Re: [R] truncated normal

From: Berwin A Turlach <berwin_at_maths.uwa.edu.au>
Date: Thu, 24 Jul 2008 14:47:46 +0800

G'day all,

On Wed, 23 Jul 2008 20:48:59 -0400
Duncan Murdoch <murdoch_at_stats.uwo.ca> wrote:

> On 23/07/2008 8:17 PM, cindy Guo wrote:

> > Yes, I know. I mean if I want to generate 100 numbers from
> > N(0,1)I((0,1),(5,10)). There are two intervals (0,1) and (5,10).
> > Then the function will give 50 numbers in the first interval and 50
> > in the other.

I can think of two strategies:

  1. Use rtnorm from the msm *package* to sample from the normal distribution truncated to the interval from your smallest lower bound and your largest upper bound; and then use rejection sampling.

I.e. for your original example N(0,1)I((0,1),(2,4)) sample from a N(0,1)I(0,4) distribution and reject all observations that are between 1 and 2, sample sufficiently many points until you have kept the required number. But this could be rather inefficient for your second example N(0,1)I((0,1),(5,10)).

2) If I am not mistaken, truncating the normal distribution to more than one interval essentially creates a mixture distribution where the components of the mixture are normal distributions truncated to a single interval. The weights of the mixture are given by the relative probability with which an observation from a normal distribution falls into each of the intervals. Thus, an alternative strategy, exemplified using your second example, would be (code untested):

samp1 <- rtnorm(100,mean=0,sd=1, lower=0,upper=1) samp2 <- rtnorm(100,mean=0,sd=1, lower=5,upper=10) p1 <- pnorm(1)-pnorm(0)
p2 <- pnorm(10)-pnorm(5) ## Take heed about Duncan's remark on

                          ## evaluating such quantities
p <- p1/(p1+p2)
ind <- runif(100) < p
finalsample <- samp1
finalsample[!ind] <- samp2[!ind]

HTH. Cheers,

        Berwin


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