R help - Apr 2008 - fft: characteristic function to distribution

The characteristic function is the inverse Fourier transform of the
distribution function. The characteristic function of a normaly
distributed random variable is exp(-t^2/2).

x=seq(-2,2,length=100)
fft(pnorm(x),inverse=T)/length(x)
exp(-x^2/2)

Why aren't the inverse fft and the mentioned function the same?
Thanks for help,
Thomas

On Wed, 30 Apr 2008, Thomas Steiner wrote:
> The characteristic function is the inverse Fourier transform of the
> distribution function. The characteristic function of a normaly
> distributed random variable is exp(-t^2/2).
The fft is a discrete Fourier transforn, not a continuous one.
Further in each case where the normalizing constants are placed and the 
units of frequecy differ from source to source.

?fft has references to exactly what it computes: please consult them.
>
> x=seq(-2,2,length=100)
> fft(pnorm(x),inverse=T)/length(x)
> exp(-x^2/2)
>
> Why aren't the inverse fft and the mentioned function the same?
> Thanks for help,
> Thomas
>
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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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Matthias Kohl was so kind and provided me the following lines in this issue:

library(distrEx)
chf <- function(t, D){
  E(D, function(x){exp(1i*t*x)}, useApply = FALSE)
}

## Normalverteilung
D <- Norm()
t <- seq(-3, 3, by = 0.05)
chf.norm <- sapply(t, chf, D = D)
chf.exakt <- exp(-t^2/2)
chf.diff <- chf.norm - chf.exakt
plot(chf.diff)
abs(chf.diff)


This is nice BUT:
* Only built-in distributions can be used
* I want to understand the fft() function, instead here you used E()
of the distrEx package.

Apart from this it was exactly what i was looking for: I know the
characteristic function and want to get the distribution. (My porposal
was to use fft(,inverse=T).)
Any help appreciated,
Thomas