MATH 5075: Time Series Analysis

Transcription

1 NAME: MATH 5075: Time Series Analysis Final For the entire test {Z t } WN(0, 1)!!! 1

2 1) Let {Y t, t Z} be a stationary time series with EY t = 0 and autocovariance function γ Y (h). Assume that a) Show that {X t } is not stationary. X t = a + bt + ct 2 + Y t. b) Show that 2 X t = (1 B) 2 X t, where B is the backshift operator, is stationary. (Compute E 2 X t and Cov( 2 X t, 2 X t+h )). c) Describe (in detail) a different procedure to eliminate the trendfunction a + bt + ct 2. Solution: a) EX t = a + bt is not constant. b) (1 B) 2 = 1 2B + B 2. Hence 2 X t = a + bt + ct 2 + Y t 2(a + b(t 1) + c(t 1) 2 + Y t 1 ) + a + b(t 2) + c(t 2) 2 + Y t 2 = 2c + Y t 2Y t 1 + Y t 2. Thus E[ 2 X t ] = 2c. Cov( 2 X t, 2 X t+h ) = Cov(2c + Y t 2Y t 1 + Y t 2, 2c + Y t+h 2Y t 1+h + Y t 2+h ) = E[(Y t 2Y t 1 + Y t 2 ) (Y t+h 2Y t 1+h + Y t 2+h )] = γ(h 2) 4γ(h 1) + 6γ(h) 4γ(h + 1) + γ(h + 2). c) Use for example least square estimation, i.e. minimize n [X t (a + bt + ct 2 )] 2 t=1 with respect to a, b, c. Take partial derivatives and set equal 0 = we get â,ˆb, ĉ. Now define W t = X t (â + ˆbt + ĉt 2 ). When the estimates are good W t Y t : the latter is stationary. 2

3 2) Let {X t, t Z} be an AR(3) process given by X t X t X t X t 3 = Z t. This process has a stationary and causal solution. Consider the MA( ) representation of X t : X t = ψ k Z t k. Compute ψ 0, ψ 1, ψ 2, ψ 3 numerically, then give a general recursive formula for ψ k. Solution: We have φ(b)x t = θ(b)z t, where φ(z) = z 1 8 z z3 and θ(z) = 1. Let ψ(z) = ψ kz k. Thus we have ψ(z) = 1 resp. ψ(z)φ(z) = 1, φ(z) i.e. ψ k z k + 1 ψ k z k+1 1 ψ k z k ψ k z k+3 = Now equate coefficients: ψ 0 = 1, ψ ψ 0 = 0 = ψ 1 = 1 4, ψ ψ ψ 0 = 0 = ψ 2 = 3 16, ψ ψ ψ ψ 0 = 0, = ψ 3 = , ψ k ψ k ψ k ψ k 3 = 0. 3

4 3) Let {X t, t Z} be an AR(2) process given by a) Show that this process is invertible. X t = Z t 1 2 Z t Z t 2. b) Consider the AR( ) representation of Z t : Z t = η k X t k. Compute η 0, η 1, η 2 numerically, then give a general recursive formula for η k. Solution: a) Solve Hence z z2 = 0 z 2 2z + 4 = 0. z 1,2 = 1 ± 3 = 1 ± i 3 = z i > 1, i = 1, 2. Since the roots are outside the complex unit circle, the process is invertible. b)we have φ(b)x t = θ(b)z t, where φ(z) = 1 and θ(z) = z z2. Let η(z) = η kz k. Thus we have η(z) = 1 θ(z) resp. θ(z)η(z) = 1, i.e. η k z k 1 η k z k ψ k z k+2 = Now equate coefficients: η 0 = 1, η η 0 = 0 = η 1 = 1 2, η η η 0 = 0 = η 2 = 0, η k 1 2 η k η k 2 = 0. 4

5 4) a) Show that the AR(4) process given by X t 5 3 X t X t X t X t 4 = Z t has a stationary and causal solution. Hint: (1 (5x)/6 + x 2 /6) 2 = 1 (5x)/3 + (37x 2 )/36 (5x 3 )/18 + x 4 /36. b) Show that the MA(1) process X t = Z t 1 2 Z t 1 is invertible. Determine the coefficients (η k ) k 0 in the AR( ) representation Solution Z t = a) We have to show that the roots of η k X t k. 1 (5z)/3 + (37z 2 )/36 (5z 3 )/18 + z 4 /36 are outside the complex unit circle. By the hint we only need the roots of 1 (5z)/6 + z 2 /6. It is trivial to show that the roots are z 1 = 2, z 2 = 3. b) 1 1 2z = 0 if z = 2. Done. Representation: Z t = 1 2 Z t 1 + X t = 1 4 Z t X t 1 + X t = 1 = 2 k X t k. 5

6 5) Assume that we have a causal and invertible AR(2) process X t = φ 1 X t 1 + φ 2 X t 2 + Z t. a) Denote by ρ(h) the autocorrelation function. Determine ρ(0) and ρ(1) in terms of φ 1, φ 2 and then give a recursive formula for ρ(h), h 2. b) Denote by α(h) the partial autocorrelation function. Determine α(h) for h 1 in terms of φ 1 and φ 2. Solution: a) Of course ρ(0) = 1. The equation multiplied with X t 1 yields: X t X t 1 = φ 1 X 2 t 1 + φ 2X t 2 X t 1 + Z t X t 1 E γ(1) = φ 1 γ(0) + φ 2 γ(1) : γ(0) ρ(1) = φ 1 + φ 2 ρ(1). Thus Now multply the equation with X t h : ρ(1) = φ 1 1 φ 2 X t X t h = φ 1 X t 1 X t h + φ 2 X t 2 X t h + Z t X t h E γ(h) = φ 1 γ(h 1) + φ 2 γ(h 2) : γ(0) ρ(h) = φ 1 ρ(h 1) + φ 2 ρ(h 2). b) To get α(1) solve the predicton equation Thus γ(0)φ 11 = γ(1). α(1) = φ 11 = γ(1) γ(0) = ρ(1) = φ 1 1 φ 2. For the AR(2) process α(2) = φ 22 = φ 2 and α(k) = φ kk = 0 if k 3. 6

7 6) Let X t be the MA(3) process X t = Z t Z t 1 Z t 2 + 2Z t 3. a) Compute the best linear predictor ˆX n+h based on X n and X n 1 for every h 1. b) Compute the mean squared prediction error E ˆX n+1 X n+1 2. Solution: First note that a) Solve the equation system This is trivial. Then Obviously (φ (h) b) Let (φ (1) 21, φ(h) 22 21, φ(1) 22 ( γ(0) = 6 + 1/4 γ(1) = 2 γ(2) = 0 γ(3) = 2 γ(k) = 0 for k 4. ) ( φ (h) 21 φ (h) 22 ) = ( γ(h) γ(h + 1) ˆX n+h = φ (h) 21 X n + φ (h) 22 X n 1. ) = (0, 0) if h 4. Hence then the best forecast is 0. ) = (a, b) = ( 0.36, 0.11). E[X n+1 ax n bx n 1 ] 2 = γ(0)(1 + a 2 + b 2 ) 2γ(1)(a ab) 2bγ(2) = ) 7

8 7) Let {X t µ} be the AR(1) process X t µ = 0.8(X t 1 µ) + Z t. Based on the sample X 1,...,X 400, compute the length of an approximate 95% confidence interval for µ. (I want to see the numerical value.) Solution: The CI is where τ 2 = h Z γ(h). We have [ X τ, X τ], X t µ = 0.8 k Z t k. Now compute γ(h), or use the result we derived in class for MA( ) process: τ 2 = Var(Z 0 ) (sum of coefficients) 2. Thus the length is = 1. τ 2 = ( 0.8 k ) 2 = 25. 8

9 8) Let {X t } be the MA(1) process X t = Z t Z t 1. Assume EZ 4 t = 1. Based on the sample X 1,..., X 400, compute the length of an approximate 95% confidence interval for γ(0). (I want to see the numerical value.) Solution: The CI is where [ˆγ τ, ˆγ τ], τ 2 = h ZCov(X 2 0, X2 h ). Note that Cov(X 2 0, X2 h ) = E[X2 0, X2 h ] γ(0)2, with γ(0) = 5/4. We have E[X0X 2 h] 2 = E[(Z Z2 1 + Z 0 Z 1 )(Zh Z2 h 1 + Z h Z h 1 )] 41/16 if h = 0, = 33/16 if h = 1, 25/16 else. Hence τ 2 = 1 + 2(8/16) = 2. Thus the length is = 2/5. 9

10 6000 Level 9) Let {X t } be a stationary process with mean µ. Show that P sp{1,x1,...,x n}x n+h = µ + P sp{y1,...,y n}y n+h where {Y t } = {X t µ}. Solution: The projection P sp{1,x1,...,x n}x n+h is characterized by X n+h P sp{1,x1,...,x n}x n+h X i = 0 i = 1,, n and X n+h P sp{1,x1,...,x n}x n+h 1 = 0. We have to show that X n+h µ P sp{y1,...,y n}y n+h X i = 0 i = 1,, n and X n+h µ P sp{y1,...,y n}y n+h 1 = 0. or equivalently Y n+h P sp{y1,...,y n}y n+h Y i + µ = 0 i = 1,, n and Y n+h P sp{y1,...,y n}y n+h 1 = 0. Using the linearity of the inner product, it only remains to show that Y n+h P sp{y1,...,y n}y n+h 1 = 0. And this is clear since EY i = 0. 10