If by "problem" you mean the problem of determining how, in general,
you should proceed perhaps you need an introductory guide on R
and time series such as Cowpertwait's book.
On Thu, Jul 23, 2009 at 10:37 PM, Hongwei Dong wrote:
> Hi, Gabor, it seems ARIMA model does not have that problem. For ex
Hi, Gabor, it seems ARIMA model does not have that problem. For example:
set.seed(123)
y<-ts(c(1:20))
x = ts(rnorm(20))
z = ts(rnorm(20))
tt<-ts(cbind(x, lag(x,-1),lag(x,-2),z))
fit <- arima(y[1:15],order=c(1,0,0),xreg=tt[(1:15),])
fit
pred <- predict(fit, n.ahead=5,tt[(16:20),])
pred
What do you
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$Y[i] <- with(tz, 2*Y[i-1] + 3*z[i] +4* x[i] + 5*x[i-1] + rnorm(1))
}
# keep copy of tz to compare later to simulated Y's
tz.orig <- tz
# NA out Y's that are to be p
What I want R to do is to use the estimated Y at t-1 to be the lag(Y,-1) in
the forecast equation for time t. Is there anyway I can realize this with R?
For example, when the Y value for year 18 is forecast, the estimated Y for
year 17 is used, not the actual Y for year 17 already in the data.
Than
You can't remove Y since its in the rhs of your model.
On Thu, Jul 23, 2009 at 8:25 PM, Hongwei Dong wrote:
> Thanks, Gabor. Here are the problems I'm trying to solve.
> FIRST, I run this to simulate a 20 years time series process. The data from
> 1-15 years are used to estimate the model, and thi
Thanks, Gabor. Here are the problems I'm trying to solve.
*FIRST*, I run this to simulate a 20 years time series process. The data
from 1-15 years are used to estimate the model, and this model is used to
predict the year from 16-20. The following script works.
set.seed(123)
tt <- ts(cbind(Y = 1:2
Best thing is to test it out on simulated data for which you
already know the answer.
library(dyn)
set.seed(123)
DF <- data.frame(x = rnorm(10), z = rnorm(10))
DF$Y <- 0
for(i in 2:10) {
DF$Y[i] <- with(DF, 2*Y[i-1] + 3*z[i] +4* x[i] + 5*x[i-1] + rnorm(1))
}
DF.zoo <- do.call(merge, lapply(DF, z
Hi, Gabor, got it. Thanks a lot.
I have one more question about how the "predict" function works here,
especially for the lag(Y,-1) part.
In my model, I assume I know predictors x and z in the next two years, and
use them to predict Y. For each forecast step at time t, the lag(Y,-1) in
the model s
Please provide your code in a reproducible form (as requested previously).
Here we fit with first 9 points and then add a point for prediction. (Of
course your model can only predict the current value of Y so you may
have to rethink your model even aside from the implementation if you
really want
Hi, Gabor and Other R users,
I'm re-posting my script and the results I got.
here is the dynamic model I used to estimate in-sample model (1996-2006)
and it works:
fit<-dyn$lm(Y~lag(Y,-1)+z+x+lag(x,-1)+lag(x,-2)+lag(x,-3)+lag(x,-4))
Then I used this model to do out sample forecast with t
Here is an example closer to yours.
> library(dyn)
> set.seed(123)
> x <- zooreg(rnorm(10))
> y <- zooreg(rnorm(10))
> L <- function(x, k = 1) lag(x, k = -k)
> mod <- dyn$lm(y ~ L(y) + L(x, 0:2))
> mod
Call:
lm(formula = dyn(y ~ L(y) + L(x, 0:2)))
Coefficients:
(Intercept) L(y) L(x, 0:
Use dyn.predict like this:
> library(dyn)
> x <- y <- zoo(1:5)
> mod <- dyn$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 wrote:
> I have a dynamic time series model like this:
> dyn$lm( y ~ lag(y,-1) + x + lag
I have a dynamic time series model like this:
dyn$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
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