Simulate from an ARIMA Model with constant in differenced-form

[ac_1]

Hi Tom,

How do I simulate from the following ARIMA + constant models, d=1 (in differenced-form), in RATS?
(i) ARIMA(1,1,0) + c
(ii) ARIMA(0,1,1) + c
(iii) ARIMA(1,1,1) + c

Regards
Amarjit

[TomDoan]

I'm not sure what you mean by +C. From your description it sounds like you simulate the ARIMA process, then add C to the result.

Simulating an ARIMA process itself is relatively straightforward---put in the model with EQUATION and do a SIMULATE putting zeros in for the pre-sample values. Just create an extra 50 or so data points for a burn-in and extract the information after that.

[ac_1]

I'm not sure what you mean by +C. From your description it sounds like you simulate the ARIMA process, then add C to the result.

Simulating an ARIMA process itself is relatively straightforward---put in the model with EQUATION and do a SIMULATE putting zeros in for the pre-sample values. Just create an extra 50 or so data points for a burn-in and extract the information after that.

I can follow the examples viewtopic.php?t=3090, but they are for d=0, and not including a constant.

The question is how to model those 3 ARIMA models including a constant (constant as in the output from BOXJENK), but in differenced form: d=1.
(i) ARIMA(1,1,0) including a constant
(ii) ARIMA(0,1,1) including a constant
(iii) ARIMA(1,1,1) including a constant

[TomDoan]

You can do the whole thing in one go by inputting an equation which incorporates the differencing into the AR:

Code: Select all

equation(ar=1,ma=1,const,variance=1.0) yeq y
set y 1 500 = 0.0
set u 1 500 = 0.0
*
* Coefficients are in order:
*  constant
*  AR(1) (here 1.0 to make this an ARIMA(0,1,1)
*  MA(1)
*
associate(resids=u) yeq
# 0.3 1.0 0.4
uforecast(simulate,equation=yeq) y 2 150
graph
# y 51 150

However, you can take a simulation without the differencing and just do an ACCUMULATE instruction to integrate it up. And the constant just requires adding the constant to the simulation of the AR or MA:

Code: Select all

set u 1 nburn+nobs = %ran(sigma)
set ma 1 nburn+nobs = C+u+theta*u{1}
set(first=0.0) ar 1 nburn+nobs = C+rho*ar{1}+u

acc ma / arima011
aaa ar / arima110

[ac_1]

In BOXJENK Output there is a CONSTANT which is the mean (mu) of the model or E(y(t)). Shouldn't BOXJENK CONSTANT be relabeled MEAN instead?

Using RATS ExampleThree.xls data, estimating an AR(1) model, comparing LINREG with BOXJENK the CONSTANT's are different and the mean's are the same.

Code: Select all

linreg rate 1959:02 1996:02
# constant rate{1}
comp mu = %beta(1)/(1-%beta(2))
disp mu

boxjenk(constant,ar=||1||,NOMAXL) rate 1959:02 1996:02

How is Linear Regression - Estimation by Least Squares, different to Box-Jenkins - Estimation by LS Gauss-Newton?


Further, if I simulate in-levels an ARMA model as e.g. y = 3.0 + 0.7*y(t-1) + 0.6*y(t-2) + e(t), the mu = (c1/(1.0 - phi1)) which is shown as a CONSTANT in BOXJENK, yet I cannot use BOXJENK output results for a forecast by-hand. To convert BOXJENK CONSTANT to be used for forecasting I would c1 = mu*(1.0 - phi1). Why does BOXJENK not display the constant (maybe as-well) which is used for forecasting, the mean is not?

If I integrate-up y(t) using the levels simulation series and estimate using BOXJENK with diff=1 I get the same results.

But how do I write the model? Is it

ARIMA(1,1,1)\\y(t) = 3.0 + 0.7*y(t-1) + e(t) + 0.6*e(t-1)
Or
ARIMA(1,0,1)\\d(y(t)) = 3.0 + 0.7*d(y(t-1)) + e(t) + 0.6*e(t-1)

Code: Select all

*===============================
* y = c1 + phi1*y(t-1) + e(t) + theta1*e(t-1)

seed 123

compute sigma=1.0

compute nburn=100
compute nobs=200

set e 1 nburn+nobs = %ran(sigma)

compute c1 = +3.0
compute phi1 = +0.7
compute theta1 = +0.6
compute mean1 = (c1/(1.0 - phi1))
disp 'mean' mean1

set(first=0.0) y 1 nburn+nobs = c1 + phi1*y{1} + e + theta1*e{1}


spgraph(vfields=1,hfields=2,footer="")
graph(head="ARMA(1,1)\\y(t) = 3.0 + 0.7*y(t-1) + e(t) + 0.6*e(t-1)") 1
# y nburn+1 nburn+nobs
@bjident(method=yule,diffs=0,report,number=15) y nburn+1 nburn+nobs
spgraph(done)


boxjenk(constant,ar=||1||,diff=0,ma=||1||,MAXL) y nburn+1 nburn+nobs


* integrate-up
acc Y / arima111

spgraph(vfields=2,hfields=2,footer="")
graph(head="ARIMA(1,1,1)\\y(t) = 3.0 + 0.7*y(t-1) + e(t) + 0.6*e(t-1)") 1
# arima111 nburn+1 nburn+nobs
@bjident(method=yule,diffs=1,report,number=15) arima111 nburn+1 nburn+nobs
spgraph(done)

boxjenk(const,diff=1,ar=||1||,ma=||1||,MAXL) arima111 nburn+1 nburn+nobs

[TomDoan]

In BOXJENK Output there is a CONSTANT which is the mean (mu) of the model or E(y(t)). Shouldn't BOXJENK CONSTANT be relabeled MEAN instead?

No. Because (a) the process may have no mean (if it's integrated) and (b) if there are any other shift variables, it's not the mean anyway.

Using RATS ExampleThree.xls data, estimating an AR(1) model, comparing LINREG with BOXJENK the CONSTANT's are different and the mean's are the same.

Code: Select all

linreg rate 1959:02 1996:02
# constant rate{1}
comp mu = %beta(1)/(1-%beta(2))
disp mu

boxjenk(constant,ar=||1||,NOMAXL) rate 1959:02 1996:02

How is Linear Regression - Estimation by Least Squares, different to Box-Jenkins - Estimation by LS Gauss-Newton?

Because the Box-Jenkins model is a structural model, while the LINREG is the equivalent reduced form model. The estimated coefficients have different meanings.

From the User's Guide: "The advantages of the form chosen by RATS is that the estimated coefficient on the constant has a simple interpretation as the mean of the differenced process. It also generalizes better to intervention models and the like."

The description of BOXJENK explains how DEFINE transforms the structural ARIMA model into something usable for forecasting. Note, for instance, that the output from BOXJENK describes the AR terms as AR{1}, AR{2},... NOT as Y{1} and Y{2},... since they aren't.

The rest of your question seems to be a repetition of the confusion between the two.