similar to: using histogram to find cdf

Dear all, I have the cdf of the following power fuction distribution: F(y)=(y/350)^a               ,0<y<350, where " a " is some parameter with range a>0. I want to use it as the argument of the discretize function of the actuar package. So I think I need to define this function to R so that if I entered a=1, I get the following F(y)=(y/350) and if I entered a=4.5, I get the

I first answered to the email below in r-help, but as I did not see any response, and it looks like a bug/unwanted behavior, I am also posting here. I have observed this in RGui, whereas it seems not to happen in RStudio. Similar to OP, I sometimes have a problem with functions using the progress bar. Frequently, the console is cleared after x iterations when the progress bar is called in a

Hi all, I've a question concerning the R 3.3.1 version. I have a long code that I used to run on versions earlier to the 3.3.1 version, and when I copied the code to the R console, I can still see the code while the loop is executing , along with the output printed after each iteration of the loop. Now, on the 3.3.1 version, after I copy the code to the console, it disappears and I only see

Hi all, I've a question concerning the R 3.3.1 version. I have a long code that I used to run on versions earlier to the 3.3.1 version, and when I copied the code to the R console, I can still see the code while the loop is executing , along with the output printed after each iteration of the loop. Now, on the 3.3.1 version, after I copy the code to the console, it disappears and I only see

Hi all, I want the y-axis label to be ( in symbols) g(sigma given alpha) where given is the conditional sign. I've tried ylab=expression(g(sigma|alpha))) but it gave me g(|(sigma,alpha)) where the sigma and alpha are in greek but the conditional sign is misplaced (before the bracket) Any help would be appreciated. Maram. [[alternative HTML version deleted]]

Hi all, I've a linear equation of the form: 0.95=2 ( [3+ln(x/3)]^-13  + 4 [3+2ln(x/3)]^-13  + [3+4ln(x/3)]^-13 ) and I want to solve it for x, can I do this using R? Thanks in advance. Maram. [[alternative HTML version deleted]]

Dear All, Attached are the codes of a histogram & a kernel density estimate and the output they produced. In fact the variable q is a vector of 1000 simulated values; that is I generated 1000 samples from the pareto distribution, from each sample I calculated the value of q ( a certain fn in the sample observations), and thus I was left with 1000 values of q and I don't know the

Good Afternoon, I'm trying to create a cdf plot, with the following code. It works well, but I have little doubt, if you can help solve. When I create the plot, like the graph line would still not appear with point #cdf x<-table(Dataset$Apcode) View(s) hist(s) *plot(ecdf(x))* x<-1 37607 2 26625 3 5856 4 25992 5 30585 6 16064 7 9850 .. ... .. 186 52 -- View this message in

Hi all, is there any package in R capable of smooth approximation of CDF basing on given sample? (Thus, I am not speaking about ecdf) In particular, I expect very much that the approximation should subject to the property: f(x0)<=f(x1) for x0<x1, where x0 and x1 belong to range of the sample given. Polynomial approximation could be OK for me as well. P.S.

Hi r-users, I would like to overlap 2 ecdf plots.  I tried this below and it gives me two plots of ecdf but just both just in black. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs="i", yaxs="i") plot(ecdf(datobs)) lines(ecdf(gam_sum_gen)) Then I try to add colors etc and also the legend but fail. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs="i", yaxs="i")

Good day everyone, I want to assess the error when fitting a Gram-Charlier CDF to some data 'ws', that is, I want to calculate: Err = |ecdf(ws) - GCh_ser(ws)| The problem is, I cannot get the F(x) values from the ecdf. 'Summary(ecdf())' returns some of the x-axis values, but how do you get the F(x) values? Thank you for any help you can provide. Regards, Augusto

Hi, I would like to overlap the cdf curve for observed and generated data  Here is my code: plot(cdf,main ="CDF of the sum for winter season-Hume",cex.axis=1.2,xlab="Rainfall (mm)", xaxs="i",yaxs="i",col=c("black","red"), lty=c(1,1),ylab="Cumulative probability", xlim=c(0,800),lwd=1) lines(ecdf(datobs))

Dear all, I would like for a given vector to calculate the pdf and the cdf and then plot these two graphs. Is there in R any toolkit that can help me do that fast? Regards Alex

Hi all, I'm using a certain  procedure to calculate the value of some variable(Bayes risk),B. So I got the values B1, B2, ........, B1000, each under certain input values and using a long procedure. Now, I want to put the values I got in a nummerical vector and find their minimum value. I think c( ) should work.For example if I have only 10 values I could have used

Dear all, I want to use the histogtam as a density estimator, with the binwidths calculated using scott's formula which is binwidth = 3.49*ST.dev.*n^(-1/3) for the following data  (30 data points) 12-9-3-6-1-23-21-7-18-16-15-4-19-22-20-2-3-18-8-10-1-7-5-4-11-12-3-9-19-7 so first,I' ve tried this manually, and substituted in the above formula and I got st.dev.=7.02745 and thus the

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

Hi all, I've a trivial question. If (q) is a continous variable,actually a vector of 1000 values. how to calculate the probability that q is greater than a specific value, i.e. P(q>45)?? Thanks Maram [[alternative HTML version deleted]]

I have no idea to be honest. I have never used the package, I simply did a search for it. Hopefully a more experienced user can help --- On Sun, 7/12/09, maram salem <marammagdysalem at yahoo.com> wrote: > From: maram salem <marammagdysalem at yahoo.com> > Subject: Re: [R] (no subject) > To: "John Kane" <jrkrideau at yahoo.ca> > Received: Sunday, July 12,

Hi Group, I've a vector of 1000 numeric values for which I want to draw a histogram. I've read this vector into R with no variable name.I mean only the 1000 values, which makes V1 the name of the variable by default?? Then I tried > hist(V1, breaks = "Sturges", +      freq = NULL, probability = !freq, +      include.lowest = TRUE, right = TRUE, +      density = NULL, angle =

Dear UseRs, maybe is a silly question: how can I get Empirical CDF values from an object created with ecdf()?? Using print I obtain: Empirical CDF Call: ecdf(t) x[1:57] = 4.1, 4.4, 4.5, ..., 491.3, 671.27 Thanks in advance. Regards, Vito Diventare costruttori di soluzioni Became solutions' constructors "The business of the statistician is to catalyze the scientific