TS Module 9 Non-stationary IMA processes.

(The attached PDF file has better formatting.)

Background: Starting points for non-stationary time series

Cryer and Chan use non-stationary processes that have a fixed starting point. If the non-stationary time series has no starting point, its mean and variance are not defined.

Illustration: Infinite ARIMA(0,1,1) process

Suppose Y_t is an ARIMA(0,1,1) process = an IMA(1,1) process:

 or

If the time series has no starting point, then Y_t = å_t + (1 – è) å_t-1 + (1 – è) å_t-2 + …

Y_t has no mean and a non-finite variance. Each entry in the non-stationary time series is the sum of an infinite number of random variables that do not die out.

Jacob: The expected value of each å_t is zero. Doesn’t this imply that the mean of Y_t is zero?

Rachel: If the current value of Y_t is 1, the expected values of all future observations is also 1. If the current value of Y_t is –1, the expected values of all future observations is also –1. If a time series has a mean, then the expected value as t ➝ ∞ is this mean. In this time series, the expected value in the limit depends on the current value, so the process is not stationary.

This is a simplified explanation. A more rigorous analysis shows that the process has no mean. Suppose one asks: “What is the mean of the real number line?” One might say: “Start at zero. The probability of a value k is the same as the probability of the value –k, so the two values offset each other ➾ the mean of the real number line is zero.” This reasoning is not correct. One could also start at any point m and argue that the points m + k and m – k offset each other ➾ the mean of the real number line is m. The correct answer it that the real number line has no mean. Similarly, a random walk has no mean.

To examine the pattern of non-stationary ARIMA processes, assume they begin at some time t = –m, so

➾ Y_t = å_t + (1 – è) å_t-1 +  (1 – è) å_t-2 + … + (1 – è) å_-m – è å_-m-1 

This is not really a restrictive condition. Most commonly, we forecast future values of a time series based on historical observations. Before the first observation, we assume all values of the time series are zero. We model the evolution of the mean and variance of the process.

**Question 1.2: ARIMA(0,1,1) process

Suppose

with è = 0.4 and ó² = 4 for t > 0 and Y_t = 0 for t ≤ 0.

What is the variance of Y₂?

A.      5.440

B.      6.080

C.      7.520

D.     8.160

E.      20.320

Answer 1.2: B

Write the ARIMA(0,1,1) process in the long form and determine the variance of each random variable.

           Y₂ = Y₁ + å₂ – 0.4 å₁ = å₂ + (1 – 0.4) å₁ – 0.4 å₀

           The variance of Y₂ is 4 + (1 – 0.4)² × 4 + 0.4² × 4 = 6.080

Jacob: If we don’t specify that Y_t = 0 for t < 0, what is the variance of Y_t?

Rachel: The process is not stationary and has no variance.

Jacob: If we know Y₁ and forcast Y₂, what is the variance of the forecast of Y₂?

Rachel: The one period ahead forecast has only one random error term, so the variance is 4.

See Cryer and Chan, P94: equation 5.2.7, for IMA(1,1) process:

 =

Jacob: Do we always assume the time series starts at time 0?

Rachel: Cryer and Chan assume it starts at Period –m. Final exam problems often assume it starts at Period 0 or Period 1. They can be solved by first principles, as in this exercise.

**Question 1.3: IMA(1,1) process

An IMA(1,1) process has è = 0.4 and Y_t = 0 for t < –20.

What is ñ(Y_t, Y_t-1) for t = 50?

A.      –0.02

B.      –0.05

C.      +0.02

D.     +0.05

E.      +0.99

Answer 1.3: E

(See Cryer and Chan, chapter 5, page 94, equation 5.2.8)

Jacob: What is the intuition for this result?

Rachel: Each Y_t is the sum of t+20 random error terms. For t = 50, the correlation of Y_t and Y_t-1 has 70 terms which overlap and one term which differs. The two elements are almost identical, whereas the range of possible values of Y_t is broad. As N ➝ ∞, the correlation of successive elements of a non-stationary IMA time series ➝ 1.

**Question 1.4: IMA(2,2) process

Which of the following is an IMA(2, 2) process?

A.      Y_t = 2 Y_t-1 – Y_t-2 + e_t – è₁ e_t-1 – è₂ e_t-2

B.      Y_t = Y_t-1 – Y_t-2 + e_t – è₁ e_t-1 – è₂ e_t-2

C.      Y_t = Y_t-1 – 2 Y_t-2 – e_t – è₁ e_t-1 – è₂ e_t-2

D.     Y_t = Y_t-1 – Y_t-2 – e_t – è₁ e_t-1 – è₂ e_t-2

E.      Y_t = Y_t-1 + Y_t-2 + e_t – è₁ e_t-1 – è₂ e_t-2

Answer 1.4: A

(See Cryer and Chan chapter 5, page 94, equation 5.2.9)

IMA(2,2) process:

Y_t = 2 Y_t-1 – Y_t-2 + e_t – è₁ e_t-1 – è₂ e_t-2

➾ Y_t – Y_t-1 = Y_t-1 – Y_t-2 + e_t – è₁ e_t-1 – è₂ e_t-2

➾ ∇Y_t = ∇Y_t-1 + e_t – è₁ e_t-1 – è₂ e_t-2

➾ ∇² Y_t = e_t – è₁ e_t-1 – è₂ e_t-2

**Question 1.5: ARIMA processes

           A time series is zero for periods before t = –m.

           For t > –m, the value is Y_t = å_t + (1 – è) å_t-1 +  (1 – è) å_t-2 + … + (1 – è) å_-m – è å_-m-1 

What type of ARIMA process is this time series?

A.      ARIMA(0,1,1)

B.      ARIMA(1,1,0)

C.      ARIMA(1,1,1)

D.     ARIMA(1,2,1)

E.      ARIMA(2,2,1)

Answer 1.5: A

See equations 5.2.5 and 5.2.6 at the bottom of page **.

           Equation 5.2.5 is an IMA process: Y_t = Y_t-1 + å_t – è å_t-1

           Equation 5.2.6 writes the IMA process in the form given above.

**Question 1.6: Drift of IMA(1,1) process

Let Y_tʹ be an IMA(1,1) process with a drift of zero.

Let Y_t = Y_tʹ + â₀ + â₁ × t

           â₀ and â₁ are scalars, not random variables.

           t is the index for time.

What is the drift of Y_t?

A.      â₀

B.      â₁

C.      â₀ + â₁ × t

D.     1 + â₀

E.      1 + â₁

Answer 1.6: B

           â₀ + â₁ × t is a regression line with a slope (= drift) of â₁

           The drift of Y_t is the drift of Y_t  + â₁ = â₁

Intuition: The drift of Y_t is the mean of ∇Y_t = 0 + â₀ + â₁ × t – (â₀ + â₁ × (t-1) ) = â₁