similar to: simple random number generation

Displaying 20 results from an estimated 11000 matches similar to: "simple random number generation"

2009 Mar 10
6
Pseudo-random numbers between two numbers
I would like to generate pseudo-random numbers between two numbers using R, up to a given distribution, for instance, rnorm. That is something like rnorm(HowMany,Min,Max,mean,sd) over rnorm(HowMany,mean,sd). I am wondering if dnorm(runif(HowMany, Min, Max), mean, sd) is good. Any idea? Thanks. -james
2004 Aug 06
3
Bug in qnorm or pnorm?
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))
2011 Sep 03
3
question with uniroot function
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,
2011 Aug 04
2
random value generation with added constraint.
Hi I am looking at generating a random dataset of say 100 values fitting in a normal distribution of a given mean and SD, I am aware of rnorm function. However i am trying to build into this function one added constraint that all the random value generated should also obey the constraint that they only take values between say X to X+25 How do i do this in R? Any help would be highly appreciated,.
2012 Jun 18
3
(1-1e-100)==1 true?
Hi, This problems has bothered me for the lase couple of hours. > 1e-100==0 [1] FALSE > (1-1e-100)==1 [1] TRUE How can I tell R that 1-1e-100 does not equal to 1, actually, I found out that > (1-1e-16)==1 [1] FALSE > (1-1e-17)==1 [1] TRUE The reason I care about this is that I was try to use qnorm() in my code, for example, > qnorm(1e-100) [1] -21.27345 and if I want to
2001 Jul 02
2
Shapiro-Wilk test
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
2019 Jun 21
4
Calculation of e^{z^2/2} for a normal deviate z
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()
2010 Oct 03
2
sampling from normal distribution
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]]
2011 Feb 03
1
random sequences for rnorm and runif
Dear R experts, For a fixed seed, the first random number produced by rnorm and runif has the same rank within the distribution, which I find useful. The following ranks differ, however. > set.seed(123) > runif(4) [1] *0.2875775* 0.7883051 *0.4089769* 0.8830174 > set.seed(123) > pnorm(rnorm(4)) [1] 0.2875775 0.4089769 0.9404673 0.5281055 I noticed that rnorm seems to
2008 Apr 13
4
R equivalent of erfcinv in matlab
I am converting some matlab code into R that use inverse of the complementary error function, erfcinv and did not find an equivalent in R, is there such a function in some contributed modules? Thanks.
2006 Oct 27
3
Power of test
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
2003 Mar 31
2
point-biserial correlation
Dear list, has anyone written a package/function in R for computing a point- biserial resp. biserial correlation? Thanks in advance Bernd
2006 Jan 31
1
approximation to ln \Phi(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
2019 Jun 23
2
Calculation of e^{z^2/2} for a normal deviate z
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
2010 Nov 12
4
dnorm and qnorm
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:
2012 Apr 24
2
Some Help Needed
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
2019 Jun 21
4
Calculation of e^{z^2/2} for a normal deviate z
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
2005 Feb 21
4
rnorm??
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)
2011 May 30
1
Error in minimizing an integrand using optim
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
2009 Feb 06
1
16 digits and beyond? R64-bit a solution?
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