Command meaning

[Bird]

When I run the ARIMA model, I find out that it has seasonal lag. Therefore, I use the seasonal ARIMA. I have a question with it. For example, I read in a SARIMA guiding material, they logarize a time series and find out that the new time series has seasonal factor s=6. They then use the Gen function in eviews to generate a new series by typing: New series=d(log(variable(A),0,6)

I do not understand what it means with the function d(log(variable(A),0,6)? What does eviews really do with it? And after that I will use the new time series to forecast but it will give the result respective with the values of new time series. How can I have the result respective with the value of the original time series?

Thank you so much for your help!

[EViews Gareth]

Are you sure that that is a) an EViews command, or b) you've copied it correctly?

It is not a valid EViews command.

[Bird]

Are you sure that that is a) an EViews command, or b) you've copied it correctly?

It is not a valid EViews command.

I copied exactly that example: D(A,0,6)) (with time series A=log(xyz), you can generate 2 times: First one: typing Gen and write the command: A=log(xyz) and Second one: B=D(a,0,6))

I did it and I think it is a valid one. Thank you!

[EViews Gareth]

ok, you misplaced a parenthesis in your original post.


D(a,0,6) = (1-L^6)*a = a-a(-6)

where L is the lag operator.

[Bird]

ok, you misplaced a parenthesis in your original post.


D(a,0,6) = (1-L^6)*a = a-a(-6)

where L is the lag operator.

Thank you. But I wonder what is the difference between the command D(a,6) and the command D(a,0,6)? I think D(a,6) =(1-L^6)*a = a-a(-6) either.

BTW, the result forecasted will be respective with the value of new time series D(log(xyz),0,6). So how can I change it to the value of the original time series(say:xyz)?

Thank so much for your help!

[EViews Gareth]

No, D(a,6) = (1-L)^6

If you're doing a forecast from an equation, the forecast gives you the option to forecast the original series, xyz.

[Bird]

No, D(a,6) = (1-L)^6

If you're doing a forecast from an equation, the forecast gives you the option to forecast the original series, xyz.

Thank you so much for your help. To be honest, I am not familiar with Eviews. Could you show me how can I choose the option to forecast the original series, xyz? I normally choose Quick-Estimate Equation then typing e.g seriesxyz c ar(..) ma(...) there. So since the original time series is seasonal, I have to generate it to the new serie as above. Now I should type
seriesA c ar(...) ma(..) with the seriesA is the new series. ANd the result will give the value that are different much from the original one.

BTW: With D(a,0,6) what does 0 stand for? If I change it to D(a,1,6) what willl it mean?

I am sorry to ask you so much and somehow silly! Thank again for your help!

[EViews Gareth]

I don't understand your first question, in fact I'm not sure I understand where the seasonality comes in.

But to answer your last question, the 0 stands for the lag operation. D(a,i,j) = (1-L)^i * (1-L^j) * a. Thus:

d(a,0,6) = a-a(-6).

d(a,1,6) = (a-a(-6)) - (a(-1)-a(-7))

[Bird]

I mean when i logarize the time series and take difference it by the command d(a,0,6) (I do this for the time series to become stationary) the value of the new series now is much smaller in comparison with the original one. After that I use the new series to make a forecast. And I got the result. The forecasted result may be around 0.2 or 0.3 for example. How can I use this small value since the original value is in fact very big?

Thanks a lot for your help!

[EViews Gareth]

ok, say you have an equation of the following form:

d(log(a),1,6) c x1 x2 x3


Then when you forecast from this equation it will ask you whether you wish to forecast values of d(log(a),1,6), or values of a.

In your case you should ask for values of a to be forecasted.

[Bird]

I find it already. Thank you very much for your help.