sol3 - HOMEWORK #3 SOLUTIONS IE121 7-19. f ( x) = e − λ...

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Unformatted text preview: HOMEWORK #3 SOLUTIONS IE121 7-19. f ( x) = e − λ λx x! n L (λ ) = ∏ i =1 n n e λ e = xi ! xi −λ − nλ n λ n ∑ xi i =1 ∏x ! i i =1 ln L(λ ) = − nλ ln e + ∑ x i ln λ − ∑ ln x i ! i =1 i =1 d ln L(λ ) 1 = −n + dλ λ n ∑x i =1 i n i ≡0 −n+ ∑x i =1 λ =0 ∑x i =1 n i = nλ ˆ λ= 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 0≤y≤a ∂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. σ 3.5 = = 1.429 , µ X = 75.5 psi σ X = n 6 75.75 − 75.5 1.429 X −µ P ( X ≥ 75.75) = P ≥ σ / 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 ( σ ) + (1 − 2a + a 2 )( σ ) n n 1 2 2 2 = a 2σ 2 σ 2 2aσ 2 a 2σ 2 + − + n2 n2 n2 n1 2 = (n2 a 2 + n1 − 2n1a + n1a 2 )( σ ) n1n2 2 ∂V ( X ) = ( σ )(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 ∑ (σ 2 + µ 2 + σ 2 + µ 2 − 2 µ 2 ) i =1 n −1 Therefore, k = = k ( n − 1)2σ 2 1 2 ( n −1) ...
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This homework help was uploaded on 02/26/2008 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .

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sol3 - HOMEWORK #3 SOLUTIONS IE121 7-19. f ( x) = e − λ...

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