similar to: Poission distribution

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,

(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

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

Dear users, In a study with recurrent events: My objective is to get estimates of survival (obtained through a Cox model) by rank of recurrence and by treatment group. With the following code (corresponding to a model with a global effect of the treatment=rx), I get no error and manage to obtain what I want : data<-(bladder)

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

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

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, >> >>>

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. _

Hello R-help. I am trying to make a stacked barplot where the color of the sections of each bar depend on another variable. > myData[1:11,] score percent marker cellType Malignant 1 0 100.00000 ESR1 (ER) Bladder.M(5) TRUE 2 0 80.00000 PAX8 Bladder.M(5) TRUE 3 1 20.00000 PAX8 Bladder.M(5) TRUE 4 0 100.00000 ESR1 (ER) Brain.N(3) FALSE 5 0

A quick question! The number of episodes per year of otitis media follows a Possion distribution with lambda = 1.6 episodes per year. Wouldn't the probability of getting 3 or more episodes of otitis media in the first 2 years of life be: > ppois(q=3, lambda=1.6*2, lower.tail = TRUE, log.p = FALSE) [1] 0.6025197 I am confused with the lambda and 3 or more.. thx much

It would be nice if the 00check.log file also included this part of the output: Running ?bladder.R? Comparing ?bladder.Rout? to ?bladder.Rout.save? ... OK Running ?book1.R? Comparing ?book1.Rout? to ?book1.Rout.save? ... OK Running ?book2.R? Comparing ?book2.Rout? to ?book2.Rout.save? ... OK etc. The survival package has enough test scripts that it exceeds my terminal's scroll

Hello, In package survival should be a dataset bladder1. http://stat.ethz.ch/R-manual/R-devel/library/survival/html/bladder.html I can not open it (not found). Both bladder and bladder2 are there. Thanks, Petra

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

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

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

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) +

> On Apr 16, 2018, at 3:20 PM, David Winsemius <dwinsemius at comcast.net> wrote: > >> >> On Apr 16, 2018, at 2:58 PM, Therneau, Terry M., Ph.D. via R-devel <r-devel at r-project.org> wrote: >> >> A user asked me about this and I can't figure it out. >> >> tmt% R >> R Under development (unstable) (2018-04-09 r74565) --

Dear volunteer: I am a graduate student of medcine in china.And now,I am devoting myself to constructing a nomograms of bladder cancer.I want to do it with R-project.However, I do not know how to construct a nomograms with R-project.I want to get yours help,thank you! I wish you can tell me the operating procedure of the R-project. And I apologize for my english,it is poor,sorry!

version .62: --------------------------------------------- > ?qpois The Poisson Distribution dpois(x, lambda) ppois(q, lambda) qpois(p, lambda) rpois(n, lambda) Arguments: x: vector of (positive) quantiles. p: vector of probabilities. n: number of random values to return. lambda: vector of positive means.