similar to: Bionomial and possion

Hi - I have a likelihood function that involves sums of two possions: L = a*dpois(Xi,theta1)*dpois(Yi,theta2)+b*(1-c)*a*dpois(Xi,theta1+theta3)*dpois(Yi,theta2) where a,b,c,theta1,theta2,theta3 are parameters to be estimated. (Xi,Yi) are observations. However, Xi and Yi are usually big (> 20000). This causes dpois to returns 0 depending on values of theta1, theta2 and theta3. My first

Hi all, I am wondering whether there is a small bug in the binom.test function of the ctest library (I'm using R 1.6.0 on windows 2000, but Splus 2000 seems to have the same behaviour). Or perhaps I've misunderstood something. the command binom.test(11,100,p=0.1) and binom.test(9,100,p=0.1) give different p-values (see below). As 9 and 11 are equidistant from 10, the mean of the

Full_Name: Petr Savicky Version: 2.4.0 OS: Fedora Core release 2 Submission from: (NULL) (62.24.91.47) the error is > binom.test(0.56*10000,10000) Error in binom.test(0.56 * 10000, 10000) : 'x' must be nonnegative and integer while > binom.test(5600,10000) yields correct result. The same error occurrs for > binom.test(0.57*10000,10000)

Hi. Here is a piece of cpp code. It works, but I do not understand the rational for the use of "R::dpois" to call the function dpois since in the examples I have always found directly "dpois" or "Rcpp::dpois" that both do not work in my code. Could anyone be so patient to explain me why should it be like that? Thaks a lot, Enrico #include <Rcpp.h> using

Hello all, I am trying to get a better understanding of the underlying code for the stats::dpois function in R and, specifically, what happens under the hood when it is called. I finally managed to track down the C course at: https://github.com/wch/r-source/blob/trunk/src/nmath/dpois.c. It would seem that the dpois C function is taking a double for each of the x and lambda arguments so I am a bit

I'm very new to R and am trying to create my first loop. I have: x <-c(0:200) A <- dpois(x,exp(4.5355343)) B <- dpois(x,exp(4.5355343 + 0.0118638)) C <- dpois(x,exp(4.5355343 -0.0234615)) D <- dpois(x,exp(4.5355343 + 0.0316557)) E <- dpois(x,exp(4.5355343 + 0.0004716)) F <- dpois(x,exp(4.5355343 + 0.056437)) G <- dpois(x,exp(4.5355343 + 0.1225822)) and would like to

Dear all, Even though one of R users answered my question, I cannot understand, so I re-ask this question. I am trying to use "do.call", but I don't think I totally understand this function. Here is an simple example. -------------------------------------------- > B <- matrix(c(.5,.1,.2,.3),2,2) > B [,1] [,2] [1,] 0.5 0.2 [2,] 0.1 0.3 > x <- c(.1,.2) >

Dear all, I am trying to use "do.call", but I don't think I totally understand this function. Here is an simple example. -------------------------------------------- > B <- matrix(c(.5,.1,.2,.3),2,2) > B [,1] [,2] [1,] 0.5 0.2 [2,] 0.1 0.3 > x <- c(.1,.2) > X <- cbind(1,x) > X x [1,] 1 0.1 [2,] 1 0.2 > > lt <-

Hello all, I am trying to understand the different results I am getting from the following 3 commands: chisq.test(c(62,50), p = c(0.512,1-0.512), correct = F) # p-value = 0.3788 binom.test(x=62,n=112, p= 0.512) # p-value = 0.3961 2*(1-pbinom(62,112, .512)) # p-value = 0.329 Well, the binom.test was supposed to be "exact" and give the same results as the pbinom, while the chisq.test

Hello everybody. Does anyone else find the last test in the following sequence odd? Can anyone else reproduce it or is it just me? > binom.test(100,200,0.13)$p.value [1] 2.357325e-36 > binom.test(100,200,0.013)$p.value [1] 6.146546e-131 > binom.test(100,200,0.0013)$p.value [1] 1.973702e-230 > binom.test(100,200,0.00013)$p.value [1] 0.9743334 (R 1.5.1, Linux RedHat 7.1) --

Hi, I noticed this problem on my home desktop running FC4 and again on my laptop running FC5. Both have previously compiled and passed make check-all on 2.3.1 svn revisions from 10 days ago or so. On both these machines, make check-all is consistently failing (4 out of 4 attempts on the FC 4 desktop and 3 out of 3 on the FC 5 laptop) in the p-r-random-tests tests. This is with both default

R. 1.1.0 The example below is self explanatory. ## 1 ## # works fine > binom.test((50*.64),50,.5,alt='g') ... Exact binomial test ... ## 2 ## # WHAT ! ? > binom.test((50*.65),50,.5,alt='g') Error in binom.test((50 * 0.65), 50, 0.5, alt = "g") : x must be an

Hi All, My dependent variable is a ratio that takes a value of 0 (zero) for 95% of the observations and positive non-integer values for the other 5%. What model would be appropriate? I'm thinking of fitting a GLM with a Poisson ~. Now, becuase it takes non-integer values, using the glm function with Poisson family issues warning messages. Warning messages: 1: In dpois(y, mu, log = TRUE) :

Hi Folks, The recent correspondence about "strange fisher.test result", and especially Peter Dalgaard's reply on Tue 03 April 2007 (which I want to investigate further) led me to take a close look at the code for binom.test(). I now have a query! The code for the two-sided case computes the p-value as follows: if (p == 0) (x == 0) else if (p == 1) (x == n)

No, since I'd like to test null: p <= p0 alternative: p > p0. and my understanding is that binom.test tests null: p = p0 (can only be a "simple" null hypothesis according to help(binom.test)) alternative: p > p0 (or p < p0 or p != p0). Thanks, Mirko. > -----Urspr?ngliche Nachricht----- > Von: Douglas Bates [mailto:bates at stat.wisc.edu] >

Hi, I was hoping to get some advice regarding the testing of interactions, when one factor is modelled as a random effect... I have a model with binomial error structure where the response variable is the proportion of time spent at the main sett (animals were tracked for 28 consecutive days in each season, and were recorded either at the main sett or an outlier sett, so the response variable is

The nice new log=TRUE option in dpois appears to mess up the case where lambda=0 (I was trying to calculate the likelihood of a saturated model). Because the behavior is now always to calculate the probability in terms of exp(log(prob)), there's a test for lambda<=0 which really needs to be only lambda<0. dpois(0:5,0) ought to give 1 0 0 0 0 but gives NaNs instead. Here's

A question about the inner workings of glm() and dpois(): Suppose I call glm(y ~ x, family=poisson, weights = w) where y contains NON-INTEGER (but still nonnegative) values. (a) Does glm() still correctly maximise the weighted Poisson loglikelihood ? (i.e. the function given by the same formal expression as the weighted loglikelihood of independent Poisson variables Y_i except that the

i read a book from WOOD, there's an example which is talking about the pollutant. library(gamair) library(mgcv) y<-gam(death~s(time,bs="cr",k=200)+s(pm10median,bs="cr")+s(so2median,bs="cr")+s(o3median,bs="cr")+s(tmpd,bs="cr"),data=chicago,family=Possion) lag.sum<-function(a,10,11) {n<-length(a) b<-rep(0,n-11) for(i in 0:(11-10))

Dear all, I would like to model the relationship between y and x. y is binary variable, and x is a count variable which may be possion-distribution. I think it is better to divide x into intervals and change it to a factor before calling glm(y~x,data=dat,family=binomail). I try to use rpart. As y is binary, I use "class" method and get the following result. >