similar to: .Random.seed (PR#911)

>From basic statistics principle,we know,given several i.i.d Gaussian RVs with zero or nonzero mean,the sum of square of them is a central or noncentral Chi-distributed RV.However if these Gaussian RVs have different variances,what does the sum of square of them obey? Thanks in advance.

Sorry for the contents not relating to R. Assume there are N i.i.d zero-mean complex gaussian random variables(RVs),as w(i),0<=i<N} with known variance,from which one can generate another N RVs,as R(0)=sum over i {w(i)*w'(i)} R(1)=sum over i {w(i+1)*w'(i)} ... up to R(N-1)= w(N-1)w'(i) where w'(i) is the complex conjugate of w(i). (from viewpoint of signal

a warning message appears when i use the chisq.test ,but it doesnt appear everytime, why? "Warning message: Chi-squared approximation may be incorrect in: chisq.test(matrix(c(20.1, 18.5, 2.6, 32.9, 23.5, 5.4), nc = 2)) " why does the warning message appear, please? Thank you very much here is the data which I have tried appear and not appear warning message have warning message

Dear R People, This question has nothing to do with R directly, but it is a simulation question. I need to generate a random variable distributed as gamma(\alpha,\beta), with the additional proviso that it must be a function of random variable(s) which do not depend on \alpha, \beta. In this case, I have \alpha = (T-1)/2, where T is a positive integer. So, it seems reasonable to first simulate

I have written an application in perl, to run multiple rsyncs, retrying those which fail. I have started getting this error on very large modules since i moved to the latest CVS version. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ToolSyncProgress Succeeded Getting big/tools/vlsi_local/etc from willy at 1004767211:2001:11:3:6:0:11. ToolSyncProgress Getting big1/cadappl1/hpux

Hi all, I try to use kernel 2.6.36.1 in HVM but building the drivers xen-blkfront, xenfs, platform-pci. I used Xen 4.0.2-rc1-pre and a DomU is a ubuntu 10.04 and I try with ubuntu 10.10. In boot I receive error "Cannot find a root partition /dev/xvda2" and a prompt (initramfs) appear. In dmesg (initramfs) appear: blkfront: xvda: barriers enabled (tag) xvda: xvda1 vbd vbd-51712: 16

Hi all, I try to use kernel 2.6.36.1 in HVM but building the drivers xen-blkfront, xenfs, platform-pci. I used Xen 4.0.2-rc1-pre and a DomU is a ubuntu 10.04 and I try with ubuntu 10.10. In boot I receive error "Cannot find a root partition /dev/xvda2" and a prompt (initramfs) appear. In dmesg (initramfs) appear: blkfront: xvda: barriers enabled (tag) xvda: xvda1 vbd vbd-51712: 16

Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as

I got some R code that I don't understand. Question as comment in code //where is t comming from, what is phi inverse rAC <- function(name, n, d, theta){ #generic function for Archimedean copula simulation illegalpar <- switch(name, clayton = (theta < 0), gumbel = (theta < 1), frank = (theta < 0), BB9 = ((theta[1] < 1) | (theta[2] < 0)), GIG = ((theta[2] < 0) |

Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with

Hi. I have made 100 experiments of an M/M/1 queue, and for each one I have calculated both, mean and variance of the queue size. Now, a professor has told me that variance is usually chi-squared distributed. Is there a way in R that I can find the parameter that best fits a chi-square to the variance data? I know there's fitdistr()m but this function doesn't handle chi-square. I believe

I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of

Hi, I have a function that generates a set of data but I am having problems determining the parameters using the nls fitting procedure. #### "MH"<-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){ #variables mu, sig, chi, Ms #input: field and diameter #all in CGS rho <- 5 kb <- 1.38e-16 t <- 300 length.d<-length(diameter) length.H<-length(field)

I need to write a method for calculating the fee for medical procedures with respect to area. I have all the tables and data but its hard to give that here. Can someone take a look at this ''stub'' method and help me out? def calculate_fee # strip out the first 3 numbers of the user entered zip code # use those 3 numbers to identify row in [zipcodes table] (between upper and

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in context:

. HI there, I'd like to show demonstrate how the chi-squared distribution works, so I've come up with a sample data frame of two categorical variables y<-data.frame(gender=sample(c('Male', 'Female'), size=100000, replace=TRUE, c(0.5, 0.5)), tea=sample(c('Yes', 'No'), size=100000, replace=TRUE, c(0.5, 0.5))) And I'd like to create a list of 100

Hi, I have calculated chi-square goodness of fit test,Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frauds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) lambda<- mean(No_of_Frauds) # Chi-Squared

Dear all, I met a problem while doing SEM by sem package. I got a negative chi-square of null model. Because the theoretical value of chi-square cannot be negative, I checked the source code of sem.R in sem package and I found the Chi-square of null model was computed by the following expression: result$chisqNull <- (N - 1) * (sum(diag(S %*% diag(1/diag(S)))) + log(prod(diag(S)))) I think

Hi! This is going to be a real newbie question, but I can't figure it out. I'm trying to plot densities of various functions of chi-square. A simple chi-square plot I can do with dchisq(). But e.g. chi.sq/degrees of freedom I only know how to do using density(rchisq()/df). For example: plot(1, type="n", xlab="", ylab="", xlim=c(0,2), ylim=c(0,7)) for (i