similar to: more on acf mis-feature (PR#1177)

Full_Name: Hiroyuki Kawakatsu Version: 1.3.1 OS: win2k Submission from: (NULL) (130.158.140.105) the lag labels in the acf plot is screwed when the ts frequency>1. try, for example, #acf mis-feature x <- ts(rnorm(200), frequency=4); acf(x,lag.max=24); h. > version _ platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32

hi, in 1.9.1, the return value from ARMAacf(pacf=TRUE) is not named by lags, contrary to ?ARMAacf. the simple fix is to move names(Acf) <- down after if(pacf), with an appropriate starting lag as pacf=TRUE appears to start at lag 1 (whereas pacf=FALSE starts at lag 0). for consistency, one could argue to append 1 for lag 0 for pacf=TRUE (or start pacf=F at lag 1). however, given the

I need to modify the graph of the autocorrelation. I tried to do it through plot.acf but with no success. 1. I would like to get rid of the lag zero 2. I would like to have numbers on the x-axis only at lags 12, 24, 36, 48, 60, ... Could anybody help me in this? Any help will be appreciated Thank you for your attention Stefano [[alternative HTML version deleted]]

Dear all, I have one Model (M3) fitted using the lme package, and I have checked the correlation structure of within-group errors using plot(ACF (M3,maxLag=10),alpha=0.05) But now I am not sure how to interpret this plot for the empirical autocorrelation function. The problem is that I am used to see/interpret diagrams in which all the autocorrelation Lags, except lag-1, are inside the

Hi list, I have the following code to compute the acf of a time series acfresid <- acf(residfit), where residfit is the series when I type acfresid at the prompt the follwoing is displayed Autocorrelations of series ?residfit?, by lag 0.0000 0.0833 0.1667 0.2500 0.3333 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333 1.000 -0.015 0.010 0.099 0.048 -0.014 -0.039 -0.019 0.040 0.018

Dear all, I have an annual time-series of population numbers and I would like to estimate the auto-correlation. Can I use acf() function and judge whether auto-correlation is significant by the plots? The acf array, eg: Autocorrelations of series 'x$log.s.r', by lag 0 1 2 3 4 5 6 7 8 9 10 11 12 1.000 0.031 -0.171

this one is not a false alarm like my previous message. i have cut and paste the code below so if anyone could run it would be appreciated. basically, my question is why the horizontal axis of the acf plot is labelled with such huge numbers when the labels should be 1 through 10 since may lag.max = 10 ? i looked at the cdoe of acf but it was pretty much beyond me. i think it has something to

Hi List, I'm trying to calculate the autocorrelation coefficients for a time series using acf at various lags. This is working well, and I can get the coefficients without any trouble. However, I don't seem to be able to obtain the significance of these coefficients from the returned acf object, largely because I don't know where I might find them. It's clear that the acf

When I run the "acf()" function using the "acf(ts.union(mdeaths, fdeaths))" example, the "acf()" function calls the "acf.plot()" function to generate this plot... http://members.cox.net/ddebarr/images/acf_example.png The plot in the lower right-hand corner is labeled "fdeaths & mdeaths", but the negative lags appear to belong to "mdeaths

I have used acf() and pacf() in R to get the acf and pacf values at max/lag=20 but the output did not show the values associated with lag numbers. lag numbers is shown in decimals. -- Rashid Ameer View my recent publication at * http://www.emeraldinsight.com/fwd.htm?id=aob&ini=aob&doi=10.1108/17538391211282854 * Details for my works are available directly at

Hi, P-values in Durbin-Watson test obtained through the use of functions available in packages "lmtest" and "car" are different. The difference is quite significant. function "dwtest" in "lmtest" is much faster than "burbinwatson" in "car". Actually, you can take a nap while the latter trying to calculated Durbin-Watson test. My question

Hi all, I'm new with R (and S), and relatively new to statistics (I'm a computer scientist), so I ask sorry in advance if my question is silly. My problem is this: I have a (sample of a) discrete time stochastic process {X_t} and I want to estimate Pr{ X_t | X_{t-l_1}, X_{t-l_2}, ..., X_{t-l_k} } where l_1, l_2, ..., l_k are some fixed time lags. It will be enough for me to compute

Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e", using the function "lag"? If so, how? If not, of what use is the function "lag"? I get the same answer from y~x as y~lag(x), whether using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider the following: > set.seed(1) > x <- rep(c(rep(0, 4), 9), len=9) > y <-

Ok I made Jarque-Bera test to the residuals (merv.reg$residual) library(tseries) jarque.bera.test(merv.reg$residual) X-squared = 1772.369, df = 2, p-value = < 2.2e-16 And I reject the null hypotesis (H0: merv.reg$residual are normally distributed) So I know that: 1 - merv.reg$residual aren't independently distributed (Box-Ljung test) 2 - merv.reg$residual aren't indentically

Dear all, I know that my question is somewhat special but I tried several times to solve the problems on my own but I am unfortunately not able to compute the following test statistic using the quantreg package. Well, here we go, I appreciate every little comment or help as I really do not know how to tell R what I want it to do^^ My situation is as follows: I have a data set containing a

I am a beginner in using R and I need help in the interpretation of AR result by R. I used 12 observations for my AR(2) model and it turned out the intercept showed 5.23 while first and second AR coefficients showed 0.40 and 0.46. It is because my raw data are in million so it seems the intercept is too small and it doesn't make sense. Did i make any mistake in my code? My code is as follows:

Hi I'm getting the following error that do not make sense to me, what am Idoing wrong ? > acf(Recsim[1,], lag.max=1) Error in acf(Recsim[1, ], lag.max = 1) : 'lag.max' must be at least 1 Regards EJ

I would like to get the result of acf min of lag 2 and max of lag 50. When I use time series ( acf, lag.max = 50, type="covariance"), I got lag 0 to lag 50. How do I get lag 2 to lag 50? Sincerely, Stephen

Full_Name: Ian McLeod Version: 2.3.1 OS: Windows Submission from: (NULL) (129.100.76.136) > There is a simple bug in acf as shown below: > > z <- 1 > acf(z,lag.max=1,plot=FALSE) > Error in acf(z, lag.max = 1, plot = FALSE) : > 'lag.max' must be at least 1 > This is certainly a bug. There are two problems: (i) the error message is wrong since lag.max is

Dear all, given two numeric vectors x and y, the ACF(x) at lag k is cor(x(t),x(t+k)) while the CCF(x,y) at lag k is cor(x(t),y(t-k)). See below for a simple example. > set.seed(1) > x <- rnorm(10) > y <- rnorm(10) > x [1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078 -0.8204684 0.4874291 0.7383247 0.5757814 -0.3053884 > y [1] 1.51178117 0.38984324