[理工] [機率] variance
設 X1 為 uniform 分佈 0 < x1 < 2
X2 為 Exp 分佈, 參數為 1/3 -> 所以E[X2] = 1/(1/3) = 3
X3 ~ G(2,2)
令 X = 0.3*X1 + 0.3*X2 + 0.4*X3
VAR(X) = ?
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 163.18.104.233
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抱歉,X小改,實際上題目是給fx(x)
實在不好打,簡化打出的話是 fx(x) = 0.3*fx1(x1) + 0.3*fx2(x2) + 0.4*fx3(x3)
注 X1 X2 X3是我分析完給的變數,實際上是給方程式,所以不知道是否indep
※ 編輯: ofd168 來自: 163.18.104.233 (01/18 21:10)
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01/18 21:33, , 6F
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但 E(X^2) 該如何算呢??
※ 編輯: ofd168 來自: 163.18.104.233 (01/18 21:40)
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E[X] = E[0.3*X1 + 0.3*X2 + 0.4*X3]
= 0.3*E[X1] + 0.3*E[X2] + 0.4*E[X3]
E[X^2] = ∫x^2*fx(x)dx
= 0.3*∫x^2*fx1(x)dx + 0.3*∫x^2*fx2(x)dx + 0.4*∫x^2*fx3(x)dx
= 0.3*E[X1^2] + 0.3*E[X2^2] + 0.4*E[X3^2]
= 0.3*(VAR(X1)+E[X1]^2) + 0.3*(VAR(X2)+E[X2]^2)
+ 0.4*(VAR(X3)+E[X3]^2)
VAR(X) = E[X^2] - E[X]^2
= { 0.3*VAR(X1) + 0.3*VAR(X2) + 0.4*VAR(X3)
+ 0.3*E[X1]^2 + 0.3*E[X2]^2 + 0.4*E[X3]^2}
- {0.3*E[X1]^2 + 0.3*E[X2]^2 + 0.4*E[X3]^2}
= 0.3*VAR(X1) + 0.3*VAR(X2) + 0.4*VAR(X3)
= 0.3*(1/3) + 0.3*(9) + 0.4*(2)
= 0.1 + 2.7 + 0.8 = 3.8
這樣對嗎?? 我是這樣想的
※ 編輯: ofd168 來自: 163.18.104.233 (01/18 22:28)
※ 編輯: ofd168 來自: 163.18.104.233 (01/18 22:29)
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懂了!! 謝謝k大
※ 編輯: ofd168 來自: 163.18.104.233 (01/18 23:25)