search for: psi1

...he main programme. For instance, is there any smart way for the following programme that will lessen time? --------------------------------------------------------------------------------------------------------------- dose<-seq(0,10,.5) alpha1<--4 beta1<-0.5 alpha2<--6 beta2<--.7 psi1<-function(alpha1,beta1,alpha2,beta2,d){ z1<-exp(alpha1+beta1*d) z2<-exp(alpha2+beta2*d) z1/((1+z1)*(1+z2)) } psi2<-function(alpha2,beta2,d) { z2<-exp(alpha2+beta2*d) z2/(1+z2)...

UseRs, I used the optim function valor.optim <- optim(c(1,1,1),logexp1,method ="BFGS",control=list(fnscale=-1),hessian=T); and I want to calculate the derivates, psi1<-valor.optim$par[1] psi2<-valor.optim$par[2] psi3<-valor.optim$par[3] a0=exp(psi1); a1=exp(psi2)/(20+exp(psi2)+exp(psi3)); a2=exp(psi3)/(20+exp(psi2)+exp(psi3)) deriv.psi1<-numericDeriv(a0,c("psi1","psi2","psi3")); d...

...----------------------------------------------------------------------- library(cubature) dose<-c(2,3,5) y0<-c(2,1,0) y1<-c(1,1,1) y2<-c(0,1,2) lf<-function (x) { v<-1 for (i in 1:length(dose)) { psi0<-1/((1+exp(x[1]+x[2]*dose[i]))*(1+exp(x[3]+x[4]*dose[i]))) psi1<-exp(x[1]+x[2]*dose[i])/((1+exp(x[1]+x[2]*dose[i]))*(1+exp(x[3]+x[4]*dose[i]))) v<-v*(psi0^y0[i])*(psi1^y1[i])*((1-psi0-psi1)^y2[i]) } return(v) } adaptIntegrate(lf, lowerLimit = c(-20, 0,-20,0), upperLimit = c(0,10,0,10)) --------------------------------------...

...e more specific, I wish to draw vertical lines at d=c(5.0,5.5,6) and they should go till p=c(0.12,0.60,0.20) . I haven't found any way out, though made several attempts. Please run the following commands first if you are interested in! ####################################################### psi1<-function(alpha1,beta1,alpha2,beta2,d){ exp(alpha1+beta1*d)/((1+exp(alpha1+beta1*d))*(1+exp(alpha2+beta2*d))) } alpha1<--3.5 beta1<-1 alpha2<--6 beta2<-0.72 d<-seq(0.5,10,0.01) plot(d,psi1(alpha1,beta1,alpha2,beta2,d),type="l",...

Hi all, I am using the package ?Segmented? to estimate logistic regression models with unknown breakpoints (see Muggeo 2003 Statistics in Medicine 22:3055-3071). In the documentation it suggests that it might be possible to include several variables with breakpoints in the same model: ?Z = a vector or a matrix meaning the (continuous) explanatory variable(s) having segmented relationships with

...tell me what is wrong with my model? Thank you. Here is the code: model.ram1 <- specifyModel() UNIT -> Y1, ty,0.3 UNIT -> Z1, tz1,-0.1 UNIT -> Z2, tz2,0.1 CF -> Y1, lamy,0.5 CF -> Z1, lamz1,0.85 CF -> Z2, lamz2,0.2 UNIT -> CF, k Y1 <-> Y1, psi3, NA Z1 <-> Z1, psi1, NA Z2 <-> Z2, psi2, NA CF <-> CF,vCF1,NA sem.m1<-sem(model=model.ram1,S=S2,N=500,fixed.x="UNIT",raw=T) -- I.J. Williams Ph.D. Student in Education Measurement and Statistics Statistics MS Mathematics BS Rutgers University [[alternative HTML version deleted]]

...,2] time.WC <- X[,3] olm <- lm(y~0+ group + time.KV + time.KW + time.WC) os <- segmented(olm, seg.Z= ~ time.KV + time.KW + time.WC, psi=list(time.KV=c(300,500), time.KW=c(450,600), time.WC=c(300,450))) The error message is as follows: Error in `rownames<-`(`*tmp*`, value = c("psi1.time.KV", "psi2.time.KV", : Length of 'dimnames' [1] not equal to array extent Zusätzlich:Warnmeldung: In cbind(initial, psi, sqrt(vv)) : number of rows of result is not a multiple of vector length (arg 2) However, when I take out one break.point from any of the gro...

Hello all! I want to estimate parameters in a MIMIC model. I have one latent variable (ksi), four reflexive indicators (y1, y2, y3 and y4) and four formative indicators (x1, x2, x3, x4). Is there a way to do it in R? I know there is the SEM library, but it seems not to be possible to specify formative indicators, that is, observed exogenous variables which causes the latent variable. Thanks,

...ernative hypothesis: less segments <- segmented(lmfit, seg.Z=~HRRPUA_kWm2,psi=c(475,550)) summary(segments) ***Regression Model with Segmented Relationship(s)*** Call: segmented.glm(obj = lmfit, seg.Z = ~HRRPUA_kWm2, psi = c(475, 550)) Estimated Break-Point(s): Est. St.Err psi1.HRRP 430.2 4.087 psi2.HRRP 484.6 3.077 t value for the gap-variable(s) V: 0 0 Meaningful coefficients of the linear terms: Estimate Std. Error t value Pr(>|t|) (Intercept) -38.6993 274.7666 -0.141 0.8891 HRRPUA_kWm2 1.4297 0.7472 1.914 0.0668 . U1.HRRP 42.288...

...Bentler CFI = 0.902 SRMR = 0.0682 BIC = 51.4 SYNTAX rm(sem.enf.rq) mdl.rq <- specify.model() enf -> law2, NA, 1 enf -> law3, lam2, 1 enf -> law4, lam3, 1 enf <-> enf, psi1, 0.6 law2 <-> law2, theta1, 0.3 law3 <-> law3, theta2, 0.3 law4 <-> law4, theta3, 0.5 gender -> enf, a1, 0.2 incomex -> enf, a2, 0.2 oftdrnkr...

Hi there, as a student sociology, I'm starting to learn about SEM. The course I follow is based on LISREL, but I want to use the SEM-package on R parallel to it. Using LISREL, I found it to be very usable to be able to see the total direct and total indirect effects (standardized and unstandardized) in the output. Can I create these effects using R? I know how to calculate them

Dear All, I'm a beginner user in R and I would like to make a quadratic and plateau model in R. Can you help please with an example? Thanks so much -- Ahmed M. Attia Assistant Lecturer El-Khattara farm Station Agronomy Dept., Zgazig Univ., Egypt Visiting Scientist Haskell Agricultural laboratory Agronomy and Horticultural Dept., Univ. of Nebraska-Lincoln ahmedatia at zu.edu.eg

Dear R list members, Since today seems to be the day for optimization questions, I have one that has been puzzling me: I've been doing some work on sem, my structural-equation modelling package. The models that the sem function in this package fits are essentially parametrizations of the multinormal distribution. The function uses optim and nlm sequentially to maximize a multinormal