similar to: seemingly unrelated regression

Hello everyone, I would like to a linear regression with the following code. lm(a[,"fquamsci"]~., data=a) a is a list with class "mts" "ts" , and "fquamsci" is the name of the response variable in a. I would like to do a linear regression of "fquamsci" to the rest of the variables. But it turns out the "fquamsci" is also included in

Hi all, I am trying to use mle() to find a self-defined function. Here is my function: test <- function(a=0.1, b=0.1, c=0.001, e=0.2){ # omega is the known covariance matrix, Y is the response vector, X is the explanatory matrix odet = unlist(determinant(omega))[1] # do cholesky decomposition C = chol(omega) # transform data U = t(C)%*%Y WW=t(C)%*%X beta = lm(U~W)$coef Z=Y-X%*%beta

Good morning reader, I have encountered a, probably, simple issue with respect to the *formulae* of a *regression model* I want to use in my research. I’m researching alliances as part of my study Business Economics (focus Strategy) at the Vrije Universiteit in Amsterdam. In the research model I use a moderating variable, I’m looking for confirmation or help on the formulation of the model.

Dear all, I am trying to estimate a system of equations with a Seemingly Unrelated Regression. However because of the characteristics of the data I'd like to do it with a negative binomial estimation. Would anybody know how to implement a Seemingly Unrelated Negative Binomial (SUNB) estimation in R? Thank you in advance for your help. Best regards, Elisa Lanzi -- Elisa Lanzi PhD student in

Hello, I would like to fit a linear regression and when I use summary(), I got the following result: Call: lm(formula = weight ~ group - 1) Residuals: Min 1Q Median 3Q Max -1.0710 -0.4938 0.0685 0.2462 1.3690 Coefficients: Estimate Std. Error t value Pr(>|t|) groupCtl 5.0320 0.2202 22.85 9.55e-15 *** groupTrt 4.6610 0.2202 21.16 3.62e-14

Hi All, I am trying to make a loop for a function and I am using the following codes. "p" and "var" are some matrix obtained before. I would like to apply the function "gpdlow" for i in 1:12 and get the "returnlow" for i in 1:12. But when I ask for "returnlow" there are warnings and it turns out some strange result. for (i in 1:12){ gpdlow

Hi, I would like to generate a series in the following form (0.8^1, 0.8^2, ..., 0.8^600) Could anyone tell me how can I achieve that? I am really new to R. -- View this message in context: http://www.nabble.com/sequences-tf4019146.html#a11414836 Sent from the R help mailing list archive at Nabble.com.

Hello, I am using the following codes to plot a histogram and density line for x. For the density line, I just want it to show the two tails, eg, for x larger than 0.05 ans smaller than -0.05 hist (x, seq(-0.1,0.1,0.01),freq = FALSE) lines (density(x,bw="SJ"), x> 0.05 & x< (-0.05), col = "red") But is does not work, can anyone give me some advice? -- View this

Hello, I would like to perform a linear regression and the data is a list.e.g lm(list$abc~., data=list) or lm(abc~., data=list), which would give the same result. The problem is I would like to call the response variable in a more general form. What I try to achieve is sth like lm(list$(paste("a","b","c"))~., data=list), but it does not work. Could anyone give me

Hello everyone, Do we have package in R which will estimates the parameter in Seemingly Unrelated Spatial Regression Model? Any help would be appreciated. Thanks Arun -- View this message in context: http://www.nabble.com/Seemingly-unrelated-model-with-spatial-processes-tp22478913p22478913.html Sent from the R help mailing list archive at Nabble.com.

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

Hello to everyone, I found some problems using R in the estimation of systems of nonlinear equations like SURE (Seemingly Unrelated Regression Equations) with mutual parameters as the following system example: EQ1 PQ1=P1*G1+B1*(Y-P1*G1-P2*G2-P3*G3) EQ2 PQ2=P2*G2+B2*(Y-P1*G1-P2*G2-P3*G3) where G1,G2,G3,B1,B2 are the coefficients to estimate. command "nls" accept only single

I am having problem using output of lm() function for further analysing using systemfit package. Basicaly, the problem s following - I generate several formulas using lm() > fo1 <- lm(r98[,2] ~ f98[,1] + f98[,2] + ... + f98[,43]) > fo2 <- lm(r98[,1] ~ f98[,1] + f98[,2] + ... + f98[,43]) and than I want to estimate a general model using package systemfit. > fitsur <-

Hi, I am trying to write a function with the following codes and I would like it to return the values for "alpha beta para parab " seperately. Then I would like to use this funstion for "variable" with factor "a" and "b". But the result turns out to be a matrix with element like "Numeric,2" ... I guess they are just the values for

Hi, I have a series of return data, a and b are factors. I would like to build a matrix which contains each vector of "returns". I am thinking about something as following, but I guess there should be a sensible way of doing this. returns <- split(return, list(regimef, assetf)) cbind(returns[[1]], returns[[2]],...,returns[[n]]) Could anyone give me some advice? Many thanks. --

I have got a list named "filtered", I would like to construct alist named "fdata" as following: fdata <- cbind(matrix(unlist(filtered),ncol=28), myregime) If I try names(filtered), it gives all the correct name for each vector, but if I try names(fdata), it appears "filtered[[1]]" "filtered[[2]]" ..., How can I keep the name in "fdata"?

Hello, I would like to use the cbind() function to construct a matrix used in the middle of a function as following for (i in 1:1000) { b[i] <- function(cbind(a[[1]][[i]], a[[2]][[i]],a[[3]][[i]],...a[[67]][[i]])) } Is there an easy way of achieving this rather than "cbind" every column? -- View this message in context: http://www.nabble.com/cbind%28%29-tf4724534.html#a13507633

Dear all, For an introductory course on glm?s I would like to create an example to show the difference between glm and transformation of the response. For this, I tried to create a dataset where the variance increases with the mean (as is the case in many ecological datasets): poissondata=data.frame( response=rpois(40,1:40), explanatory=1:40) attach(poissondata) However, I have run into

Dear all, I am having difficulties using the seemingly-simple function lag. I have a dataframe with several weather variables (maxitemp, windspeed, rainfall etc), and the response variable (admissions). The dataset is fairly large (1530 observations). I simply want to model the response against a lag of a couple of the explanatory variables, say maxitemp and rainfall. I would like to look at

Hi, I want to import some data from Access and I am using the following codes: testdb <- file.path("c/../db1") channel <- odbcConnect("testdb") sqlFetch(channel,"tbl",colnames = TRUE, rownames = FALSE) It comes out the error message: 1: [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver Manager] Data source name not found and no default