Explanation Given: Consider following expectation: K ( a , b ) = E [ ( Y − a − b X ) 2 ] Calculation: Coefficients that minimize this expectation will give us a line that best fits to the probability distribution. ( a 0 , b 0 ) is a critical part of K(a,b) if ∂ K ∂ a ( a 0 , b 0 ) = 0 ∂ K ∂ b ( a 0 , b 0 ) = 0 And ( a 0 , b 0 ) is local minimum if K(a,b) is continuous function of a and b and ∂ 2 K ∂ a 2 ( a 0 , b 0 ) > 0 ∂ 2 K ∂ b 2 ( a 0 , b 0 ) > 0 Assume ∂ K ∂ a ( a 0 , b 0 ) = 0 The partial derivative, ∂ K ∂ a ( a 0 , b 0 ) = E [ − 2 ( Y − a 0 − b 0 X ) ] = − 2 E [ Y ] + 2 a 0 + 2 b 0 E [ X ] = − 2 μ y + 2 a 0 + 2 b 0 μ x ∂ K ∂ a ( a 0 , b 0 ) = 0 is equivalent to μ y − a 0 − b 0 μ x = 0 Assume ∂ K ∂ b ( a 0 , b 0 ) = 0 The partial derivative, ∂ K ∂ b ( a 0 , b 0 ) = E [ − 2 X ( Y − a 0 − b 0 X ) ] = − 2 E [ X Y ] + 2 a 0 E [ X ] + 2 b 0 E [ X 2 ] Covariance: C o v ( X , Y ) = E ( X , Y ) − μ x μ y Thus, E ( X , Y ) = C o v ( X , Y ) + μ x μ y Since, C o v ( X , Y ) = ρ σ x σ y Using this, E ( X , Y ) = ρ σ x σ y + μ x μ y Where, σ 2 x = E ( x 2 ) − μ x 2 Using this, E ( x 2 ) = σ 2 x + μ x 2 On substitution of σ 2 x + μ x 2 for E ( x 2 ) and C o v ( X , Y ) + μ x μ y for E ( X , Y ) in following formula ∂ K ∂ b ( a 0 , b 0 ) = E [ − 2 X ( Y − a 0 − b 0

Pearson eText for Probability and Statistical Inference -- Instant Access (Pearson+)

10th Edition

ISBN: 9780137538461

Author: Robert Hogg, Elliot Tanis

Publisher: PEARSON+

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Chapter 4.2, Problem 5E

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To find: The best fit line.

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Pearson eText for Probability and Statistical Inference -- Instant Access (Pearson+)

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The best fit line.

Solution Summary: The author explains that the best fit line is ly, which minimizes this expectation.

To find: The best fit line.

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Chapter 4 Solutions