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Unformatted text preview: 1 Time series models for the random component & Let Y t = D t +X t & where D t =Deterministic component X t = Random (stochastic) component & Note: The covariance of X t and X t+k is cov(X t , X tk ) = E(X t X tk )  E(X t )E(X t+k ) & Def: A stochastic process {X t } is stationary if (1) E(X t )= & , does not dependent on time (2) Cov(X t , X tk ) = ¡ x (k) depends only on lag (separation) k not time t. 2 Theoretical ACF & The ACF of a stationary time series with autocovariance function { & x (k) } is & ¡ x (k) = & x (k) , k = 0,1,2, & x (0) where & x (0) = cov(X t , X t ) = var(X t ) = ¢ 2  ¡ x (k) £ 1 i. e. 1 £ ¡ x (k) £ 1 3 White noise {Z t } & Def : A process {Z t } is called white noise or a white noise process, if {Z 1 , Z 2 , … } are uncorrelated each with mean & z =0 and variance ¡ z 2 Eg_1 Show that a white noise time series is stationary . Sketch the correlogram at lags k=0,1,2, …. 4 Solution : Example_1 & Cov(Z t , Z tk ) = E(Z t Z tk ) – E(Z t )E(Z tk ) = 0 since the Z t ’s are uncorrelated & Hence, & & ¡ & ¢ £ = = = ,.... 2 , 1 , , ) ( 2 k k k σ γ & ¡ & ¢ £ = = = ,.... 2 , 1 , , 1 ) ( k k k ρ 5 Examples of stationary time series & Eg_2 : A time {X t } satisfies the model & X t = Z t – Z t1 where {Z t } is white noise. Show that {X t } is stationary. Hint: Use the results (a) var( a +a 1 X 1 +a 2 X 2 ) = a 1 2 var(X 1 )+ 2a 1 a 2 cov(X 1 , X 2 )+ a 2 2 var(X 2 ) (b) cov( a +a 1 X 1 +a 2 X 2 , b + b 1 Y 1 + b 2 Y 2 ) = a 1 b 1 cov(X 1 ,Y 1 ) + a 1 b 2 cov(X 1 ,Y 2 ) + a 2 b 1 cov(X 2 ,Y 1 )+ a 2 b 2 cov(X 2 ,Y 2 ). 6 Example_2: Solution & (1) E(X t ) = E(Z t – Z t1 ) = E(Z t )–E(Z t1 ) = 0 since {Z t } is white noise & (2) Cov(X t , X t+K ) = cov(Z t – Z t1 , Z t+k – Z t+k1 ) = cov (Z t, , Z t+k )  cov (Z t , Z t+k1 ) cov (Z t1, , Z t+k ) + cov (Z t1 , Z t+k1 ) Case 1: k = 0 Cov(X t , X t ) =var(X t ) = var(Z t – Z t1 ) = var(Z t ) 2cov(Z t , Z t1 )+var(Z t1 ) = & z 2 – 2(0)+ & z 2 = 2 & z 2 7 Example_2: Solution(continued) & Case 2: k =1 & Cov(X t , X t+K ) = cov(Z t – Z t1 , Z t+k – Z t+k1 ) = cov (Z t, , Z t+k )  cov (Z t , Z t+k1 )  cov (Z t1, , Z t+k ) + cov (Z t1 , Z t+k1 ) = cov (Z t, , Z t+1 )  cov (Z t , Z t )  cov (Z t1, , Z t+1 ) + cov (Z t1 , Z t ) = 0  & z 2 0 + 0 = & z 2 Case 3: k =2 : Cov(X t , X t+K ) = cov(Z t – Z t1 , Z t+k – Z t+k1 ) = cov (Z t, , Z t+k )  cov (Z t , Z t+k1 )  cov (Z t1, , Z t+k ) + cov (Z t1 , Z t+k1 ) = cov (Z t, , Z t+2 )  cov (Z t , Z t+1 )  cov (Z t1, , Z t+2 ) + cov (Z t1 , Z t+1 ) = 0  0  0 + 0 = 0 8 Example_2: Solution(continued) & Hence & & ¡ & & ¢ £ = = = = ,.... 3 , 2 , 1 , , 2 ) ( 2 2 k k k k z z x σ σ γ & ¡ & ¢ £ = = = = ,.... 3 , 2 , 1 , 2 / 1 , 1 ) ( k k k k x ρ 9 Moving Average Processes, MA(q) & Def: Let {Z t } be white noise. A process {X t } called a moving average process of order,...
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This note was uploaded on 10/19/2009 for the course MATH 611 taught by Professor Jsdkasj during the Spring '09 term at Kansas.
 Spring '09
 jsdkasj
 Covariance, Variance

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