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Unformatted text preview: STAT 410 Homework #7 Fall 2011 (due Friday, October 21, by 3:00 p.m.) 0. Warmup: 4.2.3 By Chebyshev’s Inequality, P (  W n – μ  ≥ ε ) ≤ 2 2 W g G n = 2 g p n b → 0 as n → ∞ for all ε > 0. Therefore, ¡ W P n → . 1. Suppose P ( X n = i ) = 6 3 + + n i n , for i = 1, 2, 3. Find the limiting distribution of X n . F X n ( x ) = g g g g g G g g g g g ¡ ¢ ≥ < ≤ + + < ≤ + + < 3 1 3 2 6 3 2 2 1 6 3 1 x x n i n x n i n x . () g g g g g G g g g g g ¡ ¢ ≥ < ≤ < ≤ < ∞ = → 3 1 3 2 3 2 2 1 3 1 1 F X lim x x x x x n n . Then X X D n → , where P ( X = i ) = 3 1 , for i = 1, 2, 3. 2. Let X n have p.d.f. f n ( x ) = n n x 2 1 1 1 + + , for 0 < x < 1, zero elsewhere. Find the limiting distribution of X n . F X n ( x ) = g g g g G g g g g ¡ ¢ ≥ < ≤ + + < 1 1 1 2 1 1 2 2 x x n n x x x . () g g g G g g g ¡ ¢ ≥ < ≤ < ∞ = → 1 1 1 F X lim x x x x x n n . Then X X D n → , where X has a Uniform distribution over ( 0, 1 ) . 3. 4.3.2 Let Y 1 denote the minimum of a random sample of size n from a distribution that has p.d.f. () ( ) ¢ = x e x f , θ < x < ∞ , zero elsewhere. Let Z n = n ( Y 1 – θ ) . Investigate the limiting distribution of Z n . F Y 1 ( x ) = ( ) θ x n e 1 , x > θ . ( Recall problems 8 & 9 of Homework #7. ) F Z n ( z ) = P ( n ( Y 1 – θ ) ≤ z ) = P ( Y 1 ≤ n z + θ ) = 1 – e – z , z > 0. Therefore, the limiting distribution of Z n is Exponential with mean 1. ( Exponential distribution with mean 1 is same as Gamma distribution with α = 1, β = 1. ) 4. 4.3.3 Let Y n denote the maximum ( the last order statistic ) of a random sample of size n from a distribution of the continuous type that has c.d.f. from a distribution of the continuous type that has c....
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This note was uploaded on 01/26/2012 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Monrad
 Statistics

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