# Re: [R] simulate zero-truncated Poisson distribution

From: Peter Dalgaard <p.dalgaard_at_biostat.ku.dk>
Date: Mon 02 May 2005 - 05:35:22 EST

(Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> writes:

> On 01-May-05 Glazko, Galina wrote:
> >
> > Dear All
> >
> > I would like to know whether it is possible with R to
> > generate random numbers from zero-truncated Poisson
> > distribution.
> >
> > Galina
.....
> (B) This has a deeper theoretical base.
>
> Suppose the mean of the original Poisson distribution (i.e.
> before the 0's are cut out) is T (notation chosen for intuitive
> convenience).
>
> The number of events in a Poisson process of rate 1 in the
> interval (0,T) has a Poisson distribution with mean T.
> The time to the first event of a Poisson process of rate 1
> has the negative exponential distribution with density exp(-t).
>
> Conditional on the first event lying in (0,T), the time
> to it has the conditional distribution with density
>
> exp(-t)/(1 - exp(-T)) (0 <= t <= T)
>
> and the PDF (cumulative distribution) is
>
> F(t) = (1 - exp(-t))/(1 - exp(-T))
>
> If t is randomly sampled from this distribution, then
> U = F(t) has a uniform distribution on (0,1). So, if
> you sample U from runif(1), and then
>
> t = -log(1 - U*(1 - exp(-T)))
>
> you will have a random variable which is the time to
> the first event, conditional on it being in (0,T).
>
> Next, the number of Poisson-process events in (t,T),
> conditional on t, simply has a Poisson distribution
> with mean (T-t).
>
> So sample from rpois(1,(T-t)), and add 1 (corresponding to
> the "first" event whose time you sampled as above) to this
> value.
>
> The result is a single value from a zero-truncated Poisson
> distribution with (pre-truncation) mean T.
>
> Something like the following code will do the job vectorially:
>
> n<-1000 # desired size of sample
> T<-3.5 # pre-truncation mean of Poisson
> U<-runif(n) # the uniform sample
> t = -log(1 - U*(1 - exp(-T))) # the "first" event-times
> T1<-(T - t) # the set of (T-t)
>
> X <- rpois(n,T1)+1 # the final truncated Poisson sample
>
> The expected value of your truncated distribution is of course
> related to the mean of the pre-truncated Poisson by
>
> E(X) = T/(1 - exp(-T))

There must be an easier way... Anything wrong with

rtpois <- function(N, lambda)

qpois(runif(N, dpois(0, lambda), 1), lambda)

rtpois(100,5)

?

```--
O__  ---- Peter Dalgaard             Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics     2200 Cph. N
(*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard@biostat.ku.dk)             FAX: (+45) 35327907

______________________________________________
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help