Dr. Hackney STA Solutions pg 164

# Dr. Hackney STA Solutions pg 164 - Second Edition 10-7...

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Second Edition 10-7 10 . 19 a. The variance of ¯ X is Var ¯ X = E( ¯ X - μ ) 2 = E ± 1 n X i X i - μ ! 2 = 1 n 2 E X i ( X i - μ ) 2 + 2 X i>j ( X i - μ )( X j - μ ) = 1 n 2 ² 2 + 2 n ( n - 1) 2 ρσ 2 ³ = σ 2 n + n - 1 n ρσ 2 b. In this case we have E X i>j ( X i - μ )( X j - μ ) = σ 2 n X i =2 i - 1 X j =1 ρ i - j . In the double sum ρ appears n - 1 times, ρ 2 appears n - 2 times, etc. . so n X i =2 i - 1 X j =1 ρ i - j = n - 1 X i =1 ( n - i ) ρ i = ρ 1 - ρ ² n - 1 - ρ n 1 - ρ ³ , where the series can be summed using (1 . 5 . 4), the partial sum of the geometric series, or using Mathematica. c. The mean and variance of X i are E X i = E[E( X i | X i - 1 )] = E ρX i - 1 = ··· = ρ i - 1 E X 1 and Var X i = VarE( X i | X i - 1 ) + EVar( X i | X i - 1 ) = ρ 2 σ 2 + 1 = σ 2 for σ 2 = 1 / (1 - ρ 2 ). Also, by iterating the expectation E X 1 X i = E[E( X 1 X i | X i - 1 )] = E[E( X 1 | X i - 1 )E( X i | X i - 1 )] = ρ E[ X 1 X i - 1 ] , where we used the facts that X 1 and X i are independent conditional on X i - 1 . Continuing with the argument we get that E X 1 X i = ρ i - 1 E X 2 1 . Thus, Corr( X 1 ,X i ) = ρ i - 1 E X 2 1 - ρ i - 1 (E X 1 ) 2 Var X 1 Var X i
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