similar to: Simulating SVAR Data

Hello! I'm doing a svar and when I make the estimation the next error message appears: In SVAR(x, Amat = amat, Bmat = bmat, start = NULL, max.iter = 1000, : The AB-model is just identified. No test possible. Could you help me to interpret it please. Also I have the identification assumption that one of my shocks is exogenous relative to the contemporaneous values of the other variables

Hello folks! I'm using R-Package {vars} and I'm trying to estimate an A-Model. I have serious problems regarding the restrictions. 1) My A-Matrix needs (!) to have the following form: # 1 NA NA NA # 0 1 NA NA # 0 0 1 NA # 0 0 0 1 That is done in R by: A_Matrix <- diag(4) # main diagonal = 4 restrictions A_Matrix [1, 2] <- NA # A_Matrix [1, 3] <- NA #

How to determine restricted variable in SVAR and SVEC? There are some values which set to be zero and others set to be NA.. How to determine values that set to be 0? Thanks Regards, Arif _________________________________________________________________ Facebook. k-basics.aspx?ocid=PID23461::T:WLMTAGL:ON:WL:en-id:SI_SB_2:092009 [[alternative HTML version deleted]]

Hello, I'm working on an structural VAR using the var command to estimate and the svar command on the resultant object (package: vars). I want to constrain coefficients to equal one another, but that value to be estimated. So for the A matrix, I want A[2,1]=A[1,2] to be my constraints. Can this be done with this package? If so, how? If not, is there another package that it might be done with

Dear R-experts, Here below my R code. I get a NaN response for gam with mgcv library. How to solve that problem? Many thanks. ######################################################### library(mgcv) ? y=c(23,24,34,40,42,43,54,34,52,54,23,32,35,45,46,54,34,36,37,48) x1=c(0.1,0.3,0.5,0.7,0.8,0.9,0.1,0.7,0.67,0.98,0.56,0.54,0.34,0.12,0.47,0.52,0.87,0.56,0.71,0.6)

I use code ( actually its code from vars package and its directly below ) to do a sequence of lm calls and the data I use from the matrix depends on restrictions. for(i in 1:K){ datares <- datasub[, which(x$restrictions[i, ] == 1),drop=FALSE] y <- yendog[, i] lmres <- lm(y ~ -1 + ., data=datares) # x$varresult[[i]] <- lmres # x$resid[, i] <-

Hello With due respect, have a nice time. I would like to ask some command in R. It is regarding variable selection in linear regression. In R, there is one rebuild function called "step" which selecting variables according to AIC. let say i have data [y, x1,x2,x3,x4] we start with y~b0 i compute the partial F test and choose the variable with maximum partial F to enter the

Hi R-users, I would like to ask question related to error output. If an error comments come out, then the program will automatically stop. I want to ask , how I can still continue the program even though there is an error comment? var=VAR(Canada,p=3,type="const") for (j in 1:nrow(com)) { mat=ma { for (i in 1:ncol(com)) { y=which(mat==com[j,i]) mat[y]=NA }

Dear R users, Biostatistics Denmark would like to annouce the availability of the new utility: sas2R --- a SAS to R parser. For almost 40 years SAS has been the primary tool for statisticians worldwide and its easy-to-learn syntax, unsurpassed graphical system, powerful macro language and recent graphical user interfaces have made SAS the number one statistical software choice for both beginners

I am trying to understand what is the difference between linear and polynomial kernel: linear: u'*v polynomial: (gamma*u'*v + coef0)^degree It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree = 1 should be identical to linear kernel, however it gives me significantly different results for very simple data set, with linear kernel

Hi, this is possibly a bug report (I'm not sure if this is a feature). It's related to the --keep-dirlinks option, when combined with --delete . I have the following directory structure on server A: ls -lR software software: total 238 drwxr-xr-x 2 biolord bioinf 1024 Sep 20 10:49 EMBOSS/ lrwxrwxrwx 1 biolord bioinf 6 Feb 5 2003 MSE -> EMBOSS/ lrwxrwxrwx 1

Dear R users The following function is R code for the main compuations in the article: M. Eliasziw, S Lorraine Young, M Gail Woodbury and Karen Fryday-Field (1994): Statistical Methodology for the Concurrent Assessment of Intrarater and Intrarater Reliability: Using Goniometric Measurements as an Example. Physical Therapy 74 (8); 777-788 The function gives the estimated inter- and intrarater

Hi everyone, I'm working with the Vector Autoregressive Model (VAR), and it seems like the ur.ca package provides the best function for this purpose. My problem is, that I don't know how to extract a certain value from the output without using the variables names. I was hoping that this could be done by using numerical value, e.g. [1], instead of writing for instance $Liabilities. The

Dear Researches, I am using RF (in regression way) for analize several metrics extract from image. I am tuning RF setting a loop using different range of mtry, tree and nodesize using the lower value of MSE-OOB mtry from 1 to 5 nodesize from1 to 10 tree from 1 to 500 using this paper as refery Palmer, D. S., O'Boyle, N. M., Glen, R. C., & Mitchell, J. B. O. (2007). Random Forest Models

Hi R-users, I am trying to estimate function parameters using optim(). My count observations follows a Poisson like distribution. The problem is that I wanna express the lambda coefficient, in the passion likelihood function, as a linear function of other covariates (and thus of other coefficients). The codes that I am using (except data frame) are the following (FYI the parameters need to be

All -- I have been trying to work with the 'Party' package using R v2.15.1 and have cobbled together a (somewhat) functioning code from examples on the web. I need to run a series of unbiased, conditional, cForest tests on several subsets of data which I have made into a loop. The results ideally will be saved to an output file in matrix form. The two questions regarding the script in

Dear friends, I have written the following lines in R console wich already exist in pdf file systemfit: data( "GrunfeldGreene" ) library( "plm" ) GGPanel <- plm.data( GrunfeldGreene, c( "firm", "year" ) ) greeneSur <- systemfit( invest ~ value + capital, method = "SUR", + data = GGPanel ) greenSur I have obtained the following incomplete

Hi all, I'm trying to fit a model using the shorthand coeff[factor] instead of coding dummy variables. Is there a way to keep this notation when specifying constraints? See example below: x = runif(200) b0 = c(rep(0,100),runif(100)) b1 = 1 fac <- as.factor(rep(c(0,1), each=100)) y = b0+b1*x+rnorm(200,sd=0.05) ## No constraint - works nls(y~b0[fac]+b1*x,

I have the following function, a MSE calc based on some Multivariate normals: MV.MSE<-function(n,EP,X,S){ (dmvnorm(X,mean=rep(0,2),I+S+EP)-dmvnorm(X,mean=rep(0,2),I+S))^2 + 1/n*(dmvnorm(X,mean=rep(0,2),1+S+EP/2)*det(4*pi*EP)^-.5- (dmvnorm(X,mean=rep(0,2),I+S+EP ))^2)} I can get the MV.MSE for given values of the function e.g

Several ways: 1. Read ?randomForest, especially the `Value' section. 2. Look at str(myforest.rf). 3. Look at print.randomForest. If the forest has 100 trees, then the mse and rsq are vectors with 100 elements each, the i-th element being the mse (or rsq) of the forest consisting of the first i trees. So the last element is the mse (or rsq) of the whole forest. HTH, Andy > From: David