similar to: (1-1e-100)==1 true?

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) ,

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

You may want to look into using the log option to qnorm e.g., in round figures: > log(1e-300) [1] -690.7755 > qnorm(-691, log=TRUE) [1] -37.05315 > exp(37^2/2) [1] 1.881797e+297 > exp(-37^2/2) [1] 5.314068e-298 Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit manipulation of the log values. qnorm()

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,

I agree with many the sentiments about the wisdom of computing very small p-values (although the example below may win some kind of a prize: I've seen people talking about p-values of the order of 10^(-2000), but never 10^(-(10^8)) !). That said, there are a several tricks for getting more reasonable sums of very small probabilities. The first is to scale the p-values by dividing the

Hi, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being

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

Hello If i want to resampl from the tails of normal distribution , are these commans equivelant??   upper tail:qnorm(runif(n,pnorm(b),1))  if b is an upper tail boundary   or   upper tail:qnorm((1-p)+p(runif(n))  if p is the probability of each interval (the observatins are divided to intervals)   Regards [[alternative HTML version deleted]]

Hello, Well, try it: p <- .Machine$double.eps^seq(0.5, 1, by = 0.05) z <- qnorm(p/2) pnorm(z) # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16 #[11] 1.110223e-16 p/2 # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15

Hi, I am working with some extremely small p-values and I want to capture the corresponding quantiles. I see the help file it says: 'qnorm' is based on Wichura's algorithm AS 241 which provides precise results up to about 16 digits. What happen after the 16th digits? If I am running R in a server 64-bit, can that improve the chances that beyond 16th digits to still have

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

how R implement qnorm() I wonder anyone knows the mathematical process that R calculated the quantile? The reason I asked is soly by curiosity. I know the probability of a normal distribution is calculated through integrate the Gaussian function, which can be implemented easily (see code), while the calculation of quantile (or Zα) in R is a bit confusing as it requires inverse error function (X

I am using pnorm() with the log.p=T argument to get approximations to ln \Phi(x) and qnorm with the log.p=T argument to get estimates of \Phi^{-1}(exp(x)). What approximations are used in these two functions (I noticed in the source pnorm.c it doesn't look like Abramowitz and Stegen) and where can I find the citation? Thanks, Richard Morey

Dear list, I have a problem on persp() x <- u1data #first coloum in attached data y <- u2data #second coloum in attached data f <- function(x,y){qgev(pnorm(rhoF*qnorm(pnorm((qnorm(y)-rho2*qnorm(x)/sqrt(1-rho2^2)))) +sqrt(1-rhoF^2)*qnorm(0.95)),-0.3935119, 0.4227890, 0.2701648)} z <- outer(x,y,f) persp(x,y,z) The R will display: "Error in persp.default(x, y,

What would be the R formulae for a two-sided test? I have a formula for a one-sided test: powertest <- function(a,m0,m1,n,s){ t1 = -qnorm(1-a) num = abs(m0-m1) * sqrt(n) t2 = num/s pow = pnorm(t1 + t2) } Would you pls let me know if you know of? Thank you, ej

>>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: >>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: > include/Rmath.h declares a set of 'logspace' functions for use at the C > level. I don't think there are core R functions that call

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)

I'm familiar with the quantile() command, but what if I have a specific number that I want to know its location in a vector? I know that in known distributions, (for example the normal distribution), there is pnorm and qnorm, but how can I do it with unknown vector? thanks in advance _________________________________________________________________ Walla! Mail - [1]Get

I am wondering whether there is a bug in rnorm. When generating rnorm(1000000) and counting the cases > 4 and the cases < (-4) I get rather unexpectedly low counts for the latter. The problem goes away when using qnorm(runif(1000000)). Fritz Scholz, PhD Applied Statistics Group Boeing Phantom Works fritz.scholz at pss.boeing.com 425-865-3623 Tu/We 206-542-6545 (most likely)