{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dr. Hackney STA Solutions pg 78

# Dr. Hackney STA Solutions pg 78 - 5-12 Solutions Manual for...

This preview shows page 1. Sign up to view the full content.

5-12 Solutions Manual for Statistical Inference c. S 2 n σ 2 in probability. By a), S n = S 2 n σ 2 = σ in probability. By b), σ/S n 1 in probability. 5.33 For all > 0 there exist N such that if n > N , then P ( X n + Y n > c ) > 1 - . Choose N 1 such that P ( X n > - m ) > 1 - / 2 and N 2 such that P ( Y n > c + m ) > 1 - / 2. Then P ( X n + Y n > c ) P ( X n > - m, + Y n > c + m ) P ( X n > - m ) + P ( Y n > c + m ) - 1 = 1 - . 5.34 Using E ¯ X n = μ and Var ¯ X n = σ 2 /n , we obtain E n ( ¯ X n - μ ) σ = n σ E( ¯ X n - μ ) = n σ ( μ - μ ) = 0 . Var n ( ¯ X n - μ ) σ = n σ 2 Var( ¯ X n - μ ) = n σ 2 Var ¯ X = n σ 2 σ 2 n = 1 . 5.35 a. X i exponential(1). μ X = 1, Var X = 1. From the CLT, ¯ X n is approximately n(1 , 1 /n ). So ¯ X n - 1 1 / n Z n(0 , 1) and P ¯ X n - 1 1 / n x P ( Z x ) . b. d dx P ( Z x ) = d dx F Z ( x ) = f Z ( x ) = 1 2 π e - x 2 / 2 . d dx P ¯ X n - 1 1 / n x = d dx n i =1 X i x n + n W = n i =1 X i gamma( n, 1) = d dx F W ( x n + n ) = f W ( x n + n ) · n = 1 Γ( n ) ( x n + n ) n - 1 e - ( x n + n ) n. Therefore, (1 / Γ( n ))( x n + n ) n - 1 e - ( x n + n ) n 1 2 π e - x 2 / 2 as n → ∞ . Substituting x = 0 yields n ! n n +1 / 2 e - n
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}