You are right. Specifically, I need to predict the mean and median time to
failure from a coxph model and several parametric models using new data.
Thanks.
El lun., 5 nov. 2018 a las 7:11, Therneau, Terry M., Ph.D. (<
thern...@mayo.edu>) escribió:
> First, type='expected' gives the expected cumu
I am trying to predict follow-up time using several survival models, both
parametric and semi-parametric. I achieve it for semi parametric models
using predict.coxph function in R from survival package using type =
"expected" as indicated in help. However, for parametric models, this
option doesn't
Pareto in this form? I assume that you would like
> survreg to solve for some parameters -- how do you map them onto the beta
> and s values that survreg will attempt to optimize? I have not yet grasped
> what it is that you want survreg to DO.
>
> Terry T.
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Thanks Terry, I use the following formula for density:
[image: f_X(x)= \begin{cases} \frac{\alpha
x_\mathrm{m}^\alpha}{x^{\alpha+1}} & x \ge x_\mathrm{m}, \\ 0 & x <
x_\mathrm{m}. \end{cases}]
Where *x*m is the minimum value for x. I get this fórmula in
https://en.wikipedia.org/wiki/Pareto_distri
Thanks Terry but the error persists. See:
> library(foreign)> library(survival)> library(VGAM) > mypareto <-
> list(name='Pareto',+ init= function(x, weights,parms){+
> alpha <- length(x)/(sum(log(x)))#this is a MLE for alpha+
> c(media <-(
Hi, I want to perform a survival analysis using survreg procedure from
survival library in R for a pareto distribution for a time variable, so I
set the new distribution using the following sintax:
library(foreign)
library(survival)
library(VGAM)
mypareto <- list(name='Pareto',
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