\ (\\text{not required to be}\\ 1)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "列和是 1/2+1/4+0=3/4、1/2+1/2+1/3=4/3、0+1/4+2/3=11/12（N4 到 N6），都不是 1；它們的總和 3 等於行數。列和不是條件分布，沒有任何要求；P 仍然是合法的轉移矩陣（只是不是雙隨機）。",
          "en": "The column sums are 1/2+1/4+0=3/4, 1/2+1/2+1/3=4/3 and 0+1/4+2/3=11/12 (N4 to N6), none equal to 1; their total 3 equals the number of rows. A column sum is not a conditional distribution and carries no requirement; P is still a valid transition matrix (merely not doubly stochastic)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "這一步只說明列和不在責任清單上；若把列和當行和，會誤判本矩陣不合格（MKO 指出的誤解）。",
            "en": "This step only shows that column sums are not among the obligations; mistaking column sums for row sums would wrongly reject this matrix (the misconception named by the MKO)."
          }
        },
        "leaf_basis": "numeric_evaluation + finite_sum_rules"
      },
      {
        "id": "S05",
        "uses": [
          "G2"
        ],
        "before_tex": "\\mu_0=(\\tfrac12,\\tfrac12,0)",
        "after_tex": "\\mu_0\\ge0,\\ \\ \\tfrac12+\\tfrac12+0=1",
        "relation": "implies",
        "rule": "prob_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "初始律的三個分量非負，和為 1（N7），所以 μ_0 是 {1,2,3} 上的機率向量。MKO 指出：時刻律與路徑機率需要初始律，所以這是 S08 之後計算的必要輸入。",
          "en": "The three entries of the initial law are nonnegative and sum to 1 (N7), so μ_0 is a probability vector on {1,2,3}. The MKO notes that time laws and path probabilities need an initial law, so this is a required input for the computations from S08 on."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prob_axioms",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "初始律與 P 是兩份獨立資料；P 合格不代表初始律已知（S15）。",
            "en": "The initial law and P are two independent pieces of data; a valid P does not mean the initial law is known (S15)."
          }
        },
        "leaf_basis": "prob_axioms + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "G2",
          "L_one_step"
        ],
        "before_tex": "\\mu_{n+1}(j)=P(X_{n+1}=j)",
        "after_tex": "\\sum_{i}\\mu_n(i)P_{ij}\\ \\ (\\mu_{n+1}=\\mu_nP)",
        "relation": "theorem_application",
        "rule": "L_one_step",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_one_step（S06.1、S06.2 證明）：下一時刻的分布是現在的分布（行向量）右乘 P。這是 MKO 段落 1 的 row-vector convention μP。",
          "en": "Lemma L_one_step (proved in S06.1, S06.2): the law at the next time is the current law (a row vector) multiplied on the right by P. This is the row-vector convention μP of MKO paragraph 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_one_step",
            "matrix_product_def"
          ],
          "explanation": {
            "zh-Hant": "依矩陣乘法定義，(μP)_j=Σ_i μ(i)P_{ij}，與引理的和式一致；若誤寫成 Pμ（列向量），會用到列而不是行。",
            "en": "By the definition of matrix product, (μP)_j=Σ_i μ(i)P_{ij}, which matches the sum in the lemma; writing Pμ (a column vector) by mistake would use columns instead of rows."
          }
        },
        "leaf_basis": "L_one_step + matrix_product_def"
      },
      {
        "id": "S06.1",
        "uses": [
          "prob_axioms",
          "finite_sum_rules"
        ],
        "before_tex": "P(X_{n+1}=j)",
        "after_tex": "\\sum_{i=1}^{3}P(X_n=i,\\ X_{n+1}=j)",
        "relation": "equivalent",
        "rule": "prob_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "事件 {X_n=1}、{X_n=2}、{X_n=3} 兩兩不交且聯集為全空間，所以 {X_{n+1}=j} 被分成三個不交部分；機率的可加性給出這個和。",
          "en": "The events {X_n=1}, {X_n=2}, {X_n=3} are pairwise disjoint with union the whole space, so {X_{n+1}=j} splits into three disjoint parts, and additivity of probability gives this sum."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "prob_axioms",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "分割與可加性都是等式，雙向成立。",
            "en": "The partition and additivity are equalities, so the step holds both ways."
          }
        },
        "leaf_basis": "prob_axioms + set_operations_def"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "p_markov_property",
          "p_cond_prob"
        ],
        "before_tex": "P(X_n=i,\\ X_{n+1}=j)",
        "after_tex": "\\mu_n(i)P_{ij}",
        "relation": "equivalent",
        "rule": "p_cond_prob",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分兩種情況。若 μ_n(i)>0：條件機率給 P(X_n=i,X_{n+1}=j)=P(X_n=i)P(X_{n+1}=j∣X_n=i)=μ_n(i)P_{ij}。若 μ_n(i)=0：交集包含於 {X_n=i}，機率為 0，而 μ_n(i)P_{ij}=0 也為 0。兩種情況都得到同一公式，代回 S06.1 證得引理 L_one_step。",
          "en": "Two cases. If μ_n(i)>0: conditional probability gives P(X_n=i,X_{n+1}=j)=P(X_n=i)P(X_{n+1}=j∣X_n=i)=μ_n(i)P_{ij}. If μ_n(i)=0: the intersection lies inside {X_n=i}, so its probability is 0, and μ_n(i)P_{ij}=0 as well. Both cases give the same formula; substituting into S06.1 proves Lemma L_one_step."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "p_cond_prob",
            "p_markov_property",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩邊在每種情況下都是同一個數；μ_n(i)=0 的情況不能用條件機率，必須另外處理。",
            "en": "Both sides are the same number in each case; the case μ_n(i)=0 cannot use conditional probability and is handled separately."
          }
        },
        "leaf_basis": "p_cond_prob + case_analysis + mul_zero"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S06",
          "L_prob_vector"
        ],
        "before_tex": "\\mu_n\\ \\text{probability vector}",
        "after_tex": "\\mu_{n+1}=\\mu_nP\\ \\text{probability vector}",
        "relation": "theorem_application",
        "rule": "L_prob_vector",
        "conditions": [
          "P\\ \\text{row-stochastic}"
        ],
        "explanation": {
          "zh-Hant": "引理 L_prob_vector（S07.1 證明）：機率向量乘行隨機矩陣仍是機率向量。證明中用的是行和為 1，而不是列和；這正是 S03 的資格要先於步進計算的原因。",
          "en": "Lemma L_prob_vector (proved in S07.1): a probability vector times a row-stochastic matrix is again a probability vector. The proof uses row sums equal to 1, not column sums; this is why the qualification of S03 must precede any step computation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_prob_vector",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "需要行隨機；列和為 1 對這個結論既不必要也不充分。",
            "en": "Needs row-stochasticity; column sums equal to 1 are neither necessary nor sufficient for this conclusion."
          }
        },
        "leaf_basis": "L_prob_vector + finite_sum_rules"
      },
      {
        "id": "S07.1",
        "uses": [
          "finite_sum_rules",
          "distributive"
        ],
        "before_tex": "\\sum_j\\sum_i\\mu(i)P_{ij}",
        "after_tex": "\\sum_i\\mu(i)\\sum_jP_{ij}=\\sum_i\\mu(i)\\cdot1=1",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [
          "\\sum_jP_{ij}=1"
        ],
        "explanation": {
          "zh-Hant": "有限雙重和可以交換求和次序，再用分配律把 μ(i) 提出：Σ_j Σ_i μ(i)P_{ij}=Σ_i μ(i)(Σ_j P_{ij})（N8 對本矩陣驗證這個恆等式）。每行和為 1，所以等於 Σ_i μ(i)=1。非負性：非負數之積與和非負。這證得引理 L_prob_vector。",
          "en": "A finite double sum may be summed in either order, and distributivity factors out μ(i): Σ_j Σ_i μ(i)P_{ij}=Σ_i μ(i)(Σ_j P_{ij}) (N8 checks this identity for this matrix). Each row sums to 1, so this equals Σ_i μ(i)=1. Nonnegativity: products and sums of nonnegative numbers are nonnegative. This proves Lemma L_prob_vector."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "交換有限和與提出公因子都是恆等變形；最後一步用到行和條件。",
            "en": "Swapping finite sums and factoring are identities; the final step uses the row-sum condition."
          }
        },
        "leaf_basis": "finite_sum_rules + distributive + mul_one"
      },
      {
        "id": "S08",
        "uses": [
          "S05",
          "S06",
          "G1"
        ],
        "before_tex": "\\mu_1=\\mu_0P",
        "after_tex": "\\mu_1=(\\tfrac38,\\ \\tfrac12,\\ \\tfrac18)",
        "relation": "equivalent",
        "rule": "L_one_step",
        "conditions": [],
        "explanation": {
          "zh-Hant": "μ_1(1)=1/2·1/2+1/2·1/4+0·0=3/8（N9）；μ_1(2)=1/2·1/2+1/2·1/2+0·1/3=1/2（N10）；μ_1(3)=1/2·0+1/2·1/4+0·2/3=1/8（N11）。和為 1，與 S07 一致。",
          "en": "μ_1(1)=1/2·1/2+1/2·1/4+0·0=3/8 (N9); μ_1(2)=1/2·1/2+1/2·1/2+0·1/3=1/2 (N10); μ_1(3)=1/2·0+1/2·1/4+0·2/3=1/8 (N11). They sum to 1, consistent with S07."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_one_step",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個分量都是精確計算，可由 N9 到 N11 重算。",
            "en": "Each entry is an exact computation, recomputed by N9 to N11."
          }
        },
        "leaf_basis": "L_one_step + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S06",
          "G1"
        ],
        "before_tex": "\\mu_2=\\mu_1P",
        "after_tex": "\\mu_2=(\\tfrac{5}{16},\\ \\tfrac{23}{48},\\ \\tfrac{5}{24})",
        "relation": "equivalent",
        "rule": "L_one_step",
        "conditions": [],
        "explanation": {
          "zh-Hant": "再套一次 L_one_step：μ_2(1)=3/8·1/2+1/2·1/4+1/8·0=5/16（N12）；μ_2(2)=3/8·1/2+1/2·1/2+1/8·1/3=23/48（N13）；μ_2(3)=3/8·0+1/2·1/4+1/8·2/3=5/24（N14）。和為 15/48+23/48+10/48=1（N15）。",
          "en": "Apply L_one_step again: μ_2(1)=3/8·1/2+1/2·1/4+1/8·0=5/16 (N12); μ_2(2)=3/8·1/2+1/2·1/2+1/8·1/3=23/48 (N13); μ_2(3)=3/8·0+1/2·1/4+1/8·2/3=5/24 (N14). The sum is 15/48+23/48+10/48=1 (N15)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_one_step",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數計算；總和為 1 是 S07 的獨立核對。",
            "en": "Exact rational computation; the total of 1 is an independent check of S07."
          }
        },
        "leaf_basis": "L_one_step + numeric_evaluation + frac_rules"
      },
      {
        "id": "S10",
        "uses": [
          "G2",
          "L_path"
        ],
        "before_tex": "P(X_0=i_0,X_1=i_1,X_2=i_2)",
        "after_tex": "\\mu_0(i_0)P_{i_0i_1}P_{i_1i_2}",
        "relation": "theorem_application",
        "rule": "L_path",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_path（S10.1、S10.2 證明）：有限路徑機率是初始質量乘上逐步轉移因子（MKO 段落 2）。",
          "en": "Lemma L_path (proved in S10.1, S10.2): a finite path probability is the initial mass times the successive transition factors (MKO paragraph 2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_path"
          ],
          "explanation": {
            "zh-Hant": "需要 Markov 性（G2）；若只知道每步的條件機率而不知歷史無關，乘法公式沒有根據。",
            "en": "Needs the Markov property (G2); knowing each one-step conditional probability without independence from the past would not justify the product formula."
          }
        },
        "leaf_basis": "L_path + p_markov_property"
      },
      {
        "id": "S10.1",
        "uses": [
          "p_markov_property",
          "p_cond_prob"
        ],
        "before_tex": "P(X_0=i_0,X_1=i_1)",
        "after_tex": "\\mu_0(i_0)P_{i_0i_1}",
        "relation": "equivalent",
        "rule": "p_cond_prob",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 μ_0(i_0)>0：P(X_0=i_0,X_1=i_1)=P(X_0=i_0)P(X_1=i_1∣X_0=i_0)=μ_0(i_0)P_{i_0i_1}。若 μ_0(i_0)=0：兩邊都是 0。這是兩點路徑的公式。",
          "en": "If μ_0(i_0)>0: P(X_0=i_0,X_1=i_1)=P(X_0=i_0)P(X_1=i_1∣X_0=i_0)=μ_0(i_0)P_{i_0i_1}. If μ_0(i_0)=0: both sides are 0. This is the formula for two-point paths."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "p_cond_prob",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "與 S06.2 相同的兩種情況處理。",
            "en": "The same two-case treatment as in S06.2."
          }
        },
        "leaf_basis": "p_cond_prob + case_analysis + mul_zero"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "p_markov_property",
          "p_cond_prob"
        ],
        "before_tex": "P(X_0=i_0,X_1=i_1,X_2=i_2)",
        "after_tex": "P(X_0=i_0,X_1=i_1)\\,P_{i_1i_2}=\\mu_0(i_0)P_{i_0i_1}P_{i_1i_2}",
        "relation": "equivalent",
        "rule": "p_markov_property",
        "conditions": [],
        "explanation": {
          "zh-Hant": "令 H={X_0=i_0,X_1=i_1}。若 P(H)>0：P(H∩{X_2=i_2})=P(H)P(X_2=i_2∣H)，而 Markov 性給 P(X_2=i_2∣H)=P_{i_1i_2}（只依賴現在狀態 i_1）。若 P(H)=0：三點事件包含於 H，機率為 0，而由 S10.1，μ_0(i_0)P_{i_0i_1}=P(H)=0，公式也給 0。再代入 S10.1，證得引理 L_path。",
          "en": "Let H={X_0=i_0,X_1=i_1}. If P(H)>0: P(H∩{X_2=i_2})=P(H)P(X_2=i_2∣H), and the Markov property gives P(X_2=i_2∣H)=P_{i_1i_2} (depending only on the current state i_1). If P(H)=0: the three-point event lies inside H and has probability 0, and by S10.1 μ_0(i_0)P_{i_0i_1}=P(H)=0, so the formula also gives 0. Substituting S10.1 proves Lemma L_path."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "p_markov_property",
            "p_cond_prob",
            "S10.1"
          ],
          "explanation": {
            "zh-Hant": "兩種情況下兩邊同值；P(H)=0 的情況必須單獨處理，因為條件機率不定義。",
            "en": "Both sides agree in both cases; the case P(H)=0 must be handled separately because the conditional probability is undefined."
          }
        },
        "leaf_basis": "p_markov_property + p_cond_prob + case_analysis"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G1",
          "G2"
        ],
        "before_tex": "P(X_0=1,X_1=2,X_2=3)",
        "after_tex": "\\tfrac12\\cdot\\tfrac12\\cdot\\tfrac14=\\tfrac{1}{16}",
        "relation": "equivalent",
        "rule": "L_path",
        "conditions": [],
        "explanation": {
          "zh-Hant": "μ_0(1)=1/2、P_{12}=1/2、P_{23}=1/4，乘積 1/16（N16）。注意 P_{23} 是第 2 行第 3 列（從 2 到 3），不是第 3 行第 2 列。",
          "en": "μ_0(1)=1/2, P_{12}=1/2, P_{23}=1/4, with product 1/16 (N16). Note that P_{23} is row 2, column 3 (from 2 to 3), not row 3, column 2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_path",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "代入引理的精確計算。",
            "en": "An exact computation substituting into the lemma."
          }
        },
        "leaf_basis": "L_path + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "G1",
          "G2"
        ],
        "before_tex": "P(X_0=2,X_1=3,X_2=3)",
        "after_tex": "\\tfrac12\\cdot\\tfrac14\\cdot\\tfrac23=\\tfrac{1}{12}",
        "relation": "equivalent",
        "rule": "L_path",
        "conditions": [],
        "explanation": {
          "zh-Hant": "μ_0(2)=1/2、P_{23}=1/4、P_{33}=2/3，乘積 1/12（N17）。",
          "en": "μ_0(2)=1/2, P_{23}=1/4, P_{33}=2/3, with product 1/12 (N17)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_path",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "代入引理的精確計算。",
            "en": "An exact computation substituting into the lemma."
          }
        },
        "leaf_basis": "L_path + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S10",
          "G1",
          "G2"
        ],
        "before_tex": "P(X_0=1,X_1=3,X_2=j),\\ P(X_0=3,X_1=i,X_2=j)",
        "after_tex": "0\\ \\ (P_{13}=0,\\ \\mu_0(3)=0)",
        "relation": "implies",
        "rule": "L_path",
        "conditions": [],
        "explanation": {
          "zh-Hant": "P_{13}=0，所以任何以 1→3 開頭的路徑機率為 0；μ_0(3)=0，所以任何從 3 出發的路徑機率為 0。乘積中只要有一個因子為 0，路徑就不可能。",
          "en": "P_{13}=0, so every path beginning 1→3 has probability 0; μ_0(3)=0, so every path starting at 3 has probability 0. A single zero factor in the product makes the path impossible."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_path",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "零因子推出乘積為零；逆否命題（乘積非零推出每個因子非零）也成立但此處不需要。",
            "en": "A zero factor makes the product zero; the contrapositive (a nonzero product has nonzero factors) also holds but is not needed here."
          }
        },
        "leaf_basis": "L_path + mul_zero"
      },
      {
        "id": "S14",
        "uses": [
          "S10",
          "S09",
          "G1"
        ],
        "before_tex": "\\mu_2(3)=\\sum_{i_0,i_1}\\mu_0(i_0)P_{i_0i_1}P_{i_13}",
        "after_tex": "\\tfrac{1}{16}+\\tfrac{1}{16}+\\tfrac{1}{12}=\\tfrac{5}{24}",
        "relation": "equivalent",
        "rule": "prob_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交叉核對：{X_2=3} 依 (X_0,X_1) 分割，μ_2(3) 等於所有結尾為 3 的路徑機率之和。非零的只有 (1,2,3):1/16、(2,2,3):1/2·1/2·1/4=1/16、(2,3,3):1/12；其他都含 μ_0(3)=0 或 P_{13}=0。和為 5/24（N18），與 S09 一致。也可用 P² 核對：μ_2(3)=1/2·(P²)_{13}+1/2·(P²)_{23}=1/2·1/8+1/2·7/24=5/24（N19）。",
          "en": "Cross-check: {X_2=3} is partitioned by (X_0,X_1), so μ_2(3) is the sum of the probabilities of all paths ending at 3. The only nonzero ones are (1,2,3): 1/16, (2,2,3): 1/2·1/2·1/4=1/16, (2,3,3): 1/12; all others contain μ_0(3)=0 or P_{13}=0. The sum is 5/24 (N18), agreeing with S09. A check through P² also works: μ_2(3)=1/2·(P²)_{13}+1/2·(P²)_{23}=1/2·1/8+1/2·7/24=5/24 (N19)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "prob_axioms",
            "L_path",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩種算法（逐步分布與路徑求和）給出同一個數；這是獨立的一致性核對，不是新的假設。",
            "en": "Two methods (step-by-step laws and the path sum) give the same number; this is an independent consistency check, not a new assumption."
          }
        },
        "leaf_basis": "prob_axioms + L_path + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "G3",
          "S06",
          "S08"
        ],
        "before_tex": "\\mu_0'=(0,0,1)",
        "after_tex": "\\mu_0'P=(0,\\tfrac13,\\tfrac23)\\neq\\mu_1=(\\tfrac38,\\tfrac12,\\tfrac18)",
        "relation": "implies",
        "rule": "L_one_step",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一個 P，初始律換成 μ_0'=(0,0,1)，一步後的分布就是 P 的第 3 行 (0,1/3,2/3)（N20），與 S08 的 μ_1 不同。所以 P 單獨一般不能決定完整的無條件時刻律或路徑律；這不表示每個個別機率都無法確定，例如 P_{13}=0 使任何以 1→3 開頭的路徑對所有初始律都為零。",
          "en": "With the same P but the initial law μ_0'=(0,0,1), the law after one step is row 3 of P, (0,1/3,2/3) (N20), which differs from μ_1 of S08. Thus P alone generally does not determine the complete unconditional marginal or path law. This does not make every individual probability undetermined: P_{13}=0 forces every path beginning 1→3 to have probability zero for all initial laws."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_one_step",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "一個反例就否定「P 單獨決定時刻律」。",
            "en": "One counterexample refutes 'P alone determines the time laws'."
          }
        },
        "leaf_basis": "L_one_step + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S03",
          "S08",
          "S09",
          "S11",
          "S12",
          "S15"
        ],
        "before_tex": "\\text{S03 to S15}",
        "after_tex": "P\\ \\text{row-stochastic};\\ \\mu_1,\\mu_2;\\ \\text{path probabilities}\\ \\tfrac1{16},\\ \\tfrac1{12};\\ \\mu_0\\ \\text{required}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：P 是行隨機矩陣（列和不必為 1）；μ_1=(3/8,1/2,1/8)、μ_2=(5/16,23/48,5/24)；兩條路徑的機率為 1/16 與 1/12；這些都需要初始律。本稿不宣稱平穩、不可約、遍歷或長期行為（MKO 假設 a2）。",
          "en": "Summary: P is row-stochastic (column sums need not be 1); μ_1=(3/8,1/2,1/8) and μ_2=(5/16,23/48,5/24); the two path probabilities are 1/16 and 1/12; all of these need the initial law. No stationarity, irreducibility, ergodicity or long-term behaviour is asserted (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S09",
            "S11",
            "S12"
          ],
          "explanation": {
            "zh-Hant": "只合取已證的有限步結論。",
            "en": "Only conjoins finite-step conclusions already proved."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "P 的三行和皆為 1、元素非負，是行隨機矩陣；列和 3/4、4/3、11/12 不受限制。由 μ_{n+1}=μ_nP 得 μ_1=(3/8,1/2,1/8)、μ_2=(5/16,23/48,5/24)；路徑機率 P(X_0=1,X_1=2,X_2=3)=1/16、P(X_0=2,X_1=3,X_2=3)=1/12，與路徑求和交叉核對一致。換成初始律 (0,0,1) 會得到不同的 μ_1，所以時刻律需要初始律。",
      "en": "The three rows of P are nonnegative and sum to 1, so P is row-stochastic; the column sums 3/4, 4/3, 11/12 are unrestricted. From μ_{n+1}=μ_nP, μ_1=(3/8,1/2,1/8) and μ_2=(5/16,23/48,5/24); the path probabilities are P(X_0=1,X_1=2,X_2=3)=1/16 and P(X_0=2,X_1=3,X_2=3)=1/12, consistent with the path-sum cross-check. With the initial law (0,0,1) a different μ_1 results, so time laws require an initial law."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "P 的列和不是 1，所以 P 不是轉移矩陣",
          "en": "The columns of P do not sum to 1, so P is not a transition matrix"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "資格條件是行非負且行和為 1（S02、S03）；列和沒有要求（S04）。",
          "en": "The qualification is nonnegative rows summing to 1 (S02, S03); there is no requirement on column sums (S04)."
        }
      },
      {
        "claim": {
          "zh-Hant": "只要知道 P，就能算出 P(X_1=2)",
          "en": "Knowing P alone, one can compute P(X_1=2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "初始律 (1/2,1/2,0) 給 1/2，初始律 (0,0,1) 給 1/3（S08、S15）。",
          "en": "The initial law (1/2,1/2,0) gives 1/2 while (0,0,1) gives 1/3 (S08, S15)."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(X_0=1,X_1=2,X_2=3)=P_{12}P_{23}=1/8",
          "en": "P(X_0=1,X_1=2,X_2=3)=P_{12}P_{23}=1/8"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "漏了初始質量 μ_0(1)=1/2；正確值是 1/16（S11）。1/8 是條件機率 P(X_1=2,X_2=3∣X_0=1)。",
          "en": "The initial mass μ_0(1)=1/2 is missing; the correct value is 1/16 (S11). 1/8 is the conditional probability P(X_1=2,X_2=3∣X_0=1)."
        }
      },
      {
        "claim": {
          "zh-Hant": "μ_1≠μ_0 且 μ_2≠μ_1，所以 μ_n 一定收斂到某個平穩分布",
          "en": "μ_1≠μ_0 and μ_2≠μ_1, so μ_n must converge to some stationary distribution"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限步計算不能推出長期收斂；MKO 假設 a2 明示不宣稱平穩、遍歷或長期收斂。",
          "en": "Finite-step computations do not imply long-term convergence; MKO assumption a2 explicitly asserts no stationarity, ergodicity or long-term convergence."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "matrix_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。",
          "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "mko_markov_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-markov-chain",
          "sha256": "713cef8fb3c1efc7627452592ab30fb69b9ebb7992c47aad78b9ae322ee7fd55",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "獨立來源追加式勘誤（使用者已採用於本地修訂；未合併或發布）：固定狀態次序後，P 是每行非負且精確和為 1 的指定轉移核。在 Pr(X_n=i)>0 時，P_{ij} 才與比值定義的 Pr(X_{n+1}=j∣X_n=i) 相同；零機率狀態的核行仍可指定，但不由 0/0 決定。本輪只處理單一時間齊次有限轉移核；完整的無條件時刻律與路徑律另需初始分布。原始引文與 canonical MKO 保留。",
          "en": "Independent append-only source erratum (owner-adopted in this local revision; not merged or published): with a fixed state order, P is a prescribed transition kernel whose rows are nonnegative and sum exactly to 1. When Pr(X_n=i)>0, P_ij agrees with the ratio-defined Pr(X_(n+1)=j ∣ X_n=i). A kernel row at a null state can still be prescribed, but is not determined by 0/0. This slice uses a single finite time-homogeneous kernel; full unconditional marginal and path laws additionally require an initial distribution. The original citation and canonical MKO are retained."
        }
      },
      {
        "id": "p_markov_property",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "以 P 為核、μ_0 為初始律的 Markov 鏈：P(X_0=i)=μ_0(i)；對每個 n、每個狀態 j，以及每個由 X_0 到 X_n 的取值所定義、機率為正且 X_n=i 的歷史事件 H，P(X_{n+1}=j∣H)=P_{ij}。特別地，P(X_n=i)>0 時 P(X_{n+1}=j∣X_n=i)=P_{ij}。",
          "en": "A Markov chain with kernel P and initial law μ_0: P(X_0=i)=μ_0(i); for every n, every state j and every history event H, fixing the values of X_0 through X_n with X_n=i and P(H)>0, P(X_{n+1}=j∣H)=P_{ij}. In particular, P(X_{n+1}=j∣X_n=i)=P_{ij} whenever P(X_n=i)>0."
        }
      },
      {
        "id": "p_cond_prob",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "條件機率：P(B)>0 時 P(A∣B)=P(A∩B)/P(B)，因此 P(A∩B)=P(B)P(A∣B)。",
          "en": "Conditional probability: for P(B)>0, P(A∣B)=P(A∩B)/P(B), hence P(A∩B)=P(B)P(A∣B)."
        }
      },
      {
        "id": "L_one_step",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "對每個 n 與狀態 j：μ_{n+1}(j)=Σ_i μ_n(i)P_{ij}，即 μ_{n+1}=μ_nP（行向量乘矩陣）。",
          "en": "For every n and state j: μ_{n+1}(j)=Σ_i μ_n(i)P_{ij}, that is μ_{n+1}=μ_nP (row vector times matrix)."
        }
      },
      {
        "id": "L_prob_vector",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "若 μ 是機率向量且 P 行隨機，則 μP 非負且分量和為 1。",
          "en": "If μ is a probability vector and P is row-stochastic, then μP is nonnegative and its entries sum to 1."
        }
      },
      {
        "id": "L_path",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2"
        ],
        "statement": {
          "zh-Hant": "對所有狀態 i_0、i_1、i_2：P(X_0=i_0,X_1=i_1,X_2=i_2)=μ_0(i_0)P_{i_0i_1}P_{i_1i_2}。",
          "en": "For all states i_0, i_1, i_2: P(X_0=i_0,X_1=i_1,X_2=i_2)=μ_0(i_0)P_{i_0i_1}P_{i_1i_2}."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03.1",
        "lhs": "1/2+1/2+0",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S03.2",
        "lhs": "1/4+1/2+1/4",
        "rhs": "1"
      },
      {
        "id": "N3",
        "step": "S03.3",
        "lhs": "0+1/3+2/3",
        "rhs": "1"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "1/2+1/4+0",
        "rhs": "3/4"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "1/2+1/2+1/3",
        "rhs": "4/3"
      },
      {
        "id": "N6",
        "step": "S04",
        "lhs": "0+1/4+2/3",
        "rhs": "11/12"
      },
      {
        "id": "N7",
        "step": "S05",
        "lhs": "1/2+1/2+0",
        "rhs": "1"
      },
      {
        "id": "N8",
        "step": "S07.1",
        "vars": [
          "m1",
          "m2",
          "m3"
        ],
        "degree": 1,
        "lhs": "(m1*(1/2)+m2*(1/4)+m3*0)+(m1*(1/2)+m2*(1/2)+m3*(1/3))+(m1*0+m2*(1/4)+m3*(2/3))",
        "rhs": "m1+m2+m3"
      },
      {
        "id": "N9",
        "step": "S08",
        "lhs": "1/2*1/2+1/2*1/4+0*0",
        "rhs": "3/8"
      },
      {
        "id": "N10",
        "step": "S08",
        "lhs": "1/2*1/2+1/2*1/2+0*1/3",
        "rhs": "1/2"
      },
      {
        "id": "N11",
        "step": "S08",
        "lhs": "1/2*0+1/2*1/4+0*2/3",
        "rhs": "1/8"
      },
      {
        "id": "N12",
        "step": "S09",
        "lhs": "3/8*1/2+1/2*1/4+1/8*0",
        "rhs": "5/16"
      },
      {
        "id": "N13",
        "step": "S09",
        "lhs": "3/8*1/2+1/2*1/2+1/8*1/3",
        "rhs": "23/48"
      },
      {
        "id": "N14",
        "step": "S09",
        "lhs": "3/8*0+1/2*1/4+1/8*2/3",
        "rhs": "5/24"
      },
      {
        "id": "N15",
        "step": "S09",
        "lhs": "5/16+23/48+5/24",
        "rhs": "1"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "1/2*1/2*1/4",
        "rhs": "1/16"
      },
      {
        "id": "N17",
        "step": "S12",
        "lhs": "1/2*1/4*2/3",
        "rhs": "1/12"
      },
      {
        "id": "N18",
        "step": "S14",
        "lhs": "1/2*1/2*1/4+1/2*1/2*1/4+1/2*1/4*2/3",
        "rhs": "5/24"
      },
      {
        "id": "N19",
        "step": "S14",
        "lhs": "1/2*(1/2*0+1/2*1/4+0*2/3)+1/2*(1/4*0+1/2*1/4+1/4*2/3)",
        "rhs": "5/24"
      },
      {
        "id": "N20",
        "step": "S15",
        "lhs": "0*1/2+0*1/2+1*1/3",
        "rhs": "1/3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 到 G3 指定的單一時間齊次有限轉移矩陣與初始律，只算有限步結果（MKO 假設 a1、a2）。不宣稱平穩分布、不可約、遍歷或長期收斂。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the single time-homogeneous finite transition matrix and initial law of G1 to G3, computing finite-step outcomes only (MKO assumptions a1, a2). No stationary distribution, irreducibility, ergodicity or long-term convergence is asserted. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "S=(1,2,3)",
        "固定次序的狀態",
        "the states in fixed order"
      ],
      [
        "P=(P_{ij})",
        "轉移矩陣：行為現在、列為下一步",
        "the transition matrix: rows current, columns next"
      ],
      [
        "\\mu_n",
        "時刻 n 的分布（行向量）",
        "the law at time n (a row vector)"
      ],
      [
        "\\mu_0=(\\tfrac12,\\tfrac12,0)",
        "初始律",
        "the initial law"
      ],
      [
        "(X_n)",
        "Markov 鏈",
        "the Markov chain"
      ],
      [
        "\\mu_0'",
        "比較用的另一個初始律 (0,0,1)",
        "an alternative initial law (0,0,1) for comparison"
      ]
    ],
    "md_error_controls": [
      {
        "title": "行和，不是列和 / Row sums, not column sums",
        "zh": "MKO 指出的誤解之一是把列和當行和。本例列和為 3/4、4/3、11/12，若誤用列和會錯誤地拒絕這個合法矩陣；反過來，一個列和都為 1 而某行和不為 1 的矩陣不是轉移矩陣。S07.1 的證明顯示，μP 仍是機率向量靠的正是行和為 1。",
        "en": "One misconception named by the MKO is to use column sums in place of row sums. Here the column sums are 3/4, 4/3, 11/12, so using them would wrongly reject this valid matrix; conversely, a matrix whose columns sum to 1 but some row does not is not a transition matrix. The proof of S07.1 shows that μP stays a probability vector precisely because rows sum to 1."
      },
      {
        "title": "初始律與完整的無條件分布 / Initial laws and complete unconditional distributions",
        "zh": "MKO 的另一個誤解是在沒有初始律時聲稱時刻分布與路徑機率。S15 中同一個 P 配上 (1/2,1/2,0) 與 (0,0,1) 兩個初始律，得到不同的一步分布 (3/8,1/2,1/8) 與 (0,1/3,2/3)。P 給指定轉移核；完整無條件路徑律還要初始質量，但某些個別值可先確定，例如含 P_{13}=0 的路徑機率對任意初始律都是 0。",
        "en": "Another misconception named by the MKO is to claim time laws and path probabilities without an initial law. In S15 the same P with initial laws (1/2,1/2,0) and (0,0,1) gives different one-step laws (3/8,1/2,1/8) and (0,1/3,2/3). P supplies the prescribed transition kernel. The full unconditional path law additionally needs initial mass, although some individual values are already fixed: every path product containing P_{13}=0 is zero for any initial law."
      },
      {
        "title": "行向量乘在左邊，索引不要倒置 / The row vector multiplies on the left; do not transpose indices",
        "zh": "一步律是 μP：(μP)_j=Σ_i μ(i)P_{ij}，對行索引 i 求和。若誤算成 Pμ^T，就變成對列求和，例如第 1 分量會變成 1/2·1/2+1/2·1/2+0·0=1/2 而不是 3/8。路徑中的 P_{23} 是從 2 到 3，不是 P_{32}=1/3。",
        "en": "The one-step law is μP: (μP)_j=Σ_i μ(i)P_{ij}, summing over the row index i. Computing Pμ^T by mistake sums along rows of P against μ, and the first entry would become 1/2·1/2+1/2·1/2+0·0=1/2 instead of 3/8. In a path, P_{23} means from 2 to 3, not P_{32}=1/3."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "S14 的 P² 交叉核對用到 (μ_0P)P=μ_0(P²)（矩陣乘法結合律），本稿未證明結合律；主要結論不依賴它。",
        "en": "The P² cross-check in S14 uses (μ_0P)P=μ_0(P²) (associativity of matrix multiplication), which is not proved here; the main conclusions do not depend on it."
      },
      {
        "zh-Hant": "狀態分類、平穩分布與首達時間（MKO 的 formal next）不在範圍內。",
        "en": "State classification, stationary distributions and first-passage times (the MKO's formal next) are out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "6f5cd3a61fc98aa7b2e492e044393761f66e3b4905bd398b1b0096fd7ef4ca68",
    "original_source_sha256": "7bf7d81f8e1bef72de3c74bbacf391a9a6ee30e2667b8069d65cf1aa1f8f28c7",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M061",
      "BR-M062",
      "BR-M063"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}