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If I have a discrete time series $x(t_i)$, and each of the $x(t_{i})$ are normally distributed, i.e., come from a Gaussian distribution with mean zero and variance one, would a windowed finite Fourier transform of $x(t_0)$ through $x(t_{N-1})$ also be Gaussian distributed? In other words, would the real and imaginary parts of:

$$y(f) = \sum_{t=0}^{N-1}exp(-i 2 \pi f t) x(t) a(t)$$

also have Gaussian distributions? a(t) is a window function that decays to 0 at $t=0$ and $t=N-1$.

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Please do not silently modify important aspects of your post, as this can make some answers already posted look fully or partly non pertinent. I went back to the previous version. –  Did Aug 15 '12 at 16:39

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Yes, provided one assumes that the family $(x(t))_{0\leqslant t\leqslant N-1}$ is gaussian (recall that, as soon as $N\geqslant2$, this asks strictly more than each $x(t)$ being normally distributed). Then $y(f)$, the real part of $y(f)$ and the imaginary part of $y(f)$ are all linear combinations of a gaussian family hence they are normally distributed (complex-valued for $y(f)$ and real-valued for the others).

Edit: A standard way to ensure that a family is gaussian is to assume that each random variable is normally distributed and that the family is independent.

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Yes it is correct. As long as $a(t)$ are deterministic the resulting $y(f)$ is Gaussian. Just came to my mind. Let $X_i$ be correlated normally distributed RVs. When we take the fourier transform how is the correlation pattern in the frequency domain between frequency bins? –  Seyhmus Güngören Aug 15 '12 at 9:43

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