Displaying 20 results from an estimated 333 matches for "armas".
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areas
2005 Jun 14
1
using forecast() in dse2 with an ARMA model having a trend component
(My apologies if this is a repeated posting. I couldn't find any trace
of my previous attempt in the archive.)
I'm having trouble with forecast() in the dse2 package. It works fine
for me on a model without a trend, but gives me NaN output for the
forecast values when using a model with a trend. An example:
# Set inputs and outputs for the ARMA model fit and test periods
2009 Mar 06
0
modifying a built in function from the stats package (fixing arima) (CONCLUSIONS)
Thanks a lot to everybody that helped me out with this.
Conclusions:
(1)
In order to edit arima in R:
>fix(arima)
or alternatively:
>arima<-edit(arima)
(2)
This is not contained in the "Introduction to R" manual.
(3)
A "productive" fix of arima is attached (arma coefficients printed out and
error catched so that it doesn't halt parent loops to search for
2003 May 16
3
ARMA.predict?
Hi there,
Does anyone know how to predict ARMA? It doesn?t have either predict or forecast methods. I found couple of packages called fbasic and fseries at http://www.itp.phys.ethz.ch/econophysics/R/, which has ?arma.predict? in it, but it doesn?t seem to be working. Any help in this regard would be appreciated. Thanks in advance.
Regards
Skanda Kallur
"Prediction is very difficult,
2011 Dec 05
1
RcppArmadillo compilation error: R CMD SHLIB returns status 1
Dear all,
running the example by D. Eddebuettel (http://dirk.eddelbuettel.com/blog/2011/04/23/) I get an error message. Specifically, the R code I was taking from the above example is
### BEGIN EXAMPLE ###
suppressMessages(require(RcppArmadillo))
suppressMessages(require(Rcpp))
suppressMessages(require(inline))
code <- '
arma::mat coeff = Rcpp::as<arma::mat>(a);
arma::mat
2009 Jul 15
2
storing lm() results and other objects in a list
to clean up some code I would like to make a list of arbitrary length
to store?various objects for use in a loop
sample code:
############ BEGIN SAMPLE ##############
# You can see the need for a loop already
linearModel1=lm(modelSource ~ .,mcReg)
linearModel2=step(linearModel1)
linearModel3=lm(modelSource ~ .-1,mcReg)
linearModel4=step(linearModel3)
#custom
linearModel5=lm(modelSource ~ .
2007 Oct 22
1
Newbie help: Data in an arma fit
I'd like to fit an ARMA(1,1) model to some data (Federal Reserve Bank
interest rates) that looks like:
...
30JUN2006, 5.05
03JUL2006, 5.25
04JUL2006, N <---- here!
05JUL2006, 5.25
...
One problem is that holidays have that "N" for their data. As a test, I
tried fitting ARMA(1,1) with and without the holidays deleted. In other
words, I fit the above data
2005 Dec 23
1
dse package problems
I am having problems with the package dse. I just installed R 2.2.1
and reinstalled all packages. I am running Windows XP Pro with all
updates.
Below there are two examples of error messages generated when trying
to execute some simple programs. The code was taken directly from the
package documentation.
Any help on this will be greatly appreciated.
Merry Christmas
Fernando
2004 Oct 25
1
output processing / ARMA order identification
Dear R users,
I need to fit an ARMA model. As far as I've seen, EACF (extended ACF)
is not available in R.
1. Let's say I fit a series of ARMA models in a loop. Given the
code/output included below, how do I pull 'Model' and 'Fit' (AIC)
from each summary() so that I can combine them into an array/data
frame to be sorted by AIC?
2. Apart from EACF, are you aware perhaps
2009 Oct 22
1
arima crashes too
Another pathological test.
arima does not crash for that series that crashes arma:
arima(c(2.01, 2.22, 2.09, 2.17, 2.42), order=c(1,0,0))
However, arima crashes for this:
arima(c(1.71, 1.78, 1.95, 1.59, 2.13), order=c(1,0,0))
arima seems pretty consistent in its crashing behaviour, since crashing for
one series means crashing for all affine series:
lets.crash.arima <- c(71, 78, 95, 59,
2006 Nov 07
1
Comparison between GARCH and ARMA
Dear all R user,
Please forgive me if my problem is too simple.
Actually my problem is basically Statistical rather
directly R related. Suppose I have return series ret
with mean zero. And I want to fit a Garch(1,1)
on this.
my is r[t] = h[i]*z[t]
h[t] = w + alpha*r[t-1]^2 + beta*h[t-1]
I want to estimate the three parameters here;
the R syntax is as follows:
#
2008 Aug 20
2
arma: what is the meaning of Pr(>|t|)?
In the summary of the output of arma, there's a number Pr(>|t|), however, I
don't know what is its meaning - at least, it doesn't _seem_ to be a
Student's t distribution.
Reproducible test case:
x <- c(0.5, sin(1:9))
reg <- arma(x, c(1,0))
summary(reg)
<output>
Call:
arma(x = x, order = c(1, 0))
Model:
ARMA(1,0)
Residuals:
Min 1Q Median 3Q
2011 Oct 12
0
ARMA and prediction
Hello,
I am running an ARMA model to run forecast for changes in S&P 500 prices.
My ARMA calculations look as follows
armacal <- arma( spdata, order = c(0,4), lag = list(ma = c(1,2,4)) )
Output:
Call:
arma(x = spdata, order = c(0, 4), lag = list(ma = c(1, 2, 4)) )
Coefficient(s):
ma1 ma2 ma4 intercept
-0.073868 0.058020 -0.081292 0.007082
All's
1999 Nov 14
1
bug in arma.sim (PR#322)
Dear Sir,
I think I found a bug in the function arma.sim, which is defined in
the help page of the function filter:
arma.sim <- function(n, ar = NULL, ma = NULL, sigma = 1.0)
{
x <- ts(rnorm(n+100, 0, sigma^2), start = -99)
if(length(ma)) x <- filter(x, ma, sides=1)
if(length(ar)) x <- filter(x, ar, method="recursive")
as.ts(x[-(1:100)])
}
I am using R
2009 Apr 29
1
arma model with garch errors
Dear R experts,
I am trying to estimate an ARMA 2,2 model with garch errors.
I used the following code on R 2.9.
#library
library(fGarch)
#data
data1<-ts(read.table("C:/Users/falcon/Desktop/Time
Series/exports/goods1.csv"), start=c(1992,1), frequency=12)
head(data1)
#garch
garchFit(formula.mean= ~arma(2,2),formula.var=~garch(1,1), data=data1)
but get this error:
>
2011 May 08
1
ARMA
Hello,Could somebody tell me what is the difference between theese 3 calls of functionsarma(x,order=c(1,0)), arima(x,order=c(1,0,0)) ar(x,order=1)?I expected same residuals of theese three models,but unexpectably for the first two R requiredinitial value of something (what?)...Thanks in advance!
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2004 Feb 12
0
How to predict ARMA models?
Hi all,
I am fitting an ARMA(1,(1,4)) model.
y(t) = a*y(t-1) + e(t) + b1*e(t-1) + b4*e(t-4)
> arma1.14 <- arma(series, lag=list(ar=1, ma=c(1,4)),
+ include.intercept = F, qr.tol = 1e-07)
works fine:
Coefficient(s):
ar1 ma1 ma4
0.872 -0.445 0.331
I want to forecast 50 periods.
I could not find a 'predict' function for ARMA models.
I
2006 Aug 14
1
ARMA(1,1) for panel data
Dear List,
I am new to TS-Modeling in R. I would like to fit an ARMA(1,1) model
for a balanced panel, running Y on a full set of unit and year dummies
using an arma(1,1) for the disturbance:
y_it=unit.dummies+yeardummies+e_it
where: e_it=d*e_it-1+u_it+q*u_it-1
How can I fit this model in R? arma() does not seem to take covariates
(or I don't understand how to specify the function so that
2008 Mar 21
1
tseries(arma) vs. stats(arima)
Hello,
The "arma" function in the "tseries" package allows estimation of models
with specific "ar" and "ma" lags with its "lag" argument.
For example: y[t] = a[0] + a[1]y[t-3] +b[1]e[t-2] + e[t] can be estimated
with the following specification : arma(y, lag=list(ar=3,ma=2)).
Is this possible with the "arima" function in the
2010 Aug 23
1
Fitting a regression model with with ARMA error
Hi,
I want to fit a regression model with one independent variable. The error
part should be fitted an ARMA process.
For example,
y_t = a + b*x_t + e_t where e_t is modelled as an ARMA process.
Please let me know how do I do this in R. What code should I use?
TIA
Aditya
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2013 Apr 08
0
Maximum likelihood estimation of ARMA(1,1)-GARCH(1,1)
Hello
Following some standard textbooks on ARMA(1,1)-GARCH(1,1) (e.g. Ruey
Tsay's Analysis of Financial Time Series), I try to write an R program
to estimate the key parameters of an ARMA(1,1)-GARCH(1,1) model for
Intel's stock returns. For some random reason, I cannot decipher what
is wrong with my R program. The R package fGarch already gives me the
answer, but my customized function