similar to: Product integral in R

Hello all, I'm hoping to find a way to evaluate the following sort of integral in R. \int_a^b \int_{g(y)}^Inf f(x,y) dx dy. The integral has no closed form and so must be evaluated numerically. The "adapt" package provides for multidimensional integration but does not appear to allow the limits of integration to be a function. I need to evaluate a number of integrals of this

Dear R-Friends, How can I calculate a double integral like \int_a^b \int_c^y g(x, y) dx dy where a, b, c are constants, g(x, y), e.g., g(x, y) = tan(x + y). I tried to nested integrate() and adapt(), but none of them working, seemingly due to the limits can not be specified constants. Best regards, C. Joseph Lu Department of Statistics National Cheng-Kung University Voice:

Dear R-users, I'm wondering how to calculate this double integral in R: int_a^b int_c^y g(x, y) dx dy where g(x,y) = exp(- alpha (y - x)) * b Thanks for answering! Cheers, Alui [[alternative HTML version deleted]]

Hi, I have a function that I need to do double integration: \int^T_0 \int^t_0 N(\delta / \sigma \sqrt(u)) (1-N(\delta / \sigma \sqrt(u))) du dt where N(x) is a standard normal probability of x. I start off by writing an inner integral into a function. Meaning \int^t_0 N(\delta,\sigma \sqrt(u)) (1-N(\delta,\sigma \sqrt(u))) du. Then calling integrate function on this function. This

Dear R gurus, To start, let me confess to not being an experienced programmer, although I have used R fairly extensively in my work as a graduate student in statistics. I wish to find the root of a function of two variables that is defined by an integral which must be evaluated numerically. So the problem I want to solve is of the form: Find k such that f(k)=0, where f(y) = int_a^b g(x,y)

Hi all, I am trying to draw a Kaplan-Meier curve and I found online that Kaplan - Meier estimates are computed with a function called km in the event package. Is there an update for that because when I choose to download packages in R,. there is no package called event, even though I have selected all the repositories. Thanks in advance, Eleni [[alternative HTML version deleted]]

Hello, I am not sure how to do this in R. Any suggestion would be appreciated. I have a vector of values from where I build an empirical CDF. For example: > x <- seq(1,100) > x <- sample(x,1000,replace=T) > quantile(x,probs=seq(0,1,.05)) 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50% 55% 1.00 5.00 10.00 16.00 20.00 25.00 31.00 36.00 41.00

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

Is it possible to calculate Kaplan-Meier for left-truncated, right-censored data using survival5? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

Hello All, Learning to use the odfWeave package. I really like the package. It has good documentation, makes some very nice looking tables, and seems to have lots of options for customizing output. There are a few things I'd like to do that don't seem to be covered in the documentation though. So I'm not sure if they're possible or not. Here's a list of some things I'd

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

Hello, I have a question about the function lifetab in package KMsurv. The description of the output value surv says "the estimated survival function at the start of the intervals". Are these estimates the ones calculated via Kaplan-Meier probability of survival ? Thanks in advance! -- View this message in context:

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

Hi I have been asked to provide Weibull parameters from a paper using Kaplan Meir survival analysis. This is something I am not familiar with. The survival analysis in R works nicely and is the same as commercial software (only the graphs are superior in R). The Weibull does not and produces an error (see below). Any ideas why this error should occur? My approach may be spurious.

Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time <- cumsum(rexp(1000)/10) status <- rbinom(1000, 1, 0.5) ## kaplan meier estimates fit <- survfit(Surv(time,

As I have previously asked, in response to a similar question: Is this a homework problem? cheers, Rolf Turner rolf at math.unb.ca

Hello, it seems that the main results of survival analysis with package survival are shown only as side effects of the print method. If I compute e.g. a Kaplan-Meier estimate by > km.survdur<-survfit(s.survdur) then I can simply print the results by > km.survdur Call: survfit(formula = s.survdur) n events median 0.95LCL 0.95UCL 100.0 58.0 46.8 41.0 79.3 Is

Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is just "1 minus kaplan-Meier survival"? Under what circumstance, you should use cumulative incidence vs KM survival? If the relationship is just CI = 1-survival, then what difference it makes to use one vs. the other? And in R how I can draw a cumulative incidence plot. I know I can make a Kaplan-Meier

Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to include enough detail. I have a question about survival curves I have been battling off and on for a few months. No one local seems to be able to help, so I turn here. The issue seems to either be how R calculates Kaplan-Meier Plots, or something with the underlying statistic itself that I am misunderstanding. Basically,