第5講 回帰分析

説明変数が 2個の重回帰モデル

p=2 のときの重回帰モデルを考える.

yi=β0+β1xi1+β2xi2+ϵi=α0+β1(x1x¯1)+β2(x2x¯2)+ϵi
ただし,
α0=β0+β1x¯1+β2x¯2
ϵiN(0, σ)
ここで,
y=(y1y2yn)
β=(α0β1β2)
X=(1x11x¯1x12x¯21x21x¯1x22x¯21xn1x¯1xn2x¯2)
ϵ=(ϵ1ϵ2ϵn)
とすると,
y=Xβ+ϵ
ϵN(0,σ2In)
と書ける.

回帰式の推定

β^=(XX)1Xy
XX=(111x11x¯1x21x¯1xn1x¯1x12x¯2x22x¯2xn2x¯2)(1x11x¯1x12x¯21x21x¯1x22x¯21xn1x¯1xn2x¯2)=(ni(xi1x¯1)i(xi2x¯2)i(xi1x¯1)i(xi1x¯1)2i(xi1x¯1)(xi2x¯2)i(xi2x¯2)i(xi1x¯1)(xi2x¯2)i(xi2x¯2)2)=(n000S11S120S12S22)

|XX|=|n000S11S120S12S22|=n|S11S12S12S22|=n(S11S22S122)

(XX)1=(n000S11S120S12S22)1=1n(S11S22S122)(S11S22S122000nS22nS120nS12nS11)

Xy=(111x11x¯1x21x¯1xn1x¯1x12x¯2x22x¯2xn2x¯2)(y1y2yn)=(iyii(xi1x¯1)yii(xi2x¯2)yi)=(iyii(xi1x¯1)yiy¯i(xi1x¯1)i(xi2x¯2)yiy¯i(xi2x¯2))=(iyii(xi1x¯1)(yiy¯)i(xi2x¯2)yiy¯))=(ny¯S1yS2y)
β^=1n(S11S22S122)(S11S22S122000nS22nS120nS12nS11)(ny¯S1yS2y)=1n(S11S22S122)(n(S11S22S122)y¯n(S22S1yS12S2y)n(S12S1y+S11S2y))=1S11S22S122((S11S22S122)y¯S22S1yS12S2yS12S1y+S11S2y)
よって,
α^0=y¯

(β^1β^2)=1n(S11S22S122)(S22S1yS12S2yS12S1y+S11S2y)=1n(S11S22S122)(S22S12S12S11)(S1yS2y)=(S11S12S12S22)1(S1yS2y)