search for: minuslogl

Hi I have the following formula (I hope it is clear - if no, I can try to do better the next time) h(x, a, b) = integral(0 to pi/2) ( ( integral(D/sin(alpha) to Inf) ( ( f(x, a, b) ) dx ) dalpha ) and I want to do an mle with it. I know how to use mle() and I also know about integrate(). My problem is to give the parameter values a and b to the

...39;s sufficiently interested I can post the package somewhere -- with the package installed, it's only a few steps to reproduce the example. I end up, deep in the process of trying to compute a likelihood profile, with the following situation: I want to evaluate "call": call mle2(minuslogl = function (lmu = NULL, ltheta = NULL) { if (!is.null(parameters)) { pars <- unlist(as.list(match.call())[-1]) for (i in seq(along = parameters)) { assign(vars[i], mmats[[i]] %*% pars[vpos[[i]]]) } } arglist1 <- lapply(arglist1, eval, envir = da...

...sing mle() from the stats4 package. Here's what I would have thought would work: -------------- library(stats4) ## simulate values r = rnorm(1000,mean=2) ## very basic neg. log likelihood function mll <- function(mu,logsigma) { -sum(dnorm(r,mean=mu,sd=exp(logsigma),log=TRUE)) } mle(minuslogl=mll,start=list(mu=1,logsigma=0)) r2 = rnorm(1000,mean=3) ## second "data set" with(list(r=r2), mle(minuslogl=mll,start=list(mu=1,logsigma=0)) ) ------------- but this doesn't work -- it fits to the original data set, not the new one --- presumably because mll() picks up i...

Thank you very much. now, i call mle(minuslogl=loglik, start=start, method <<- method, fixed=list()) in the mle.wrap() function, and the profile.mle() worked. however, it created a variable named "method" in user workspace. if there had been a variable with same name, then the value of that variable would be destroyed. Is th...

...in lm, nls, nlme, etc., that would allow the log-likelihood function to be evaluated with different sets of data. (Peter's advice was to use closures, writing a function-to-generate-likelihood-functions such as fn0 <- function(x) { function(mu,sd) { -sum(dnorm(x,mu,sd,log=TRUE) } } mle(minuslogl=fn0(x),...) My feeling is that this will be somewhat mysterious to the intermediate R users who are my target audience.) I have three thoughts on how to allow different data sets to be substituted in the same objective function, and I'm not sure which is best. 1. passed in ... as in opt...

...optim.method = "BFGS", ## optim method optim.lower = numeric(length(initial.para)) + 0.00001, optim.upper = numeric(length(initial.para)) + Inf, ...) { require(stats4) ## Create a string with the log likelihood definition minusLogLikelihood.txt <- paste("function( ", paste(names(initial.para), collapse = ", "), " ) {", "isi <- eval(", deparse(sub...

Hi there, When I do bootstrap on a maximum likelihood estimation, I try the following code, however, I get error: Error in minuslogl(alpha = 0, beta = 0) : object "x" not found It seems that mle() only get data from workspace, other than the boot.fun(). My question is how to pass the data to mle() in my case. I really appreciated to any suggestions. Best wishes, Jinsong #-----------code start here--------------- x...

Hi I am trying to find a parametric bootstrap confidence interval and when I used the boot function I get zero bias and zero st.error? What could be my mistake? Thank you and take care. Laila [[alternative HTML version deleted]]

...fint(m1)) ---------------- for what it's worth, the logic appears to be OK in mle in the stats4 library: -------------- library(stats4) mfun <- function(a,b,s) { if (a<0 || b<0 || s<0) stop("bounds violated") -sum(dnorm(y,a*x/(1+a*b*x),sd=s,log=TRUE)) } m2 = mle(minuslogl=mfun, start=list(a=1,b=1,s=0.1), method="L-BFGS-B",lower=c(0.002,0.002,0.002)) confint(m2) m2b = mle(minuslogl=mfun, fixed=list(b=0),start=list(a=1,s=0.1), method="L-BFGS-B",lower=c(0.002,0.002,0.002)) ## set boundary slightly above zero to avoid ## boundary ca...

Dear all, I'm using igraph for some analysis about the network I have. I have a question about the function "power.law.fit". I wonder if there is any test for checking whether the "power.law.fit" is good for the input, i.e., under which situation, could we use this function to get a reliable result. I'm afraid even I input a random graph without any property of

...cEcon}*. I have documented some of my attempts below ((a) package name (b) usage (c) my attempt and corresponding error). In all humility I apologise for any bad coding, and ask if anyone can *direct me in finding these estimators*. Yours sincerely. *(1a) mle{stats4} (b) Usage: mle(minuslogl, start = formals(minuslogl), method = "BFGS", fixed = list(), ...) (For this I use the negative of the log-likelihood function,bn)* *(c) >mle(start=list(psi=1,alpha=0),fn, method="BFGS",fixed=list(c=c))* Error in optim(start, f, method = method, hessian = TRUE, ...)...

...nable expectation since it *is* specified as a list). The fundamental problem is that optim() loses names of the parameter vector somewhere. Example: x = runif(200) y = 1+x+x^2+rnorm(200,sd=0.05) fn <- function(a,b,z=2,c,d) { -sum(dnorm(y,mean=a+c*x+d*x^2,sd=exp(b),log=TRUE)) } m1 = mle(minuslogl=fn,start=list(a=1,b=1,c=1,d=1)) ## fails with "missing argument" warning, about wrong argument m1 = mle(minuslogl=fn,start=list(a=1,b=1,c=1,d=1),fixed=list(z=2)) ## works m2 = mle(minuslogl=fn,start=list(a=1,d=1,c=1,b=1),fixed=list(z=2)) ## fails -- coeffs returned in wrong order TO DO:...

...;-function(alfa,beta) + {n<-24 + x<-data2 + -n*log(alfa)-n*log(beta)+alfa*sum(x^beta)-(beta-1)*sum(log(x))} > est<-mle(minuslog=ll, start=list(alfa=10, beta=1)) There were 50 or more warnings (use warnings() to see the first 50) > summary(est) Maximum likelihood estimation Call: mle(minuslogl = ll, start = list(alfa = 10, beta = 1)) Coefficients: Estimate Std. Error alfa 0.002530163 0.0006828505 beta 0.641873010 0.0333072184 -2 log L: 511.6957 > library(stats4) > ll<-function(alfa,beta) + {n<-24 + x<-data2 + -n*log(alfa)-n*log(beta)+alfa*sum(x^beta)-(beta-1)*...

...he CRAN. When doing the check, there is an example that has an error running in the 32 bits version. The problem comes from the mle function, using it with a lower constrain. In 64 bits version it works fine but when I put it in the R 32 bits it fails. (same numbers, all equal!) The call is: *mle(minuslogl = p.est,start = beta,method = "L-BFGS-B",lower=llim*reduction)* lower = -0.01570427 The optimizer (optim function in 32 bits) display: -0.015704 -loglik 48.690236 -0.015704 -loglik 48.690236 -0.017704 -loglik 1.#QNAN0 And it is not respecting the lower constrain. Could anyone explain...

...hat worked for a single pair of interactions, when I try to evaluate two pairs of interactions( flowers*gopher, flowers*rockiness) my computer runs out of memory, and the larger desktop I use just doesn't go anywhere after about 20 minutes. Is it really that big a calculation? to start: mle2(minuslogl = Lily_sum$seedlings ~ dnbinom(mu = a, size = k), start = list(a = 10, k = 1)) then: i2<-interaction(Lily_sum$flowers, Lily_sum$gopher) i3<-interaction(Lily_sum$flowers, Lily_sum$rockiness) mle2(Lily_sum$seedlings ~ dnbinom(mu = a, size = k), start=list(a=10,k=1) ,parameters=list(a~i3+i2+L...

...f <- list() f$IP <- mle(...) f$NE <- mle(...) but when I say: > summary(f) I get: Length Class Mode IP 0 mle list NE 0 mle list I don't get the output I would have, i.e. the one from > summary(f$IP) summary(f$IP) Maximum likelihood estimation Call: mle(minuslogl = IPNeglogPoisL, method = "L-BFGS-B", fixed = list(), control = list(maxit = 1e+08, factr = 1e-20)) Coefficients: Estimate Std. Error a 1242.0185506 44.92341097 b 0.8802538 0.01685811 -2 log L: 145.3509 What I want to do is something like: AICs <- AIC(logLik(f)) an...

...LE: > nLL <- function(lambda) -sum(stats::dpois(y, lambda, log=TRUE)) *#they > define the Poisson negative loglikelihood* > fit0 <- mle(nLL, start = list(lambda = 5), nobs = NROW(y)) * #they > estimate the Poisson parameter using mle* > fit0 *#they call the output* Call: mle(minuslogl = nLL, start = list(lambda = 5), nobs = NROW(y)) Coefficients: lambda 11.545 * #this is their estimated Lambda Vallue.* *Now my question is in a Poisson distribution the Maximum Likelihood estimator of the mean parameter lambda is the sample mean, so if we calculate the sample mean of tha...

Dear all, I'd like to find a minimum of (-loglik) function which is a function of k parameters. I'd like to run the minimization algorithm for the different subsets of the parameters and assign the fixed values to the complementary subset. How should I define my (-loglik) function such that it can be passed to the optim or other optimization function? Much thanks for any suggestions.

...if i try using the method mle() i get stock and i don't know, how to make it work. can anybody help me? thank you very much, indeed. Nadja I tried so fare ll<-function(lambda,alfa) {n<-24 x<-data2 -n*alfa*log(lambda)+n*log(gamma(alfa))-(alfa-1)*sum(log(x))+lambda*sum(x) est<-mle(minuslogl=ll,start=list(lambda=29827.51,alfa=0.4954725)) summary(est) NaN's are produced with optim, i just don't know how to avoid this!

...mmary of "a"? > x <- 0:10 > y <- c(26, 17, 13, 12, 20, 5, 9, 8, 5, 4, 8) > LL <- function(ymax=15, xhalf=6) + -sum(stats::dpois(y, lambda=ymax/(1+x/xhalf), log=TRUE)) > a <- mle2(LL, fixed=list(xhalf=6)) > summary(a) Maximum likelihood estimation Call: mle2(minuslogl = LL, fixed = list(xhalf = 6)) Coefficients: Estimate Std. Error z value Pr(z) ymax 19.2881 1.7115 11.269 < 2.2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 -2 log L: 60.39244 Tom -- View this messa...