Q.7.39 - A First Course in Probability

Question

The best quadratic predictor of$Y$with respect to $X$ is a + b$X + c{X}_2$, where a, b, and c are chosen to minimize $E\left[\right(Y - (a + bX + c{X}_2)\left)2\right]$. Determine $a$, $b$, and $c$. 

Step-by-Step Solution

Answer

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

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

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

Step 1: Given Information

The best quadratic predictor of $Y$ with respect to $X$ is $a + bX + c{X}_2$. 

Step 2: Explanation

This exercise is a very special case of the previous exercise where $X_1=X$ and $X_2=X^2$. In that case, we have

$Cov\left(X_1,X_2\right)=Cov\left(X,X^2\right)=E\left(X^3\right)-E\left(X\right)E\left(X^2\right)$

$Cov\left(Y,X_1\right)=Cov(Y,X)=E\left(XY\right)-E\left(X\right)E\left(Y\right)$

$Cov\left(Y,X_2\right)=Cov\left(Y,X^2\right)=E\left(X^{2}Y\right)-E\left(X^2\right)E\left(Y\right)$

$Var\left(X_1\right)=Var\left(X\right)$$Var\left(X_2\right)=Var\left(X^2\right)$

Step 3: Final Answer

Using the previous exercise, 

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

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

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