{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-iid-sequence-01", "mko_id": "mko-iid-sequence", "title": { "zh-Hant": "獨立同分布序列：由 Bernoulli(1/3) 柱集乘積公式證明 IID", "en": "IID sequence: proving IID from the Bernoulli(1/3) cylinder product formula" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-iid-sequence-01", "source": { "mko_id": "mko-iid-sequence", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "bac79da4634be201a10f08ec4489946a636db5ebc2e628c0b1830b743f4ddbd0", "english_sha256": "9fc20d5d2102056dfc01584c16ff07f2a621a4acee305e717a7c68822a5dfa48" }, "mode": "theorem_proof", "audience_prerequisites": [ "prob_axioms", "power_laws", "distributive", "finite_sum_rules", "nat_induction", "mul_comm" ], "goal": { "zh-Hant": "設 p=1/3，柱集證書規定每個有限 0-1 word w（長度 n、含 k 個 1）的機率為 p^k(1−p)^{n−k}。證明：此族一致且每層總和為 1；由此（配合引用的 Kolmogorov 存在定理）得到 {0,1}^ℕ 上的機率；每個 X_i 的完整邊際都是 Bernoulli(1/3)；每個有限子族都獨立。再以三個對照說明：檢查過的 prefix、pairwise 獨立與相同期望都不足以宣稱無限 IID。", "en": "Let p=1/3, and let the cylinder certificate give each finite binary word w (length n, with k ones) the probability p^k(1−p)^{n−k}. Prove that this family is consistent with total mass 1 at every level; obtain from it (with the cited Kolmogorov existence theorem) a probability on {0,1}^ℕ; show that every X_i has the complete marginal Bernoulli(1/3) and that every finite subfamily is independent. Then use three contrasts to show that a checked prefix, pairwise independence and equal expectations are each insufficient for an infinite IID claim." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "p=1/3，為 [0,1] 中的精確有理數（MKO 假設 a1）；q(1)=p=1/3、q(0)=1−p=2/3。", "en": "p=1/3, an exact rational in [0,1] (MKO assumption a1); q(1)=p=1/3 and q(0)=1−p=2/3." } }, { "id": "G2", "statement": { "zh-Hant": "符號乘積律證書：對每個 n≥1 與每個 word w=(w_i)_{i=1}^{n}∈{0,1}^n，P(X_i=w_i 對所有 1≤i≤n)=c(w)=p^{k(w)}(1−p)^{n−k(w)}，k(w) 為 w 中 1 的個數（MKO 假設 a2）。", "en": "Symbolic product-law certificate: for every n≥1 and every word w=(w_i)_{i=1}^{n}∈{0,1}^n, P(X_i=w_i for all 1≤i≤n)=c(w)=p^{k(w)}(1−p)^{n−k(w)}, where k(w) is the number of ones in w (MKO assumption a2)." } }, { "id": "G3", "statement": { "zh-Hant": "定義：(X_i)_{i≥1} 是 IID，若 (a) 同分布：所有 X_i 有相同的完整邊際分布；(b) 獨立：對每個有限的相異指標集 J 與值 (a_j)_{j∈J}，P(X_j=a_j 對所有 j∈J)=Π_{j∈J} P(X_j=a_j)。", "en": "Definition: (X_i)_{i≥1} is IID if (a) identically distributed: all X_i have the same complete marginal law; and (b) independent: for every finite set J of distinct indices and values (a_j)_{j∈J}, P(X_j=a_j for all j∈J)=Π_{j∈J} P(X_j=a_j)." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2" ], "before_tex": "c(101)", "after_tex": "\\left(\\tfrac13\\right)^2\\left(\\tfrac23\\right)^1=\\tfrac{2}{27}", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "先讀懂證書：w=101 長度 n=3，含 k=2 個 1，所以 c(101)=p^2(1−p)^1=(1/3)^2(2/3)=2/27（N1）。公式只依賴 1 的個數，所以 110、011 也是 2/27（N9）。", "en": "First read the certificate: w=101 has length n=3 and k=2 ones, so c(101)=p^2(1−p)^1=(1/3)^2(2/3)=2/27 (N1). The formula depends only on the number of ones, so 110 and 011 also get 2/27 (N9)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation", "G2" ], "explanation": { "zh-Hant": "精確有理數求值，可獨立重算；兩邊是同一數。", "en": "An exact rational evaluation, independently recomputable; both sides are the same number." } }, "leaf_basis": "numeric_evaluation + power_laws" }, { "id": "S02", "uses": [ "G1", "G2" ], "before_tex": "c(w)=p^{k(w)}(1-p)^{n-k(w)}", "after_tex": "c(w)=\\prod_{i=1}^{n}q(w_i)", "relation": "equivalent", "rule": "L_product_form", "conditions": [], "explanation": { "zh-Hant": "依引理 L_product_form（在 S02.1 證明），柱集機率是各位置機率 q(w_i) 的乘積。這個乘積形式是之後一致性與邊際計算的關鍵。", "en": "By Lemma L_product_form (proved in S02.1), the cylinder probability is the product of the position probabilities q(w_i). This product form is the key to the consistency and marginal computations that follow." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_product_form" ], "explanation": { "zh-Hant": "兩式對每個 w 相等，雙向成立。", "en": "The two expressions are equal for every w, so it holds both ways." } }, "leaf_basis": "mul_comm + mul_assoc + power_laws" }, { "id": "S02.1", "uses": [ "G1", "mul_comm", "power_laws" ], "before_tex": "\\prod_{i=1}^{n}q(w_i)", "after_tex": "p^{k(w)}(1-p)^{n-k(w)}", "relation": "equivalent", "rule": "power_laws", "conditions": [], "explanation": { "zh-Hant": "乘積中 w_i=1 的 k(w) 個位置各貢獻 q(1)=p，w_i=0 的 n−k(w) 個位置各貢獻 q(0)=1−p。依乘法交換與結合律把相同因子集中，得 p^{k(w)}(1−p)^{n−k(w)}。S02.1 證得引理 L_product_form。", "en": "In the product, the k(w) positions with w_i=1 each contribute q(1)=p, and the n−k(w) positions with w_i=0 each contribute q(0)=1−p. Commutativity and associativity gather equal factors, giving p^{k(w)}(1−p)^{n−k(w)}. S02.1 proves Lemma L_product_form." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mul_comm", "mul_assoc", "power_laws" ], "explanation": { "zh-Hant": "重排有限乘積不改變其值；兩向都只是重排。", "en": "Rearranging a finite product does not change its value; both directions are rearrangements." } }, "leaf_basis": "mul_comm + mul_assoc + power_laws" }, { "id": "S03", "uses": [ "S02" ], "before_tex": "c(w0)+c(w1)", "after_tex": "c(w)", "relation": "equivalent", "rule": "L_consistency", "conditions": [], "explanation": { "zh-Hant": "依引理 L_consistency（在 S03.1 證明），把長度 n+1 的 word 對最後一個位置求和，回到長度 n 的機率。這正是 Kolmogorov 存在定理要求的一致性。", "en": "By Lemma L_consistency (proved in S03.1), summing words of length n+1 over the last position gives back the probability at length n. This is exactly the consistency required by the Kolmogorov existence theorem." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_consistency" ], "explanation": { "zh-Hant": "恆等式，雙向成立。", "en": "An identity, so it holds both ways." } }, "leaf_basis": "distributive + add_comm + mul_one" }, { "id": "S03.1", "uses": [ "S02", "distributive" ], "before_tex": "c(w)q(0)+c(w)q(1)", "after_tex": "c(w)\\big((1-p)+p\\big)=c(w)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "由乘積形式（S02），c(wa)=c(w)q(a)。所以 c(w0)+c(w1)=c(w)(1−p)+c(w)p=c(w)((1−p)+p)=c(w)·1=c(w)（N3 以多項式恆等式核對）。S03.1 證得引理 L_consistency。", "en": "By the product form (S02), c(wa)=c(w)q(a). So c(w0)+c(w1)=c(w)(1−p)+c(w)p=c(w)((1−p)+p)=c(w)·1=c(w) (N3 checks this as a polynomial identity). S03.1 proves Lemma L_consistency." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "add_inverse", "mul_one" ], "explanation": { "zh-Hant": "分配律與 (1−p)+p=1 都是恆等式，雙向成立。", "en": "Distributivity and (1−p)+p=1 are identities, so the step is two-way." } }, "leaf_basis": "distributive + add_inverse + add_comm + mul_one" }, { "id": "S04", "uses": [ "S03", "G1" ], "before_tex": "\\sum_{w\\in\\{0,1\\}^n}c(w)", "after_tex": "1\\quad\\text{for every}\\ n\\ge1", "relation": "induction_step", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "對 n 歸納。基底 n=1：c(0)+c(1)=2/3+1/3=1（N8）。歸納步：長度 n+1 的 word 恰是長度 n 的 w 接上 0 或 1，所以 Σ_{|v|=n+1} c(v)=Σ_{|w|=n}(c(w0)+c(w1))=Σ_{|w|=n} c(w)=1（S03 與歸納假設）。例 n=3：8 個 word 的總和 8/27+12/27+6/27+1/27=1（N2、N10）。所以每一層都是機率分布。", "en": "Induct on n. Base n=1: c(0)+c(1)=2/3+1/3=1 (N8). Step: the words of length n+1 are exactly the words w of length n followed by 0 or 1, so Σ_{|v|=n+1} c(v)=Σ_{|w|=n}(c(w0)+c(w1))=Σ_{|w|=n} c(w)=1 (S03 and the induction hypothesis). Example n=3: the 8 words sum to 8/27+12/27+6/27+1/27=1 (N2, N10). So every level is a probability distribution." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "nat_induction", "finite_sum_rules" ], "explanation": { "zh-Hant": "全稱結論來自歸納法，不是來自 n=3 的數值核對；N2 只是一個實例。", "en": "The universal conclusion comes from induction, not from the numerical check at n=3; N2 is one instance only." } }, "leaf_basis": "nat_induction + finite_sum_rules + numeric_evaluation" }, { "id": "S05", "uses": [ "S03", "S04" ], "before_tex": "(\\mu_n)_{n\\ge1}\\ \\text{consistent, each a probability}", "after_tex": "\\exists!\\ P\\ \\text{on}\\ \\{0,1\\}^{\\mathbb{N}}\\ \\text{with cylinder probabilities}\\ c(w)", "relation": "theorem_application", "rule": "kolmogorov_ext", "conditions": [ "\\text{consistency (S03)}", "\\text{total mass 1 (S04)}" ], "explanation": { "zh-Hant": "令 μ_n(w)=c(w)。S04 給出每個 μ_n 是 {0,1}^n 上的機率，S03 給出一致性。依引用的 Kolmogorov 存在定理，{0,1}^ℕ 上存在唯一機率使柱集機率為 c(w)；X_i 是第 i 個座標。本稿只核對定理的前提，不證明定理本身。", "en": "Let μ_n(w)=c(w). S04 shows each μ_n is a probability on {0,1}^n, and S03 gives consistency. By the cited Kolmogorov existence theorem there is a unique probability on {0,1}^ℕ with cylinder probabilities c(w); X_i is the i-th coordinate. This guide checks only the theorem's hypotheses and does not prove the theorem itself." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "kolmogorov_ext" ], "explanation": { "zh-Hant": "定理需要每層是機率且族一致，兩者都已證；定理本身是引用（citation_verified）。", "en": "The theorem needs each level to be a probability and the family to be consistent, both proved; the theorem itself is cited (citation_verified)." } }, "leaf_basis": "kolmogorov_ext" }, { "id": "S06", "uses": [ "S05", "G2", "prob_axioms" ], "before_tex": "P(X_j=a_j\\ \\forall j\\in J),\\ n=\\max J", "after_tex": "\\sum_{w\\in\\{0,1\\}^n,\\ w_j=a_j\\ (j\\in J)}c(w)", "relation": "equivalent", "rule": "prob_axioms", "conditions": [ "J\\ \\text{finite, nonempty}" ], "explanation": { "zh-Hant": "取 n=max J。事件「X_j=a_j 對所有 j∈J」是長度 n、在 J 上取指定值的 word 的柱集之聯集；不同 word 的柱集互斥。依機率的有限可加性，其機率等於這些 c(w) 之和。", "en": "Take n=max J. The event 'X_j=a_j for all j∈J' is the union of the cylinders of the words of length n that take the prescribed values on J, and different words give disjoint cylinders. By finite additivity of probability, its probability is the sum of these c(w)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "prob_axioms", "set_operations_def" ], "explanation": { "zh-Hant": "事件分解是集合等式，可加性給出機率相等；兩邊同值。", "en": "The decomposition of the event is a set equality, and additivity gives equal probabilities; both sides have the same value." } }, "leaf_basis": "prob_axioms + set_operations_def + finite_sum_rules" }, { "id": "S07", "uses": [ "S06", "S02" ], "before_tex": "\\sum_{w_J=a}\\prod_{i=1}^{n}q(w_i)", "after_tex": "\\prod_{j\\in J}q(a_j)", "relation": "equivalent", "rule": "L_marginal", "conditions": [], "explanation": { "zh-Hant": "用乘積形式（S02）改寫和中的 c(w)，再依引理 L_marginal（在 S07.1、S07.2 證明）：自由位置的和全部化為因子 q(0)+q(1)=1，只剩指定位置的因子。", "en": "Rewrite each c(w) in the sum by the product form (S02), then apply Lemma L_marginal (proved in S07.1 and S07.2): summing over each free position produces a factor q(0)+q(1)=1, and only the factors of the prescribed positions remain." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_marginal" ], "explanation": { "zh-Hant": "引理是等式，雙向成立。", "en": "The lemma is an equality, so the step holds both ways." } }, "leaf_basis": "distributive + finite_sum_rules + nat_induction" }, { "id": "S07.1", "uses": [ "S02" ], "before_tex": "J=\\{1,2,\\text{up to }n\\}", "after_tex": "\\sum=\\prod_{i=1}^{n}q(a_i)", "relation": "induction_base", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "對自由位置的個數 f=n−|J| 歸納。基底 f=0：所有位置都被指定，和中只有一個 word w=a，其值為 Π_{i=1}^{n} q(a_i)=Π_{j∈J} q(a_j)。", "en": "Induct on the number f=n−|J| of free positions. Base f=0: every position is prescribed, the sum has the single word w=a, and its value is Π_{i=1}^{n} q(a_i)=Π_{j∈J} q(a_j)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "nat_induction" ], "explanation": { "zh-Hant": "只是單項和，無需其他前提。", "en": "A one-term sum; nothing else is needed." } }, "leaf_basis": "nat_induction + finite_sum_rules" }, { "id": "S07.2", "uses": [ "S07.1", "S02", "distributive" ], "before_tex": "f\\ \\text{free positions},\\ i\\ \\text{one of them}", "after_tex": "\\sum=\\big(q(0)+q(1)\\big)\\cdot\\big(\\text{sum with}\\ f-1\\ \\text{free positions}\\big)=\\prod_{j\\in J}q(a_j)", "relation": "induction_step", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "歸納步：設 f≥1，取一個自由位置 i。按 w_i 的值把和分成兩組；每項含因子 q(w_i)，依分配律提出：和=Σ_{b∈{0,1}} q(b)·T=(q(0)+q(1))T=T，其中 T 是去掉位置 i 後、其餘位置的乘積在同樣約束下的和，只有 f−1 個自由位置。依歸納假設 T=Π_{j∈J} q(a_j)。S07.1 與 S07.2 合起來證得引理 L_marginal。", "en": "Step: let f≥1 and pick a free position i. Split the sum by the value of w_i; each term has the factor q(w_i), which distributivity pulls out: sum=Σ_{b∈{0,1}} q(b)·T=(q(0)+q(1))T=T, where T is the sum, under the same constraints, of the products over the other positions, with only f−1 free positions. By the induction hypothesis T=Π_{j∈J} q(a_j). S07.1 and S07.2 together prove Lemma L_marginal." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07.1", "distributive", "nat_induction" ], "explanation": { "zh-Hant": "提出因子需要乘積形式（S02）：q(w_i) 不依賴其他位置；若機率不是乘積形式，這一步就失敗（見 S11）。", "en": "Pulling out the factor needs the product form (S02): q(w_i) does not depend on the other positions; without the product form this step fails (see S11)." } }, "leaf_basis": "nat_induction + distributive + finite_sum_rules + add_inverse" }, { "id": "S08", "uses": [ "S06", "S07", "G1" ], "before_tex": "J=\\{i\\}", "after_tex": "P(X_i=1)=\\tfrac13,\\ P(X_i=0)=\\tfrac23\\quad\\text{for every}\\ i", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "取 J={i}：P(X_i=a)=q(a)，即 P(X_i=1)=1/3、P(X_i=0)=2/3，對任意 i 相同；1/3+2/3=1（N8）。這是完整的邊際分布（兩個值的機率），不只是期望 E[X_i]=1/3。所以 (a) 同分布成立。", "en": "Take J={i}: P(X_i=a)=q(a), so P(X_i=1)=1/3 and P(X_i=0)=2/3, the same for every i; 1/3+2/3=1 (N8). This is the complete marginal law (the probabilities of both values), not just the mean E[X_i]=1/3. So (a), identical distribution, holds." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S07", "forall_intro" ], "explanation": { "zh-Hant": "i 是任意的，證明沒有對 i 加條件，所以對所有 i 成立；不是逐一核對有限多個 i。", "en": "i is arbitrary and the proof puts no condition on it, so it holds for all i; this is not a check of finitely many i." } }, "leaf_basis": "forall_intro + forall_elim" }, { "id": "S09", "uses": [ "S06", "S07", "S08", "G3" ], "before_tex": "J\\ \\text{finite},\\ (a_j)_{j\\in J}", "after_tex": "P(X_j=a_j\\ \\forall j\\in J)=\\prod_{j\\in J}P(X_j=a_j)", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "先分情況。若 J=∅，沒有座標限制的事件是整個 Ω，機率為 1；右邊依空乘積慣例也為 1，所以獨立性等式成立，不取 max∅。若 J 為非空有限集，才依 S06 取 n=max J；由 S06、S07 左邊等於 Π_{j∈J} q(a_j)，再由 S08 每個 q(a_j)=P(X_j=a_j)，所以聯合機率等於邊際之積。兩個分支涵蓋所有有限 J，因此 (b) 對每個有限子族成立，不只是成對。", "en": "First split cases. If J is empty, the event with no coordinate constraints is all of Ω and has probability 1; the right side is also 1 by the empty-product convention. The independence equality holds without taking max of the empty set. If J is finite and nonempty, S06 takes n=max J. Then S06 and S07 give Π_{j∈J} q(a_j) on the left, and S08 identifies each q(a_j) with P(X_j=a_j), giving the product of marginals. These cases cover every finite J, so (b) holds for every finite subfamily, not just pairs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S07", "S08", "forall_intro" ], "explanation": { "zh-Hant": "量化範圍是所有有限 J（MKO 段落 1）；空集由 Ω 的機率與空乘積均為 1 的分支處理；非空集才用 S06 的 max J 論證，所以沒有遺漏空子族。", "en": "The quantifier ranges over all finite J (MKO paragraph 1); the empty case uses probability 1 for Ω and product 1 for the empty family; only the nonempty case uses max J in S06, so the empty subfamily is not omitted." } }, "leaf_basis": "forall_intro + eq_replace" }, { "id": "S10", "uses": [ "S09", "S08" ], "before_tex": "P(X_2=1,\\ X_5=0)", "after_tex": "\\tfrac13\\cdot\\tfrac23=\\tfrac29", "relation": "implies", "rule": "forall_elim", "conditions": [], "explanation": { "zh-Hant": "實例：J={2,5}、a_2=1、a_5=0。由 S09 得 (1/3)(2/3)=2/9。直接由 S06 驗算：n=5，自由位置 1、3、4 共 2^3=8 個 word，和為 (1/3)(2/3)((2/3)+(1/3))^3=2/9（N4）。", "en": "Instance: J={2,5}, a_2=1, a_5=0. S09 gives (1/3)(2/3)=2/9. Direct check from S06: n=5, the free positions 1, 3, 4 give 2^3=8 words, summing to (1/3)(2/3)((2/3)+(1/3))^3=2/9 (N4)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "numeric_evaluation" ], "explanation": { "zh-Hant": "只是一個實例，不取代 S09 的全稱證明。", "en": "One instance only; it does not replace the universal proof of S09." } }, "leaf_basis": "forall_elim + numeric_evaluation" }, { "id": "S11", "uses": [ "G3" ], "before_tex": "Z_1,Z_2\\ \\text{IID Bernoulli}(\\tfrac13),\\ Z_i=Z_1\\ (i\\ge3)", "after_tex": "P(Z_1=1,Z_3=1)=\\tfrac13\\neq\\tfrac19", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "對照（prefix）：令 Z_1、Z_2 為獨立 Bernoulli(1/3)，其後每個 Z_i 都等於 Z_1。長度 2 的 prefix 滿足乘積公式，每個 Z_i 的邊際也是 Bernoulli(1/3)；但 P(Z_1=1,Z_3=1)=P(Z_1=1)=1/3，而 P(Z_1=1)P(Z_3=1)=1/9（N5），所以 J={1,3} 的獨立性失敗。檢查過的有限 prefix 不能證明無限序列 IID。", "en": "Contrast (prefix): let Z_1, Z_2 be independent Bernoulli(1/3) and let every later Z_i equal Z_1. The length-2 prefix satisfies the product formula and every Z_i has marginal Bernoulli(1/3); but P(Z_1=1,Z_3=1)=P(Z_1=1)=1/3, while P(Z_1=1)P(Z_3=1)=1/9 (N5), so independence fails for J={1,3}. A checked finite prefix cannot prove that the infinite sequence is IID." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "quantifier_negation" ], "explanation": { "zh-Hant": "一個失敗的有限子族就否定 (b)；這也顯示 S09 必須對所有 J 證明。", "en": "One failing finite subfamily refutes (b); it also shows why S09 must be proved for all J." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S12", "uses": [ "G3" ], "before_tex": "U_1,U_2\\ \\text{IID Bernoulli}(\\tfrac12),\\ U_3=U_1\\oplus U_2", "after_tex": "\\text{pairwise independent, but}\\ P(U_1=U_2=U_3=1)=0\\neq\\tfrac18", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "對照（pairwise）：U_3 是 U_1、U_2 的模 2 和。四種 (U_1,U_2) 各有機率 1/4。對任意 a、b，P(U_1=a,U_3=b)=P(U_1=a,U_2=a⊕b)=1/4=(1/2)(1/2)（N7），(U_2,U_3) 同理，且三者邊際都是 Bernoulli(1/2)。但 U_1=U_2=1 推出 U_3=0，所以 P(U_1=U_2=U_3=1)=0≠1/8（N6）。pairwise 獨立加同分布仍不是 IID。", "en": "Contrast (pairwise): U_3 is the sum of U_1 and U_2 modulo 2. The four values of (U_1,U_2) each have probability 1/4. For any a, b, P(U_1=a,U_3=b)=P(U_1=a,U_2=a⊕b)=1/4=(1/2)(1/2) (N7); likewise for (U_2,U_3), and all three marginals are Bernoulli(1/2). But U_1=U_2=1 forces U_3=0, so P(U_1=U_2=U_3=1)=0≠1/8 (N6). Pairwise independence plus identical distribution is still not IID." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "quantifier_negation" ], "explanation": { "zh-Hant": "三個成對都核對了，失敗的是三元子族；(b) 量化每個有限子族。", "en": "All three pairs are checked; the failure is in the three-element subfamily, and (b) quantifies over every finite subfamily." } }, "leaf_basis": "quantifier_negation + prob_axioms + numeric_evaluation" }, { "id": "S13", "uses": [ "S08", "G3" ], "before_tex": "V\\equiv\\tfrac13", "after_tex": "E[V]=E[X_1]=\\tfrac13,\\ \\text{but}\\ P(V=1)=0\\neq\\tfrac13", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "對照（相同期望）：常數 V≡1/3 的期望與 X_1 相同，都是 1/3，但 P(V=1)=0，而 P(X_1=1)=1/3。所以期望相同不推出分布相同；(a) 要比較完整邊際（S08）。", "en": "Contrast (equal expectations): the constant V≡1/3 has the same mean 1/3 as X_1, but P(V=1)=0 while P(X_1=1)=1/3. So equal expectations do not give equal laws; (a) compares complete marginals (S08)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "G3" ], "explanation": { "zh-Hant": "一個值的機率不同就足以否定同分布。", "en": "A single value with a different probability suffices to refute identical distribution." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S14", "uses": [ "S05", "S08", "S09", "S11", "S12", "S13" ], "before_tex": "\\text{S05, S08, S09, S11 to S13}", "after_tex": "(X_i)_{i\\ge1}\\ \\text{IID Bernoulli}(\\tfrac13)", "relation": "implies", "rule": "mko_iid_def", "conditions": [], "explanation": { "zh-Hant": "在 S05 的機率下，每個 X_i 的完整邊際都是 Bernoulli(1/3)（S08），每個有限子族都獨立（S09），所以 (X_i) 是 IID；這依賴的是對所有 n 與所有 word 成立的符號證書，而不是某段 prefix。三個對照（S11 到 S13）說明較弱的檢查都不足夠。", "en": "Under the probability of S05, every X_i has complete marginal Bernoulli(1/3) (S08) and every finite subfamily is independent (S09), so (X_i) is IID; this rests on the symbolic certificate valid for all n and all words, not on any prefix. The three contrasts (S11 to S13) show that the weaker checks are insufficient." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S05", "S08", "S09", "mko_iid_def" ], "explanation": { "zh-Hant": "目標 G：(X_i) 是 IID Bernoulli(1/3)。足夠條件 H：機率存在（S05，引用）、(a)（S08）、(b)（S09）；H⇒G 由定義 G3。H 由 G1、G2 正向證成，只有 S05 依賴引用的定理。", "en": "Goal G: (X_i) is IID Bernoulli(1/3). Sufficient condition H: the probability exists (S05, cited), (a) (S08) and (b) (S09); H⇒G by the definition G3. H is proved forwards from G1 and G2; only S05 relies on the cited theorem." } }, "leaf_basis": "and_intro_elim + def_unfold" } ], "result": { "zh-Hant": "柱集證書 c(w)=(1/3)^{k}(2/3)^{n−k} 等於各位置機率之積，族一致且每層總和為 1；配合引用的 Kolmogorov 存在定理，它在 {0,1}^ℕ 上定出唯一機率。對任何有限 J，P(X_j=a_j, j∈J)=Π q(a_j)，所以每個 X_i 的完整邊際是 Bernoulli(1/3)，每個有限子族獨立：(X_i) 是 IID。檢查過的 prefix、pairwise 獨立與相同期望各有反例，不能代替證書。", "en": "The cylinder certificate c(w)=(1/3)^{k}(2/3)^{n−k} equals the product of the position probabilities, the family is consistent and has total mass 1 at every level; with the cited Kolmogorov existence theorem it determines a unique probability on {0,1}^ℕ. For every finite J, P(X_j=a_j, j∈J)=Π q(a_j), so every X_i has complete marginal Bernoulli(1/3) and every finite subfamily is independent: (X_i) is IID. A checked prefix, pairwise independence and equal expectations each have counterexamples and cannot replace the certificate." }, "negative_controls": [ { "claim": { "zh-Hant": "長度至多 2 的 word 都符合乘積公式，所以整個序列是 IID", "en": "All words of length at most 2 satisfy the product formula, so the whole sequence is IID" }, "expected": "reject", "reason": { "zh-Hant": "S11 的 Z 序列滿足此條件，但 P(Z_1=1,Z_3=1)=1/3≠1/9。", "en": "The sequence Z of S11 satisfies this, yet P(Z_1=1,Z_3=1)=1/3≠1/9." } }, { "claim": { "zh-Hant": "兩兩獨立且同分布的隨機變數就是 IID", "en": "Pairwise independent, identically distributed random variables are IID" }, "expected": "reject", "reason": { "zh-Hant": "S12 的模 2 和例子：P(U_1=U_2=U_3=1)=0≠1/8。", "en": "The modulo-2 sum example of S12: P(U_1=U_2=U_3=1)=0≠1/8." } }, { "claim": { "zh-Hant": "每個 X_i 的期望都是 1/3，所以它們同分布", "en": "Every X_i has mean 1/3, so they are identically distributed" }, "expected": "reject", "reason": { "zh-Hant": "常數 1/3 的期望也是 1/3，但分布不同（S13）；同分布要比較完整邊際。", "en": "The constant 1/3 also has mean 1/3 but a different law (S13); identical distribution compares complete marginals." } }, { "claim": { "zh-Hant": "word 110 與 101 的 1 位置不同，所以機率不同", "en": "The words 110 and 101 have their ones in different places, so their probabilities differ" }, "expected": "reject", "reason": { "zh-Hant": "公式只依賴 1 的個數：兩者都是 (1/3)^2(2/3)=2/27（N9）。", "en": "The formula depends only on the number of ones: both are (1/3)^2(2/3)=2/27 (N9)." } } ], "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": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "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": "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": "nat_induction", "kind": "prerequisite", "statement": { "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。", "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "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": "mul_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "乘法結合律：(uv)w=u(vw)。", "en": "Associativity of multiplication: (uv)w=u(vw)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "mul_one", "kind": "prerequisite", "statement": { "zh-Hant": "乘法單位元：1·u=u。", "en": "Multiplicative identity: 1·u=u." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "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": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "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": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "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": "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": "mko_iid_def", "kind": "cited", "source": { "mko_id": "mko-iid-sequence", "sha256": "bac79da4634be201a10f08ec4489946a636db5ebc2e628c0b1830b743f4ddbd0", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "本輪用 Bernoulli(p) 無限 product-law／cylinder symbolic certificate 建立 IID 序列：任意有限 0-1 word 的機率為 p 的一數次方乘 (1-p) 的零數次方。有限 prefix 只能稱 finite IID family，不能獨自證無限序列。", "en": "This slice uses a symbolic Bernoulli(p) infinite product-law cylinder certificate: each finite binary word has probability p raised to its number of ones times (1−p) raised to its number of zeros. A finite prefix can only certify a finite IID family; by itself it does not certify the infinite sequence." } }, { "id": "kolmogorov_ext", "kind": "cited", "source": { "reference": "Rick Durrett, Probability: Theory and Examples, 5th edition, Appendix A.3, Theorem A.3.1 and Example A.3.2", "original_reference": "Kolmogorov existence (extension) theorem; e.g. P. Billingsley, Probability and Measure, 3rd ed., Wiley 1995, Section 36", "url": "https://sites.math.duke.edu/~rtd/PTE/PTE5_011119.pdf#page=472", "locator": "Kolmogorov extension for countable real coordinate products", "checked_date": "2026-10-19", "verification": "verified_replacement", "verification_note": "For binary coordinates, embed {0,1} into R. The extension is supported on {0,1}^N since the complement is a countable union of null coordinate events; restrict the measure." }, "needs_citation": false, "statement": { "zh-Hant": "若對每個 n 給定 {0,1}^n 上的機率 μ_n，且族一致（刪去最後一個座標時 μ_{n+1} 的像是 μ_n），則 {0,1}^ℕ（配乘積 σ 代數）上存在唯一的機率，其前 n 個座標的分布為 μ_n。", "en": "If for every n a probability μ_n on {0,1}^n is given and the family is consistent (deleting the last coordinate maps μ_{n+1} to μ_n), then there is a unique probability on {0,1}^ℕ (with the product σ-algebra) whose law on the first n coordinates is μ_n." } }, { "id": "L_product_form", "kind": "lemma", "proved_in": [ "S02.1" ], "statement": { "zh-Hant": "c(w)=Π_{i=1}^{n} q(w_i)。", "en": "c(w)=Π_{i=1}^{n} q(w_i)." } }, { "id": "L_consistency", "kind": "lemma", "proved_in": [ "S03.1" ], "statement": { "zh-Hant": "對每個 word w：c(w0)+c(w1)=c(w)，其中 wa 表示在 w 後面接上 a。", "en": "For every word w: c(w0)+c(w1)=c(w), where wa means w followed by a." } }, { "id": "L_marginal", "kind": "lemma", "proved_in": [ "S07.1", "S07.2" ], "statement": { "zh-Hant": "對 n≥1、J⊆{1,2,直到 n} 與值 (a_j)_{j∈J}：Σ_{w∈{0,1}^n, w_j=a_j (j∈J)} Π_{i=1}^{n} q(w_i)=Π_{j∈J} q(a_j)。", "en": "For n≥1, J⊆{1, 2, up to n} and values (a_j)_{j∈J}: Σ_{w∈{0,1}^n, w_j=a_j (j∈J)} Π_{i=1}^{n} q(w_i)=Π_{j∈J} q(a_j)." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S01", "lhs": "(1/3)**2*(2/3)", "rhs": "2/27" }, { "id": "N2", "step": "S04", "lhs": "(2/3)**3+3*(1/3)*(2/3)**2+3*(1/3)**2*(2/3)+(1/3)**3", "rhs": "1" }, { "id": "N3", "step": "S03.1", "vars": [ "c", "p" ], "degree": 2, "lhs": "c*(1-p)+c*p", "rhs": "c" }, { "id": "N4", "step": "S10", "lhs": "(1/3)*(2/3)*((2/3)+(1/3))**3", "rhs": "2/9" }, { "id": "N5", "step": "S11", "lhs": "1/3", "relation": "!=", "rhs": "(1/3)*(1/3)" }, { "id": "N6", "step": "S12", "lhs": "0", "relation": "!=", "rhs": "(1/2)**3" }, { "id": "N7", "step": "S12", "lhs": "(1/2)*(1/2)", "rhs": "1/4" }, { "id": "N8", "step": "S08", "lhs": "1/3+2/3", "rhs": "1" }, { "id": "N9", "step": "S01", "lhs": "(1/3)*(1/3)*(2/3)", "rhs": "(1/3)*(2/3)*(1/3)" }, { "id": "N10", "step": "S04", "lhs": "2**3", "rhs": "1+3+3+1" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 範圍內工作：p 為 [0,1] 中的精確有理數（此處 1/3），cylinder word 為有限 0-1 序列（假設 a1）；無限 IID 的宣稱以明示的符號乘積律證書為依據（假設 a2）。對每個有限子族的獨立性與同分布給出完整證明；無限乘積測度的存在與唯一性引用 Kolmogorov 存在定理，本稿不證明。不處理大數律或中心極限定理；不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: p is an exact rational in [0,1] (here 1/3), and cylinder words are finite binary sequences (assumption a1); the infinite IID claim rests on the explicit symbolic product-law certificate (assumption a2). Independence and identical distribution of every finite subfamily are proved in full; existence and uniqueness of the infinite product measure are cited from the Kolmogorov existence theorem and not proved here. Laws of large numbers and the central limit theorem are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "p=\\tfrac13", "Bernoulli 參數", "the Bernoulli parameter" ], [ "q(1)=p,\\ q(0)=1-p", "單一位置的機率", "the one-position probabilities" ], [ "c(w)", "word w 的柱集機率 p^{k}(1−p)^{n−k}", "the cylinder probability p^{k}(1−p)^{n−k} of the word w" ], [ "k(w)", "w 中 1 的個數", "the number of ones in w" ], [ "J,\\ a=(a_j)_{j\\in J}", "有限指標集與指定值", "a finite index set and prescribed values" ] ], "md_error_controls": [ { "title": "prefix 不能證無限序列 / A prefix does not certify the infinite sequence", "zh": "MKO 指出的誤解：檢查固定 prefix 就宣稱無限 IID。S11 的 Z 序列前兩項完全符合乘積公式，但 Z_3=Z_1 使 J={1,3} 的獨立性失敗（1/3≠1/9）。證書必須對所有長度 n 與所有 word 以符號形式成立（G2）。", "en": "The MKO's misconception: declaring an infinite IID sequence after checking a fixed prefix. The first two terms of the sequence Z in S11 match the product formula exactly, yet Z_3=Z_1 breaks independence for J={1,3} (1/3≠1/9). The certificate must hold symbolically for every length n and every word (G2)." }, { "title": "pairwise 獨立不夠 / Pairwise independence is not enough", "zh": "MKO 的另一個誤解。S12 的 U_1、U_2、U_3 兩兩獨立且同分布，但三者的聯合機率 0≠1/8。獨立性要對每個有限子族核對，S09 正是對任意有限 J 證明。", "en": "Another MKO misconception. U_1, U_2, U_3 in S12 are pairwise independent and identically distributed, yet their joint probability is 0≠1/8. Independence must hold for every finite subfamily, which is what S09 proves for arbitrary finite J." }, { "title": "相同期望不是同分布 / Equal expectations are not equal laws", "zh": "MKO 段落 0：同分布要核完整邊際，不是只看平均。常數 1/3 與 Bernoulli(1/3) 期望相同，但 P(V=1)=0≠1/3（S13）。", "en": "MKO paragraph 0: identical distribution compares complete marginals, not just means. The constant 1/3 and Bernoulli(1/3) have the same mean, yet P(V=1)=0≠1/3 (S13)." } ], "md_open_items": [ { "zh-Hant": "{0,1}^ℕ 上乘積測度的存在與唯一性引用已核對的 Durrett《Probability: Theory and Examples》第 5 版附錄 A.3 定理 A.3.1 及例 A.3.2，取支撐於 {0,1} 的可數座標特例；本稿核對其前提（S03、S04），不重證定理。原 Billingsley 版次未查閱，已另存沿革。大數律與中心極限定理仍不在範圍內。", "en": "Existence and uniqueness of the measure on {0,1}^ℕ are cited from the checked Theorem A.3.1 and Example A.3.2 in Appendix A.3 of Durrett's Probability: Theory and Examples, fifth edition, specialized to countably many coordinates supported on {0,1}. The guide checks the hypotheses in S03 and S04, not the theorem's proof. The uninspected Billingsley edition is preserved separately as provenance. Laws of large numbers and the central limit theorem remain out of scope." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "ba5654672fd1ce1b24fc7ce24c4d5916ec5632235207e7fb67cdc22967ffafa9", "original_source_sha256": "49bd8455452eda4395e6e1383315934f9e90e96eda5273f92b4fdc2148171d52", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M096" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }