I am making my way through this paper here, and I understand everything up to equation 16, and I am stuck here.

Long story short, the vector $\mathbf{x}[i]$ here is an $N$ by 1 vector, consisting of the complex envelopes of signal + noise + interference across $N$ different sensors, taken at some time index $i$. We have a desired signal, an interference signal, and AWGN noise at each sensor. Naturally, the desired and interference signals at each sensor are delayed versions of each other, but the AWGN at each sensor is independent of the AWGN at another sensor.

What I do not understand, is the author's claim, where he states that since the desired and/or interference signals are just delayed copies of each other, "the circular symmetry of the complex envelopes thus implies that:"

$$ \mathbb{E} \Big[\mathbf{x}[i] \mathbf{x}^T[i] \Big] = \mathbf{0} $$

Circular symmetry of the complex envelopes? What does that mean, and why do we get the above equation from that?

  • $\begingroup$ A quick read doesn't tell me the immediate answer, but the equations you referenced seem suspect to me. I think there should at least be nonzero values along the diagonal of the correlation matrix of $\mathbf{x}$ (which is what $\mathbb{E}(\mathbf{x}\mathbf{x^T})$ gives). $\endgroup$ – Jason R Sep 25 '13 at 20:29
  • $\begingroup$ Also, note that the elements of $\mathbf{x}$ are soft symbol observations after despreading. Thus, $\mathbf{x}[i]$ refers to the observation vector for the $i$-th symbol. For a phase-shift-keyed consetellation (which is what I believe the paper considers), the expected symbol positions lie symmetrically in a circle about the origin. That may be what they are getting at by "circular symmetry of the complex envelopes." $\endgroup$ – Jason R Sep 25 '13 at 20:36
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    $\begingroup$ For a circularly symmetric or circular complex random vector $\mathbf{Z}$, $\mathrm{E}[\mathbf{Z}]=0$ and $\mathrm{E}[\mathbf{Z}\mathbf{Z}^T]=0$. These are definitions which can be proven. For information, any standard Complex Random Variables chapter from a book will have it. $\endgroup$ – Sudarsan Sep 25 '13 at 21:05
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    $\begingroup$ And as Jason says, the definition is as follows. A complex Random Vector $\mathbf{Z}$ is circularly symmetric if $\mathbf{Z}$ and $e^{j\theta}\mathbf{Z}$ have the same PDF for all $\theta$. $\endgroup$ – Sudarsan Sep 25 '13 at 21:07
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    $\begingroup$ @TheGrapeBeyond: Good point; I hadn't read closely enough to notice the $^T$ instead of $^H$. $\endgroup$ – Jason R Sep 25 '13 at 21:30

A complex random variable with dimension being 2 $Z=X+jY$ can be defined as a Random vector with components $X$ and $Y$ which take values along 2 different dimensions. The Probability Density Function of $Z$ is defined using the PDFs of $X$ and $Y$. The Expectation is defined as $\mathrm{E}(Z)=\mathrm{E}(X)+j\mathrm{E}(Y)$ and the variance is defined the usual way as $Z$ being a complex function of $X$ and $Y$ and the Variance is $\mathrm{VAR}(Z)=\mathrm{E}(|Z^2|)-|\mathrm{E}(Z)|^2$. Intuitively speaking, $Z$ is a complex function of two random variables which just assumes value $X+jY, \forall x,\forall y$. The PDF is written as follows: Write the 2-dimensional Random variable as $$\tilde{Z}=\left[\begin{array}{ccc}X \\Y\end{array}\right]$$where $\tilde{Z}$ is a Real random vector. The PDF of $Z$ is the PDF of $\tilde{Z}$.

Now with Complex Random vectors, $\mathbf{Z}=\mathbf{X}+j\mathbf{Y}$ is a vector of Complex Random variables which has a lot of interesting properties in a sense that it has more to offer than a Vector of Real Random Variables. Particularly of importance are the Covariance, Pseudocovariance(this is something new) and the Expectation. In our situation, we have a special Complex Random vector called the Symmetric Random Vector $\mathbf{Z}$.

Define the $2n$-dimensional Complex Random Vector as $$\tilde{\mathbf{Z}}=\left[\begin{array}{ccc}\mathbf{X} \\\mathbf{Y}\end{array}\right]$$ where $\mathbf{X}$ and $\mathbf{Y}$ are $n$-dimensional real random vectors. The PDF definition follows the same from Complex Random Variable. You can look up what the Pseudocovariance and Variance mean for a Complex Random Vector. I'm referring to Digital Communications by Proakis here in my definitions and notations.

Now as in the comments, if $\mathbf{Z}$ is a Symmetric i.e the PDF doesn't change on rotation, we can say that $\mathrm{E}(\mathbf{ZZ^\mathrm{T}})=0$. This is a consequence of the following. Since $\mathbf{Z}$ and $\mathbf{Z}e^{j\theta}$ have the same PDF, $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(e^{j\theta}\mathbf{Z})=e^{j\theta}\mathrm{E}(\mathbf{Z}), \forall \theta$$ If we had $\theta=\pi$ (the condition has to be satisfied for all $\pi$), we can see that $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(-\mathbf{Z})=-\mathrm{E}(\mathbf{Z})$$ which clearly says that $\mathrm{E}(\mathbf{Z})=0$. Going further, we can prove that $\mathrm{E}(\mathbf{Z}\mathbf{Z}^\mathrm{T})=0$ the same way assuming a special case of $\theta=\frac{\pi}{2}$. And hence Circularly Symmetric Complex Random Vectors are Zero-mean and Proper. The last equation left unproven here is the definition of being proper.

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  • $\begingroup$ Ok, I am sorry but there are many things that I am still not getting. i) "Z is symmetric". Z is a vector. You are saying it is symmetric to what? What does this mean? ii) "Its PDF doesnt changed upon rotation." Rotation on what? You just mean rotation on the complex plane? iii) Lastly, I am still unclear as to how any of this shows that $\mathbb{E}[Z Z^H] = \mathbb{0}$. THank you. $\endgroup$ – TheGrapeBeyond Sep 27 '13 at 1:51
  • $\begingroup$ Yes exactly, it's with respect to the rotation on the complex plane. It simply means that each Random Variable in the vector preserves its PDF upon rotation. To show $\mathrm{E}(\mathbf{Z}\mathbf{Z}^\mathrm{T})=0$, you start from the same argument. You just write down $\mathrm{E}(e^{2j\theta}\mathbf{Z}\mathbf{Z}^\mathrm{T})= e^{2j\theta}\mathrm{E}(\mathbf{Z}\mathbf{Z}^\mathrm{T})=\mathrm{E}(\mathbf{Z} \mathbf{Z}^\mathrm{T})$ and since they are the same when $\theta=\frac{\pi}{2}$ it can't be anything but our required proof. $\endgroup$ – Sudarsan Sep 27 '13 at 19:24

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