Suppose $X$ is a real-valued N-dimensional Gaussian vector, $X \sim \mathcal{N}(\mathbf{0}, C_X)$. The discrete Fourier transform can be obtained by left-multiplying with the unitary DFT matrix, i.e. $\hat{X} = W X$ (where $W$ is defined as in this page). What is the distribution of $\hat{X}$?

My attempt to answer this is as follows: Since the DFT is linear, $\hat{X}$ will also be Gaussian, and thus $\hat{X} \sim \mathcal{N}(\mathbf{0}, WC_XW^\top)$... BUT that doesn't make sense to me; $\hat{X}$ is inevitably complex valued!

On the other hand, saying $\hat{X} \sim \mathcal{CN}(\mathbf{0}, WC_XW^H)$ doesn't make sense either; even if it is correct, what does it mean to have complex valued variances? Do we not have to account for the fact that $X$ is purely real?

  • $\begingroup$ The closest I could find to an answer only mentions the element-wise variances, not the entire covariance matrix, and only addresses the specific case of zero-mean i.i.d noise. $\endgroup$ Commented Nov 11, 2023 at 17:30

1 Answer 1


As the DFT of real $X$ is conjugate symmetric, $\hat{X}$ is not N-dimensional jointly Gaussian and neither your two distributions is correct.

Representing the N-dimensional DFT by a $2N$ dimensional linear transformation $$Y =\begin{bmatrix}\hat{X}_{re}\\\hat{X}_{im}\end{bmatrix}=\begin{bmatrix}W_{re}&-W_{im}\\W_{im}&W_{re}\end{bmatrix}\begin{bmatrix}X\\0\end{bmatrix}=\begin{bmatrix}W_{re}\\W_{im}\end{bmatrix}X=\tilde{W}X$$

The covariance matrix $\tilde{W}C_X\tilde{W}^T$ is singular and the distribution is degenerate. You can find the subspace to which $\hat{X}$ is mapped by exploiting the conjugate symmetric property as follows.

Exclude the last $N/2$ lines of both $W_{re}$ and $W_{im}$, and the line corresponding to the imaginary part of DC from the matrix $\tilde{W}$ to get $\tilde{W}_{sub}$ (by $\tilde{W}_{sub}=S\tilde{W}$ with $S$ is a diagonal matrix with corrresponding 1 and 0 in its diagonal). Hence the dimensions of $\tilde{W}_{sub}$ is $(N-1)\times N$.

$\tilde{W}_{sub}C_X\tilde{W}_{sub}^T$ is non-singular and the $SY$ is a $(N-1)$-dimensional zero mean multivariate Normal with covariance $\tilde{W}_{sub}C_X\tilde{W}_{sub}^T$.

We know that the distribution of $SY$ fully characterizes $Y$, and $\hat{X}$, because the conjugate symmetric property implies that the dimension of $\hat{X}$ is at most $N-1$.

The $2N$-dimensional real $Y$, and the $N$-dimensional complex $\hat{X}$, has zero probability outside of the space of $SY$ and is impulsive from the view of such higher-than-$(N-1)$-real-dimensional density.

  • $\begingroup$ Thanks!! Could you please suggest a resource for understanding why $\hat{X}$ won't be Gaussian, and how to deal with degenerate Gaussian distributions? The wiki article goes into measure theory stuff, surely there must be a simpler way to understand it? $\endgroup$ Commented Nov 15, 2023 at 6:34
  • $\begingroup$ My understanding is still shaky, but what if we instead excluded the last N/2 lines from the original complex matrix $W$ and set the imaginary DC part to zero - is this equivalent to your method? Can we then say $\hat{X} \sim \mathcal{CN}(0, \bar{W}C_X\bar{W}^H$? Intuitively I'm trying to get at a DFT matrix that corresponds to something like what numpy.fft.rfft does. $\endgroup$ Commented Nov 15, 2023 at 6:59
  • $\begingroup$ @DangerousTim To understand why the elements of $\hat{X}$ are not jointly Gaussian, you can read Chapter 3 of "STOCHASTIC PROCESSES: THEORY FOR APPLICATIONS" of Robert Gallager. Specifically, 3.4.3 for real vectors and 3.7 for the complex model. $\endgroup$
    – AlexTP
    Commented Nov 15, 2023 at 10:36
  • $\begingroup$ For the degenerate distributions, you might have a simpler way if you can exploit the structure of the non-singular covariance matrix induced by the DFT. I don't know how. $\endgroup$
    – AlexTP
    Commented Nov 15, 2023 at 10:39
  • $\begingroup$ @DangerousTim For the question about manipulating directly the complex DFT matrix $W$, you can but this does not mean that the remaining part of $\hat{X}$ is circularly symmetric complex Gaussian like your notation because the resulting vector cannot be written as a linear transform of iid circularly symmetric gaussian random variables (note that the definition of complex gaussian vector is not standardized and I am using the definition of the aforementionned Gallager book and the Wikipedia article). $\endgroup$
    – AlexTP
    Commented Nov 15, 2023 at 10:51

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