# Hw4sol - AMS316 Homework 4 Solution 1 Since | < 1 we have Xt = and V ar(Xt = Furthermore(k = E[Xt Xt k = E Zt = Zt Zt-1 2 Zt-2(1 2 Z(1 2 i Zt-i k j

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AMS316 Homework 4 Solution 1. Since | α | < 1, we have X t = Z t (1 - α ß) = Z t + αZ t - 1 + α 2 Z t - 2 + ... and V ar ( X t ) = σ 2 Z (1 - α 2 ) Furthermore, γ ( k ) = E [ X t X t + k ] = E [ X α i Z t - i X α j Z t + k - j ] = σ 2 Z X i =0 α i α ( k + i ) = α k σ 2 Z (1 - α 2 ) Hence ρ ( k ) = α | k | k = 0 , ± 1 , ± 2 ,... and ρ ( k ) = 0 . 7 | k | . 2. The roots of equation y 2 - 1 3 y - 2 9 = 0 are 2 / 3 and -1 / 3, then the ACFs are given by ρ ( k ) = A 1 ( 2 3 ) | k | + A 2 ( - 1 3 ) | k | . Since A 1 + A 2 = ρ (0) = 1 and ρ (1) = 2 3 A 1 - 1 3 A 2 and ρ (1) = 3 / 7, solving for A 1 and A 2 gives A 1 = 16 21 and A 2 = 5 21 . 3. The roots of equation y 2 - 1 12 y - 1 12 = 0 are 1 / 3 and - 1 / 4. Then the ACFs are given by ρ ( k ) = A 1 ( 1 3 ) | k | + A 2 ( - 1 4 ) | k | . 1

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Since A 1 + A 2 = ρ (0) = 1 and ρ (1) = 1 3 A 1 - 1 4 A 2 and ρ (1) = 1 / 11, solving for A 1 and A 2 gives A 1 = 45 77 and A 2 = 32 77 . 4. (a) The root of 1 - 0 . 3 B = 0 is 1 / 0 . 3, which lies outside the unit circle, so X t is stationary. AR process is always invertible. X t = ( 1 1 - 0 . 3 B Z t ) = (1 + 0
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## This note was uploaded on 02/28/2012 for the course AMS 316 taught by Professor Xing during the Fall '09 term at SUNY Stony Brook.

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Hw4sol - AMS316 Homework 4 Solution 1 Since | < 1 we have Xt = and V ar(Xt = Furthermore(k = E[Xt Xt k = E Zt = Zt Zt-1 2 Zt-2(1 2 Z(1 2 i Zt-i k j

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