On 10/04/2009, at 11:44 PM, Achilleas Achilleos wrote:

Hi,

A very simple question.

I know that the Fourier transform of a Laplace distribution, with zero
mean and variance 2b^2, is equal to 1/(1+(tb)^2). Therefore, the Fourier
transform is a positive function (actually is always between 0 and 1).

BUT, if I use the fft(alpha) function of R, where alpha is a vector
containing 128 values of the probability density function from -2 to 2, I
get negative values as well.

Why?


Basically because there is a (considerable!) difference between the continuous Fourier transform and the discrete Fourier transform, the latter being what is calculated by fft(). (NOTE: ***NOT*** ``FFT'' --- R is case sensitive.)

        cheers,

                Rolf Turner

######################################################################
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}

______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Reply via email to