5102-Solution-Set-04

# 5102-Solution-Set-04 - SOLUTIONS FOR PROBLEM SET 4 STAT...

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Unformatted text preview: SOLUTIONS FOR PROBLEM SET 4 STAT 5102-004 (1) If X 1 ,...,X n are a random sample from a N( θ, 4) population, then ¯ X n ∼ N( θ, 4 /n ) and E( ¯ X n- θ ) 2 = 4 /n . If n ≥ 40, then E( ¯ X- θ ) 2 ≤ 1 / 10. Let Z n = 1 2 √ n ( ¯ X n- θ ). Pr( | ¯ X n- θ | ≤ . 1) = Pr( | Z n | ≤ . 05 √ n ) . Note that c ≥ 1 . 96 implies Pr( | Z n | ≤ c ) ≥ . 95 since Z n is standard normal. If n ≥ 400(1 . 96) 2 , then 0 . 05 √ n ≥ 1 . 96, and n = 1537 is the smallest integer that satisfies this condition. Thus, if n ≥ 1537, then Pr( | ¯ X n- θ | ≤ . 1) ≥ . 95. (2) A χ 2 ( n ) random variable has density f ( x | n ) = c ( n ) x n 2- 1 exp(- 1 2 x ) on the positive real line. The normalizing constant c ( n ) has no effect on the shape of the density. The exponential factor exp(- 1 2 x ) decreases in x , from 1 asymptotically toward 0 as x increases, and does not change with the parameter n . Instead, n affects the shape of the density through the polynomial factor x n 2- 1 . This factor is a decreasing function for n = 1, constant for n = 2, and increasing for n ≥ 3. x exp(-x/2) 2 4 6 8 10 1 x exp(-x/2) 2 4 6 8 10 1 x exp(-x/2) 2 4 6 8 10 1 x c ( 1 29 x 2 4 6 8 10 6 x c(2) 2 4 6 8 10 0.0 0.5 x c ( 3 29 x 2 4 6 8 10 0.00 1.25 x f(x|1) 2 4 6 8 10 6 x f(x|2) 2 4 6 8 10 0.0 0.5 x f(x|3) 2 4 6 8 10 0.00 0.25 For n ∈ { 1 , 2 } , the density is a strictly decreasing function. If n = 1, the density is unbounded as x approaches zero since lim x ↓ x- 1 2 = ∞ and lim x ↓ exp(- 1 2 x ) = 1 is bounded. If n = 2, the density is bounded with a maximum at zero. For n ≥ 3, the increasing polynomial and decreasing exponential factors produce internal maxima. Working with logarithms translates the product to a sum, preserves order, and simplifies the analysis. The function g ( x | n ) = log f ( x | n ) has derivative g ( x | n...
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5102-Solution-Set-04 - SOLUTIONS FOR PROBLEM SET 4 STAT...

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