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.
>
> Thanks in advance,
> Galina
Maybe someone has written an efficient function for this.
If not, you anyway have at least two options with basic R.
(A) "Rejection sampling": OK if the probability of 0 is small.
In that case, suppose you want n zero-truncated Poisson values.
Sample n values from rpois. Reject any which are 0, leaving
you with say r to find. Sample r from rpois, and continue
in the same way until you have all n.
This could be done by something on the following lines:
n<-1000; T<-3.5
Y<-rpois(n,T); Y0<-Y[Y>0]; r<-(n - length(Y0))
while(r>0){
Y<-rpois(r,T); Y0<-c(Y0,Y[Y>0]); r<-(n - length(Y0))
}
and Y is then the required sample of n=1000 from a Poisson
distribution with mean T=3.5, after rejecting all zeros.
(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))
Hoping this helps,
Ted.
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E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 01-May-05 Time: 17:20:09
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