similar to: Poisson distribution help requested

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, 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

> How can I draw a Hypercube sample for the variable mortality_probability so > that this variable exhibits the same pattern as the observed distribution? One simple way is to use the uniform random output of randomLHS as input to the quantile function for your desired distribution(s). For example: q <- randomLHS(1000, 3) colnames(q) <- c("A", "B",

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

Hi folks! I run the following code to get a CI for a Poisson with lambda=12.73 library(MASS) set.seed(125) x <- rpois(100,12.73) lambda_hat<-fitdistr(x, dpois, list(lambda=12))$estimate #Confidence Intervals - Normal Approx. alpha<-c(.05,.025,.01) for(n in 1:length(alpha)) { LowerCI<-mean(x)-(qnorm(1-alpha[n]/2, mean = 0, sd = 1)*sqrt(var(x)/length(x)))

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

(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

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, >> >>>

I am interested in R functions for the following integrals / sums (expressed best I can in text) - Normal: G_u(k) = Integration_{Lower limit=k}^{Upper limit=infinity} [(u -k) f(u) d(u)], where where u is N(0,1), and f(u) is the density function. Poisson: G(lambda,k) = Sum_{Lower limit=k}^{Upper limit=infinity} [(x-k) p(x, lambda)] where P(x,lambda) is the Poisson prob function with parameter

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. _

Hi all, So far I only know one way to get the confidence limit for the Poisson distribution is to use the look-up table given by the 2 parameter (the number of observation x and the confidence level, e.g. 95%) and the table is limit by the maximum number of observations (x <= 50). I know the formula to compute the CI, however, mathematically it is not easy to do it. So, anyone know an R

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,

Hi I want to write a C function for the R Code below and call it with .Call: SimPoisson <- function(lambda,tgrid,T2M) #Simulation eines Poissonprozesses { NT <- 0 Ni <- rep(0,length(tgrid)) tau <- 0 sign <- 0 if(lambda != 0) { i=1 j=1 while (1) { EVar <- rexp(1,lambda) sign <- sign + EVar if (sign > T2M) { break } tau[i] <- sign

Thanks for your answer. However, my variable is simulated from the cumulative distribution function of the Poisson distribution. So, the pattern obtained from the function "qpois" is not the same as the observed pattern (i.e., obtained from the function "ppois") set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2)

Dear R-listers, I want to reperfrom a poisson distribution test that presented in a recent-published biological research paper (Plant Physiology 2008, vol 148, pp. 1189-1200). That test is about the occurrence number of a kind of gene in separate chromosomes. For instance: The observed gene number in chromosome A is 36. The expected gene number in chromosome A is 30. Then, the authors got a

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

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.

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

1. lbeta and beta do not work properly: lbeta returns its first argument and beta gives the lbeta result. In names.c lines 245-6, the codes for these should be 2 and 3 instead of 1 and 2 2. crtl-C does not work (except the first time) on Red Hat elf Linux (which has many many other problems as well) nor on the previous version of Linux for Amiga. It worked on Slackware aout Linux and now works