推 Kevintsaitsa: 謝謝！！！您真是太用心了！ 04/18 23:54
※ 引述《Kevintsaitsa (kevin)》之銘言：
: https://i.imgur.com/F8dHteR.jpg
: 想問5.10的證明怎麼證
: 雖然今天早上已經問過一題了
: 但這完全沒有頭緒……
: 沒有詳解太痛苦了…….
From Hogg McKean Craig "Introduction to Mathematical Statistics" 6ed
(p102~p104) 再加上我自己的感想
(a)
Note that E(Y) = E(E(Y|X))=E(a0+b0X) = a0+b0E(X)
2 2
and E(XY) = E(E(XY|X))=E(XE(Y|X))=E(X(a0+b0X))=E(a0X+b0X ) =a0E(X)+boE(X )
cov(X,Y) = E(XY)-E(X)E(Y) => E(XY) = Cov(X,Y) + E(X)E(Y)
=> μ =a0+b0μ ------------------------------(1)
Y X
2 2
σ +μ μ = a0μ +b0(σ +μ ) ----(2)
XY Y X Y X X
2
(2)-μ *(1) => σ =b0σ
X XY X
2
=> b0 = σ / σ 代入(1)
XY X
=> a0 = μ -b0μ
Y X
2
(b) Var(Y|x) = ∫(y-E(Y|x)) fy|x(x,y) dy
2
= ∫[y-μ -b0(x-μ )] fxy(x,y)dy/ fx(x)
Y X
兩邊對X取期望值 =>
2
c = E(Var(Y|X)) = ∫∫[(y-μ ) -b0(x-μ ) ] fxy(x,y) dydx
Y X
2 2 2
=∫∫[(y-μ ) -2b0(y-μ )(x-μ )+b0 (x-μ )]fxy(x,y) dydx
Y Y X X
2 2 2
=E[(Y-μ )] -2b0E[(X-μ )(Y-μ )] + b0 E[(X-μ )]
Y X Y X
2 2 2 2 2
= σ -2σ /σ σ + (σ /σ ) σ
Y XY X XY XY X X
2 2 2 2 2
= σ -2σ / σ + σ /σ
Y XY X XY X
2 2 2
= σ - σ /σ
Y XY X
2 2 2
= σ - (ρ σσ )/σ
Y XY X Y X
2 2 2
= σ - ρ σ
Y XY Y
2 2
= σ (1-ρ )
Y XY
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