# Re: [R] Fit model to data and use model for data generation

From: Roberto Perdisci <roberto.perdisci_at_gmail.com>
Date: Thu 25 Jan 2007 - 15:13:46 GMT

On 1/25/07, Prof Brian Ripley <ripley@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@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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