similar to: ppois

function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: > any(diff(ppois(0:19,lambda=0.9))<0) [1] TRUE Actually, > ppois(19,lambda=0.9)<ppois(18,lambda=0.9) [1] TRUE Which could not be TRUE. Code is tested in both R 3.5.1 and Microsoft R Open 3.5.1. _

Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: > On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >> >>> any(diff(ppois(0:19,lambda=0.9))<0) >> [1] TRUE >> >> Actually, >> >>>

Full_Name: Murray H Smith Version: 0.63.3 OS: Windows NT Submission from: (NULL) (130.216.5.57) The ppois function is displaced by -0.5. Try: > ppois(-0.5,1) [1] 0.3678794 > ppois(-0.51,1) [1] 0 > ppois(0,1) [1] 0.3678794 and > par(mfrow=c(2,1)) > x<-seq(-1,5,0.01) > plot(x,ppois(x,1),type="s",ylab="F(x)",main="Poisson CDF?") >

Hello, by checking the precision of a convolution algorithm, we found the following "inexactness": We work with R Version 1.8.1 (2003-11-21) on Windows systems (NT, 2000, XP). Try the code: ## Kolmogorov distance between two methods to ## determine P(Poisson(lambda)<=x) Kolm.dist <- function(lam, eps){ x <- seq(0,qpois(1-eps, lambda=lam), by=1) max(abs(ppois(x,

The expected number of bladder cancer over next 20 years a tire industry is 1.8. Poission distribution is assumed to hold and 6 reported deaths are caused by bladder cancer among the employees. Trying to find how unusual this event is. > ppois(q=6, lambda=1.8, lower.tail = TRUE, log.p = FALSE) [1] 0.9974306 not sure if ppois is the right one to use and the parameters... thx much

>>>>> Serguei Sokol >>>>> on Tue, 4 Dec 2018 11:46:32 +0100 writes: > Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: >> On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >>> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is:

Could somebody help me to understand the syntax of R's ppois function? I'm looking to calculate the cumulative probability density of an observed value (y) given the expected mean (mu) and the level of significance (alpha). I'm coming from using SAS to do this and don't recognize the descriptions of the arguments for ppois. The definitions of lambda and p as stated in the R manuals

Reciban un cordial saludo. De manera muy comedida les pido ayuda con la función indicada abajo. La usan en un artículo de inventarios. No veo dónde estoy errado. Gracias de antemano. César Escalante C. > lamb<-50 > L<-1 > h.<-10 > p<-25 > K<-5 > integ<-function(y){integrate(function(x)ppois(x,lamb*L),0,y)$value} > G<-function(y){(h.+p)*integ(y) +

Folks, I'm fairly sure that I'm doing something stupid, but I'm getting a few really strange results from *some* of the distributions, but by no means all, when I link directly to the R shared library. I've tried this on both Windows with the precompiled Mingw binary of R-2.5.0 (compiling my code with MinGW-3.4.2), and by building R-2.5.0 on Mandriva Linux with gcc-3.4.4 and

Hello, I am performing a sensitivity analysis using a Latin Hypercube sampling. However, I have difficulty to draw a Hypercube sample for one variable. I?ve generated this variable from a Poisson distribution as follows: set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2) barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age

Hi all, I have a large data.frame, 1530 observation with 6 columns. I want to merge a 7th column, a transformation of the response variable (hospital admissions), namely trans<-sqrt(copd$admissions+0.25) trans<-data.frame(trans) And now when I do copd2<-merge(copd,trans) (copd being my original data.frame), R either crashes or is taking an extremely long time to do the computation. I

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

Make a plot in R where you compare the probability distributions for the Binomial distributions with N=1000 trials and succes probability 0.005 and the Possion Distribution with rate lambda=5. My answer is b<-binom(x, 1000, 0.005, log = FALSE) plot(b) p<-dpois(x, lambda, log = FALSE) plot(p) *Is this the right way to solve the problem?* -- View this message in context:

I have a dataset which I want to model using a Poisson distribution, with a given parameter. I would like to know what is the proper way to do a ''goodness of fit'' test using R. I know the steps I''d take if I were to do it ''manually'': grouping the numbers into classes, calculating the expected frequencies using ''ppois'', then

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 all, Firstly I appologise if this question has been answered previously, however searching of the archives and the internet generally has not yielded any results. I am looking in to the effects of summer weather conditions (temperature, humidity etc), on the incidences of a breathing disorder brought on through smoking (COPD). I am fairly new to R and completely new to the idea of

Dear all, I appologise for cluttering up the list with such a basic question, however I have been unable to find the answer I want (possibly through my poor usage of the R help system). As I am visually impaired and using assistive technology, I think I would prefer to use R from the terminal mode, i.e. by running rterm.exe found in the bin directory. I have managed to set my working

I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1) for large negative x. This expression is actually the probability that y = 1 when y is a Poisson random variable truncated at 0, hence must satisfy 0 <= h <= 1. However, when x < -18, I may get an h value that is larger than 1 while the true value should be a quantity that is smaller than but very close to 1. For

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik