TS Module 9 Non-stationary ARIMA time series

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Time series ARIMA processes practice problems

** Exercise 9.1: ARIMA Process

A time series is Y_t = 1.4Y_t-1 + 0.1Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

A.    Write the time series in terms of W_t ( Y_t).

B.    What is the ARIMA process followed by this time series?

C.    What are the coefficients of this ARIMA process?

Part A: Rewrite the ARIMA process as

Y_t – Y_t-1 = 0.4Y_t-1 + 0.1Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

= 0.4Y_t-1 – 0.4Y_t-2 + 0.4Y_t-2 + 0.1Y_t-2  – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

= 0.4Y_t-1 – 0.4Y_t-2 + 0.5Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

➾ W_t = 0.4 W_t-1 + 0.5W_t-2 + e_t + 0.3e_t-1 + 0.2e_t-2

Jacob: What is the procedure for this transformation?

Rachel: The sum of the coefficients for the Y terms are equal on both sides of the equation.

The left side has a coefficient of 1. The right side has coefficients of 1.4 + 0.1 – 0.5 = 1.

Part B: The time series is an ARIMA(2,1,2) process.

        d = 1: we took one difference ( Y_t).

        p = 2: we use W_t-1 and W_t-2.

        q = 2: we use e_t-1 and e_t-2.

Part C: The ARIMA coefficients are

ö₁ = 0.4

ö₂ = 0.5

è₁ = –0.3

è₂ = – 0.2

** Exercise 9.2: ARIMA Process

A time series is Y_t = è₀ + á₁ × Y_t-1 + á₂ × Y_t-2 + á₃ × Y_t-3 + â₁ × e_t + â₂ × e_t-1 + â₃ × e_t-2

A.    Write the time series in terms of W_t ( Y_t).

B.    What is the ARIMA process followed by this time series?

Part A: Rewrite the ARIMA process as

Y_t – Y_t-1 = è₀ + (á₁ – 1) × Y_t-1 + á₂ × Y_t-2 + á₃ × Y_t-3 + â₁ × e_t + â₂ × e_t-1 + â₃ × e_t-2

Y_t–Y_t-1 = è₀ + [(á₁ – 1) × Y_t-1 – (á₁ – 1) × Y_t-2] + [(á₁ – 1) × Y_t-2 + á₂ × Y_t-2 + á₃ × Y_t-3] + â₁ × e_t + â₂ × e_t-1 + â₃ ×e_t-2

The coefficient of Y_t-1 – Y_t-2 is (á₁ – 1).

For this to be an ARIMA(2,1,2) process, we must have

(á₁ – 1) + á₂ = –á₃

á₁ + á₂ + á₃ = 1

Jacob: What about the â coefficients?

Rachel: Any â coefficients are fine.

        If â₁ = 1, then ö₁ = –â₂ and ö₂ = –â₃.

        If â₁   1, then ö₁ = –â₂ / â₁ and ö₂ = –â₃ / â₁.

** Exercise 9.3: Combining error terms

Suppose Y_t = M_t + e_t and M_t = M_t-1 + å_t

A.    Write Y_t as a function of M_t-1 and error terms.

B.    What type of time series is M_t?

C.    What type of time series is Y_t?

D.    What is  Y_t (the first difference of Y_t)?

E.    What is the variance of Y_t?

F.    What is the variance of  Y_t?

G.    What is the covariance of  Y_t and  Y_t-1?

H.    What is ñ₁, the autocorrelation of  Y_t and  Y_t-1?

Part A: Y_t = M_t + e_t = M_t-1 + e_t + å_t

Part B: M_t is a random walk.

Part C: Y_t = M_t-1 + e_t + å_t = Y_t-1 + å_t + e_t – e_t-1. This is a random walk with a more complex error term.

Part D: ∇Y_t = Y_t – Y_t-1 = M_t-1 + e_t + å_t – ( M_t-1 + e_t-1) = å_t + e_t – e_t-1

Part E: If the random walk has no beginning, the variance is infinite, so it does not exist. If the random walk has a beginning, the variance depends on the period.

Part F: The variance of ∇Y_t = var( å_t + e_t – e_t-1 ). The three random variables are independent, so the variance = 2ó²_e + ó²_ε.

Part G: The covariance of ∇Y_t and ∇Y_t-1 is covariance ( å_t + e_t – e_t-1 , å_t-1 + e_t-1 – e_t-2 ) = –ó²_e.

Part H: The autocorrelation of ∇Y_t and ∇Y_t-1 (ñ₁) is –ó²_e / 2ó²_e + ó²_ε = –1 / (2 + ó²_ε / ó²_e ). This is equation 5.1.10 on page 90.

** Exercise 9.4: IMA(1,1) process

Each of the following time series is an IMA(1,1) process. What is the value of è for each time series?

A.    Y_t = Y_t-1 + e_t – 0.4e_t-1

B.    Y_t = Y_t-1 – e_t – 0.4e_t-1

C.    Y_t = Y_t-1 + 0.4e_t – 0.4e_t-1

D.    Y_t = Y_t-1 – 0.4e_t – 0.4e_t-1

Part A: The first difference of an IMA(1,1) is an MA(1) process.

The first difference of this time series is e_t – 0.4e_t-1, which is an MA(1) process with è = 0.4.

Part B: The first difference of this time series is – e_t – 0.4e_t-1. Use a change of the error term å_tʹ = –å_t, which gives a first difference of + e_tʹ + 0.4e_tʹ-1, which is an MA(1) process with è = –0.4.

Part C: The first difference of this time series is 0.4 e_t – 0.4e_t-1. Use a change of the error term å_tʹ = 2.5å_t, which gives a first difference of + e_tʹ – e_tʹ-1, which is an MA(1) process with è = 1.

Part D: The first difference of this time series is –0.4 e_t – 0.4e_t-1. Use a change of the error term å_tʹ = –2.5å_t, which gives a first difference of + e_tʹ + e_tʹ-1, which is an MA(1) process with è = –1.

(P93)

** Exercise 9.5: ARI(1,1) process

The time series Y_t = è₀ + á Y_t-1 + â Y_t-2 + e_t is an ARI(1,1) process.

A.    Write the time series in terms of W_t ( Y_t).

B.    What is the relation of á and â?

C.    What is the value of ö for this ARI(1,1) process?

Part A: W_t = ∇Y_t = Y_t – Y_t-1 = è₀ + (á – 1) × Y_t-1 + â Y_t-2 + e_t

Part B: If á – 1 = –â, we can write the time series as Y_t – Y_t-1 = è₀ + (–â) × (Y_t-1 – Y_t-2) + e_t

Part C: ö = – â

A.    Y_t = Y_t-1 – (1 + ö) Y_t-2 + e_t

B.    Y_t = Y_t-1 + ö Y_t-2 + e_t

C.    Y_t = Y_t-1 – ö Y_t-2 + e_t

D.    Y_t = (1 + ö) Y_t-1 – ö Y_t-2 + e_t

E.    Y_t = (1 + ö) Y_t-1 + ö Y_t-2 + e_t

** Exercise 9.6: Time series process

A time series is Y_t = è₀ + 1.75 Y_t-1 – 0.75Y_t-2 + e_t is an ARI(2,1) process.