第 27225 題：有限變異數IID中央極限定理

原題與已知條件

明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-application-boundary；資料={"mode":"finite_histogram","model":{"values":[2,3],"probabilities":[[1,5],[4,5]],"process_certificate":"infinite_iid","n":6}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-application-boundary；資料={"mode":"finite_histogram","model":{"values":[2,3],"probabilities":[[1,5],[4,5]],"process_certificate":"infinite_iid","n":6}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-central-limit-theorem-finite-variance-iid

逐步推導

source roles = [clt-application-boundary]

derived = [[mode, mean, variance, n, evidence_scope, limit_proof_claim, boundary]]

source roles = [infinite_iid]

derived = [ASSUME infinite sequence is IID with this common marginal]

CLT ORIGINAL conditional marginal ORIGINAL law mass sum0 := sum(0, 1/5)

CLT ORIGINAL conditional marginal ORIGINAL law mass sum0 = 1/5

CLT ORIGINAL conditional marginal ORIGINAL law mass sum1 := sum(1/5, 4/5)

CLT ORIGINAL conditional marginal ORIGINAL law mass sum1 = 1

source roles = [CLT ORIGINAL conditional marginal, [2, 3], [[1, 5], [4, 5]]]

derived = [[[2, 1/5], [3, 4/5]], 1]

CLT ORIGINAL conditional marginal ORIGINAL value times probability0 := mul(2, 1/5)

CLT ORIGINAL conditional marginal ORIGINAL value times probability0 = 2/5

CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0 := sum(0, 2/5)

CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0 = 2/5

CLT ORIGINAL conditional marginal exact value square0 := mul(2, 2)

CLT ORIGINAL conditional marginal exact value square0 = 4

CLT ORIGINAL conditional marginal ORIGINAL square times probability0 := mul(4, 1/5)

CLT ORIGINAL conditional marginal ORIGINAL square times probability0 = 4/5

CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0 := sum(0, 4/5)

CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0 = 4/5

CLT ORIGINAL conditional marginal ORIGINAL value times probability1 := mul(3, 4/5)

CLT ORIGINAL conditional marginal ORIGINAL value times probability1 = 12/5

CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1 := sum(2/5, 12/5)

CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1 = 14/5

CLT ORIGINAL conditional marginal exact value square1 := mul(3, 3)

CLT ORIGINAL conditional marginal exact value square1 = 9

CLT ORIGINAL conditional marginal ORIGINAL square times probability1 := mul(9, 4/5)

CLT ORIGINAL conditional marginal ORIGINAL square times probability1 = 36/5

CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1 := sum(4/5, 36/5)

CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1 = 8

CLT ORIGINAL conditional marginal ORIGINAL mean square := mul(14/5, 14/5)

CLT ORIGINAL conditional marginal ORIGINAL mean square = 196/25

CLT ORIGINAL conditional marginal ORIGINAL second minus mean square := sub(8, 196/25)

CLT ORIGINAL conditional marginal ORIGINAL second minus mean square = 4/25

source roles = [CLT ORIGINAL conditional marginal, [[2, 1/5], [3, 4/5]]]

derived = [14/5, 8, 4/25]

source roles = [finite_histogram, 14/5, 4/25, 6]

derived = [finite_histogram, 14/5, 4/25, 6, one_finite_histogram, 否, finite_histogram_does_not_prove_distributional_limit]

逐步反向核對

• S019 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S018 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S017 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL second minus mean square，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL mean square，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL square times probability1，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal exact value square1，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL value times probability1，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL square times probability0，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal exact value square0，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL value times probability0，不以小數近似替代。
• S005 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S004 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL law mass sum1，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL law mass sum0，不以小數近似替代。
• S002 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S001 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。

結論

(finite_histogram, 14/5, 4/25, 6, one_finite_histogram, 否, finite_histogram_does_not_prove_distributional_limit)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。