quiz1_sol

# quiz1_sol - 1 a Since X 1… X n are independent and...

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Unformatted text preview: 1 a) Since X 1… X n are independent and identically distributed (i.i.d.) N ( μ , σ 2 ) So, the joint distribution of X 1… X n is f ( x1 , … , x n ) = ∏ n i =1 f ( xi ) = ∏ n − ( xi − μ ) 2σ 2 2 e i =1 2π σ 2 Then, the likelihood function is, − ( xi − μ ) 2σ 2 2 2 L( x1 ,…, xn ) = Log{ f ( x1 ,…, xn )} = Log{∏ f ( xi ) = Log{∏ i =1 i =1 n n e 2πσ } n 1n 2 = − log(2πσ ) − 2 ∑ ( xi − μ )2 2 2σ i =1 Since σ 2 is known, we set Then, we get the maximum likelihood estimate of μ is, ∧ − 1n μ = x = ∑ xi # n i =1 b) Since X 1… X n are independent and identically distributed N ( μ , σ 2 ) , 1n ∑ xi E ( x ) = μ ,so n i =1 is still distributed normal distribution, and i 1n 1n 1n 1 E ( ∑ xi ) = ∑ E ( xi ) = ∑ μ = nμ = μ n i =1 n i =1 n i =1 n n 1n 1 1 Var ( ∑ xi ) = 2 Var ∑ xi = 2 n i =1 n n i =1 1 Var ( xi ) = 2 ∑ n i =1 n ∑σ 2 = n σ 2 i =1 n 1 So, the distribution of the estimate μ = x = 2 a) The trend components of B is a+bt ∧ − 1n σ2 xi is N ( μ , ) # ∑ n i =1 n The seasonal components of B is St # b) E ( xt ) = E (a + bt + st + yt ) = a + bt + st + E ( yt ) Since yt is stationary process with mean zero. So, E ( xt ) = a + bt + st The condition for stationary process is that E ( xt ) is independent of t, and cov( xs , xt ) = cov( xs + k , xt + k ) So xt is not stationary. # 3 a) cov( xt , xt + k ) = cov( zt + θ zt − 2 , zt + k + θ zt + k − 2 ) = cov( zt , zt + k ) + cov( zt ,θ zt + k − 2 ) + cov(θ zt − 2 , zt + k ) + cov(θ zt − 2 ,θ zt + k − 2 ) = cov( zt , zt + k ) + θ cov( zt , zt + k − 2 ) + θ cov( zt − 2 , zt + k ) + θ 2 cov( zt − 2 , zt + k − 2 ) If k=0, cov( xt , xt + k ) = cov( xt , xt ) = cov( zt , zt ) + θ cov( zt , zt − 2 ) + θ cov( zt − 2 , zt ) + θ 2 cov( zt − 2 , zt − 2 ) Since zt is i.i.d. Normal (0, σ 2 ) ⇒ cov( zt , zt ) = cov( zt − 2 , zt − 2 ) = σ 2 ⇒ cov( zt , zt − 2 ) = cov( zt − 2 , zt ) = 0 So, cov( xt , xt ) = σ 2 + θ 2σ 2 If k=1, cov( xt , xt + k ) = cov( xt , xt +1 ) = cov( zt , zt +1 ) + θ cov( zt , zt −1 ) + θ cov( zt − 2 , zt +1 ) + θ 2 cov( zt − 2 , zt −1 ) =0 Because of the independency of zt . If k=2 cov( xt , xt + k ) = cov( xt , xt + 2 ) = cov( zt , zt + 2 ) + θ cov( zt , zt ) + θ cov( zt − 2 , zt + 2 ) + θ 2 cov( zt − 2 , zt ) = θσ 2 If k=‐2 cov( xt , xt + k ) = cov( xt , xt − 2 ) = cov( zt , zt − 2 ) + θ cov( zt , zt − 4 ) + θ cov( zt − 2 , zt − 2 ) + θ 2 cov( zt − 2 , zt − 4 ) = θσ 2 For the other K’s, cov( xt , xt + k ) = 0 . So, γ (k ) = cov( xt , xt + k ) = { γ (k ) cov( xt , xt ) σ 2 + θ 2σ 2 θσ 2 0 k = ±2 k =0 other wise # b) ρ (k ) = ρ (k ) = { θ 1+θ 2 1 0 k = ±2 # k =0 other wise 4 γ (k ) = cov( xt , xt + k ) , for k ≥ 1 Plug the relation xt = φ xt −1 + ε t ⇒ xt + k = φ xt + k −1 + ε t + k into the above function, γ (k ) = cov( xt , xt + k ) = cov( xt , φ xt + k −1 + ε t + k ) = φ cov( xt , xt + k −1 ) + cov( xt , ε t + k ) = φ cov( xt , xt + k −1 ) = φγ (k − 1) = φ 2γ (k − 2) = = φ k γ (0) Since, ρ (k ) = γ (k ) , so γ (k ) = φ k # γ (0) ...
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