prob01stat_sol

# prob01stat_sol - X i = ± 2(4 By the LLN 1 n P n i =1 X 2 i...

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1. (1) E ( X ) = R 1 xp ( x ) dx = R 0 1 x 2 dx + R 1 0 x 2 dx = 1 3 x 3 & 0 1 + 1 3 x 3 1 0 = 0 . (2) V ar ( X ) = E ± [ X E ( X )] 2 ² = R 1 x 2 p ( x ) dx = R 0 1 x 3 dx + R 1 0 x 3 dx = 1 4 x 4 & 0 1 + 1 4 x 4 1 0 = 1 2 . (3) E ³ X 2 ´ = E ³ 1 2 X 1 + 1 2 X 2 ´ = 1 2 E ( X 1 ) + 1 2 E ( X 2 ) = = 0 . (4) Cov ( X 1 ;X 2 ) = 0 because X 1 and X 2 are independent (random draws). (5) V ar ³ X 2 ´ = V ar ³ 1 2 X 1 + 1 2 X 2 ´ = 1 4 V ar ( X 1 ) + 1 4 V ar ( X 2 ) + 2 ± 1 4 Cov ( X 1 ;X 2 ) = 1 2 ± 2 = 1 4 . (6) X 2 is a random variable because X 1 and X 2 are random variables. E ³ X 2 ´ and V ar ³ X 2 ´ are nonrandom constant (zero and 1 4 ). (7) E ³ X ´ = E ³ 1 n P n i =1 X i ´ = 1 n P n i =1 E ( X i ) = = 0 . (8) V ar ³ X ´ = V ar ³ 1 n P n i =1 X i ´ = 1 n 2 h P n i =1 V ar ( X i ) + 2 P i<j Cov ( X i ;X j ) i = 1 n ± 2 = 1 2 n . (9) The LLN states that X converges to a nonrandom number (in probability sense) as n ! 1 . In other words, the variance of X converges to 0 as n ! 1 . Since V ar ³ X ´ = 1 2 n ! 0 as n ! 1 , the result is consistent with the LLN. 2. (1) ± 2 = E ³ X 2 ´ . Replace E with 1 n P n i =1 to get a sample analogue estimator. Therefore, b ± 2 SA = 1 n P n i =1 X 2 i . (2) The log-likelihood is given by P n i =1 log p ³ X i 2 ´ = P n i =1 h 1 2 log ³ 2 ²± 2 ´ X 2 i 2 2 i . Take the FOC with respect to ± 2 : P n i =1 h 1 2 2 + X 2 i 2 4 i = 0 . Therefore, n 2 2 + 1 2 4 P n i =1 X 2 i = 0 , and we have b ± 2 ML = 1 n P n i =1 X 2 i . (3) E µ 1 n P n i =1 X 2 i = 1 n P n i =1 E ³ X 2 i ´ = 1 n P n i =1 V ar
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Unformatted text preview: ( X i ) = ± 2 . (4) By the LLN, 1 n P n i =1 X 2 i converges to E ³ X 2 i ´ , which is ± 2 . 3. (1) E ( X ) = P x =0 ; 1 xp ( x ) = 0 ± (1 & ³ ) + 1 ± ³ = ³ . (2) The log-likelihood is given by P n i =1 log p ( X i ;³ ) = P n i =1 [ X i log ³ + (1 & X i ) log (1 & ³ )] . Take the FOC with respect to ³ : 1 ± P n i =1 X i & 1 1 & ± P n i =1 (1 & X i ) = 0 . Therefore, 1 ± (1 & ± ) P n i =1 X i & n 1 & ± = 0 , and we have b ³ ML = 1 n P n i =1 X i . (3) E µ 1 n P n i =1 X i ¶ = 1 n P n i =1 E ( X i ) = ³ . (4) By the LLN, 1 n P n i =1 X i converges to E ( X i ) , which is ³ . 2...
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## This note was uploaded on 04/04/2011 for the course ECON 401 taught by Professor Staff during the Spring '08 term at University of Washington.

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