HWsolutions3

# HWsolutions3 - HOMEWORK#3 SOLUTIONS IE121 7-19 f x = e x x...

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Unformatted text preview: HOMEWORK #3 SOLUTIONS IE121 7-19. f ( x) = e - x x! n L ( ) = i =1 n n e e = xi ! xi - - n n n xi i =1 x ! i i =1 ln L( ) = - n ln e + x i ln - ln x i ! i =1 i =1 d ln L( ) 1 = -n + d n x i =1 i n i 0 -n+ x i =1 =0 x i =1 n i = n ^ = 7-26. x i =1 n i n a) Because a is less than or equal to "a" for every sample, E( a ) cannot equal "a". n n b) Yes, E( a ) is less than "a" by a factor of . As n , 1, and E( a ) a. n+1 n+1 (n +1) , because Ea (n +1) = n + 1E(a) = a . c) a n n a d) FY(y) = P(Y y)=P(X1 y, X2 y,..., Xn y) = P(X1 y)P(X2 y)...P(Xn y) = (y/a)n for 0 y a. f ( y) = FY ( y) nyn-1 = for 0ya y an =0 otherwise The maximum likelihood estimator for a is Y. To show that the mle for a is biased, need to show that E(Y) a: E( Y ) = y 0 a nyn-1 an = nyn an (n + 1) a = 0 n a. n+1 7-35. Thus, E(Y) a, and the mle of a is biased. 3.5 = = 1.429 , X = 75.5 psi X = n 6 75.75 - 75.5 1.429 X - P ( X 75.75) = P / n = P ( Z 0.175) = 1 - P ( Z 1.75) = 1 - 0.56945 = 0.43055 7-51. X ~ N (50,144) 53- 50 12 / 36 P (47 X 53) = P 47 -50 Z 12 / 36 = P (-1.5 Z 1.5) = P ( Z 1.5) - P ( Z -1.5) = 0.9332 - 0.0668 = 0.8664 7-57. V ( X ) = V [ aX 1 + (1 - a ) X 2 ] = a 2V ( X 1 ) + (1 - a ) 2V ( X 2 ) = a 2 ( ) + (1 - 2a + a 2 )( ) n n 1 2 2 2 = a 2 2 2 2a 2 a 2 2 + - + n2 n2 n2 n1 2 = (n2 a 2 + n1 - 2n1a + n1a 2 )( ) n1n2 2 V ( X ) = ( )(2n2 a - 2n1 + 2n1a ) 0 n1n2 a 0 = 2n2 a - 2n1 + 2n1a 2a ( n2 + n1 ) = 2n1 a ( n2 + n1 ) = n1 n1 a= n2 + n1 n -1 i =1 7-70. E (V ) = k [ E ( X i2+1 ) + E ( X i2 ) - 2 E ( X i X i +1 )] = k ( 2 + 2 + 2 + 2 - 2 2 ) i =1 n -1 Therefore, k = = k ( n - 1)2 2 1 2 ( n -1) ...
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• Spring '08
• Perevalov
• Maximum likelihood, Estimation theory, Likelihood function, maximum likelihood estimator, ASCII, Bias of an estimator

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HWsolutions3 - HOMEWORK#3 SOLUTIONS IE121 7-19 f x = e x x...

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