{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-finite-joint-distribution"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-finite-joint-distribution-01","title":{"zh-Hant":"有限聯合分布：驗證聯合質量表並由同一 joint law 聚合邊際與事件","en":"Finite joint distribution: validating a joint mass table and aggregating marginals and events from one joint law"},"summary":{"zh-Hant":"給定支撐 S={(0,0),(0,1),(1,0),(1,1),(2,1)} 與質量 p(0,0)=1/6、p(0,1)=1/3、p(1,0)=1/4、p(1,1)=1/12、p(2,1)=1/6。驗證 (S,p) 是二維有限聯合分布；由同一個 joint law 算出 X 與 Y 的邊際、矩形事件 P(X∈{1,2}, Y=1) 與單點質量 P(X=2, Y=0)；最後構造一個邊際相同但不同的 joint law，說明兩個邊際一般不能反推聯合分布。","en":"Given the support S={(0,0),(0,1),(1,0),(1,1),(2,1)} with masses p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, verify that (S,p) is a two-dimensional finite joint distribution; from this one joint law compute the marginals of X and Y, the rectangle event P(X∈{1,2}, Y=1) and the single-pair mass P(X=2, Y=0); finally construct a different joint law with the same marginals, showing that two marginals do not in general determine the joint distribution."},"mko_ids":["mko-finite-joint-distribution"]}],"title":"Concept derivation guides"}