hw2 stats practice - STAT 409 Homework#2(due Friday...

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STAT 409 Homework #2 Fall 2008 (due Friday, September 12, by 4:00 p.m.) From the textbook: 4.7-4 Let Y n be the number of “successes” in n independent Bernoulli trials with probability p of “success” on each trial. By Chebyshev’s Inequality, for ε > 0, ( ) 2 ° ° 1 1 Y P - - ° ° ± ² ³ ³ ´ µ < - n p p p n n , since p n n Y E = ° ° ± ² ³ ³ ´ µ and ( ) n p p n n 1 Y Var - ° ° ± ² ³ ³ ´ µ = . Here, p = 0.5, ε = 0.08. n n n 39.0625 1 0.08 25 . 0 1 0.08 5 . 0 Y P 2 - - ° ° ± ² ³ ³ ´ µ < - = . a) n = 100. n n 39.0625 1 0.08 5 . 0 Y P - ° ° ± ² ³ ³ ´ µ < - = 0.609375 . b) n = 500. n n 39.0625 1 0.08 5 . 0 Y P - ° ° ± ² ³ ³ ´ µ < - = 0.921875 . c) n = 1000. n n 39.0625 1 0.08 5 . 0 Y P - ° ° ± ² ³ ³ ´ µ < - = 0.9609375 . 4.7-6 E ( X ) = μ = 80, Var ( X ) = n 2 ± = 15 60 = 4. By Chebyshev’s Inequality, for ε > 0, ( ) 2 2 ° ± ° ² 1 X P - < - n . P ( 75 < X < 85 ) = ( ) 2 5 4 1 5 80 X P - < - = 0.84 .
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- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( ) 3 2 ³ 3 ³ x e x x f - = x > 0 θ > 0. b) Obtain the maximum likelihood estimator of θ , ³ ˆ . L ( θ ) = = ° ° ± ² ³ ³ ´ µ - n i x i i e x 1 2 3 ³ 3 ³ ln L ( θ ) = ( ) = = - + n i i n i i x n x 1 3 1 2 ³ ³ 3 ln ln ( ln L ( θ ) ) ' = = - n i i x n 1 3 ³ = 0 · ³ ˆ = = n i i n 1 3 X c) Is ³ ˆ a consistent estimator of θ ? Justify your answer . No credit will be given without proper justification . Hint 1: Find E ( X 3 ) . Hint 2: Let y = x 3 . E ( X 3 ) = ¸ - 0 2 3 3 ³ 3 ³ dx x x x e y = x 3 dy = 3 x 2 dx = ¸ - 0 ³ ³ dy y y e = ³ 1 . By WLLN, ( ) ³ 1 X E X 1 3 1 3 = = P n i i n . Since g ( x ) = 1 / x is continuous at ³ 1 , ³ ˆ = ° ± ² ³ ´ µ ° ° ± ² ³ ³ ´ µ = ³ 1 X 1 1 3 g n g P n i i = θ .
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