# TS_P2 - 1 Time series models for the random component &...

This preview shows pages 1–10. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 t-k ) = E(X t X t-k ) - 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 t-k ) = ¡ x (k) depends only on lag (separation) k not time t. 2 Theoretical ACF & The ACF of a stationary time series with auto-covariance 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 t-k ) = E(Z t Z t-k ) – E(Z t )E(Z t-k ) = 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 t-1 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 t-1 ) = E(Z t )–E(Z t-1 ) = 0 since {Z t } is white noise & (2) Cov(X t , X t+K ) = cov(Z t – Z t-1 , Z t+k – Z t+k-1 ) = cov (Z t, , Z t+k ) - cov (Z t , Z t+k-1 )- cov (Z t-1, , Z t+k ) + cov (Z t-1 , Z t+k-1 ) Case 1: k = 0 Cov(X t , X t ) =var(X t ) = var(Z t – Z t-1 ) = var(Z t ) -2cov(Z t , Z t-1 )+var(Z t-1 ) = & 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 t-1 , Z t+k – Z t+k-1 ) = cov (Z t, , Z t+k ) - cov (Z t , Z t+k-1 ) - cov (Z t-1, , Z t+k ) + cov (Z t-1 , Z t+k-1 ) = cov (Z t, , Z t+1 ) - cov (Z t , Z t ) - cov (Z t-1, , Z t+1 ) + cov (Z t-1 , Z t ) = 0 - & z 2- 0 + 0 = & z 2 Case 3: k =2 : Cov(X t , X t+K ) = cov(Z t – Z t-1 , Z t+k – Z t+k-1 ) = cov (Z t, , Z t+k ) - cov (Z t , Z t+k-1 ) - cov (Z t-1, , Z t+k ) + cov (Z t-1 , Z t+k-1 ) = cov (Z t, , Z t+2 ) - cov (Z t , Z t+1 ) - cov (Z t-1, , Z t+2 ) + cov (Z t-1 , 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,...
View Full Document

## This note was uploaded on 10/19/2009 for the course MATH 611 taught by Professor Jsdkasj during the Spring '09 term at Kansas.

### Page1 / 65

TS_P2 - 1 Time series models for the random component &...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online