TS module 20 seasonal models practice problems


TS module 20 seasonal models practice problems

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TS module 20 seasonal models practice problems

 

(The attached PDF file has better formatting.)

 

** Exercise 20.1: Seasonal moving average process

 

A time series is generated by Yt = et – Θ1 et-12 – Θ2 et-24, with ó2ε = ó2)

 


A.    What is ã0?

B.    What is ã12 = ã1 × 12?

C.    What is ñ12 = ñ1 × 12?


 

 

Part A: The variance of Yt = variance (et – Θ1 et-12 – Θ2 et-24) = ó2 + Θ12 × ó2 + Θ22 × ó2.

 

Part B: The covariance of (Yt, Yt-12) = covariance (et – Θ1 et-12 – Θ2 et-24, et-12 – Θ1 et-24 – Θ2 et-36) =

 

– Θ1 × ó2 + Θ1 × Θ2 × ó2

 

Part C: The autocorrelation for one seasonal lag = ñ12 = ñ1 × 12 =

 

(– Θ1 × ó2 + Θ1 × Θ2 × ó2 ) / (ó2 + Θ12 × ó2 + Θ22 × ó2) = (– Θ1 + Θ1 × Θ2) / (1 + Θ12 + Θ22)

 


 

** Exercise 20.2: Seasonal moving average process

 

A seasonal moving average process has a characteristic polynomial of (1 – èx)(1 – Θx12).

 

For simplicity, assume ó2e = 1, so we can ignore the ó2ε term in the covariances.

 


 

A.    What is ã0?

B.    What is ã1?

C.    What is ñ1?

D.    What is ã12?

E.    What is ñ12?

F.    What is ã11?

G.    What is ñ11?

H.    What is ã13?

I.      What is ñ13?


 

 

Part A: This time series can be written as Yt = etè et-1 – Θ et-12 + è Θ et-13 (see equation 10.1.2 on page 230). 

ã0 = 1 + è2 + Θ2 + è2 × Θ2 = (1 + è2) × (1 + Θ2).

 

Part B: ã1 = covariance (etè et-1 – Θ et-12 + è Θ et-13, et-1è et-2 – Θ et-13 + è Θ et-14 ) =

 

èè Θ2 = – è × (1 + Θ2).

 

Part C: ñ1 = ã1 / ã0 = – è × (1 + Θ2) / (1 + è2) × (1 + Θ2) = – è / (1 + è2).

 

Part D: ã12 = covariance (etè et-1 – Θ et-12 + è Θ et-13, et-12è et-13 – Θ et-24 + è Θ et-25 ) =

 

– Θ – è2 Θ = – Θ × (1 + è2).

 

Part E: ñ12 = ã12 / ã0 = – Θ × (1 + è2) / (1 + è2) × (1 + Θ2) = – Θ / (1 + Θ2) .

 

Part F: ã11 = covariance (etè et-1 – Θ et-12 + è Θ et-13, et-11è et-12 – Θ et-23 + è Θ et-24 ) = + Θ × è.

 

Part G: ñ11 = ã11 / ã0 = Θ × è / (1 + è2) × (1 + Θ2).

 

Part H: ã13 = covariance (etè et-1 – Θ et-12 + è Θ et-13, et-13è et-14 – Θ et-25 + è Θ et-26 ) = + Θ × è.

 

Part I: ñ13 = ã13 / ã0 = Θ × è / (1 + è2) × (1 + Θ2).

 

Jacob: Do exam problems give the characteristic polynomial for seasonal models?

 

Rachel: Yes. A characteristic polynomial of (1 – èx)(1 – Θx12) means Yt = etè et-1 – Θ et-12 + è Θ et-13

 


 

** Exercise 20.3: Seasonal moving average process

 

A seasonal moving average process has a characteristic polynomial of (1 – èx)(1 – Θx12), with è = 0.4, Θ = 0.5, and ó2e = 4.

 


 

A.    What is ã0?

B.    What is ã1?

C.    What is ñ1?

D.    What is ã12?

E.    What is ñ12?

F.    What is ã11?

G.    What is ñ11?

H.    What is ã13?

I.      What is ñ13?


 

 

Part A: ã0 = (1 + è2) × (1 + Θ2) × ó2ε = (1 + 0.16) × (1 + 0.25) × 4 = 5.800.

 

Part B: ã1 = – è × (1 + Θ2) × ó2ε = – 0.4 × 1.25 × 4 = -2.000.

 

Part C: ñ1è / (1 + è2) = –0.4 / 1.16 = -0.345.

 

Part D: ã12 = – Θ × (1 + è2) × ó2ε = -2.320.

 

Part E: ñ12 = – Θ / (1 + Θ2) = -0.5 / 1.25 = -0.400.

 

Part F: ã11 = + Θ × è × ó2ε = 0.4 × 0.5 × 4 = 0.800.

 

Part G: ñ11 = ã11 / ã0 = 0.13793.

 

Part H: ã13 = + Θ × è × ó2ε = 0.4 × 0.5 × 4 = 0.800.

 

Part I: ñ13 = ã13 / ã0 = 0.13793.

 

 

 


 

** Exercise 20.4: Seasonal ARMA process

 

A seasonal ARMA process Yt = Ö Yt-12 + etè et-1 has a variance of the error term ó2e.

 

This is a multiplicative seasonal ARMA(p,q) × (P,Q) process, with p = 0, q = 1, P = 1, Q = 0.

 

Let k be an integer (1, 2, 3, …).

 


 

A.    What is ã0?

B.    What is ñ12?

C.    What is ñ13?

D.    What is ñ11?

E.    What is ñ12k?

F.    What is ñ12k+1?

G.    What is ñ12k–1?


 

 

Part A: The variance of the multiplicative ARMA process is the product of the autoregressive and moving average parts.

 


 

        An MA(1) process has ã0 = ó2å (1 + è2). 

        An AR(1) process has ã0 = ó2å / (1 – ö2). 

        A seasonal AR(1) process has ã0 = ó2å / (1 – Ö2). 

        A multiplicative seasonal ARMA(p,q) × (P,Q) process has ã0 = ó2e × (1 + è2) / (1 – Ö2).


 

 

Part B: The moving average part of this process has a one-period effect. The 12 month autocorrelation stems from the autoregressive process: ñ12 = Ö

 

Part C: The 13 month autocorrelation is the product of the one-month moving average autocorrelation and the 12 month autoregressive correlation: ñ13 = – Ö è / (1 + è2)

 

Part D: The 11 month autocorrelation is the product of the moving average autocorrelation for a lag of negative one month and the 12 month autoregressive correlation: ñ13 = – Ö è / (1 + è2)

 

Jacob: How did we get the moving average autocorrelation for a lag of negative one month?

 

Rachel: A stationary ARMA process is symmetric: ñk = ñ–k

 

Part E: The autocorrelation for a lag of 12k months is Ök.

 

Part F: The autocorrelation for a lag of 12k+1 months is – Ök è / (1 + è2).

 

Part G: The autocorrelation for a lag of 12k-1 months is – Ök è / (1 + è2).

 

Jacob: For the multiplicative seasonal moving average process, we wrote the process as a combination of error terms to compute the covariance at each lag. Can we do the same with the multiplicative seasonal ARMA process?

 

Rachel: Yt = et + Ö Yt-12 can be written as Yt = et + Ö et-12 + Ö2 et-24 + Ö3 et-36 + … Combine this infinite series with the moving average piece to get a single expression of error terms. Compute the covariances to get the formulas in the textbook. Each covariance is a infinite series of Ö terms, which is rewritten as Ök / (1 – Ö2).

 


 

** Exercise 20.5: Seasonal ARMA process

 

A seasonal ARMA process Yt = Ö Yt-12 + etè et-1 has Ö = 0.4, è = 0.5, and ó2e = 4

 


 

A.    What is ã0?

B.    What is ñ12?

C.    What is ñ13?

D.    What is ñ11?

E.    What is ñ24?

F.    What is ñ23??


 

 

Part A: ã0 = ó2e × (1 + è2) / (1 – Ö2) = 4 × (1 + 0.52) / (1 – 0.42) = 6

 

Part B: ñ12 = Ö = 0.4

 

Part C: ñ13 = – Ö è / (1 + è2) = – (0.4) × 0.5 / (1 + 0.52) = -0.160.

 

Part D: ñ11 = – Ö è / (1 + è2) = – (0.4) × 0.5 / (1 + 0.52) = -0.160.

 

Part E: ñ24 = Ö2 = (0.4)2 = 0.160.

 

Part F: ñ23 = – Ö2 è / (1 + è2) = – (0.4)2 × 0.5 / (1 + 0.52) = -0.064.

 

 


 

** Exercise 20.6: Seasonal non-stationary AR(1)12 process

 

A store’s toy sales in constant dollars follow a seasonal non-stationary AR(1)12 process: Yt = Ö Yt-12 + et.

 


 

        Sales are $10,000 in January 20X1 and $100,000 in December 20X1.

        Projected sales are $110,000 in December 20X2.


 

 


 

A.    What is Ö?

B.    What are projected sales for January 20X2?

C.    What are projected sales for January 20X3?


 

 

Part A: Using December 20X1 and December 20X2, we have $110,000 = Ö × $100,000 E(et) Ö = 1.1, since the expected residual is zero.

 

Part B: The forecast for January 20X2 is 1.1 × $10,000 = $11,000.

 

Part C: The forecast for January 20X3 is 1.1 × $11,000 = $12,100.

 

(See Cryer and Chan, page 241, last line; Ö = 1.1)

 

Jacob: How do we make this time series into a stationary process?

 

Rachel: Take logarithms and first differences to make this process stationary.

 


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For the autocorrelation coefficient, isn't the definition for p(t,s) = Corr(Yt,Ys) = Cov(Yt,Ys)/ Sqrt(Var(Yt)x Var(Ys)). In the first problem why are we not taking square root in part C?


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apgarrity - 11/28/2017 2:25:06 AM

For the autocorrelation coefficient, isn't the definition for p(t,s) = Corr(Yt,Ys) = Cov(Yt,Ys)/ Sqrt(Var(Yt)x Var(Ys)). In the first problem why are we not taking square root in part C?


Hey,

aurocorrelation rho_12 = gamma_12/gamma_0

We do use a square root in the denominator. Since gamma_0 (variance of Y) is the same for Y_t and Y_(t-12), the denominator is sqrt[var(Y_t)*var(Y_(t-12)]=sqrt[var(Y_t)*var(Y_t)]=var(Y_t)=gamma_0 ‌‌ 
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