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Hello: I have a question in Math. If we want to get X's and Y's solution, X>0 and Y>0 We have two equation : 2*exp(X)+X^2+3*Y=2*exp(1)+4 3*X+4*(Y^2)=7 How I use R-project to solve above question?? THANKS YOU !!!! HLC [[alternative HTML version deleted]]

I forgot to add that I first gave a starting value for K. Nonlinear least squares won't work because my errors are not normally distributed. Any advide on my maximum likelihood function would be greatly appreciated. ---------------------------- Original Message ---------------------------- Subject: RE: [R] maximum likelihood problem From: "Ravi Varadhan" <rvaradhan at

Hi, I'm trying to estimate the parameters of a state space model of the following form measurement eq: z_t = a + b*y_t + eps_t transition eq y_t+h = (I -exp(-hL))theta + exp(-hL)y_t+ eta_{t+h}. The problem is that the distribution of the innovations of the transition equation depend on the previous value of the state variable. To be exact: y_t|y_{t-1} ~N(mu, Q_t) where Q is a diagonal

I am trying to figure out how to run maximum likelihood in R. Here is my situation: I have the following equation: equation<-(1/LR-(exp(-k*T)*LM)*(1-exp(-k))) LR, T, and LM are vectors of data. I want to R to change the value of k to maximize the value of equation. My attempts at optim and optimize have been unsuccessful. Are these the recommended functions that I should use to maximize

Dear R users, When running the program below I receive the following error message: fit <- optim(parm, objective, yt = tyield, hessian = TRUE) Error in as.vector(data) : no method for coercing this S4 class to a vector I can't figure out what the problem is exactly. I imagine that it has something to do with "tyield" being a matrix. Any help on explaining what's going on

Dear R Users on a self-written function for calculating maximum likelihood probability (plz check function code at the bottom of this message), one value, wden, suddenly jump to zero. detail info as following: w[11]=2.14 lnw =2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182 0.178 0.179... w[11]=2.14 lnw=2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182

Hey, R users I do not know how to describe my question. I am a new user for R and write the following?code for a dynamic labor economics?model and use OPTIM to get optimizations and parameter values. the following code does not work due to the?equation: ?? wden[,i]<-dnorm((1-regw[,i])/w[5])/w[5] where w[5]?is one of the parameters (together with vector a, b and other elements in vector

I am using both nlminb and optim to get MLEs from a likelihood function I have developed. AFAIK, the model I has not been previously used in this way and so I am struggling a bit to unit test my code since I don't have another data set to compare this kind of estimation to. The likelihood I have is (in tex below) \begin{equation} \label{eqn:marginal} L(\beta) = \prod_{s=1}^N \int

Dear R help, I am trying solve an MLE convergence problem: I would like to estimate four parameters, p1, p2, mu1, mu2, which relate to the probabilities, P1, P2, P3, of a multinomial (trinomial) distribution. I am using the mle2() function and feeding it a time series dataset composed of four columns: time point, number of successes in category 1, number of successes in category 2, and

Hello R Help, I am trying solve an MLE convergence problem: I would like to estimate four parameters, p1, p2, mu1, mu2, which relate to the probabilities, P1, P2, P3, of a multinomial (trinomial) distribution. I am using the mle2() function and feeding it a time series dataset composed of four columns: time point, number of successes in category 1, number of successes in category 2, and

Deal all: i want to do the probit with sample selection estimation, the following is my code: probit with sample selection can be done by stata :heckprob The heckprobll is the likelihood function shown in W.H. Greene 5th p714 ¡´ The question is the convergence is very slow compare with Stata using likellihood only. ¡´ Second i did the similar way in matlab using fminsearch , the estimated

I have constructed the function mml2 (below) based on the likelihood function described in the minimal latex I have pasted below for anyone who wants to look at it. This function finds parameter estimates for a basic Rasch (IRT) model. Using the function without the gradient, using either nlminb or optim returns the correct parameter estimates and, in the case of optim, the correct standard

Dear All,   Currently I am running the following code:   library(stats4) library(odesolve) library(rgenoud) Input<-data.frame(SUB=c(1),time=c(0.5,3,10,15),lev=c(2.05,12.08,9.02,8)) XD<-500 IT<-3 diffeqfun<-function(time, y, parms) {   if(time<=IT)      dCpdt <- (XD/IT)/parms["Vol"] -

Dear R users,   I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:     ### Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable   ## my response variable   rl <-

?s 12:13 de 13/07/2024, Christofer Bogaso escreveu: > Hi, > > I ran below code > > Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv') > head(Dat) > Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) > head(predict(Model,

Dear R forum I have following variables - EAD = 10000 LGD = 0.45 PD = 0.47 M = 3 # Equation 1 R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) + 0.24*(1-(1-exp(-50*PD))/(1-exp(-50))) b = (0.11852 - 0.05478 * log(PD))^2 K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 * qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b) RWA = K * 12.5 * EAD > RWA [1] 22845.07 #

Hi all, I've got a problem with the nls function. I have an adjustment which works when I fix one of the argument of my function (Xo=150) : *Xo*=150 f<- function (tt*,Xo*,a,b) ifelse(tt<*Xo*,a*exp(-b**Xo*),a*exp(-b*tt)) ajust<-nls(RER~f(tt,*Xo*,a,b),data=data.frame(tt=Ph2[,2*k],RER=Ph2[,2*k+1]),start=list(a=0.5,b=0.014)) But, when I use it as a "normal" parameter (and

Hi there, I have a complex math equation which does not have a closed form solution. It is - y <- (p*exp(-a*d)*(1-exp((d-p)*(a-x[1]))))/((p-d)*(1-exp(-p*(a-x[1])))) For this equation, I have all the values except for x[1]. So I need to solve this problem numerically. Can anyone suggest an optimization package that I can use to estimate the value for x[1]? Thanks in advance, Diviya

Dear R users, I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:  Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable my response variable rl <-

Hello all, I would like to get parameter estimates for different models. For one of them I give the code in example. I am estimating the parameters (i,j and k) with the nls function, which sees the error distribution as normal, I would however like to do the same as nls with the assumption that the errors are poisson distributed. Is there a way to do this with R? Are there packages designed