similar to: Use of geometric mean .. in good data analysis

If I use the Ineq library and the Gini function in this way: >Gini(c(100,0,0,0)) I obtain the result 0.75 instead of 1 (that is the perfect inequality). I think Gini's formula in Ineq is based on a formula as reported here: http://mathworld.wolfram.com/GiniCoefficient.html but in the case of perfect inequality: x_1=.......=x_n-1 =0 x_n>0 these formula are equal to 1 - 1/n, not to

Sorry to prolong a thread on something that is clearly off topic, but when Michael Meyer wrote >by using the geometric mean all asymptotic results no longer apply. that is flat our wrong. It's true that the geometric mean converges to something different that E[X], but it does indeed have an asymptotic distribution and one that makes sense in some contexts. There is no reason that

>>>>> jing hua zhao >>>>> on Mon, 24 Jun 2019 08:51:43 +0000 writes: > Hi All, > Thanks for all your comments which allows me to appreciate more of these in Python and R. > I just came across the matrixStats package, > ## EXAMPLE #1 > lx <- c(1000.01, 1000.02) > y0 <- log(sum(exp(lx))) > print(y0) ## Inf

Dear useRs, I'm happy to announce a substantial update of package actuar that bumps the version number to 2.0-0. This release focuses on additional support for continuous and discrete distributions, new functions to simulate data from compound models and mixtures, and revised and improved documentation. A slightly shortened version of the NEWS file follows: NEW FEATURES ? Support for the

Dear useRs, I'm happy to announce a substantial update of package actuar that bumps the version number to 2.0-0. This release focuses on additional support for continuous and discrete distributions, new functions to simulate data from compound models and mixtures, and revised and improved documentation. A slightly shortened version of the NEWS file follows: NEW FEATURES ? Support for the

Hello, I have to compute a multiple summation (not an integration because the independent variables a are discrete) for all the values of a function of several variables f (x_1,...,x_n), that is sum ... sum f(x_1,...,x_n) x_1 x_n have you some suggestion? Is it possible? I know that for multiple integration there is the function adapt, but it has at most n=20. In my case n depends on the

Hi Hadley, Thanks for the counterpoint. Response below. On Thu, May 16, 2019 at 1:59 PM Hadley Wickham <h.wickham at gmail.com> wrote: > The existing behaviour seems inutitive to me. I would consider these > invariants for n vector x_i's each with size m: > > * nrow(rbind(x_1, x_2, ..., x_n)) equals n > Personally, no I wouldn't. I would consider m==0 a degenerate

The existing behaviour seems inutitive to me. I would consider these invariants for n vector x_i's each with size m: * nrow(rbind(x_1, x_2, ..., x_n)) equals n * ncol(rbind(x_1, x_2, ..., x_n)) equals m Additionally, wouldn't you expect rbind(x_1[i], x_2[i]) to equal rbind(x_1, x_2)[, i, drop = FALSE] ? Hadley On Thu, May 16, 2019 at 3:26 PM Gabriel Becker <gabembecker at

Dear All, I am looking for: A software to select the lag length for a lag window spectral estimator. Also, I have a small query in the reprex given below. Background for the above, from the book by Percival and Walden: 1. We are given X_1,...,X_n which is one realization of a stochastic process. 2. We may compute the periodogram using FFT, for example by the function spectrum in R. 3. The

Suppose one wanted to consider random variables X_1,...X_n and from each subtract off the piece which is correlated with the previous variables in the list. i.e. make new variables Z_i so that Z_1=X_1 and Z_i=X_i-cov(X_i,Z_1)Z_1/var(Z_1)-...- cov(X_i,Z__{i-1})Z__{i-1}/var(Z_{i-1}) I have code to do this but I keep getting a "non-conformable array" error in the line with the covariance.

Hi, I have to draw samples from an asymmetric-Laplace-Normal distribution: f(u|y, x, beta, phi, sigma, tau) \propto exp( - sum( ( abs(lo) + (2*tau-1)*lo )/(2*sigma) ) - 0.5/phi*u^2), where lo = (y - x*beta) and y=(y_1, ..., y_n), x=(x_1, ..., x_n) -- sorry for this huge formula -- A WinBUGS Gibbs sampler and the HI package arms sampler were used with the same initial data for all parameters. I

Dear All, Suppose I have a vector X = (x_1, x_2, ...., x_n), X_sort = sort(X) = (x_(1), x_(2), ... , x(n) ), and I would like to know the original position of these ordered x_(i) in X, how can I do it? case 1: all values are unique x <- c( 3, 5, 4, 6) x.sort <- sort(x) # # I would like to obtain a vector (1, 3, 2, 4) which indicates that 3 in x is still the 1st element in x.sort, 5 is at

I have a set of n vectors, x_1, ..., x_n, of the same length. I would like to form the vector of dot products -- x_1'x_1, ..., x_n'x_n the fastest way I can think to do this is to put the vectors into a matrix and do diag(crossprod(X)) however, this seems to be very wasteful since this computes n(n+1)/2-n unnecessary dot products. Is there a better way using existing functions in R?

?Dear All I have some data expressed in geometric means and 95% confidence intervals. Can I code them in metafor as: rma(m1i=geometric mean 1, m2i=geometric mean 2, sd1i=geometric mean 1 CI /3.92, sd2i=geometric mean 2 CI/3.92.......etc, measure="MD") All of the studies use geometric means. Thanks! Edward ---------------------------- [[alternative HTML version deleted]]

Dear all, The help page on the t distribution says: The most used applications are power calculations for t-tests: Let T= (mX - m0) / (S/sqrt(n)) where mX is the `mean' and S the sample standard deviation (`sd') of X_1,X_2,...,X_n which are i.i.d. N(mu,sigma^2). Then T is distributed as non-centrally t with `df'= n-1 degrees of freedom and non-centrality

Hi all, I need to check for a difference between treatment groups in the parameter of the geometric distribution, but with a cut-off (i.e. right censored). In my experiment I stimulated animals to see whether I got a response, and stopped stimulating if the animal responded OR if I had stimulated 10 times. Since the response could only be to a stimulation, the distribution of response times

---------- Forwarded message --------- From: Sahil Sharma <sahilsharmahimalaya at gmail.com> Date: Tue, Oct 17, 2023 at 12:10?PM Subject: r-stats: Geometric Distribution To: <do-use-Contact-address at r-project.org> Hey I want to raise one issue in *r-stats **geometric distribution * function. I have found the dgeom(x,p) which denotes probability density function of geometric

Hi, I have a bunch of data which is assumed to be instances of a geometric random variable with outliers. How can I do a robust estimation of the parameter p so that the effect of outliers is minimized? As a part of the estimation process, I also need to know which are the outliers in the data. I found glmrob which does robust estimation of Poisson and binomial random variables but not geometric

Dear useRs, I'd like to announce a new package called geozoo, short for geometric zoo. It's a compilation of functions to produce high-dimensional geometric objects, including hypercubes and hyperspheres, Boy's surface, the hyper torus and a selection of polytopes. For a complete list, as well as images and movies, visit

Dear list, I have got an array with observational values t and I would like to fit a geometric distribution to it. As I understand the geometric distribution, there is only one parameter, the probability p. I estimated it by 1/mean(t). Now I plotted the estimated density function by plot(ecdf(t),do.points=FALSE,col.h="blue"); and I would like to add the geometric distribution. This