I came across this as a property, but wanted to prove it myself, but can't get through with the proof. Any help would be appreciated.

  • $\begingroup$ Hopefully you found my answer below helpful; if so would you please consider upvoting / accepting. Stackexchange sites survive by people donating their time to help each other, and that is an easy way to give back. Thanks! $\endgroup$ – owen88 Feb 25 '18 at 9:28

Is this true?

Suppose we let $Z = X + i Y$, with $X,Y \sim N(0,1)$ independent and Normally distributed, and let $W_n$ be i.i.d. draws from a uniform distribution on $\{-1,1\}$, i.e.

$$\mathbf P[ W_n = 1] = \mathbf P[W_n = -1] = \frac12,$$

(often referred to as the Rademacher distribution). Now define

$$Z_n = W_n Z, \qquad n = 1,\ldots, N.$$

Clearly the $Z_n$ are not independent, since for example

$$ \mathbf P[ Z_n = \pm z \, | \, Z_1 = z ] = 1, \qquad n = 1,\ldots, N.$$

And then computing the covariance we have for any $m \neq n$

\begin{align*} \text{Cov}(Z_m,Z_n) & = \mathbf E[ Z_m Z_n ] - \mathbf E[Z_m] \mathbf E[Z_n] \\ & = \mathbf E[ Z_m Z_n] \\ & = \mathbf E[ W_m W_n Z^2] \\ & = \mathbf E[W_m] \mathbf E[W_n] \mathbf E[Z^2] \\ & = 0, \end{align*} where in the first line we used the fact that $\mathbf E[Z_n] = 0$, and in the last line we used independence of the $W_n$, and the fact that $\mathbf E [W_n] = 0$.

  • $\begingroup$ Although X and Y are random variables in the usual sense - they have distribution functions, Z does not have a distribution function. Therefore Z is not, strictly speaking, a random variable. $\endgroup$ – herb steinberg Feb 23 '18 at 18:34
  • $\begingroup$ I think this depends on the interpretation of the poster's reference to complex Gaussian. The standard probabilistic definition is to consider a complex Gaussian to be as I described: $Z = X + iY$, (wikipedia). In this context there is a natural definition of expectation $$\mathbf E[Z] =\mathbf{E}[X] + i\,\mathbf{E}[Y].$$ This is the interpretation that I assumed was intended in the question... $\endgroup$ – owen88 Feb 23 '18 at 23:01
  • $\begingroup$ Alternatively, the poster may have been intended the case of complex Gaussian measures on $\phi \in \mathcal{C}^N$ where $$d \mu(\phi) \propto e^{- \phi A \overline \phi} d \overline \phi_1 d \phi_1 \cdots d \overline \phi_N d \phi_N,$$ with $A$ a matrix with positive Hermitian part. In this sense I agree that the notion of expectation and covariance are less well defined. But so too are the notions of independence, and correlation. $\endgroup$ – owen88 Feb 23 '18 at 23:01
  • $\begingroup$ @ herb steinberg, I'm inferring that you are really getting at the question posed being ill defined, but hopefully the comments above help to clarify why this need not be the case. $\endgroup$ – owen88 Feb 23 '18 at 23:06
  • $\begingroup$ The main point of my comment was - how can a distribution function be defined? $\endgroup$ – herb steinberg Feb 24 '18 at 1:23

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