Try this: # test data set.seed(1) y <- ts(1:10 + rnorm(10, 0, 0.1))
# fit model library(dyn) y.lm <- dyn$lm(y ~ lag(y,-1)) # use predict tail(predict(y.lm, list(y = y)), 1) # or multiply by coefficients giving same result coef(y.lm) %*% c(1, tail(y,1)) # Now try it using quantile regression library(quantreg) y.rq <- dyn$rq(y ~ lag(y,-1)) tail(predict(y.rq, list(y = y)), 1) coef(y.rq) %*% c(1, tail(y,1)) On 5/15/06, Kerpel, John <[EMAIL PROTECTED]> wrote: > All: > > > > How do I obtain one step ahead out-of-sample forecasts from a model > using "dyn" or "dynlm" ? > > > > Thanks! > > > > Best, > > > > John > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
