91影视

The best linear predictor of $Y$with respect to $X_1$and $X_2$is equal to $a+b{X}_1+c{X}_2$.

Let's suppose that $Y,X_1,X_2$are random variables $惟鈫�\mathrm{鈩�}$, where$(惟,\mathcal{F},P)$is the probability space. In that case, we have that

$E\left[{\left(Y-\left(a+b{X}_1+c{X}_2\right)\right)}^2\right]={鈭�}_{惟}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

Applying partial derivation of that expression respective to $a$

$={鈭�}_{惟}\frac{鈭�}{鈭�a}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

$=-2{鈭�}_{惟}\left(Y-a-b{X}_1-c{X}_2\right)dP$

$=-2E\left(Y-a-b{X}_1-c{X}_2\right)$

With respect to $b$,

$\frac{鈭�}{鈭�b}E\left[{\left(Y-\left(a+b{X}_1+c{X}_2\right)\right)}^2\right]=\frac{鈭�}{鈭�b}{鈭�}_{惟}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

$={鈭�}_{惟}\frac{鈭�}{鈭�b}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

$=-2{鈭�}_{惟}\left(Y-a-b{X}_1-c{X}_2\right)X_{1}dP$

$=-2E\left(\left(Y-a-b{X}_1-c{X}_2\right)X_1\right)$

$=-2{鈭�}_{惟}\left(Y-a-b{X}_1-c{X}_2\right)X_{2}dP$With respect to $c$,

$\frac{鈭�}{鈭�c}E\left[{\left(Y-\left(a+b{X}_1+c{X}_2\right)\right)}^2\right]=\frac{鈭�}{鈭�c}{鈭�}_{惟}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

$={鈭�}_{惟}\frac{鈭�}{鈭�c}{\left(Y-a-b{X}_1-c{X}_2\right)}^{2}dP$

$=-2E\left(\left(Y-a-b{X}_1-c{X}_2\right)X_2\right)$

Setting these partial derivations equal to zero gives us conditions

$E\left(Y{X}_1\right)=aE\left(X_1\right)+bE\left(X_1^2\right)+cE\left(X_1{X}_2\right)$

$E\left(Y{X}_2\right)=aE\left(X_2\right)+bE\left(X_1{X}_2\right)+cE\left(X_2^2\right)$

Implies that,

$a=E\left(Y\right)-bE\left(X_1\right)-cE\left(X_2\right)$

$b=\frac{Cov\left(X_1,X_2\right)Cov\left(Y,X_2\right)-Cov\left(Y,X_1\right)Var\left(X_2\right)}{Cov{\left(X_1,X_2\right)}^2-Var\left(X_1\right)Var\left(X_2\right)}$

$c=\frac{Cov\left(X_1,X_2\right)Cov\left(Y,X_1\right)-Cov\left(Y,X_2\right)Var\left(X_1\right)}{Cov{\left(X_1,X_2\right)}^2-Var\left(X_1\right)Var\left(X_2\right)}$