The RATS Software Forum

Sorry, I think I'm not clear. For the prevision out of sample I use Forecasts and steps, and for the prevision in sample I use FItted. But with fitted I can't use steps, so I can't define how step I want for my in sample prevision.

But I can't use fitted beacause I want to put diferent step for the in sample and not the out of sample

Hello, I try to make a prevision with an exponential smoothing but I have a probleme with the sample fit with reestimation prevision. For the prevision I do: do time = prevstart,prevend-stepsahead esmooth(trend=exponential,estimate, resids=resexo2,fitted=forecast1) ev 1960:1 * forecast(model= ?????,...

Hello, I don't understand why but with two diferent version of rats, I haven't the same results. In Rats 8.0: boxjenk(constant,ar=1, ma=1,maxl,diffs=1,regressor) pib 1960:01 * resexo # trend trend2 Box-Jenkins - Estimation by ML Gauss-Newton Convergence in 20 Iterations. Final criterion was 0.000001...

Hello,

I have an ARMA regression that is not stable, what can I do to make it stable? Differentiate it? Add a trend?

I don't understand the interpretation of the test. If the p-value is <5% ther is stability or if the p-value is >5%?

Hello, I would like to do my low rationnality test with: linreg ev 1968:1 2014:1 # constant intrasrmere restrict 2 # 1 # 1 0 # 2 # 1 1 But I don't understant if i should use a linreg like this because I make my prevision with an arma model: boxjenk(constant,ar=0, ma=3, maxl, regressors,define=eveq) ...

Hello,

I have a question about stability test (@stabtest et Cusum), I know how to do this tests with linear regression but I don't know how to do with an ARMA.

Thank you again!

I have watch the regression output and I find what was diferent. I have exactly the same think that in R with:

set ind = t<1980:1
set pibd = pib*ind
set pibsqd = pib*pib*ind
set pibsqd2 = pib*pib

linreg ev 1960:1 2014:1 resmere2
# constant pib pibsqd2 ind pibd pibsqd

I try to do a linear regression on ev with pib + pib² on the first part and pib on the second part. Sorry if I am not clear, I have some difficulty to explain myself, especially in English. Note that R has a negative outlier at entry 9, and there's nothing at all like that in the RATS output. → That...

Does the R regression include an intercept by default? I don't know. What regression are you trying to run? In R, I'm doing: regev <-lm(evts ~ pibts+(pibts+pibts2)*ind(x)) And for the residuals I have : http://img11.hostingpics.net/pics/623682residusR.png ANd for the CUSUM test: http://img11.hosting...

Thank you very much. I still have not exactly the same thing as with R but I will seek a solution thanks to your indications

Hello, I have a program under R that I try to perform on ratz but I can't get the same result. My R program: x <- 1960:2014 ind <- function(x) ifelse (x<1980, TRUE, FALSE) pibts2 <- pibts*pibts regev <-lm(evts ~ pibts+(pibts+pibts2)*ind(x)) summary(regev) res2<- residuals(regev) plot(res2, type = &q...