similar to: help with R

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

Dear all, I need to do some calculation where the code used are below. I get error message when I choose k to be large, say greater than 25. The error message is "Error in integrate(temp, lower = 0, upper = 1, k, x, rho, m) : the integral is probably divergent". Can anyone give some help on resolving this. Thanks. Hannah m <- 100 alpha <- 0.05 rho <- 0.1 F0

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,

I have the following problem. inf p= Int [dnorm(x-d) * pnorm(x) ^ (m-1)] dx -inf Given p and m, I want to find d. For example, p m d .9 2 1.81 Is there a way to solve this problem in R? Ideally I would have a function with arguments p and m, and output d. Thanks very much for any help. Bill -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing

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

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,

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

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Hi, I'm trying to calculate the value of the variable, dp, below, in the argument to the integral of dnorm(x-dp) * pnorm(x)^(m-1). This corresponds to the estimate of the sensitivity of an observer in an m-alternative forced choice experiment, given the probability of a correct response, Pc, a Gaussian assumption for the noise and no bias. The function that I wrote below gives me an error:

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]

> version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 1 minor 7.1 year 2003 month 06 day 16 language R Bug: integrate(f,lower,upper,extra_args) where f <- function(x,extra_args) { body } integrate doesn't pass the extra arguments when calling f. As a first check of this finding I integrated dnorm from

Hi all, The function h below is a function of c and it should be a monotone increasing function since the integrand is nonnegative and integral is taken from c to infinity. However, as we can see from the plot, it is not shown to be monotone. Something wrong with the usage of integrate function? Thanks so much for your help. Hanna h <- function(c){ g <- function(x){pnorm(x-8.8,

Hi all, I have some problem regarding the integration function. Can any one take a look at the following and give me some help? Thank you in advance! Hannah ## f0 is a function of r,x, and y f0 <- function(r, x, y){ pnorm((x-sqrt(r)*y)/sqrt(1-r), mean=0,sd=1, lower.tail=T)^2*dnorm(y) } ## f1 is the function of r and x after integrating out y f1 <- function(r,x)

Hi, Does anyone know what the distribution for the product of two correlated normal? Say I have X~N(a, \sigma1^2) and Y~N(b, \sigma2^2), and the \rou(X,Y) is not equal to 0, I want to know the pdf or cdf of XY. Thanks a lot in advance. yu [[alternative HTML version deleted]]

Sorry. I meant in the previous email that the function h() is a monotone decreasing function. Thanks very much. 2018-02-06 13:32 GMT-05:00 li li <hannah.hlx at gmail.com>: > Hi all, > The function h below is a function of c and it should be a monotone > increasing function since the integrand is nonnegative and integral is > taken from c to infinity. However, as we can see

Dear R-people, I have to compute C - -(pnorm(B)*dnorm(B)*B + dnorm(B)^2)/pnorm(B)^2 This expression seems to be converging to -1 if B approaches to -Inf (although I am unable to prove it). R has no problems until B equals around -28 or less, where both numerator and denominator go to 0 and you get NaN. A simple workaround I did was C <- ifelse(B > -25, -(pnorm(B)*dnorm(B)*B

Hi all, I have a function rho.f which gives a list of estimators. I have the following problems. rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give me a different answer, even though corr[4]==0.3. This prevents me from using a for loop. Can someone give me some help? Thank you very much in advance. Hannah >

Hi Hanna, your function is essentially zero outside a short interval around 9. And the help page states: "If the function is approximately constant (in particular, zero) over nearly all its range it is possible that the result and error estimate may be seriously wrong." You could try to integrate over a finite interval, say (7, 12). G?ran Brostr?m On 2018-02-06 19:40, li li wrote:

Hi r-users,   I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <- function(pars,n) {  runi    <- runif(974, min=0, max=1)

I believe that this may be more appropriate here in r-devel than in r-help. The normal hazard function, or reciprocal Mill's Ratio, may be obtained in R as dnorm(z)/(1 - pnorm(z)) or, better, as dnorm(z)/pnorm(-z) for small values of z. The latter formula breaks dowm numerically for me (running R 2.4.1 under Windows XP 5.1 SP 2) for values of z near 37.4 or greater. Looking at the pnorm