R help - Nov 2015 - Fwd: Error survreg: Density function returned an an invalid matrix

I don?t know how to write a pareto distribution in that form, I want a
pareto function for time because I have a time variable that fits that
distribution. For a weibull and lognormal it is very easy because they are
particular cases from a extreme value and gaussian distributions. I think
it is possible to write a pareto using an exponential distribution,but I?m
not sure, I tried this using :

# OPTION 1.
# Define one distribution in terms of another

library(foreign)
library(survival)
library(VGAM)


my.pareto <- survreg.distributions$exponential
my.pareto$name <- "Pareto"
my.pareto$scale <- NULL

#Using the following transformation:

my.pareto$dtrans <- function(y) min(y)*exp(y)


survregDtest(my.pareto, TRUE)
set.seed(1)
a <- rpareto(100, 1, 6)
b <- rnorm(100,5,1)
c <- rep(1,100)
base <- cbind.data.frame(a,b,c)
mod1 <- survreg(Surv(a, c) ~ b, base, dist = my.pareto)
summary(mod1)

# It works but I don?t know if it's correct

# OPTION 2.
# Using Density, distribution function, quantile function and random
generation for the Pareto(I) distribution
# from VGAM package.

my.pareto3 <- list(name='Pareto',
                 init= function(x, weights,alpha,k){
                   alpha <- length(x)/(sum(log(x))-length(x)*log(min(x)))
                   k <-min(x)
                   c(media <-(alpha*k/(alpha-1)),varianza <-
((k/alpha)^2)*(alpha/(alpha-2)))},
                 density= function (x, alpha,k)  {
                   alpha <- length(x)/(sum(log(x))-length(x)*log(min(x)))
                   k <-min(x)
                   pvec <- seq(0.1, 0.9, by = 0.1)
                   qvec <- qpareto(pvec, alpha, k)
                     cbind(ppareto(qvec, alpha, k),
                           1-ppareto(qvec, alpha,  k),
                           dpareto(x,  alpha,  k),
                           -(alpha+x)/x,
                           (alpha+1)*(alpha+2)/x^2)},
                 deviance=function(x) {stop('deviance residuals not
defined')},
                 quantile= function(alpha,k) qpareto(seq(0.1, 0.9, by 0.1),
alpha, k))

survregDtest(my.pareto3, TRUE)

mod3 <- survreg(Surv(a, c) ~ b, base, dist = my.pareto3)

# Did not work and I don't want a fixed value for alpha and k paremeters
but the function needs a default.
# I got this error:

Error in logdensity[xok] <- log(shape[xok]) + shape[xok] * log(scale[xok])
- : NAs are not allowed in subscripted assignments
6
dpareto(x, alpha, k)
5
cbind(ppareto(qvec, alpha, k), 1 - ppareto(qvec, alpha, k), dpareto(x,
alpha, k), -(alpha + x)/x, (alpha + 1) * (alpha + 2)/x^2)
4
density(z, parms)
3
derfun(y, yy, exp(vars), sd$density, parms)
2
survreg.fit(X, Y, weights, offset, init = init, controlvals = control, dist
= dlist, scale = scale, nstrat = nstrata, strata, parms = parms)
1
survreg(Surv(a, c) ~ b, base, dist = my.pareto3)



So, I don't know what else can I do.

Thanks.



2015-11-16 11:38 GMT-06:00 Therneau, Terry M., Ph.D. <therneau at
mayo.edu>:
> You are still missing the point.
> The survreg routine handles distribution of the form:
>
>   (t(y) - m)/s ~ f, where f is a distribution on the real line.
>
> Here t is an optional but fixed transform and m= X\beta.  Beta and s=scale
> are the parameters that the routine will fit.
>
> For a log-normal, t=log and f= the density of a Gaussian mean=0, sd=1.
> The distribution function is dnorm(x)
> For a Weibull,    t=log and f= the density of the least extreme value
> distribution: exp(-exp(x))
>
> How do you write a 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.
>
>
>
>
>
>
>
>
> On 11/16/2015 08:56 AM, Israel Ortiz wrote:
>
> 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 f?rmula in
> https://en.wikipedia.org/wiki/Pareto_distribution but there are a lot of
> books and sites that use the same f?rmula. This part of the code use that
> formula:
>
>  distribution <- function(x, alpha) ifelse(x > min(x) ,
> alpha*min(x)**alpha/(x**(alpha+1)), 0)
>
> Also, I support my sintax in the following post:
>
>
>
http://stats.stackexchange.com/questions/78168/how-to-know-if-my-data-fits-pareto-distribution
>
> Another option is transform my variable for time from pareto to
> exponential (but this solution it's not very elegant):
>
> If X is pareto distributed then
> [image: Y = \log\left(\frac{X}{x_\mathrm{m}}\right)]
>
> it's exponential distributed.
>
> The syntax:
>
> library(foreign)
> library(survival)
> library(VGAM)
>
> set.seed(3)
> X <- rpareto(n=100, scale = 5,shape =  1)
>
> Y <- log(X/min(X))
>
> hist(X,breaks=100)
> hist(Y,breaks=100)
> b <- rnorm(100,5,1)
> c <- rep(1,100)
> base <- cbind.data.frame(X,Y,b,c)
> mod1<-survreg(Surv(Y+1, c) ~ b, base, dist = "exponential")#
+1 it's
> because time should be > 1
>
> summary(mod1)
>
> This solution works but I don?t like it.
>
> Thanks.
>
>
>
>
> 2015-11-16 7:40 GMT-06:00 Therneau, Terry M., Ph.D. <therneau at
mayo.edu>:
>
>> The error message states that there is an invalid value for the
density.
>> A long stretch of code is not very helpful in understanding this.  What
we
>> need are the definition of your density -- as it would be written in a
>> textbook.  This formula needs to give a valid response for the range
>> -infinity to +infinity.  Or more precisely, for any value that the
>> maximizer might guess at some point during the iteration.
>>
>> Terry T.
>>
>>
>> On 11/14/2015 05:00 AM, r-help-request at r-project.org wrote:
>>
>>> Thanks Terry but the error persists. See:
>>>
>>> >library(foreign)> library(survival)> library(VGAM) >
mypareto <-
>>>> list(name='Pareto',+                 
init>>>>
>>>
>> remainder of message trucated
>>
>
>
>
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