similar to: rexp question

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Hi, Wondering if anyone could help me out with this error.Im trying to fill a matrix with random numbers taken from an exponential distribution using a loop: x.3<-matrix(rep(0,3000),nrow=1000,byrow=T)for(i in 1:1000){x[i,]<-rexp(3,rate=2/3)} I get the error message: Error in x[i, ] <- rexp(3, rate = 2/3) : incorrect number of subscripts on matrix Any ideas??? Appreciate any thoughts.

l want to create a column of 1 and 2 randomly what command should l use eg treatment strata 1 1 2 0 1 1 2 1 2 0 2 1 2 0 1 0 these should be created randomly secondly if l have

Hello, I have written a function that demonstrates the CLT by generating samples following the exponential distribution, calculating the means, plotting the histogram, and drawing the limiting normal curve as an overlay. I have the title of each histogram state the sample size and rate (1/theta) for the exponential (the output is actually 4 histograms), but I can't get the greek letter theta

I need to generate 100 I.I.D samples from an exponential distribution. I use rexp(100,parameter). Is there anyway to specify a seed to determine the first input for the uniform random number generator used to generate these exponentials? -Dhiren

In splus rexp() and rweibull() are related: > set.seed(153) > rexp(1) [1] 0.0493267 > set.seed(153) > rweibull(1, shape=1) [1] 0.0493267 (you can also try shape =2, then rweibull = sqrt(rexp) ) However, in rw0.64.1 (on Win NT) they are different > .Random.seed <- 1:4 > rexp(1) [1] 1.412030 > .Random.seed <- 1:4 > rweibull(1, shape=1) [1] 2.054032 May be rweibull

I have some data measured with a coarsely-quantized clock. Let's say the real data are q<- sort(rexp(100,.5)) The quantized form is floor(q), so a simple quantile plot of one against the other can be calculated using: plot(q,type="l"); points(floor(q),col="red") which of course shows the characteristic stair-step. I would like to smooth the quantized

Hi R community, I have some data set and construct the likelihood as follows likelihood <- function(alpha,beta){ lh<-1 d<-0 p<-0 k<-NULL data<-read.table("epidemic.txt",header = TRUE) attach(data, warn.conflicts = F) k <-which(inftime==1) d <- (sqrt((x-x[k])^2+(y-y[k])^2))^(-beta) p<-1 - exp(-alpha*d) for(i in 1:100){

Why the warning messages (2:4)? > x <- rexp(1000,0.2) > fitdistr(x,"exponential",list(rate=1)) rate 0.219824219 (0.006951308) Warning messages: 1: one-diml optimization by Nelder-Mead is unreliable: use optimize in: optim(start, mylogfn, x = x, hessian = TRUE, ...) 2: NaNs produced in: dexp(x, 1/rate, log) 3: NaNs produced in: dexp(x, 1/rate, log) 4: NaNs

Hi all, I just run into this today. Apparently rexp() sometimes gives different slightly results for the same seed on 32 bit and 64 bit machines. runif() is the same for both, so the problem seems to be in rexp(). 64 bit Linux is the same as 64 bit OSX, and R-devel gives the same results as R-3.0.2. Best, Gabor # --------------------------------------------- > options(digits=22) ;

Hello, I want to estimate the exponential parameter by using?optim?with the following input, where t contains 40% of the data and q contains 60% of the data within an interval. In implementing the code command for optim i want it to contain both the t and q data so i can obtain the correct estimate. Is there any suggestion as to how this can be done. I have tried h<-c(t,q) but it is not working

Dear Mr. for writing program about Gibbs sampling, i have a question. if i want to generate data from Exponential distribution but range of X is restricted, how can i do? regards, A.Rashidi   [[alternative HTML version deleted]]

Any one know is there any package or function to generate bivariate exponential distribution? I gusee there should be three parameters, two rate parameters and one correlation parameter. I just did not find any function available on R. Any suggestion is appreciated. -- View this message in context:

Assuming the days of raining during half a year of all states(provinces) of a country is normally distributed (mean=?, standard deviation=?) with sigma (?) equals to 2. We now have 10 data points here: 26.64, 30.65, 31.27, 33.04, 32.56, 29.10, 28.96, 26.44, 27.76, 32.27. Try to get the 95% level of CI for ?, using parametric Bootstrap method with bootstrap size B=8000. my code - what am i doing

Hello, I would like to use the parfm package: https://cran.r-project.org/web/packages/parfm/parfm.pdfhttps://cran.r-project.org/web/packages/parfm/parfm.pdf in my work. This package fits parametric frailty models to survival data. To ensure I was using it properly, I started by running some small simulations to generate some survival data (without any random effects), and analyse the data using

Dear userRs, "playing around" with combinations of replicate() and random number generating functions inside a self-defined "wrapper" function I encounterd a puzzling behaviour. The following are intentionally simple (and rather nonsense-) examples to isolate the relevant aspects. Please, note the seemingly "inconsistent" behaviour for the second call of

>May 26, 2017; 11:41am Nelly Reduan Latin Hypercube Sampling when parameters are >defined according to specific probability distributions >Hello, > I would like to perform a sensitivity analysis using a Latin Hypercube Sampling (LHS). >Among the input parameters in the model, I have a parameter dispersal distance which is defined according to an exponential probability

Hi all, I'm trying to make a little script to determine an "unknown" rate for a number of known exponential trials. My Code: #Set Trials and generate number trials=100 rand<-runif(1,0,1) vector=0 #Generate vector of 100 random exponentials and sum them for (i in 1:100) { vector<-rexp(trials,rate=rand) } sumvect=sum(vector) #Create the log likelihood function