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Unformatted text preview: STAT 409 Homework #2 Fall 2008 (due Friday, September 12, by 4:00 p.m.) From the textbook: 4.74 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 ⋅ ⋅ ≥ & & ¡ ¢ £ £ ¤ ¥ < n p p p n n , since p n n Y E = & & ¡ ¢ £ £ ¤ ¥ and ( ) n p p n n 1 Y Var & & ¡ ¢ £ £ ¤ ¥ ⋅ = . Here, p = 0.5, ε = 0.08. n n n 39.0625 1 0.08 25 . 1 0.08 5 . Y P 2 ≥ & & ¡ ¢ £ £ ¤ ¥ < = ⋅ . a) n = 100. n n 39.0625 1 0.08 5 . Y P ≥ & & ¡ ¢ £ £ ¤ ¥ < = 0.609375 . b) n = 500. n n 39.0625 1 0.08 5 . Y P ≥ & & ¡ ¢ £ £ ¤ ¥ < = 0.921875 . c) n = 1000. n n 39.0625 1 0.08 5 . Y P ≥ & & ¡ ¢ £ £ ¤ ¥ < = 0.9609375 . 4.76 E ( X ) = μ = 80, Var ( X ) = n 2 ¡ = 15 60 = 4. By Chebyshev’s Inequality, for ε > 0, ( ) 2 2 & ¡ & ¢ 1 X P ⋅ ≥ < n . P ( 75 < X < 85 ) = ( ) 2 5 4 1 5 80 X P ≥ < = 0.84 .                                     3. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( ) 3 2 & 3 & x e x x f = x > 0 θ > 0. b) Obtain the maximum likelihood estimator of θ , & ˆ . L ( θ ) = ∏ = & & ¡ ¢ £ £ ¤ ¥ n i x i i e x 1 2 3 & 3 & ln L ( θ ) = ( ) ¦ ¦ = = ⋅ ⋅ + n i i n i i x n x 1 3 1 2 & & 3 ln ln ( ln L ( θ ) ) ' = ¦ = n i i x n 1 3 & = 0 § & ˆ = ¦ = n i i n 1 3 X c) Is & 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 ) = ¨ ∞ 2 3 3 & 3 & dx x x x e y = x 3 dy = 3 x 2 dx = ¨ ∞ & & dy y y e = & 1 ....
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This note was uploaded on 10/11/2010 for the course STATS 410 taught by Professor Alexey during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
 Alexey
 Bernoulli, Probability

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