{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-strong-law-large-numbers-finite-valued-iid-01",
  "mko_id": "mko-strong-law-large-numbers-finite-valued-iid",
  "title": {
    "zh-Hant": "有限值IID的強律資格與固定n界線",
    "en": "Finite-valued IID strong-law qualification and fixed-n boundaries"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-strong-law-large-numbers-finite-valued-iid-01",
    "source": {
      "mko_id": "mko-strong-law-large-numbers-finite-valued-iid",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "14ad2e3637b7d7efdbf46bd6d640bf2637a5ab65b1f9ceafa6ff1a08565e9037",
      "english_sha256": "d7006cf0173ce35dcb02c4b0f80110b0124038bb778c5edc2903ca25a2569553"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "work"
    ],
    "goal": {
      "zh-Hant": "對齊named SLLN前提，並明示prefix不是a.s.證明。",
      "en": "Match the named SLLN hypotheses while separating prefix audits from a.s. proof."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "在一個已給定機率空間上，(X_i)_{i≥1}為無限IID序列，對任意有限不同indices的joint law都因子分解為Bernoulli(1/2)的乘積。此為全序列前提，不從有限資料估出。",
          "en": "On a supplied probability space, (X_i)_{i≥1} is an infinite IID sequence: every finite distinct-index joint law factors into Bernoulli(1/2) marginals. This is a whole-process premise, not inferred from finite data."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "work"
        ],
        "before_tex": "\\{0,1\\},\\quad p=(1/2,1/2)",
        "after_tex": "p_0,p_1\\ge0,\\quad p_0+p_1=1",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先完整驗證共同有限邊際law。",
          "en": "First validate the complete common finite marginal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "work"
          ],
          "explanation": {
            "zh-Hant": "這不單獨建立無限joint process。",
            "en": "This alone does not construct an infinite joint process."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01",
          "work"
        ],
        "before_tex": "\\mu=0(1/2)+1(1/2)",
        "after_tex": "\\mu=1/2,\\quad |X_i|\\le1\\text{ a.s.}",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有限support給共同mean與有界可積性。",
          "en": "Finite support supplies the common mean and bounded integrability."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "work"
          ],
          "explanation": {
            "zh-Hant": "boundedness綁所有i的共同law。",
            "en": "Boundedness concerns every i under the common law."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02",
          "work"
        ],
        "before_tex": "x_1,x_2,x_3,x_4=(1,0,1,1)",
        "after_tex": "\\bar x_4=3/4",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "指定一段path prefix的樣本平均只為3/4。",
          "en": "A specified path prefix has sample mean 3/4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "work"
          ],
          "explanation": {
            "zh-Hant": "prefix值不同於共同mean1/2，沒有矛盾。",
            "en": "A prefix value different from mean 1/2 is no contradiction."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03",
          "work"
        ],
        "before_tex": "S_4\\sim\\operatorname{Binomial}(4,1/2)",
        "after_tex": "P(S_4=k)=(1,4,6,4,1)_k/16\\quad(k=0,1,2,3,4)",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "利用IID完整列固定n=4 law。",
          "en": "Use IID to list the full fixed-n=4 law."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "work"
          ],
          "explanation": {
            "zh-Hant": "五格點沒有省略任何樣本和。",
            "en": "The five lattice points exhaust the possible sums."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S04",
          "work"
        ],
        "before_tex": "|S_4/4-1/2|\\ge1/4",
        "after_tex": "S_4\\in\\{0,1,3,4\\},\\quad P=10/16=5/8",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "固定n偏差事件含端點等號，精確加四項。",
          "en": "The fixed-n event includes threshold equality and sums exactly four terms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "work"
          ],
          "explanation": {
            "zh-Hant": "k=2不在事件，k=1、3在事件。",
            "en": "k=2 is excluded while k=1,3 are included."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S05",
          "work"
        ],
        "before_tex": "\\text{infinite IID common finite law}",
        "after_tex": "P\\!\\left(\\lim_{n\\to\\infty}\\frac1n\\sum_{i=1}^{n}X_i=1/2\\right)=1\\quad\\text{(named SLLN)}",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "套用明示的有限值SLLN：無限IID、有限support與共同mean均有。",
          "en": "Apply the explicit finite-valued SLLN: infinite IID, finite support and common mean are supplied."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S05",
            "work"
          ],
          "explanation": {
            "zh-Hant": "5/8不是這個極限事件機率；極限來自named theorem。",
            "en": "The value 5/8 is not the limit-event probability; the limit uses the named theorem."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06",
          "work"
        ],
        "before_tex": "P(\\lim_n\\bar X_n=1/2)=1",
        "after_tex": "\\bar X_n\\to1/2\\quad\\text{almost surely}",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "收斂事件機率一等同a.s.結論，不保證每個抽象path。",
          "en": "A probability-one convergence event gives the a.s. conclusion, not every abstract path."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06",
            "work"
          ],
          "explanation": {
            "zh-Hant": "全1事件E=交_{m≥1}{X1至Xm全為1}可測；有限prefix事件機率2^(-m)，由向下連續性得P(E)=lim 2^(-m)=0，不要求sample points本身是paths或singleton。",
            "en": "The all-ones event E is the intersection over m≥1 of measurable events{X1 through Xm all equal1}. Each finite-prefix event has probability 2^(-m); continuity from above gives P(E)=lim2^(-m)=0, without treating sample points as paths or singletons."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S07",
          "work"
        ],
        "before_tex": "X_i\\equiv c\\text{ a.s.}\\quad\\text{(boundary)}",
        "after_tex": "\\bar X_n=c\\text{ a.s. for every }n",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "零variance的有限值law仍合格；強律不需要CLT的非退化性。",
          "en": "A zero-variance finite law still qualifies; the strong law does not need CLT nondegeneracy."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S07",
            "work"
          ],
          "explanation": {
            "zh-Hant": "nonfinite support僅超出本slice，不表示一般強律一定失敗。",
            "en": "Nonfinite support is outside this slice, not a claim that every general strong law fails."
          }
        },
        "leaf_basis": "work"
      }
    ],
    "result": {
      "zh-Hant": "依named SLLN，平均a.s.趨1/2；固定n=4偏差機率5/8不證也不反證此極限。",
      "en": "By the named SLLN the mean tends a.s. to 1/2; fixed-n=4 deviation probability 5/8 neither proves nor refutes that limit."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "prefix平均3/4否定強律。",
          "en": "The prefix mean 3/4 refutes the strong law."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "極限不要求有限n等於mean。",
          "en": "A limit does not require equality at finite n."
        }
      },
      {
        "claim": {
          "zh-Hant": "強律必須σ²>0。",
          "en": "The strong law requires sigma²>0."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "常數IID序列也滿足此有限值強律。",
          "en": "Constant IID sequences satisfy this finite-valued strong law."
        }
      },
      {
        "claim": {
          "zh-Hant": "nonfinite support必違反所有SLLN。",
          "en": "Nonfinite support violates every SLLN."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "只是不屬本slice，其他強律有不同前提。",
          "en": "It is only outside this slice; other strong laws have different hypotheses."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "以明示infinite IID process certificate及normalized finite marginal law定義mu。SLLN結論為P(lim_{n→∞} n^{-1}sum_{i=1}^n X_i=mu)=1。exact families重算marginal mean、固定prefix的sample mean與固定n deviation event probability，但明示這些都不是a.s. convergence證書。non-IID與nonfinite-support只是本節scoped theorem不適用；finite prefix不證極限，wrong target不等於共同mean。",
          "en": "Assume an explicit infinite IID process and a normalized finite common marginal law with mean mu. The strong-law conclusion is P(lim_(n→infinity) n^(-1)sum_(i=1)^n X_i=mu)=1. Exact families recalculate the marginal mean, a fixed prefix sample mean and a complete fixed-n deviation-event probability. They expressly do not certify the probability-one limit. The infinite_iid certificate token is a declared process premise, not verification of infinite independence. Zero variance is allowed. Non-IID and nonfinite support are outside this finite-valued theorem scope, not counterexamples to every SLLN; a finite prefix does not establish a limit and a different target is not the common mean."
        },
        "source": {
          "mko_id": "mko-strong-law-large-numbers-finite-valued-iid",
          "sha256": "14ad2e3637b7d7efdbf46bd6d640bf2637a5ab65b1f9ceafa6ff1a08565e9037",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      },
      {
        "id": "work",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明示named theorem前提：給定機率空間上的無限IID有限值序列，其樣本平均a.s.收斂到共同mean；另假設binomial law及機率測度向下連續性。",
          "en": "Explicit named theorem prerequisite: an infinite IID finite-valued sequence on a given probability space has sample means converging a.s. to the common mean; also assume the binomial law and continuity from above of probability."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N01",
        "lhs": "(1+4+6+4+1)/16",
        "rhs": "1"
      },
      {
        "id": "N02",
        "lhs": "(1+4+4+1)/16",
        "rhs": "5/8"
      },
      {
        "id": "N03",
        "lhs": "(1+0+1+1)/4",
        "rhs": "3/4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "無限IID為前提；有限值SLLN為named theorem。",
      "en": "Infinite IID is a premise and finite-valued SLLN is a named theorem."
    },
    "md_symbols": [
      [
        "bar X_n",
        "樣本平均",
        "sample mean"
      ],
      [
        "μ",
        "共同邊際mean",
        "common marginal mean"
      ]
    ],
    "md_error_controls": [
      {
        "title": "過程與邊際 / Process and marginal",
        "zh": "有限law驗證與無限IID前提分別保留。",
        "en": "Keep finite-law validation separate from the infinite IID premise."
      },
      {
        "title": "事件層次 / Event layers",
        "zh": "固定n偏差5/8與lim事件機率1不可互換。",
        "en": "Do not interchange fixed-n deviation5/8 with probability 1 of a limit event."
      }
    ],
    "md_open_items": []
  },
  "revision": {
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    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
    "issue_ids": [
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    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
  }
}