search for: lambda

...onstrOptim(c(0.5,0.9,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.9, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible > constrOptim(c(0.3,0.5,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in f(theta, ...) : argument "lambda1" is missing, with no default I only changed the parameters, how come the lambda1 that is not missing in the first 2 cases suddently become missing? For your convenience, I put the complete code below: Best Wishes Yuchen Luo ######################################## rm(list = ls()) mat=5...

...ut my situation does not belong to any of the above cases. I have attached my code bellow and could you please help me take a look? Best Wishes Yuchen Luo rm(list = ls()) ui=matrix(c(1,-1,0,0,0,0,0,0,1,-1,0,0,0,0,0,0,1,-1),6,3) ci=c(0,-0.5,0,-2,0,-0.6) apple=expression(rr*(1.0-rec)*(1.0-(pnorm(-lambda/2.0+log(((ss+(tot/sh* 1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1000.0)*lbar)/(tot/sh* 1000.0)/lbar*exp(lambda*lambda))*pnorm(-lambda/2.0-log(((ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/lambda))+(exp(rr*(lambda*lambda/(sigmae*ss/(ss+lba...

...List of 24 $ coefficients : Named num [1:2] -17.1 3 ..- attr(*, "names")= chr [1:2] "(Intercept)" "x" $ SIMEX.estimates : num [1:6, 1:3] -1 0 0.5 1 1.5 ... ..- attr(*, "dimnames")=List of 2 .. ..$ : NULL .. ..$ : chr [1:3] "lambda" "(Intercept)" "x" $ lambda : num [1:5] 0 0.5 1 1.5 2 $ model :List of 31 ..$ coefficients : Named num [1:2] -13.27 2.32 .. ..- attr(*, "names")= chr [1:2] "(Intercept)" "x" ..$ residuals...

Suppose I have a formula like this: f <- y ~ qss(x, lambda = lambdas[1]) + qss(z, lambdas[2]) + s I?d like a function, g(lambdas, f) that would take g(c(2,3), f) and produce the new formula: y ~ qss(x, lambda = 2) + qss(z, 3) + s For only two qss terms I have been using g <- function(lambdas, f){ F <- deparse(f) F <- gsub("lambda...

...kages) biplot # Mail : joseclaudio.faria em terra.com.br #=============================================================================== # Arguments: # x Data (frame or matrix: objects in lines variables in columns) # or a object of the class 'prcomp'. # lambda.ini First eigenvalue to be considered (default is 1) # lambda.end Latest eigenvalue to be considered # (default is 2 to 2d or 3 to 3d) # center Either a logical value or a numeric vector of length equal # to the number of columns of x (TRUE is the default)....

I'm trying to write a small function (below) to compute Box & Cox transformations of x for arbitrary values of lambda. I'd like to specify a range of values for lamba (min,max,step) and am having trouble getting the for loop to work. Suggestions? Any pointers to resources for learning to write functions in R for neophyte programmers? Thanks. --Dale boxcox <- function(x,min,max,step) { lambda <-...

Hi all. I'm trying to numerically invert the characteristic function of a quadratic form following Imhof's (1961, Biometrika 48) procedure. The parameters are: lambda=c(.6,.3,.1) h=c(2,2,2) sigma=c(0,0,0) q=3 I've implemented Imhof's procedure two ways that, for me, should give the same answer: #more legible integral1 = function(u) { o=(1/2)*sum(h*atan(lambda*u)+sigma^2*lambda*u/(1+lambda^2*u^2)) - q*u/2 rho=prod((1+lambda^2*u^2)^(h/4))*exp( (1/2)*...

...ead.table("epidemic.txt",header = TRUE) + attach(data, warn.conflicts = F) + k<-97 + d <- (sqrt((x-x[k])^2 + (y-y[k])^2)) + p <- 1-exp(-alpha*d^(-beta)) + p.alpha<-1 - exp(-3*d^(-beta)) + p.beta <- 1 - exp(alpha*d^(-2)) + iterations<-1000 + mu.lambda <- c(0,0);s.lambda <- c(100,100) + prop.s <- c(0.1,0.1) + lambda <- matrix(nrow=iterations, ncol=2) + acc.prob<-0 + current.lambda <- c(0,0) + for(t in 1:iterations){ + prop.lambda <- rnorm(2,current.lambda,prop.s) + a <- p.beta/p.alpha *(...

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda v...

Could I have some suggestions as to how (various ways) I can display my confidence interval results? rm(list = ls()) set.seed(1) func <- function(d,t,beta,lambda,alpha,p.gamma,delta,B){ d <- c(5,1,5,14,3,19,1,1,4,22) t <- c(94.32,15.72,62.88,125.76,5.24,31.44,1.048,1.048,2.096,10.48) post <- matrix(0, nrow = 11, ncol = B) theta <- c(lambda,beta) beta.hat <- 2.471546 for(j in 1:B){ for(i in 1:(B-1)){ c.lambda <- rgamma(10,d+alpha...

Hello! I would like to get MLE for parameter lambda of Poisson distribution. I can use fitdistr() for this. After looking a bit into the code of this function I can see that value for lambda and its standard error is estimated via estimate <- mean(x) sds <- sqrt(estimate/n) Is this MLE? With my poor math/stat knowledge I thought that MLE...

...0.01036597923,l=list(t=w[i],gp=gp))$root + }) Error in uniroot(saeqn, lower = -5000, upper = 0.01036597923, l = list(t = w[i], : f() values at end points not of opposite sign > and here is my fonction "saeqn". > saeqn<-function(s,l) + { + + + p<- exp(-l$gp$lambda+s)*l$gp$c + + + k11<-(l$gp$mu*(l$gp$lambda^2)*l$gp$c-s*l$gp$lambda*l$gp$c*l$gp$mu+l$gp$mu*l$gp$lambda)*p + + k12 <- -l$gp$mu*l$gp$lambda-s^2+2*s*l$gp$lambda-(l$gp$lambda^2) + + k13 <-k11+k12 + + k14<-(l$gp$lambda-s)*(-l$gp$mu*s-s*l$gp$lambda+s^2+l$gp$mu*l$gp$lambda*p) + + k1<-...

Hello, I have a problem finding a root of a function, which I define like this: tuki <- function(u, x, a, lambda){ if((lambda%%1>0) & u<0) {u<-(u+0i)} f <- Re(x-(a*(u)^lambda-(1-(u))^lambda)/lambda) f } What I want to do is to find the root, but without specifying the interval within which to search for it. I can do it easily in MATLAB with fsolve() or fzero() functions. I compared results o...

Dear everybody, I have a following problem. I have a 3D array lambda <- array( dim=c(N,M,M-1)) where I have to extract the elements as follows: lambda[ 1, state[1], 1] lambda[ 1, state[1], 2] ... lambda[ 1, state[1], M-1] lambda[ 2, state[2], 1] ... lambda[ 2, state[2], M-1] ... lambda[ N, state[N], M-1] i.e. the result should be a 2D array, where the second i...

...nd line): Parameter 5 to routine DSTEIN was incorrect Mac OS BLAS parameter error in DSTEIN, parameter #0, (unavailable), is 0 In case it helps, here's the function and function call that causes the crash. n <- 4 p <- 1 f <- 2; k <- 2 lms.bcn.univar <- function(y, B, p, f, k, lambda.0, mu.0, sigma.0){ n <- length(y) D11 <- matrix(1, nrow = p, ncol = n) D1 <- rbind(cbind(D11, matrix(0, nrow = p, ncol = f)), cbind(matrix(0, nrow = f, ncol = n), diag(f))) D22 <- rbind(rep(1:0,n), rep(0:1,n)) x1 <- t(D22)[,1] x2 &...

...la performing the replacement: g <- function(e, ...) { if (length(e) > 1) { if (identical(e[[2]], as.name(names(list(...))))) { e <- eval(e, list(...)) } if (length(e) > 1) for (i in 1:length(e)) e[[i]] <- Recall(e[[i]], ...) } e } g(f, lambdas = 2:3) ## y ~ qss(x, lambda = 2L) + qss(z, 3L) + s On Fri, Oct 23, 2020 at 9:33 AM Koenker, Roger W <rkoenker at illinois.edu> wrote: > > Suppose I have a formula like this: > > f <- y ~ qss(x, lambda = lambdas[1]) + qss(z, lambdas[2]) + s > > I?d like a functio...

Hi, I am trying to write a function which defines some arguments, then uses those arguments as arguments of other function calls. It's a bit tricky to explain, so a simple example will have to suffice. I imagine this has a simple solution, but perusing through environments and other help lists has not helped. Suppose I have two functions: f1 = function(a) {     b = a + 1     b } f2 =

...find the distribution parameter for a given expectation. However I fail to understand how to define properly the functions involved and pass the parameters correctly. Can anyone help me out? Thanks, Gerrit Draisma. This what I tried: ======= > # exponential density > g <- function(x,lambda){ lambda *exp(-lambda*x) } > > # expectation with lambda=1/10 > integrate(f = function(x,lambda=1/10) {x*g(x,lambda)}, 0,Inf) 10 with absolute error < 6.7e-05 > > # *how to write this as a function?* > E <- function(lambda) { + integrate( f = function(x,th){x*g(x,...

Hi, I want to write in x axis label "fitted value of lambda" (lambda in greek letter). xlab=expression(lambda) gives the "lambda", I tryed things like xlab=paste ("fitted value of ", expression(lambda)) but I didn't get the greek letter. Thanks in advance for any hint. Antonio Olinto ---------------------------------------...

Full_Name: Version: 1.9.0 OS: Windows XP Submission from: (NULL) (134.126.93.191) I believe the noncentrality parameter, lambda, in power.anova.test is incorrect. The noncentrality paramter is defined as lambda <- (groups - 1) * n * (between.var/within.var). The n should not be there. The function pf defines the noncentrality parameter as the sum of squares of the means. Therefore, the noncentrality paramter, lambda, in...