[R] how to predict dynamic model in R

[Gabor Grothendieck]

Try this:

library(dyn)
set.seed(123)
tz <- zoo(cbind(Y = 0, x = rnorm(10), z = rnorm(10)))

# simulate values
for(i in 2:10) {
  tz$Y[i] <- with(tz, 2*Y[i-1] + 3*z[i] +4* x[i] + 5*x[i-1] + rnorm(1))
}

# keep copy of tz to compare later to simulated Y's
tz.orig <- tz

# NA out Y's that are to be predicted
tz[7:10, "Y"] <- NA

L <- function(x, k = 1) lag(x, -k)

# predict 1 ahead each iteration
for(i in 7:10) {
    # fit based on first i-1 values
    fit <- dyn$lm(Y ~ L(Y) + z + L(x, 0:1), tz, subset = seq_len(i-1))
    # get prediction for ith value
    tz[i, "Y"] <- tail(predict(fit, tz[1:i,]), 1)
}
cbind(pred = tz[7:10, "Y"], act = tz.orig[7:10, "Y"])


On Thu, Jul 23, 2009 at 9:02 PM, Hongwei Dong<pdxdong at gmail.com> wrote:
> What I want R to do is to use the estimated Y at t-1 to be the lag(Y,-1) in
> the forecast equation for time t. Is there anyway I can realize this with R?
> For example, when the Y value for year 18 is forecast, the estimated Y for
> year 17 is used, not the actual Y for year 17 already in the data.
> Thanks for you patience. I appreciate it.
> Harry
>
>
>
> On Thu, Jul 23, 2009 at 5:44 PM, Gabor Grothendieck
> <ggrothendieck at gmail.com> wrote:
>>
>> You can't remove Y since its in the rhs of your model.
>>
>> On Thu, Jul 23, 2009 at 8:25 PM, Hongwei Dong<pdxdong at gmail.com> wrote:
>> > Thanks, Gabor. Here are the problems I'm trying to solve.
>> > FIRST, I run this to simulate a 20 years time series process. The data
>> > from
>> > 1-15 years are used to estimate the model, and this model is used to
>> > predict
>> > the year from 16-20. The following script works.
>> > set.seed(123)
>> > tt <- ts(cbind(Y = 1:20, x = rnorm(20), z = rnorm(20)))
>> > L <- function(x, k = 1) lag(x, -k)
>> > tt.zoo <- as.zoo(tt)
>> > fit <- dyn$lm(Y ~ L(Y) + z + L(x, 0:1), tt.zoo[(1:15), ])
>> > fit
>> > pred <- predict(fit, tt.zoo[(16:20),])
>> > pred
>> > SECONDLY, I use similar script, but pretend that we do not know the Y
>> > data
>> > from year 16-20. We know x and z for year 16-20, and use them predict Y
>> > based on the model estimated from 1-15 years. So, in the "newdata" part,
>> > I
>> > use tt.zoo[(16:20), (2:3)] to remove the Y out. here is the script.
>> > Unfortunately, it does not work in that way.
>> > pred1 <- predict(fit, tt.zoo[(16:20),(2:3)])
>> > pred1
>> > It will be greatly appreciated if you can give me some guide on this.
>> > Thanks.
>> > Harry
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Wed, Jul 22, 2009 at 10:04 PM, Gabor Grothendieck
>> > <ggrothendieck at gmail.com> wrote:
>> >>
>> >> Use dyn.predict like this:
>> >>
>> >> > library(dyn)
>> >> > x <- y <- zoo(1:5)
>> >> > mod <- dyn$lm(y ~ lag(x, -1))
>> >> > predict(mod, list(x = zoo(6:10, 6:10)))
>> >>  7  8  9 10
>> >>  7  8  9 10
>> >>
>> >>
>> >> On Thu, Jul 23, 2009 at 12:54 AM, Hongwei Dong<pdxdong at gmail.com>
>> >> wrote:
>> >> > I have a dynamic time series model like this:
>> >> > dyn$lm( y ~ lag(y,-1) + x + lag(x,-1)+lag(x,-2) )
>> >> >
>> >> > I need to do an out of sample forecast with this model. Is there any
>> >> > way
>> >> > I
>> >> > can do this with R?
>> >> > It would be greatly appreciated if some one can give me an example.
>> >> > Thanks.
>> >> >
>> >> >
>> >> > Harry
>> >> >
>> >> >        [[alternative HTML version deleted]]
>> >> >
>> >> > ______________________________________________
>> >> > R-help at r-project.org mailing list
>> >> > https://stat.ethz.ch/mailman/listinfo/r-help
>> >> > PLEASE do read the posting guide
>> >> > http://www.R-project.org/posting-guide.html
>> >> > and provide commented, minimal, self-contained, reproducible code.
>> >> >
>> >
>> >
>
>