3 - NOTES 3 Trends of a time series Deterministic versus...

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NOTES 3 Trends of a time series : Deterministic versus Stochastic Trends Examples : y t = a t - 0 . 5 a t - 1 where a t N (0 , σ 2 a ); Purely Stochastic. y t = β 0 + β 1 t + β 2 t 2 ; Purely Deterministic y t = β 0 + β 1 t + β 2 t 2 + a t where a t N (0 , σ 2 a ); Stochastic + Deterministic Estimation of a constant mean Model: Z t = μ + X t , E ( X t ) = 0 for all t. First, we wish to estimate μ with observed time series Z 1 , Z 2 , ..., Z n . The most common estimate of μ is the sample mean ¯ Z = 1 n n X t =1 Z t . Since E ( ¯ Z ) = μ , ¯ Z is an unbiased estimate of μ . To investigate the precision of ¯ Z as an estimate of μ , we need to make further assump- tions concerning X t . Theorem Suppose { X t } is a stationary time series, then V ar ( ¯ Z ) = γ 0 n [1 + 2 n - 1 X k =1 (1 - k n ) ρ k ] = γ 0 n n - 1 X k = - n +1 (1 - | k | n ) ρ k Note that the first factor γ 0 n is the population variance assuming the observations are independent. If the { X t } series is in fact just white noise, then ρ k = 0 for k 1 and V ar ( ¯ Z ) simply reduce to γ 0 n . 1
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Proof: V ar ( ¯ Z ) = V ar ( 1 n n X t =1 Z t ) = 1 n 2 V ar ( n X t =1 Z t ) = 1 n 2 V ar ( n X t =1 X t ) = 1 n 2 [ n X t =1 V ar ( X t ) + 2 n - 1 X t =1 n X s = t +1 Cov ( X t , X s )] = 1 n 2 [ 0 + 2 n - 1 X t =1 n X s = t +1 γ t,s ] = 1 n 2 [ 0 + 2(( n - 1) γ 1 + ( n - 2) γ 2 + ( n - 3) γ 3 + . . . + γ n - 1 )] = 1 n 2 [ 0 + 2 n - 1 X k =1 ( n - k ) γ k ] = γ 0 n [1 + 2 n - 1 X k =1 (1 - k n ) ρ k ] Since ρ k = ρ - k , V ar ( ¯ Z ) = γ 0 n [1 + n - 1 X k =1 (1 - k n ) ρ k + - 1 X k = - ( n - 1) (1 - | k | n ) ρ k ] = γ 0 n n - 1 X k = - n +1 (1 - | k | n ) ρ k 2 Example: Z t = a t - 1 2 a t - 1 , ρ 1 = - 0 . 4 , ρ k = 0 , k > 1 . V ar ( ¯ Z ) = γ 0 n [1 + 2(1 - 1 n )( - 0 . 4)] = γ 0 n [1 - 0 . 8 n - 1 n ] When n is large, V ar ( ¯ Z ) 0 . 2 γ 0 n . For many stationary process, ρ k decays rapidly with increasing lag, thus X k =0 ρ k < . Under this assumption and given a large sample size n , V ar ( ¯ Z ) γ 0 n X k = -∞ ρ k . 2
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Example: Suppose ρ k = φ | k | with | φ | < 1, V ar ( ¯ Z ) = γ 0 n X k = -∞ ρ k = γ 0 n X k = -∞ φ | k | = γ 0 n (1 + 2 X k =1 φ k ) = γ 0 n (1 + 2 φ 1 - φ ) = γ 0 n ( 1 + φ 1 - φ ) Example: Suppose { X t } is a zero-mean random walk. Recall γ t,s = 2 a for 1 t s .
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