similar to: integration in R?

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 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 researchers I wish to plot a box plot without the mean line (the black line) and plot only the mean (red square). Futhermore, is it possible to add standard error and/or stadard deviation? This is an example mytest <- c(2.1,2.6,2.7,3.2,4.1,4.3,5.2,5.1,4.8,1.8,1.4,2.5,2.7,3.1,2.6,2.8) boxplot(mytest,lty = "solid") means <- mean(mytest,na.rm=TRUE) points(means, pch = 22, col

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:

Hi :) Was just wondering whether someone could help me with adjustments to trellis plots (parallel). I've got two way multivariate data. I want to make parallel plots for one of the factors, and want to color the lines according to the other factor. The first thing I manage, but with the other I'm lost :( Can only change the overall color. This is basically how far I get:

I would like to put confidence intervals on a mantel corellogram they are already calculated in the pmgram object but I am unsure how I get the x value in order to plot them? package(ecodist) X<-1:100 Y<-rnorm(1:100) Z<-rnorm(1:100) XY<-dist(data.frame(X,Y)) YX<-dist(data.frame(Y,X)) my.mgram<-mgram(XY,XZ) plot(my.mgram) print(my.mgram) > print(my.mgram) $mgram

Dear List, I am looking to plot error bars on a line using dispersion. I have values for the upper value and for the lower values, however i am unsure how to plot different values for the upper CI and the lower CI? I have been using dispersion(1:35,sim,simCI,col="red") Where there are 35 points, and simCI relates to a vector containing the lower confidence intervals, however i want

Hello all, I have an optimization routine that is giving me good results, but the results are not in the nice "model" format like "lm". How can I get optim results into a model so that I can use the clever 'fitted', 'residuals', and 'summary' functions? Using optim is the only way that I was able to make a model that 1) sums the betas to 1, 2)

Hello, I have an r function that creates the following dataframe tresults2. Notice that column 1 does not have a column heading. Tresults2: [,1] estparam 18.00000 nullval 20.00000 . . . ciWidth 2.04622 HalfInterval 1.02311 pertinent code: results<-cbind( estparam, nullval, t, pv_left, pv_right, pv_two_t, estse, df, cc, tbox, llim, ulim, ciWidth,

I have a strange problem with uniroot() function. Here is the result : > uniroot(th, c(-20, 20)) $root [1] 4.216521e-05 $f.root [1] 16.66423 $iter [1] 27 $estim.prec [1] 6.103516e-05 Pls forgive for not reproducing whole code, here my question is how "f.root" can be 16.66423? As it is finding root of a function, it must be near Zero. Am I missing something? -- View this message

This below is not solvable with uniroot to find "a": fn=function(a){ b=(0.7/a)-a (1/(a+b+1))-0.0025 } uniroot(fn,c(-500,500)) gives "Error in uniroot(fn, c(-500, 500)) : f() values at end points not of opposite sign" I read R-help posts and someone wrote a function: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html but it is not very precise. Is there any

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, I already found a conversation on the integration of a normal distribution and two suggested solutions (https://stat.ethz.ch/pipermail/r-help/2007-January/124008.html): 1) integrate(dnorm, 0,1, mean = 0, sd = 1.2) and 2) pnorm(1, mean = 0, sd = 1.2) - pnorm(0, mean = 0, sd = 1.2) where the pnorm-approach is supposed to be faster and with higher precision. I want to integrate a mixed

Hi everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

Dear Dr.Goslee and anyone may intrested in matrix manipulate, I am using your ecodist to do mantel and partial mantel test, I have locality data and shape variation data, and the two distance matrixs are given as belowings. When I run the analysis, it is always report that the matrix is not square, but I didn't know what's wrong with my data. Would you please help me on this. I am quite

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,

In looking at rootfinding for the histoRicalg project (see gitlab.com/nashjc/histoRicalg), I thought I would check how uniroot() solves some problems. The following short example ff <- function(x){ exp(0.5*x) - 2 } ff(2) ff(1) uniroot(ff, 0, 10) uniroot(ff, c(0, 10), trace=1) uniroot(ff, c(0, 10), trace=TRUE) shows that the trace parameter, as described in the Rd file, does not seem to be

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 am an R newbie and trying to grasp how the simple optimization routines in R work. Specifically, I would like some guidance on how to set up a code to calculate the internal rate of return on an investment project (http://en.wikipedia.org/wiki/Internal_rate_of_return). My main problem (I think) is that I want a generic code where N (number of periods) can be easily changed and set

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