第 19289 題：弱大數律

原題與已知條件

IID通用證書、共同有限mean與variance及exact Chebyshev tail。操作=wlln-variance-average；資料={"mean":[13,14],"variance":[14,15],"n":18,"iid_certificate":"symbolic_all_finite_subfamilies"}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — IID通用證書、共同有限mean與variance及exact Chebyshev tail。操作=wlln-variance-average；資料={"mean":[13,14],"variance":[14,15],"n":18,"iid_certificate":"symbolic_all_finite_subfamilies"}。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-weak-law-large-numbers

逐步推導

source operation=wlln-variance-average;original snapshot and mathematics retained

explicit ordered source output fields=[mean, variance, n, average_variance, certificate]

original common finite mean=13/14;POSITIVE finite variance=14/15;whole-sequence iid_certificate=symbolic_all_finite_subfamilies

ASSUME for EVERY finite subfamily the declared independence and identical law/common moments;marker declares a premise,NOT independent evidence that a real sequence satisfies it

for ANY positive integer n:sample_mean_n=(X1+...+Xn)/n;every Xi has the SAME mean and variance;distinct Xi,Xj independent

linearity gives E(sample_mean_n)=mean;centered products have zero expectation for i!=j;Var(sample_mean_n)=(n*variance)/n^2=variance/n

WLLN selected sum of diagonal variances := mul(18, 14/15)

WLLN selected sum of diagonal variances = 84/5

WLLN selected average denominator squared := mul(18, 18)

WLLN selected average denominator squared = 324

WLLN selected average variance := div(84/5, 324)

WLLN selected average variance = 7/135

逐步反向核對

• S006 · premise_and_result_audit — 以精確有理數重算 WLLN selected average variance，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 WLLN selected average denominator squared，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 WLLN selected sum of diagonal variances，不以小數近似替代。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(13/14, 14/15, 18, 7/135, independence_gives_variance_sigma_squared_over_n)

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。