TS Module 7 Stationary mixed processes practice problems
(The attached PDF file has better formatting.)
Time Series Practice Problems ARMA Means and Variances
ARMA processes have autoregressive parameters öj, moving average parameters èj, a constant ä or è0, and a standard error ó2. We derive
Means and variances of the Yt terms and an autocorrelation function.
Forecasts with their standard errors and confidence intervals.
The practice problems here cover means and variances of the time series. Other sets of practice problems show how to work more complex items.
The textbook shows all the relations tested on the final exam and used in the student projects. As you review the modules, work through the derivations of each result. The relations follow directly from first principles.
The final exam problems emphasize the intuition. An exam problem might give the mean or variance of the time series and back into a coefficient. If you understand how autoregressive and moving average parameters affect the means, forecasts, variances, and standard errors, the problems are straight-forward.
*Question 7.1: Mean of ARMA Process
An ARMA(p,q) process is Yt = 3 + åt + moving average parameters èj = 0.5j for j = 1 to 4 and autoregressive parameters ök = 0.366k for k = 1 to 2. Note that the constant term is 3. What is the mean of this time series?
A. 2
B. 4
C. 6
D. 8
E. 10
Answer 7.1: C
The constant term is called è0 in the Cryer and Chan textbook. Other authors use ä or á. The constant term causes a displacement, not a change in the time series pattern.
Intuition: If all observed terms are the mean and all residuals are zero, the forecast of the next term is the mean. Substitute ì for the Y terms and 0 for the å terms, giving a linear equation in one unknown ì.
For this exercise, ì = 3 + ì × (0.3661 +0.3662) + 0 × (terms with è)
The mean ì of this process is ì = 3 / (1 – ∑öj) = 3 / (1 – 0.366 – 0.3662) = 5.999
The textbook has two ways of writing an ARMA process:
Using terms of (Yt – ì), with no constant term.
Using terms of Yt, with a constant term.
They are equivalent.
*Question 7.2: Mean of ARMA Process
An ARMA(p,q) process is Yt = 3 + åt + moving average parameters èj = 0.5j for j = 1 to 4 and autoregressive parameters ök = 0.366k for k = 1 to 2. All values for t < 10 were the mean, and all residuals were zero.
If the actual value y10 = 3, what is the expected value of y11?
A. 3.0
B. 3.3
C. 3.6
D. 6.0
E. 6.4
Answer 7.2: E
The residuals for t < 10 are zero (by the assumptions).
The expected value for t = 10 is the mean; work out the residual for t = 10.
The moving average coefficient = the negative of the moving average parameter.
The mean ì of this ARMA process is ì = 3 / (1 – ∑öj) ➾
μ = 3 / (1 – 0.366 – 0.3662) = 5.999
The residual for y10, or å10, is 3 – 6 = –3.
The expected value of y11 = 3 + 0.366 × 3 + 0.3662 × 6 – 0.5 × (–3) = 6.402
*Question 7.3: Mean of ARMA Process
An ARMA(p,q) process has moving average parameters èj = 0.5j for j = 1 to 4 and autoregressive parameters ök = 0.366k for k = 1 to 2. The parameter ä = 5. What is the mean of this time series?
A. 7
B. 8
C. 9
D. 10
E. 11
Answer 7.3: D
The formula relating ì and ä for an autoregressive process is ì = ä / (1 – ∑öj) ➾
μ = 5 / (1 – 0.366 – 0.3662) = 9.999
To verify, we assume all past values are the mean of 10 and all past residuals are zero. The next value is expected to be 5 + 0.366 × 10 + 0.3662 × 10 = 10.000 .
*Question 7.4: Mean of ARMA Process
An ARMA(p,q) process has moving average parameters èj = 0.5j for j = 1 to 4 and autoregressive parameters ök = 0.366k for k = 1 to 2. The parameter ä = 5. Assume that all values for t < 10 were the mean, and all residuals were zero.
If y10 = 9, what is the expected value of y11? We worked out the mean in the previous problem, which gives the values for y9 and y8. The residuals for t ≤ 9 are zero (by the assumptions). The expected value for t = 10 is the mean, from which we work out the residual for t = 10. The convention in the textbook is that the moving average coefficient is the negative of the moving average parameter.
A. 8
B. 9
C. 10
D. 11
E. 12
Answer 7.4: C
The residual for y10, or å10, is 9 – 10 = –1.
The expected value of y11 = 5 + 0.366 × 9 + 0.3662 × 10 – 0.5 × (–1) = 10.005
*Question 7.5: ARMA Mean
We use an ARMA(3,1) model to forecast a time series:
The mean of this time series is 9. What is the constant term ä of the time series?
A. –2.70
B. –2.25
C. –1.20
D. +2.25
E. +2.70
Answer 7.5: E
Cryer and Chan use the variable è0 instead of ä. The discussion forum postings use è0, ä, or á. The final exam problems may derive the constant term, the mean, or one autoregressive parameter. The solution is straight-forward.
The mean is the constant term δ divided by (1 – ∑φj). The autoregressive coefficients affect the mean; the moving average coefficients do not affect the mean.
ì = ä / (1 – ∑öj) ➾ ä = ì × (1 – ∑öj) = 9 × (1 – 0.3 – 0.2 – 0.2) = 2.700
*Question 7.6: AR(1) Process
We use a time series process to model a sequence of values:
What is the mean of this process?
A. 3
B. 6
C. –6
D. 12
E. The time series is not stationary and has no mean
Answer 7.6: E
If the absolute value of the φ1 coefficient is more than or equal to one, the time series has no mean (is not stationary). For higher order processes, the conditions for stationarity are more complex.
Jacob: Why does the time series have no mean? Suppose ö1 is a negative number less than or equal to –1. Why can’t we use the formula ì = ä / (1 – ö1)?
Rachel: Suppose ö1 is –2 and y1 is zero. The formula ì = ä / (1 – ö1) = 6 / (1 – –2) = 2.
The expected values of the next entries are +6, –6, +18, –30, +66, and so forth. The expected values assume å = 0.
The values oscillate about 2, but they move away from 2; they don’t approach 2.
Jacob: What if ö1 = –1, as in the practice problem here?
Rachel: The formula gives ì = ä / (1 – ö1) = 6 / (1 – –1) = 3.
The expected values of the next entries are +6, 0, +6, 0, +6, 0, and so forth.
The values oscillate about 3, but they don’t converge or diverge. The series has no mean.
Take heed: The mean ì is the expected value as t ➝ ∞. Even if an error term å moves the time series away from this mean for a few periods, later values come back to the mean. In this time series, if a random fluctuation moves the process away from 3, later terms do not revert to 3.
*Question 7.7: AR(1) Process
We use an time series process to model a sequence of values:
What is ã0, the variance of yt?
A. 4
B. 6
C. 8
D. 12
E. The time series is not stationary and has no finite variance.
Answer 7.7: E
The φ1 coefficient is greater in absolute value than one. The uncertainty in the forecasts increases without bound as the number of periods ahead increases. The variance of yt is like the variance of the forecast for an infinite number of periods ahead.
Jacob: Is the variance of Yt the same as the variance of the error term?
Rachel: Distinguish the variance of the error term and the variance of the Y terms.
Suppose we know the past terms of the time series and we must estimate the next value.
The value is stochastic, so we do not know it with certainty.
We estimate the distribution of this value.
The distribution is normal with a variance ó2.
Suppose we do not know any past terms of the time series and we estimate the next value.
The mean of the distribution depends on the past values.
For an AR(1) process, it depends on one past value.
For other ARMA processes, it may depend on several past values.
Each unknown value adds uncertainty and increases the variance of the estimate.
We combine the various sources of uncertainty.
If the variances of the sources are independent, we add them. Some sources of uncertainty are correlated, such as autoregressive and moving average parameters for the same period.
In an ARMA(1,1) process, the uncertainty is perfectly correlated. We add the autoregressive and moving average coefficients (i.e., subtract the parameters).
Think of the variances in a continuum.
The variance of the one period ahead forecast is the variance of the error term.
The variance of the l period ahead forecast as l is the variance of Yt.
In between are the variance of other forecasts.
*Question 7.8: AR(1) Process
We use an AR(1) process (autoregressive of order 1) to model a time series:
, with σ2 = 21.
What is ã0, the variance of yt?
A. 2.1
B. 7
C. 21
D. 25
E. 49
Answer 7.8: D
21 / (1 – 0.42) = 25.000
This is not the variance of the forecast, which assumes we know the past values. This is the variance of the time series value itself.
Each Yt is the sum of two random variables: εt and 0.4yt-1. In an autoregressive process, yt depends on the lagged values, but the residuals are independent of the lagged values.
You can solve this problem by the formula in the textbook or by intuition. The intuition is that the variance is the same for all elements of the time series. The variance of yt equals the variance of yt-1. Form an equation for the relation of the variances from the time series equation Var(yt) = 0.42 × Var(yt-1) + 0 + ó2
Jacob: In the regression analysis course, the variance of Y is the variance of the error term. Why is that not true here?
Rachel: In the regression analysis course, Y is a function of X, which is not a random variable. In an autoregressive time series, Yt is a function of Yt-1, which is a random variable.
*Question 7.9: Moving Average Model of Period N
A time series follows a moving average model of period N:
What is the mean of this time series?
A. ½
B. 1
C. 1 + 1/N
D. 2
E. The time series is not stationary and does not have a mean
Answer 7.9: B
For a moving average process, μ = ä = è0 = 1 (in this problem).
The mean does not depend on the moving average parameters.
*Question 7.10: Moving Average Model of Period N
A time series follows a moving average model of period N:
What is the variance of this time series?
A. 1
B. (N + 1)/N = 1 + 1/N
C. (N + 1)2/N2
D. 2
E. The time series is not stationary and does not have a variance
Answer 7.10: B
For a moving average process, the residuals are independent and have the same variance.
The variance of the sum of the residuals is the sum of the variances.
The variance of áY, where á is a scalar and Y is a random variable, is á2 × variance(Y).
A moving average process is stationary if the sum of the moving average parameters is finite.
The variance is σ2 × (1 + N × 1/N2) = 1 × (1 + 1/N) = 1 + 1/N