Time series Mod15: MA(1) forecasts practice problems


Time series Mod15: MA(1) forecasts practice problems

Author
Message
NEAS
Supreme Being
Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)Supreme Being (5.7K reputation)

Group: Administrators
Posts: 4.2K, Visits: 1.2K

Time series Mod15: MA(1) forecasts practice problems

(The attached PDF file has better formatting.)

** Exercise 15.1: MA(1) Process

An MA(1) model of 100 observations has

ì = 50, è = 0.5, å99 = 6, y100 = 57, and ó2å = 16.

What was the estimate for Period 100, based on the observations through period 99?

What is the one period ahead forecast (for Period 101)?

What is the variance of the one period ahead forecast?

What is the two periods ahead forecast (for Period 102)?

What is the variance of the two periods ahead forecast?

Part A:

The estimate for period t is

ìè × åt-1, so the estimate for period 100 = 50 – 0.5 × 6 = 47.

Part B:

The estimate for period 100 was 47 and the actual value was 57, so the residual in period 100 = 10.

The estimate for period 101 is 50 – 0.5 × 10 = 45.

Part C:

The value in period t is

ì + åtè × åt-1. ì and è are parameters, not random variables. In period T+1, the value of åt is known, so its variance is zero.

The variance of YT+1 (the one period ahead forecast) = variance (

ì + åT+1è × åT) = 0 + ó2t + è2 × 0 = ó2t = 16.

Part D:

The fitted value is the best estimate, so the expected residual in Period 101 is zero, and the forecast for Period 102 is

ì = 50.

Part E:

In period T+2, the values of

åt+1 and åt+2 are both unknown, so their variances are ó2t. The residuals in different periods are independent, so the variance of the sum of the sum of the variances. The variance of the two periods ahead forecast = variance (ì + åT+2è × åT+1) = 0 + ó2t + è2 × ó2t = (1 + è2) × ó2å = (1 + 0.52) × 16 = 20.

** Exercise 15.2: MA(1) Process

An MA(1) process is yt

ì = åt – 0.5 åt-1 , with ì = 50, åT-1 = 6, yT = 57, and ó2t = 16.

What was the fitted value for Period T?

What is the residual for Period T?

What is the one period ahead forecast?

What is the variance of the one period ahead forecast?

What is the 95% confidence interval for the one period ahead forecast, assuming a z-value of 1.96 standard deviations?

What is the two periods ahead forecast?

What is the variance of the two periods ahead forecast?

What is the 95% confidence interval for the two periods ahead forecast?

Part A:

The estimate for period t is

ìè × åt-1, so the fitted value for period T is T = 50 – 0.5 × 6 = 47.

Part B:

The residual for Period T is 57 – 47 = 10.

Part C:

The one period ahead forecast (Period T+1) is 50 – 0.5 × 10 = 45.

Part D:

The value in period t is

ì + åtè × åt-1.

ì

and è are parameters, not random variables. In period T+1, the value of åt is known, so its variance is zero.

The variance of YT+1 (the one period ahead forecast) = variance (

ì + åT+1è × åT) = 0 + ó2t + è2 × 0 = ó2t = 16.

Part E:

The standard deviation of the forecast is 160.5 = 4, so the 95% confidence interval is

(45 – 1.96 × 4, 45 + 1.96 × 4) = (37.16, 52.84).

Jacob:

Do we always use 1.96 standard deviations for the 95% confidence interval?

Rachel:

We generally do not know

ó2t. We estimate its value from the observed values and we use t-values instead of z-values for the 95% confidence interval. The regression analysis course deals with t-values and confidence intervals; the time series course gives you the figures (usually a z-value).

Part F:

The fitted value is the best estimate, so the expected residual in Period 101 is zero, and the forecast for Period T+2 is

ì = 50.

Part G:

In period T+2, the values of

åt+1 and åt+2 are both unknown, so their variances are ó2t. The residuals in different periods are independent, so the variance of the sum of the sum of the variances. The variance of the two periods ahead forecast = variance (ì + åT+2è × åT+1) = 0 + ó2t + è2 × ó2t = (1 + è2) × ó2å = (1 + 0.52) × 16 = 20.

Part H:

The standard deviation of the forecast is 200.5 = 4.472, so the 95% confidence interval is

(50 – 1.96 × 4.472, 50 + 1.96 × 4.472) = (41.23, 58.77).


Attachments
GO
Merge Selected
Merge into selected topic...



Merge into merge target...



Merge into a specific topic ID...





Reading This Topic


Login
Existing Account
Email Address:


Password:


Social Logins

  • Login with twitter
  • Login with twitter
Select a Forum....











































































































































































































































Neas-Seminars

Search