similar to: Some Help Needed

Dear all, [system and version information below] I am trying to modify a C function for finding the root of an expression. The function is to be called from R as .Call with input parameters: f: expression for which we will find the root guesses: interval for the solution stol: tolerance rho: environment The original functions I use are: SEXP mkans(double x) { SEXP ans; PROTECT(ans =

Dear all, [system and version information below] I am trying to modify a C function for finding the root of an expression. The function is to be called from R as .Call with input parameters: f: expression for which we will find the root guesses: interval for the solution stol: tolerance rho: environment The original functions I use are: SEXP mkans(double x) { SEXP ans; PROTECT(ans =

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 value

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

Hi, Our department has detected a bug in the implementation of the Kinderman-Ramage generator for normal random variates in version 1.7.0, which can be seen from the below R session. (Consecutive calls for chisq.test(...) always gives p-values very close to 0.) We have already encountered this bug in version 1.6.2 The error is in file R-1.7.0/src/nmath/snorm.c Here is a patch for this file to

Hello, I have a bigger problem in calculating the Maximum Likelihood Estimator regarding the degree of freedom of a multivariate skew-t copula. First of all I would like to describe what this is all about, so that you can understand my problem: I have 2 time series with more than 3000 entries each. I would like to calculate a multivariate skew-t Copula that fits this time series. Notice:

Hi all, I want to draw a histgram for each row of a matrix and compare them. However the plot I got does not have the same y range and x range, which makes it difficult to make the comparison. Is there a easy way to fix the x range and y range in a xy plot for several plots, instead of specifying them for each plot. The following is my code for generalizing the matrix and draw the histogram.

Hello, could You please help: I am looking for a way to formulate test accuracy measures such as test sensitivity, specificity, predictive values, and correct classification rate using p/d/qnorm. The tests' primary values follow a bimodal distribution, which is modelled by a mixture of two normal distributions: p * dnorm ((x - u1) / s1) / s1 + (1 - p) * dnorm ((x - u2) / s2) / s2)

Hi all, I want to solve the following equation for x with rho <- 0.5 pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))==0.05 Is there a function in R to do this? Thank you very much! Hannah [[alternative HTML version deleted]]

Dear all, I have the following program for a multiple comparison procedure. There are two functions for the two steps. First step is to calculate the critical values, while the second step is the actual procedure [see below: program with two functions]. This work fine. However, However I want to put them into one function for the convenience of later use [see below: program with one

hello, I'm trying to improve the speed of my calculation but didn't get to a satisfying result. It's about the numerical Integration of a bivariate normal distribution. The code I'm currently using x <- qnorm(seq(.Machine$double.xmin,c(1-2*.Machine$double.eps),by=0.01), mean=0,sd=1) rho <- 0.5 integral <- function(rho,x1){

Dear All I am trying to find a uniroot of a function within another function (see example) but I am getting an error message (f()values at end points not of opposite sign). I was wondering if you would be able to advise how redefine my function so that I can find the solution. In short my first function calculates the intergrale which is function of "t" , I need to find the uniroot of

If this is posted elsewhere I cannot find it. I need to perform multiple integration where some of the variables are in the bounds of the other variables. I was trying to use R2Cuba function but cannot set the upper and lower bounds. My code so far is : int <- function(y){ u2 = y[1] z2 = y[2] u1 =y[3] z1 = y[4] ff <- u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)) return(ff) }

Dear all, Suppose I have a very long expression e. Lets assume, for simplicity, that it is e = expression(u1+u2+u3) Now I wish to replace u2 with x and u3 with 1. I.e. the 'new' expression, after replacement, should be: > e expression(u1+x+1) My question is how to do the replacement? I have tried using: > e = parse(text=gsub("u2","x",e)) > e =

Dear all, I am using the function 'D' in the 'stats' package to perform symbolic derivation. This works very well and it is much faster than e.g. Mathematica (at least for my purposes). First, I would like to thank the development team for this excellent function. However, I run into trouble in some cases, particularly when I am to do some operations on long expressions

Dear all, I wrote the following function previously. It worked fine with the old mvtnorm package. Somehow with the updated package, I got a error message when trying to use the function. I need some help. It is sort of urgent. Can anyone please take a look. The function is the following. Thank you very much! Hannah library(mvtnorm) cc_f <- function(m, rho,

Dear all, I would like to solve the following problem, which can be done with optimal control theory or dynamic programming: max(x,y) a*u1+b*u2+c*f1(u2) s.t. 0<u1<x, 0<u2<f2(x,u2), x'=f3(u1,u2,x) which can be rewritten if optimal control theory should be applied as H=a*u1+b*u2+c*f1(u2)+lambda*(x') s.t. 0<u1<x, 0<u2<f2(x,u2) The maximum principle

Hi, in the code below, I am drawing 1000 samples from two beta distributions, each time using the same random number seed. Using set.seed(80) produces results I expect, in that the differences between the distributions are very small. Using set.seed(20) produces results I can't make sense of. Around half of the time, it behaves as with set.seed(80), but around half of the time, it behaves

OS: CentosOS 7 I have installed samba + openldap + smbldap-tools + pam by: yum --enablerepo=extras install -y epel-release yum install -y smbldap-tools yum install -y samba openldap openldap-clients openldap-servers migrationtools yum install -y nss-pam* I know that smbldap-tools is a dead project, but I'm interested in it and would like research on it. I create users and

I'm trying to implement a recursive function using integrate, and I suspect I need a Vectorize somewhere, but I can't suss it out. Any help would be appreciated. I've tried traceback() and various debugging ideas to no avail (most likely due to my inexperience with these tools.) Here's what I have. Nk <- function(m, C) { if (length(m) > 1) { rho <- C[1, -1]