LINREG2

¯ x(44 1 n j n j 1 x j y j \$" ¯ x \$ \$ 1 n j n j

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Unformatted text preview: ¯ X , (44) 1 n j n j ' 1 X j Y j ' \$ " . ¯ X % \$ \$ 1 n j n j ' 1 X 2 j . (45) To solve these normal equations, substitute in (45). Then we get ˆ α ' ¯ Y & ˆ β . ¯ X 23 1 n j n j ' 1 X j Y j ' ( ¯ Y & ˆ β ¯ X ) 1 n j n j ' 1 X j % ˆ β 1 n j n j ' 1 X 2 j ' ¯ Y . ¯ X & ˆ β ¯ X 2 % ˆ β 1 n j n j ' 1 X 2 j ' ¯ X . ¯ Y % ˆ β 1 n j n j ' 1 X 2 j & ¯ X 2 hence 1 n j n j ' 1 X j Y j & ¯ X . ¯ Y ' \$ \$ 1 n j n j ' 1 X 2 j & ¯ X 2 . (46) Equation (46) can also be written as 1 n j n j ' 1 ( X j & ¯ X )( Y j & ¯ Y ) ' \$ \$ 1 n j n j ' 1 ( X j & ¯ X ) 2 , (47) because 1 n j n j ' 1 ( X j & ¯ X )( Y j & ¯ Y ) ' 1 n j n j ' 1 X j Y j & ¯ X . Y j & X j . ¯ Y % ¯ X . ¯ Y ' 1 n j n j ' 1 X j Y j & ¯ X . 1 n j n j ' 1 Y j & ¯ Y . 1 n j n j ' 1 X j % ¯ X . ¯ Y ' 1 n j n j ' 1 X j Y j & ¯ X . ¯ Y (48) and similarly 1 n j n j ' 1 ( X j & ¯ X ) 2 ' 1 n j n j ' 1 X 2 j & ¯ X 2 . (49) Moreover, 1 n j n j ' 1 ( X j & ¯ X )( Y j & ¯ Y ) ' 1 n j n j ' 1 ( X j & ¯ X ) Y j & 1 n j n j ' 1 ( X j & ¯ X ) ¯ Y ' 1 n j n j ' 1 ( X j & ¯ X ) Y j & ( ¯ X & ¯ X ) ¯ Y ' 1 n j n j ' 1 ( X j & ¯ X ) Y j (50) 24 The result (1) now follows from (44) and (46) through (50). Proof of Proposition 1. Recall from (1) that \$ \$ ' ' n j ' 1 ( X j & ¯ X ) Y j ' n j ' 1 ( X j & ¯ X ) 2 . (51) Substitute model (2) in (51). Then \$ \$ ' ' n j ' 1 ( X j & ¯ X )( " % \$ X j % U j ) ' n j ' 1 ( X j & ¯ X ) 2 ' " ' n j ' 1 ( X j & ¯ X ) % \$ ' n j ' 1 ( X j & ¯ X ) X j % ' n j ' 1 ( X j & ¯ X ) U j ' n j ' 1 ( X j & ¯ X ) 2 ' \$ . ' n j ' 1 ( X j & ¯ X ) X j ' n j ' 1 ( X j & ¯ X ) 2 % ' n j ' 1 ( X j & ¯ X ) U j ' n j ' 1 ( X j & ¯ X ) 2 ' \$ % ' n j ' 1 ( X j & ¯ X ) U j ' n j ' 1 ( X j & ¯ X ) 2 , (52) where the last step follows from the fact that similar to (50), j n j ' 1 ( X j & ¯ X ) 2 ' j n j ' 1 ( X j & ¯ X )( X j & ¯ X ) ' j n j ' 1 ( X j & ¯ X ) X j . (53) Now take the mathematical expectation at both sides of (52). Then, E [ \$ \$ ] ' \$ % E ' n j ' 1 ( X j & ¯ X ) U j ' n j ' 1 ( X j & ¯ X ) 2 ' \$ % ' n j ' 1 ( X j & ¯ X ) E ( U j ) ' n j ' 1 ( X j & ¯ X ) 2 ' \$ , (54) because taking the mathematical expectation of a constant ( β ) does not effect that constant, and taking the mathematical expectation of a linear function of random variables is equal to taking the linear function of the mathematical expectation of these random variables. The last conclusion in (54) follows from assumption II, and the second step in (54) can be taken because 25 we have assumed that the X j 's are non-random (assumption IV). Next consider . We have already established that Substituting the right- ˆ α ˆ α ' ¯ Y & ˆ β . ¯ X . hand side of (52) for in this equation yields ˆ β \$ " ' ¯ Y & \$ % ' n j ' 1 ( X j & ¯ X ) U j ' n j ' 1 ( X j & ¯ X ) 2 ....
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¯ X(44 1 n j n j 1 X j Y j \$" ¯ X \$ \$ 1 n j n j 1 X 2...

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