第 19183 題：抽樣分布

原題與已知條件

共同有限樣本空間、逐theta模型law與單一theta-independent total statistic。操作=sampling-pushforward-at-parameter；資料={"parameter_count":2,"outcome_count":4,"laws":[[[1,15],[2,15],[3,15],[9,15]],[[2,19],[3,19],[4,19],[10,19]]],"statistic":[0,0,8,9],"parameter":1}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 共同有限樣本空間、逐theta模型law與單一theta-independent total statistic。操作=sampling-pushforward-at-parameter；資料={"parameter_count":2,"outcome_count":4,"laws":[[[1,15],[2,15],[3,15],[9,15]],[[2,19],[3,19],[4,19],[10,19]]],"statistic":[0,0,8,9],"parameter":1}。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-sampling-distribution

逐步推導

source operation=sampling-pushforward-at-parameter;original snapshot and mathematics retained

explicit ordered source output fields=[parameter, law, total_mass, model_qualified, certificate]

sampling parameter0 full law mass := sum(1/15, 2/15, 1/5, 3/5)

sampling parameter0 full law mass = 1

sampling:original parameter SLOT=0;same outcome carrier Fin(4);ALL aligned law masses=[1/15, 2/15, 1/5, 3/5]

full sum=1;all masses nonnegative=是;qualified probability law;zero-mass outcomes retained

sampling parameter1 full law mass := sum(2/19, 3/19, 4/19, 10/19)

sampling parameter1 full law mass = 1

sampling:original parameter SLOT=1;same outcome carrier Fin(4);ALL aligned law masses=[2/19, 3/19, 4/19, 10/19]

full sum=1;all masses nonnegative=是;qualified probability law;zero-mass outcomes retained

sampling:parameter SLOT carrier Fin(2);common outcome carrier Fin(4);ONE total original statistic array=[0, 0, 8, 9]

T takes only the observation index,not the unknown parameter;SAME function at all parameter slots;its sampling law may still vary with the slot

sampling parameter1 statistic value0 full fibre mass := sum(2/19, 3/19)

sampling parameter1 statistic value0 full fibre mass = 5/19

FIXED original parameter SLOT=1;statistic value=0;FULL original fibre outcome indices=[0, 1];aligned law masses=[2/19, 3/19]

pushforward mass at value0=5/19;retain this row even if mass0

sampling parameter1 statistic value8 full fibre mass := sum(4/19)

sampling parameter1 statistic value8 full fibre mass = 4/19

FIXED original parameter SLOT=1;statistic value=8;FULL original fibre outcome indices=[2];aligned law masses=[4/19]

pushforward mass at value8=4/19;retain this row even if mass0

sampling parameter1 statistic value9 full fibre mass := sum(10/19)

sampling parameter1 statistic value9 full fibre mass = 10/19

FIXED original parameter SLOT=1;statistic value=9;FULL original fibre outcome indices=[3];aligned law masses=[10/19]

pushforward mass at value9=10/19;retain this row even if mass0

sampling parameter1 complete pushforward mass := sum(5/19, 4/19, 10/19)

sampling parameter1 complete pushforward mass = 1

fixed parameter SLOT=1;ALL increasingly ordered statistic values=[0, 8, 9];disjoint complete fibres=[[0, 1], [2], [3]]

pushforward law=[[0, 5/19], [8, 4/19], [9, 10/19]];total mass=1;ANY set of statistic values has preimage equal to the union of its fibres

逐步反向核對

• S014 · 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。
• S013 · premise_and_result_audit — 以精確有理數重算 sampling parameter1 complete pushforward mass，不以小數近似替代。
• S012 · 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。
• S011 · premise_and_result_audit — 以精確有理數重算 sampling parameter1 statistic value9 full fibre mass，不以小數近似替代。
• S010 · 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。
• S009 · premise_and_result_audit — 以精確有理數重算 sampling parameter1 statistic value8 full fibre mass，不以小數近似替代。
• S008 · 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。
• S007 · premise_and_result_audit — 以精確有理數重算 sampling parameter1 statistic value0 full fibre mass，不以小數近似替代。
• S006 · 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。
• S005 · 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。
• S004 · premise_and_result_audit — 以精確有理數重算 sampling parameter1 full law mass，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 — 以精確有理數重算 sampling parameter0 full law mass，不以小數近似替代。
• 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

(1, [[0, 5/19], [8, 4/19], [9, 10/19]], 1, 是, pushforward_sum_over_estimator_fibres)

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。