2
$\begingroup$

From this previous post, the real and imaginary parts of the Fourier transform of a zero mean Gaussian are uncorrelated (and i.i.d. Gaussians)

This some how seems counter-intuitive. It seems if the real part has a large intensity, then the imaginary part should be small. I read through the proof linked from the previous post, but I don't have a good intuitive feel from this proof.

Can you provide an intuitive explanation of why this is true?

$\endgroup$
3
$\begingroup$

Recall the following properties of the Fourier transform:

  • If $x(t)$ is an even function, then its Fourier transform $X(\omega)$ is purely real.

  • If $x(t)$ is an odd function, then its Fourier transform $X(\omega)$ is purely imaginary.

Thus, we can think of the real and imaginary parts of the Fourier transform of a zero-mean Gaussian random process $x(t)$ as the Fourier transforms of two separate inputs $x_e(t)$ and $x_o(t)$: the components of the process that have even and odd symmetry, respectively. We split the process into these components as follows:

$$ x_e(t) = \frac{x(t)+x(-t)}{2} $$

$$ x_o(t) = \frac{x(t)-x(-t)}{2} $$

Note that $x(t) = x_e(t) + x_o(t)$. Moving through this step by step,

  • Your observation was that $\text{Re}\{X(\omega)\}$ and $\text{Im}\{X(\omega)\}$ (the real and imaginary parts of the process's DFT) are uncorrelated.

  • Using the Fourier transform properties mentioned above, we can deduce that $\text{Re}\{X(\omega)\} = X_e(\omega)$ and $\text{Im}\{X(\omega)\} = X_o(\omega)$; the real and imaginary parts of $X(\omega)$ are none other than the Fourier transforms of $x_e(t)$ and $x_o(t)$, respectively.

  • Therefore, your observation is equivalent to saying that $X_e(\omega)$ and $X_o(\omega)$ are uncorrelated.

  • Since the Fourier transform is a one-to-one mapping between the time and frequency domains, I posit that the lack of correlation between $X_e(\omega)$ and $X_o(\omega)$ would imply a lack of correlation between $x_e(t)$ and $x_o(t)$ as well.

What is the correlation between $x_e(t)$ and $x_o(t)$? Simple:

$$ \begin{align} \mathbb{E}(x_e(t)x_o(t)) &= \mathbb{E}\left(\left(\frac{x(t)+x(-t)}{2}\right)\left(\frac{x(t)-x(-t)}{2}\right)\right) \\ &= \mathbb{E}\left(\frac{1}{4}\left(x^2(t) - x^2(-t)\right)\right) \\ &= \frac{1}{4} \left(\mathbb{E}(x^2(t)) - \mathbb{E}(x^2(-t))\right) \\ &= \frac{1}{4} \left(\sigma^2 - \sigma^2\right) \\ &= 0 \end{align} $$

As expected, the even and odd components are uncorrelated. So, to summarize, I would say the following:

  • The real and imaginary components of a Fourier transform correspond to the individual Fourier transforms even and odd components of the input function.

  • For a zero-mean Gaussian random process, these even and odd components are uncorrelated.

  • Therefore, their Fourier transforms (the real and imaginary components that you asked about) are also uncorrelated.

Edit: To address your followup:

  • If $x(t)$ is Gaussian, then its even and odd components $x_e(t)$ and $x_o(t)$ are as well, due to the property that any weighted sum of Gaussian random variables is also Gaussian.

  • If $x_e(t)$ and $x_o(t)$ are Gaussian random processes, then their Fourier transforms $X_e(\omega)$ and $X_o(\omega$) are as well. This follows from the same property as the previous statement; if you look at the transform, you're computing a weighted sum of a bunch of Gaussian random variables.

  • If $X_e(\omega)$ and $X_o(\omega)$ are Gaussian, and they are uncorrelated with one another (as described above), then they are also independent. This is a property of the Gaussian distribution.

$\endgroup$
  • $\begingroup$ Thanks! As a follow-up, why are they also i.i.d. Gaussian? $\endgroup$ – klurie Apr 30 '14 at 5:30
  • 1
    $\begingroup$ I added some more info to the post. $\endgroup$ – Jason R Apr 30 '14 at 13:39
  • $\begingroup$ I think the first added bullet is a bit short. The sum of two independent Gaussian variables is Gaussian, but you're perhaps a bit quick in assuming x(-t) and x(t) are independent. (They are, because the autocorrelation function of x is a delta function) $\endgroup$ – MSalters May 2 '14 at 16:39
  • $\begingroup$ @MSalters: I don't make that assumption. Check out the part above where I show that $x_e(t)$ and $x_o(t)$ are uncorrelated. I don't assume that $x(t)$ is independent of $x(-t)$ there. The expectation operator is linear, so the order of expectation and subtraction can be swapped. $\endgroup$ – Jason R May 2 '14 at 16:43
  • $\begingroup$ @JasonR: The added bullet states that the sum of two Gaussians is Gaussian, therefore x(t) is Gaussian implies xe(t) is Gaussian. That makes sense: xe(t) = x(t) + x(-t), the sum of two Gaussians. But the real theorem is that the sum of two independent Gaussians is Gaussian. Thus to prove xe(t) is Gaussian that way means proving x(-t) is independent. $\endgroup$ – MSalters May 2 '14 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.