[R] Fit model to data and use model for data generation

[Prof Brian Ripley]

On Wed, 24 Jan 2007, Stephen D. Weigand wrote:

>
> On Jan 24, 2007, at 10:34 AM, Benjamin Otto wrote:
>
>> Hi,
>>
>> Suppose I have a set of values x and I want to calculate the
>> distribution of
>> the data. Ususally I would use the "density" command. Now, can I use
>> the
>> resulting "density-object" model to generate a number of new values
>> which
>> have the same distribution? Or do I have to use some different
>> function?
>>
>> Regards,
>>
>> Benjamin
>>
>> --
>> Benjamin Otto
>> Universitaetsklinikum Eppendorf Hamburg
>> Institut fuer Klinische Chemie
>> Martinistrasse 52
>> 20246 Hamburg
>>
>
> You could sample from the x's in the density object with probability
> given by the y's:

That gives a discrete distribution, which may well matter for small 
samples.

Since density() is returning an equal-weighted mixture of (by default)
normal distributions, all you need to do is

x.new <- rnorm(n, sample(x, size = n, replace=TRUE), bw)

where bw is the bandwidth used by density (d$bw in this example).
(This is known as a 'smoothed bootstrap' in some circles.)


> ### Create a bimodal distribution
> x <- c(rnorm(25, -2, 1), rnorm(50, 3, 2))
> d <- density(x, n = 1000)
> plot(d)
>
> ### Sample from the distribution and show the two
> ### distributions are the same
> x.new <- sample(d$x, size = 100000, # large n for proof of concept
>                 replace = TRUE, prob = d$y/sum(d$y))
> dx.new <- density(x.new)
> lines(dx.new$x, dx.new$y, col = "blue")

BTW, lines(density(x.news), col = "blue") works here, and you do need to 
remember that a kde is biased.  But my solution matches better than yours.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595