✔ 最佳答案
σ12=E(x12)–E(x1)2σ22=E(x22)–E(x2)2ρσ1σ2 =E[(x1 –E(x1)) ( x2 –E(x2))]= E(x1 x2)–E[x1 E( x2)]–E[E(x1)x2] + E(x1) E( x2)= E(x1 x2)–E(x1) E( x2)For x =ω1 x1 +ω2 x2total variance of x= E[(ω1 x1 +ω2 x2)2] –E(ω1 x1 +ω2 x2) 2= E(ω1 2x12 +2ω1ω2 x1 x2 +ω22 x22)–[E(ω1 x1)+E(ω2 x2)] 2= E(ω1 2x12 +2ω1ω2 x1 x2 +ω22 x22)–E(ω1 x1) 2–2 E(ω1 x1) E(ω2 x2)–E(ω2 x2) 2= E(ω1 2x12 +2ω1ω2 x1 x2 +ω22 x22)–[E(ω1 x1)+E(ω2 x2)] 2= E(ω1 2x12) + 2 E(ω1ω2x1 x2) + E(ω22 x22)–E(ω1 x1) 2–2 E(ω1 x1) E(ω2 x2)–E(ω2 x2) 2=ω1 2 E(x12) + 2ω1ω2 E(x1 x2) +ω22 E(x22)–E(ω1 x1) 2–2ω1ω2 E(x1) E( x2)–ω2 E(x2) 2=ω1 2 [ E(x12) –E(x1) 2] + 2ω1ω2 [E(x1 x2)–E(x1) E( x2)] + ω2 2 [ E(x22) –E(x2) 2]=ω1 2σ12 + 2ω1ω2σ1σ2ρ+ ω2 2σ22
2011-02-24 14:35:36 補充：
上標和下標格式顯示不出來，比較難看，不好意思。
2011-02-24 14:43:35 補充：
σ1^2=E(x1^2)–E(x1)^2
σ2^2=E(x2^2)–E(x2)^2
ρσ1σ2 =E[(x1 –E(x1)) ( x2 –E(x2))]
= E(x1 x2)–E[x1 E( x2)]–E[E(x1)x2] + E(x1) E( x2)
= E(x1 x2)–E(x1) E( x2)
For x =ω1 x1 +ω2 x2
2011-02-24 14:43:46 補充：
total variance of x
= E[(ω1 x1 +ω2 x2)^2] –E(ω1 x1 +ω2 x2) ^2
= E(ω1^2 x1^2 +2ω1ω2 x1 x2 +ω2^2 x2^2)–[E(ω1 x1)+E(ω2 x2)] ^2
= E(ω1^2 x1^2 +2ω1ω2 x1 x2 +ω2^2 x2^2)–E(ω1 x1) ^2–2 E(ω1 x1) E(ω2 x2)–E(ω2 x2) ^2
= E(ω1 2x1^2 +2ω1ω2 x1 x2 +ω2^2 x2^2)–[E(ω1 x1)+E(ω2 x2)] ^2
2011-02-24 14:43:51 補充：
= E(ω1 2x1^2) + 2 E(ω1ω2x1 x2) + E(ω2^2 x2^2)–E(ω1 x1) ^2–2 E(ω1 x1) E(ω2 x2)–E(ω2 x2) ^2
=ω1^2 E(x1^2) + 2ω1ω2 E(x1 x2) +ω2^2 E(x2^2)–E(ω1 x1)^2–2ω1ω2 E(x1) E( x2)–ω2 ^2E(x2)^2
=ω1^2 [ E(x1^2) –E(x1)^2] + 2ω1ω2 [E(x1 x2)–E(x1) E( x2)] + ω2^2 [ E(x2^2) –E(x2)^ 2]
=ω1^2 σ1^2 + 2ω1ω2σ1σ2ρ+ ω2^2 σ2^2
2011-02-25 10:58:23 補充：
字數已超轉往意見欄
2011-02-25 18:21:00 補充：
There is no such linear relationship unless ω1, ω2, σ1 and σ2 are constant.
When every random variable involved is doubled, the total variance becomes 4 times the original value but the coefficient of correlation is still the same. This is not a linear relationship at all.
