similar to: Systems of equations in glm?

Dear R users/developers I requested help to solve the problem of formulating Multivariate Sample selection model by using Full Information Maximum Likelihood (FIML)estimation method. I could not get any response. I formulated the following code of FIML to analyse univariate sample selection problem. Would you please advise me where is my problem library (sem) library(nrmlepln) Selection

Hi there, I want to estimate simultaneous equation model with panel data. The model looks as follows Y1=a0+a1*X1+a2*X2 Y2=b0+b1*X2+b2*X1 X1=Z1-(Y1/a1) X2=Z2-(Y2/b1) I In this model Y1, Y2, X1 and X2 are endogenous variables; Z1, Z2 are exogenous variables and a0, a1, a2, b0, b1 and b2 are parameters. Could any one please help me how to estimate this model in R. Thanking you in anticipation

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

I cannot figure out how, using R, I can paste indexes or characters to the variable names which are used within loops. I will explain this with a simple example: Immagine I have a huge series of variables, each one taken two times, say x1 x2 y1 y2 z1 z2..... Now, immagine that I want to compute a variable from the difference of each couple, say dx=x1-x2, dy=y1-y2, dz=z1-z2... In Stata, for

Hi, I am having trouble plotting the graph I need given the follow kind of data > xxx <- data.frame( "x"=c(1,2,3,4,5), "y1"=c(2,4,3,5,6), "y2"=c(3,4,6,3,1), "y3"=c(1,3,5,7,3), "z1"=c(1,NA,3,5,NA), "z2"=c(2,NA,4,6,NA) ) > xxx x y1 y2 y3

I'd like to model the relationship between m responses Y1, ..., Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. My understanding is that although lm() handles vector Y's on the left-hand side of the model formula, it really just fits m separate lm models. What should

Hi, I want to know if it's possible to pass a vector v=c(x,y,...) to a function f(x,y,...) so that each vector element corresponds to a formal argument of the function. For python programmers: f(*v). Specifically, what I'm trying to achieve is: given a list of coordinates l=list(c(x1,y1,z1), c(x2,y2,z2),...) I would like to obtain the corresponding elements in some array A (3-dim in this

Hi all,I had a question last week on asking for a function that will help me draw three different circles on x,y,z axis based on polar coordinates (Each X,Y,Z circle are coming from three independent measurements of 1-360 polar coordinates). It turned out that there ?is no such function in R and thus I am trying to write my own piece of code that hopefully I will be able to share. I have spent

hello, I want to simulate 200 times the mean of a joint probability (y1) and 200 times the mean of another joint distribution (y2), that is I'm expecting to get 200 means of y1 and 200 means of y2. y1 and y2 are probabilities that I calculate from the marginal prob. (z1 and z2 respectively) multiple by the conditional prob. (x1 and x2 respectively), which I generaterd from the binomial

Please look at what I see in your code below (run-on code mush) to understand part of why it is important for you to send your email as plain text as the Posting Guide indicates. You might find [1] helpful. [1] https://wiki.openstack.org/wiki/MailingListEtiquette -- Sent from my phone. Please excuse my brevity. On June 25, 2017 2:42:26 PM EDT, Alaios via R-help <r-help at r-project.org>

Dear R-users, I am facing problem with integrating in R a likelihood function which is a function of four parameters. It's giving me the result at the end but taking more than half an hour to run. I'm wondering is there any other efficient way deal with. The following is my code. I am ready to provide any other description of my function if you need to move forward.

Dear useRs, I'm trying to simply fill in the area under a curve using RGL. Here' the set up: x <- c(0.75,75.75,150.75,225.75,300.75,375.75,450.75,525.75,600.75,675.75, 0.5,50.5,100.5,150.5,200.5,250.5,300.5,350.5,400.5,450.5, 0.25,25.25,50.25,75.25,100.25,125.25,150.25,175.25,200.25,225.25) y <- c(0.05,4.91,9.78,14.64,19.51,24.38,29.24,34.11,38.97,43.84,

Hi, can anyone help me fit betabinomial model to the following dataset where each iD is a cluster in itself , if i use package aod 's betabinom model it gives an estimate of zero to phi(the correlation coeficient ) and if i fix it to the anova type estimate obtained from icc( in package aod) then it says system is exactly singular. And when i try to fit my loglikelihood by

Consider the following scrap of code: > x<- ts(1:50,start=c(1,11),freq=12) > y <- aggregate(x,nfreq=4) > c(y) [1] 6 15 24 33 42 51 60 69 78 87 96 105 114 123 132 141 > y Error in rep.int("", start.pad) : invalid number of copies in rep.int() > tsp(y) [1] 1.833333 5.583333 4.000000 So we can aggregate into quarters, but we cannot print it using

Consider the following scrap of code: > x<- ts(1:50,start=c(1,11),freq=12) > y <- aggregate(x,nfreq=4) > c(y) [1] 6 15 24 33 42 51 60 69 78 87 96 105 114 123 132 141 > y Error in rep.int("", start.pad) : invalid number of copies in rep.int() > tsp(y) [1] 1.833333 5.583333 4.000000 So we can aggregate into quarters, but we cannot print it using

I have a set of data in the form x1, y1, z1 x1, y2, z2 ... x1, yN, zN x2, y1, z(N+1) x2, y2, z(N+2) ... x2, yN, z(2N) ...and so on... xM, yN, val(M*N) I have been trying to figure out how to get R to use this data in a persp plot. So far the only thing that I can figure out to do is to break the data file into three different files. The first file contains the x-coordinate data: x1 x2

Hi, I am having an issue when creating a postscript file from RGL window. It seems to cut off some of the axis labels. Here is the code I am using. I created a 3D plot using RGL_0.71 with R 2.5 on Windows XP. z1<-c(5,4,1,4.5,2,3,2,1,1) z2<-c(6,8,7,7.5,5,3.5,4,1,1) z3<-c(3,2,4,7,3,4.5,6,2,3) x1<-seq(1,9) x2<-seq(1,9) x3<-seq(10,18) y1<-seq(8,0) y2<--1*y1

I am trying to build up a quoted or character expression representing a component in a list in order to reference it indirectly. For instance, I have a list that has data I want to pull, and another list that has character vectors and/or lists of characters containing the names of the components in the first list. It seems that the way to do this is as evaluating expressions, but I seem to

Full_Name: Raymond Salvador Version: R 1.6.2 OS: Windows ME Submission from: (NULL) (131.111.93.195) The mean function applied on individual components of several correlation matrices gives a wrong result (gives the first value instead of the mean). Here there is a simple example x1 <- rnorm(10,1,1) y1 <- rnorm(10,1,1) z1 <- cbind(x1,y1) w1 <- cor(z1) x2 <- rnorm(10,1,1) y2

Hi I have a dataset which is like this: ID X1 Y1 Z1 X2 Y2 Z2 1 32 127 85 31.7 121 85 2 37.9 153 88 34.6 137 90 3 36.8 134 88 36.1 155 97 4 31.4 136 87 29 113 70 I would like draw the arrow or segment line for each ID from (x1, y1, z1) to (x2, y2, z2) Is there a way to do this kind of 3-D graph. Thank you -- View this message in context: