R help - Jan 2007 - Fit model to data and use model for data generation

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

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:

### 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")

Hope this helps,

Stephen
Rochester, Minnesota, USA

On 1/25/07, Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:
 > 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)
Prof. Ripley,
  I didn't understand why you used
sample(x, size = n, replace=TRUE)
I though the mixture should be computed using all the points in x as
means, like in
x.new <- rnorm(n, x, bw)

Could you explain why you propose
x.new <- rnorm(n, sample(x, size = n, replace=TRUE), bw)
instead?

Could you also briefly say in what sense kde is biased?

thank you very much,
best regards,
Roberto
> 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
>
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