Displaying 9 results from an estimated 9 matches for "possions".
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passions
2012 Aug 28
1
Bionomial and possion
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?*
--
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2011 Jun 21
5
please help for mgcv package
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))
2007 Jun 15
2
method of rpart when response variable is binary?
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.
>
2009 Oct 27
1
Poisson dpois value is too small for double precision thus corrupts loglikelihood
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,
the...
2006 Sep 20
1
ppois
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
2009 Apr 11
0
question related to fitting overdispersion count data using lmer quasipoisson
Dear R-helpers:
I have a question related to fitting overdispersed count data using lmer.
Basically, I simulate an overdispsed data set by adding an observation-level
normal random shock
into exp(....+rnorm()).
Then I fit a lmer quasipoisson model.
The estimation results are very off (see model output of fit.lmer.over.quasi
below).
Can someone kindly explain to me what went wrong?
Many thanks in
2009 Apr 11
0
Sean / Re: question related to fitting overdispersion count data using lmer quasipoisson
Hey Buddy,
Hope you have been doing well since last contact.
If you have the answer to the following question, please let me know.
If you have chance to travel up north. let me know.
best,
-Sean
---------- Forwarded message ----------
From: Sean Zhang <seanecon@gmail.com>
Date: Sat, Apr 11, 2009 at 12:12 PM
Subject: question related to fitting overdispersion count data using lmer
2002 May 13
1
GLM questions
Hi
I'm doing a glm analysis and I have two doubts (at least :)
1) When I run the function it gives a lot of warnings (see below) what
they mean ? (may be I'm ignorant about this analysis ...)
glm.poisson<-glm(log(Jkij+1)~fac.ano+fac.tri+fac.icesr+fac.mat+fac.ano:fac.icesr+fac.ano:fac.tri,family=poisson())
warnings()
40: non-integer x = 1.252763
41: non-integer x = 1.864785
42:
2010 Nov 03
2
multivariate Poisson distribution
Hello, from a search of the archives and functions, I am looking for information on creating random correlated counts from a multivariate Poisson distribution. I can not seem to find a function that does this. Perhaps, it has not yet been created. Has anyone created an R package that does this.
thanks,
Jourdan Gold
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