econ483midterm2010sol

# econ483midterm2010sol - Solution Econ 483 Midterm Spring...

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Unformatted text preview: Solution: Econ 483 Midterm, Spring 2010 1. (6) E V ar 1 2 3 R1 + 3 R 2 1 2 3 R1 + 3 R2 2 = 1 E (R1 ) + 2 E (R2 ) = 1 r1 + 3 r2 . 3 3 3 = V ar 1 3 R1 + V ar 2 3 R2 + 2Cov 1 2 3 R1 ; 3 R2 = 12 91 + 42 92 + 4 9 12 . 2. (1) E (X ) = V ar (Y ) = V ar ( (2) b R1 R2 x 0:5 1 (0 x 2) dx = 0 0:5xdx = 1. h1 i V ar (X ) = E (X E (X ))2 = E X 2 2X + 1 = E X 2 0 2E (X )+E (1) = 1 Cov (X; U ) + 1X + U) = 2V 1 ar (X ) + V ar (U ) + 2 = 4:22 V ar (X ) + 36. R2 0 1 0:5x2 dx 2+1 = 3 . 1 = Pn = = Xi X ( 0 + 1 Xi + Ui ) Pn 2 X i=1 Xi Pn Pn Pn 2 X X X Ui 0 1 i=1 Xi i=1 Xi i=1 Xi Pn 2 + Pn 2 + Pn 2 X X X i=1 Xi i=1 Xi i=1 Xi Pn X Ui i=1 Xi 1 + Pn 2: Xi X i=1 i=1 of (3) E b 1 jX = E 1. 1 + (4) b 1 !p (5) Therefore, b 1 is unbiased. 1 Pn i=1 Pn i=1 1 Xi Xi X Ui X 2 jX ! = 1+ Pn i=1 Xi Pn X E (Ui jX ) Xi X 2 = 1 for any value i=1 + Cov (X; U ) = V ar (X ) as n ! 1 because Cov (X; U ) = 0. Therefore, b 1 is consistent. (A): de…nition of variance (B): unbiasedness or from (3) (C): from (2) and conditional on X , any function of X is nonrandom (D): properties of variance and conditional on X , any function of X is nonrandom (E): homoskedasticity and random sampling (6) e 1 = + 2 Pn i=1 Pn i=1 Xi X Ui 2 Xi X V ar (2Ui jX ) V ar e 1 jX = Pn X i=1 Xi fore, se e 1 jX = 2se b 1 jX . 1 6= 1 2 d = 4V ar b 1 jX . Similarly, V ar e 1 jX i + P=1 n Pn Xi Xi X Ui X 2 i=1 = b1 d = 4V ar b 1 jX . There- 1 ...
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