similar to: transform one probability distribution into another

Hi all, Using the exponential distribution to exemplify: The dexp function is the PDF (1) and pexp is the CDF (2), that is obtained integrating the PDF. How can I get the qexp and the rexp? Considering that I have the PDF, how this two are mathematically related to the PDF? (1) ke^{-kx} (2) 1-e^{kx} Thanks in advance.

>May 26, 2017; 11:41am Nelly Reduan Latin Hypercube Sampling when parameters are >defined according to specific probability distributions >Hello, > I would like to perform a sensitivity analysis using a Latin Hypercube Sampling (LHS). >Among the input parameters in the model, I have a parameter dispersal distance which is defined according to an exponential probability

Hello, Our professor asked us to do probability plotting using weibull paper, exponential paper, normal, log-normal paper, etc. I know I can create Q-Q plot for normal dist. and see if all te points are on one line. How do I go about other distributions? I tried generating different samples and use the general qq function. However, I could not do it since I don't know the population

Thank you very much Rob for your answer. I have some difficulties to understand how to apply my agent-based model to each parameter combination generated by the LHS, in particular when parameters are defined by probability distributions. Indeed, I have multiple parameters in my model: parameters which are defined by a single value (like ?temperature", "pressure?) and parameters which are

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

I have some data measured with a coarsely-quantized clock. Let's say the real data are q<- sort(rexp(100,.5)) The quantized form is floor(q), so a simple quantile plot of one against the other can be calculated using: plot(q,type="l"); points(floor(q),col="red") which of course shows the characteristic stair-step. I would like to smooth the quantized

Hi list, I have a group of data. It looks like they follow a exponential distribution. In R, how can I esimate lamda, that is the rate in pexp, of the distribution and can I use Kolmogorov-Smirnov for hypothesis testing in such a situation? I have read the "8.2 Examing the distribution of a set of data" of "An Introduction to R" but I did not find any clues on this issue.

Please note that building R on Windows is documented in "R Installation and Administration", including links to external software. Particularly there is a link to texinfo which is part of Rtools. The documentation is maintained and it is a sufficient source of information for building R on Windows. https://cran.r-project.org/doc/manuals/r-release/R-admin.html

Dear helpers, As the title says, my question is not directly related to R. I find, however, that there are many people who are both knowledgeable and kind in this email list, and so decided to give it a try. I do stochastic simulations. Parameter values used in simulation often come from the observations of the real word phenomena. Parameter values are often given as "rates" (of

Hi, guys, I am trying to sample from a truncated normal/gamma distribution. But only the far end of the distribution (where the probability is very low) is left. e.g. mu = - 4; sigma = 0.1; The distribution is Normal(mu,sigma^2) truncated on [0,+Inf]; How can I get a sample? I tried to use inverse CDF method, but got Inf as answers. Please help me out. Also, pls help me on the similar

Hi Is there any implementation for AES encryption algorithm in R? Thanks Bander

Dear all, Sorry for bringing up an old issue: >pexp(50, 0.5) [1] 1 In some cases, pexp() gives CDF=1. I read some discussion in 2002 saying it has been patched. However it's not working in "R2.4.1Patched". Could anyone help me out? Thanks a lot, Jeann _________________________________________________________________ Fine Dining & Fancy Food. Check Out This

Hi, I have a question on generating random variables based on zipf-mandelbrot distribution. So when I execute the following lines: ZM = lnre ("zm", alpha = 2/3, B=0.1) zmsample = rlnre (ZM, n =100) zmsample It generates 100 random values based on a zipf-mandelbrot distribution as below. But how do I make sure the generated random number is within the range of 1 - 6000 only? Can I

Hi, I have normally distributed random terms u~N(0,1). I want to get gamma distributed random terms g~(scale,shape) with E(g)=1=shape/scale and var(g)=theta=1/scale=1/shape. How can I reach my goal? The following way doesn't work: use the distribution function of u to get U(0,1)- distributed random terms, then take the quantile function of the gamma distribution with shape and scale. The

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik