Displaying 20 results from an estimated 10000 matches similar to: "tsboot"
2010 Mar 01
1
p-values from bootstrapping of time series (tsboot)
Does anyone know how p-values can be generated if tsboot (stationary
bootstrap) for time series is performed?
That would be of great help. Thanks a lot for your comments.
Markus
[[alternative HTML version deleted]]
2010 Feb 03
0
Package np update (0.30-6) adds nonparametric entropy test functionality...
Dear R users,
Version 0.30-6 of the np package has been uploaded to CRAN. See
http://cran.r-project.org/package=np
Note that the cubature package is now required in addition to the boot package. The recent updates in 0.30-4 through 0.30-6 provides additional functionality in the form of five new functions that incorporate frequently requested nonparametric entropy-based testing methods to the
2010 Feb 03
0
Package np update (0.30-6) adds nonparametric entropy test functionality...
Dear R users,
Version 0.30-6 of the np package has been uploaded to CRAN. See
http://cran.r-project.org/package=np
Note that the cubature package is now required in addition to the boot package. The recent updates in 0.30-4 through 0.30-6 provides additional functionality in the form of five new functions that incorporate frequently requested nonparametric entropy-based testing methods to the
2012 Sep 17
2
Problem with Stationary Bootstrap
Dear R experts,
I'm running the following stationary bootstrap programming to find the parameters estimate of a linear model:
X<-runif(10,0,10)
Y<-2+3*X
a<-data.frame(X,Y)
coef<-function(fit){
fit <- lm(Y~X,data=a)
return(coef(fit))
}
result<- tsboot(a,statistic=coef(fit),R = 10,n.sim = NROW(a),sim = "geom",orig.t = TRUE)
Unfortunately, I got this
2011 Feb 08
0
tsboot fails on Seasonal Mann-Kendall (seaKen function, wq package)
Dear R-users,
tsboot fails when I try to perform a block bootstrap on seaKen
(package wq):
these commands:
require(wq)
require(datasets)
boot.block.sen <- function(data){seaKen(data)[[1]]}
tsboot(sunspot.month, boot.block.sen, R=1999, l=12, sim="fixed")
return:
Error in seaKen(data) : x must be a 'ts'
Any suggestion on how might I change seaKen in order to use it with
2007 Nov 22
0
Problem with tsboot
I'm trying to bootstrap some regression coefficients so that I can
estimate confidence intervals, but boot is not producing results. Can
anybody suggest what I'm doing wrong here?
> SpecPress <- ts(rnorm(501))
> Returns <- ts(rnorm(501))
> BootData <- data.frame(cbind(SpecPress, Returns))
> boot.specpress <- function(data, indices, maxit=20){
+ data <-
2010 Aug 01
0
BCa-intervals not defined in boot.ci() for tsboot() -> package: boot
Hello everybody,
when I create an object of class boot with the function tsboot() from the package boot and try to compute several types of confidence intervals with boot.ci("object of class boot created with tsboot") I obtain the warning message, that "BCa-intervals are not defined for time-series bootstraps".
Does that hold in general? Or is it just not defined in
1999 Dec 09
1
tsboot
Fritz,
I have slightly adapted (didn't work before) "tsboot" from the "boot"
library to the current time series conventions of R. The following patch
will do that. I suggest to apply this patch to the file
"boot/R/bootfuns.q" of the "boot" library at CRAN.
best
Adrian
--- bootfuns.orig.q Thu Dec 9 10:07:23 1999
+++ bootfuns.q Thu Dec 9 10:06:51 1999
1999 Dec 09
1
tsboot
Fritz,
I have slightly adapted (didn't work before) "tsboot" from the "boot"
library to the current time series conventions of R. The following patch
will do that. I suggest to apply this patch to the file
"boot/R/bootfuns.q" of the "boot" library at CRAN.
best
Adrian
--- bootfuns.orig.q Thu Dec 9 10:07:23 1999
+++ bootfuns.q Thu Dec 9 10:06:51 1999
2018 Nov 04
3
[BUG?] sftp is echoing back prompt and commands in batch mode
Il giorno dom 4 nov 2018 alle ore 01:45 Ben Lindstrom
<mouring at offwriting.org> ha scritto:
>
> I don't see it as a bug.
Yes, this is why I put a question mark in the subject.
> As if I'm writing a batch script I want to see the echo of the command and the output so if there is a failure I know where the failure is.
I see: you have a single batch file with no
2010 Jun 28
0
Forecast Package in R: auto.arima function
Hey,
I have a few doubts with regard to the usage of the auto.arima function from
the forecast package in R.
*Background:*
I have a set of about 50 time-series for which I would like to estimate the
best autroregressive model. (I want to estimate the coefficients and order
of p). Each of the series is non-stationary and are also have a non-normal
distribution. The data is non-seasonal.
My
2018 Nov 02
2
[BUG?] sftp is echoing back prompt and commands in batch mode
Thanks, Iain.
I am willing to hear from other users whether anyone else sees this as a
bug before filing it.
--
Vincenzo Romano
Il giorno ven 2 nov 2018, 20:03 Iain Morgan <imorgan at nas.nasa.gov> ha
scritto:
> If you truly intend this as a bug report, you should file it at
> bugzilla.mindrot.org.
>
> On Fri, Nov 02, 2018 at 12:25:22 +0100, Vincenzo Romano wrote:
> >
2011 Jul 18
1
XEN and vcpus
HI.
Got a question.
When I assign a number of vcpus to a DomU, Are they dedicated to that VM?
For example: If I have 16 vcpus available and 8 DomUs, should I assign 2
vcpus for each one? or could I assign more vcpus (i.e. 6) and the hypervisor
takes care of it?
Thanks
--
----------------------------------------------------------
Luis R. Romano
Estudiante de Ingeniería en Computación.
FCEFyN -
2007 Nov 26
3
Time Series Issues, Stationarity ..
Hello,
I am very new to R and Time Series. I need some help including R codes
about the following issues. I' ll really appreciate any number of
answers...
# I have a time series data composed of 24 values:
myinput = c(n1,n2...,n24);
# In order to make a forecasting a, I use the following codes
result1 = arima(ts(myinput),order = c(p,d,q),seasonal = list(order=c(P,D,Q)))
result2 =
2003 Jul 08
1
Questions about corARMA
Hi,
I'm a new member here in the list. I am a graduate from University of Georgia. Recently in doing analysis using lme on a dataset, I found several questions:
1. How to express the equation when the correlation structure is very complicated. For exmaple, if the fixed is y(t)=0.03x1(t)+1.5x2(t)(I omitted "hat" and others). And the model with corARMA(p=2,q=3) is proper. What will be
2010 Feb 17
0
adf.test help
Hi,
I am trying to test whether a series is return series stationary, but
before proceeding I wanted to make sure I understand correctly how to
use the adf.test function and interpret its output... Could you please
let me know whether I am correct in my interpretations?
ex: I take x such as I know it doesn't have a unit root, and is
therefore stationary
1/
> x <- rnorm(1000)
>
2007 Dec 08
2
time series tests
Hi all,
Can anyone clear my doubts about what conclusions to take with the following what puts of some time series tests:
> adf.test(melbmax)
Augmented Dickey-Fuller Test
data: melbmax
Dickey-Fuller = -5.4075, Lag order = 15, p-value = 0.01
alternative hypothesis: stationary
Warning message:
p-value smaller than printed p-value in: adf.test(melbmax)
2008 Jun 26
1
stationary "terminology" time series question
This is not exactly an R question but the R code below may make my
question more understandable.
If one plots sin(x) where x runs from -pi to pi , then the curve hovers
around zero obviously. so , in a"stationary in the mean" sense,
the series is stationary. But, clearly if one plots the acf, the
autocorrelations at lower lags are quite high and, in the "box jenkins"
2006 Feb 15
1
Generating random walks
Hello, here is another question, how do I generate
random walk models in R? Basically, I need an AR(1)
model with the phi^1 value equal to 1:
Yt = c + Yt-1 + E
where E is random white noise.
I tried using the arima.sim command:
arima.sim(list(ar=c(1)), n = 1000, rand.gen = rnorm)
but got this error since the model I am generating is
not stationary:
Error in arima.sim(list(ar = c(1)), n =
2013 Apr 30
1
ADF test --time series
Hi all,
I was running the adf test in R.
CODE 1:
adf.test(data$LOSS)
Augmented Dickey-Fuller Test
data: data$LOSS
Dickey-Fuller = -1.9864, Lag order = 2, p-value = 0.5775
alternative hypothesis: stationary
CODE 2:
adf.test(diff(diff(data$LOSS)))
Augmented Dickey-Fuller Test
data: diff(diff(data$LOSS))
Dickey-Fuller = -6.9287, Lag order = 2, p-value = 0.01
alternative