{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","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-covariance"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-covariance-01","title":{"zh-Hant":"協方差：對稱性、仿射規則與「零協方差不蘊含獨立」","en":"Covariance: symmetry, the affine rule, and why zero covariance does not imply independence"},"summary":{"zh-Hant":"在同一個精確有限 joint law 上：(1) 證明 Cov(Y,X)=Cov(X,Y)；(2) 證明 Cov(aX+b,cY+d)=ac·Cov(X,Y)；(3) 證明 X、Y 獨立時 Cov(X,Y)=0，因此非零協方差可否證獨立；(4) 對 X 均勻取 −1、0、1 且 Y=X² 的 joint law，算出 E[X]、E[Y]、E[XY] 與 Cov(X,Y)=0，並證明 X、Y 不獨立。","en":"On one exact finite joint law: (1) prove Cov(Y,X)=Cov(X,Y); (2) prove Cov(aX+b,cY+d)=ac·Cov(X,Y); (3) prove that independence of X and Y gives Cov(X,Y)=0, so a nonzero covariance refutes independence; (4) for the joint law with X uniform on −1, 0, 1 and Y=X², compute E[X], E[Y], E[XY] and Cov(X,Y)=0, and prove that X and Y are not independent."},"mko_ids":["mko-covariance"]}],"title":"概念推導導讀"}