Hi,
you might try this:
set.seed(100)
m <- 10
size.a <- 10
prob.a <- 0.3
prior.constant = 0
draw1 = rbinom( m , size.a, prob.a )
beta.draws <- function(draw, size.a, prior.constant, n) {
rbeta(n, prior.constant + draw, prior.constant + size.a - draw)
}
bdraws <- sapply(draw1, beta.draws, size.a = size.a, prior.constant =
prior.constant, n = 10000)
beta.post <- apply(bdraws, 2, function(x) c(post.mean = mean(x),
post.median = median(x)) )
beta.post
[,1] [,2] [,3] [,4] [,5]
[,6] [,7] [,8]
post.mean 0.2017118 0.1996809 0.2991173 0.10069613 0.3001924
0.2991149 0.4033310 0.2003104
post.median 0.1804893 0.1791630 0.2845427 0.07505278 0.2858155
0.2844503 0.3961419 0.1790511
[,9] [,10]
post.mean 0.3013020 0.1990232
post.median 0.2886199 0.1786447
best
Víctor H Cervantes
2008/9/17 Juancarlos Laguardia <[EMAIL PROTECTED]>:
> I have a problem in where i generate m independent draws from a binomial
> distribution,
> say
>
> draw1 = rbinom( m , size.a, prob.a )
>
>
> then I need to use each draw to generate a beta distribution. So, like
> using a beta prior, binomial likelihood, and obtain beta posterior, m many
> times. I have not found out a way to vectorize draws from a beta
> distribution, so I have an explicit for loop within my code
>
>
>
> for( i in 1: m ) {
>
> beta.post = rbeta( 10000, draw1[i] + prior.constant , prior.constant +
> size.a - draw1[i] )
>
> beta.post.mean[i] = mean(beta.post)
> beta.post.median[i] = median(beta.post)
>
> etc.. for other info
>
> }
>
> Is there a way to vectorize draws from an beta distribution?
>
> UC Slug
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
