Hello members!
This question had been posted by Thomas Steiner in May, but I couldn't
locate any followups thereafter, see:
https://stat.ethz.ch/pipermail/r-help/2008-May/161483.html
I want to raise up this question in a (hopefully) even simpler way:
Given a random variable X, it's characteristi
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
Thank you Prof Ripley for your answer.
> > 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.
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 e
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 functio
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