The best linear predictor of Ywith respect toX1and X2is equal to a+bX1+cX2, where a, b, and care chosen to minimizeEY-a+bX1+cX22 Determine a, b, and c.

Short Answer

Expert verified

a=E(Y)-bEX1-cEX2

b=CovX1,X2CovY,X2-CovY,X1VarX2CovX1,X22-VarX1VarX2

c=CovX1,X2CovY,X2-CovY,X2VarX1CovX1,X22-VarX1VarX2

Step by step solution

01

Given Information

The best linear predictor of Ywith respect to X1and X2is equal to a+bX1+cX2.

02

Explanation

Let's suppose that Y,X1,X2are random variables Ω, where(Ω,F,P)is the probability space. In that case, we have that

EY-a+bX1+cX22=ΩY-a-bX1-cX22dP

Applying partial derivation of that expression respective to a

aEY-a+bX1+cX22=aΩY-a-bX1-cX22dP

=ΩaY-a-bX1-cX22dP

=-2ΩY-a-bX1-cX2dP

=-2EY-a-bX1-cX2

03

Explanation

With respect to b,

bEY-a+bX1+cX22=bΩY-a-bX1-cX22dP

=ΩbY-a-bX1-cX22dP

=-2ΩY-a-bX1-cX2X1dP

=-2EY-a-bX1-cX2X1

04

Explanation

=-2ΩY-a-bX1-cX2X2dPWith respect to c,

cEY-a+bX1+cX22=cΩY-a-bX1-cX22dP

=ΩcY-a-bX1-cX22dP

=-2EY-a-bX1-cX2X2

05

Explanation

Setting these partial derivations equal to zero gives us conditions

E(Y)=a+bEX1+cEX2

EYX1=aEX1+bEX12+cEX1X2

EYX2=aEX2+bEX1X2+cEX22

06

Final Answer

Implies that,

a=E(Y)-bEX1-cEX2

b=CovX1,X2CovY,X2-CovY,X1VarX2CovX1,X22-VarX1VarX2

c=CovX1,X2CovY,X1-CovY,X2VarX1CovX1,X22-VarX1VarX2

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