Chapter 7: Q.7.39 (page 362)

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 $E [(Y - (a + b X + c X_{2})) 2]$ . Determine $a$ , $b$ , and $c$ .

Short Answer

Expert verified

$a = E (Y) - b E (X_{1}) - c E (X_{2})$

$b = \frac{Cov (X_{1}, X_{2}) Cov (Y, X_{2}) - Cov (Y, X_{1}) Var (X_{2})}{Cov {(X_{1}, X_{2})}^{2} - Var (X_{1}) Var (X_{2})}$

$c = \frac{Cov (X_{1}, X_{2}) Cov (Y, X_{1}) - Cov (Y, X_{2}) Var (X_{1})}{Cov {(X_{1}, X_{2})}^{2} - Var (X_{1}) Var (X_{2})}$

Step by step solution

Given Information

The best quadratic predictor of $Y$ with respect to $X$ is $a + b X + c X_{2}$ .

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 (X_{1}, X_{2}) = Cov (X, X^{2}) = E (X^{3}) - E (X) E (X^{2})

$Cov (Y, X_{1}) = Cov (Y, X) = E (X Y) - E (X) E (Y)$

$Cov (Y, X_{2}) = Cov (Y, X^{2}) = E (X^{2} Y) - E (X^{2}) E (Y)$

$Var (X_{1}) = Var (X)$ $Var (X_{2}) = Var (X^{2})$

Final Answer

Using the previous exercise,

$a = E (Y) - b E (X_{1}) - c E (X_{2})$

$b = \frac{Cov (X_{1}, X_{2}) Cov (Y, X_{2}) - Cov (Y, X_{1}) Var (X_{2})}{Cov {(X_{1}, X_{2})}^{2} - Var (X_{1}) Var (X_{2})}$

$c = \frac{Cov (X_{1}, X_{2}) Cov (Y, X_{1}) - Cov (Y, X_{2}) Var (X_{1})}{Cov {(X_{1}, X_{2})}^{2} - Var (X_{1}) Var (X_{2})}$

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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 $E [(Y - (a + b X + c X_{2})) 2]$ . Determine $a$ , $b$ , and $c$ .

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Explanation

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The best quadratic predictor ofYwith respect to Xis a + bX+cX2, where a, b, and c are chosen to minimize E[(Y−(a+bX+cX2))2]. Determine a, b, and c.

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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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 $E [(Y - (a + b X + c X_{2})) 2]$ . Determine $a$ , $b$ , and $c$ .