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
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