A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

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 $E [(Y 鈭� (a + b X + c X_{2})) 2]$ . 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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91影视

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

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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91影视

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

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Decision Maths

Pure Maths

Mechanics Maths

Discrete Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

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