Displaying 20 results from an estimated 3000 matches similar to: "glm() with non-integer responses"
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,
theta2 and theta3.
My first
2012 Jul 05
3
Maximum Likelihood Estimation Poisson distribution mle {stats4}
Hi everyone!
I am using the mle {stats4} to estimate the parameters of distributions by
MLE method. I have a problem with the examples they provided with the
mle{stats4} html files. Please check the example and my question below!
*Here is the mle html help file *
http://stat.ethz.ch/R-manual/R-devel/library/stats4/html/mle.html
http://stat.ethz.ch/R-manual/R-devel/library/stats4/html/mle.html
2015 Nov 03
1
Fwd: Rcpp sugar dpois
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
2018 May 31
3
Understanding the sequence of events when calling the R dpois function
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
2011 Nov 14
7
Very simple loop
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
2018 Jan 17
1
mgcv::gam is it possible to have a 'simple' product of 1-d smooths?
I am trying to test out several mgcv::gam models in a scalar-on-function regression analysis.
The following is the 'hierarchy' of models I would like to test:
(1) Y_i = a + integral[ X_i(t)*Beta(t) dt ]
(2) Y_i = a + integral[ F{X_i(t)}*Beta(t) dt ]
(3) Y_i = a + integral[ F{X_i(t),t} dt ]
equivalents for discrete data might be:
1) Y_i = a + sum_t[ L_t * X_it * Beta_t ]
(2) Y_i
2011 Aug 08
3
on "do.call" function
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)
>
2010 Feb 06
1
Canberra distance
Hi the list,
According to what I know, the Canberra distance between X et Y is : sum[
(|x_i - y_i|) / (|x_i|+|y_i|) ] (with | | denoting the function
'absolute value')
In the source code of the canberra distance in the file distance.c, we
find :
sum = fabs(x[i1] + x[i2]);
diff = fabs(x[i1] - x[i2]);
dev = diff/sum;
which correspond to the formula : sum[ (|x_i - y_i|) /
2011 Aug 08
1
problem in do.call function
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 <-
2012 Jul 03
2
EM algorithm to find MLE of coeff in mixed effects model
I have a general question about coefficients estimation of the mixed model.
I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b follows
N(0,\psi) #i.e. bivariate normal
where b is the latent variable, Z and X are ni*2 design matrices, sigma is
the error variance,
Y are longitudinal data, i.e. there are ni
2008 May 23
1
maximizing the gamma likelihood
for learning purposes and also to help someone, i used roger peng's
document to get the mle's of the gamma where the gamma is defined as
f(y_i) = (1/gammafunction(shape)) * (scale^shape) * (y_i^(shape-1)) *
exp(-scale*y_i)
( i'm defining the scale as lambda rather than 1/lambda. various books
define it differently ).
i found the likelihood to be n*shape*log(scale) +
2009 Sep 19
1
Poisson Regression - Query
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) :
2009 Oct 30
3
Fast optimizer
Hi,
I'm using optim with box constraints to MLE on about 100 data points.
It goes quite slow even on 4GB machine. I'm wondering if R has any
faster implementation? Also, if I'd like to impose
equality/nonequality constraints on parameters, which package I should
use? Any help would be appreciated. Thank you.
rc
2007 Nov 19
2
All nonnegative integer solution
Dear all,
Is there any method in R to find all possible nonnegative integer
solutions to the linear equation with unit coefficients as follow:
X1+X2+...+Xk=N
Thank you,
Amin Zollanvari
2010 Oct 08
2
R: Why this deosn't work?, matrix, rounding error?
Hello
Why this works:
ncota <- 1
nslope <- 29
resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4)
But this doesn't?
ncota <- 1
sini <- 0.1; sfin <- 1.5; spaso <- 0.05; nslope <- 1+((sfin-sini)/spaso)
resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4)
I guess the problem is that the division gives a noninteger number.
How can I get the second one work?
I
2000 Feb 25
1
lambda==0 in dpois() (PR#459)
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
2001 Mar 05
1
Canberra dist and double zeros
Canberra distance is defined in function `dist' (standard library `mva') as
sum(|x_i - y_i| / |x_i + y_i|)
Obviously this is undefined for cases where both x_i and y_i are zeros. Since
double zeros are common in many data sets, this is a nuisance. In our field
(from which the distance is coming), it is customary to remove double zeros:
contribution to distance is zero when both x_i
2001 Mar 05
1
Canberra dist and double zeros
Canberra distance is defined in function `dist' (standard library `mva') as
sum(|x_i - y_i| / |x_i + y_i|)
Obviously this is undefined for cases where both x_i and y_i are zeros. Since
double zeros are common in many data sets, this is a nuisance. In our field
(from which the distance is coming), it is customary to remove double zeros:
contribution to distance is zero when both x_i
2008 Apr 25
2
force glm estimates to be nonnegative
Is there a way to force certain formula parameters to be nonnegative?
What I want to do is to estimate student capacity over time, namely by
> capacity ~ Student + Student:Day
I add this formula to a glm call and obtain negative learning slope estimates (Student:Day) in some cases.
However, I don't want to allow for that. In such a case, glm should solve
> capacity ~ Student
and
2010 Apr 25
1
function pointer question
Hello,
I have the following function that receives a "function pointer" formal parameter name "fnc":
loocv <- function(data, fnc) {
n <- length(data.x)
score <- 0
for (i in 1:n) {
x_i <- data.x[-i]
y_i <- data.y[-i]
yhat <- fnc(x=x_i,y=y_i)
score <- score + (y_i - yhat)^2
}
score <- score/n