Considering a random variable $x$ that takes in values from a complex domain. Its real and imaginary components are totally uncorrelated. I am following this link and also studying this document.

In the wikipedia link, the pdf for a single observation is given and $k =2$ for bivariate gaussian as my assumption is that the real and imaginary are totally uncorrelated and are gaussian respectively. There is a 2 in the denominator in the pdf but the log likelihood for the complex case does not have any 2 in the denominator. The log-likelihood for $k=2$ would be $$-N \ln| \Sigma| - (x_n) {|\Sigma|}^{-1}(x_n)^H - k N \ln \pi$$ where $\mu=0$ as I am assuming zero mean r.v $x$

  • Confusion 1:

    I was thinking that the variance of $x$ was $$\Sigma = \begin{bmatrix}\sigma_1^2 & 0\\ 0& \sigma_2^2\end{bmatrix}\,\text.$$ So for $N$ samples (observation), the joint pdf would turn out to be $$P_x(x_1,x_2,...,x_N) = \prod_{n=1}^N\frac{1}{\pi \sigma^2_x} \exp \bigg(\frac{-{({x_n})}^H ({x_n})}{\sigma^2_x} \bigg)$$ where $\sigma^2_x = [\sigma_1^2,\sigma_2^2]$. Would there be a 2 in the denominator of the pdf and what is the correct pdf expression?

  • Confusion 2:

    In the document, the expression for the pdf for complex case looks different from the wikipedia link. Are they the same but maybe I am missing some link between them? Which pdf should I use?

  • Confusion 3:

    When simulating the r.v in Matlab with zero mean and variance 1, I am halving the variance because the real and imaginary part's variance should add up to 1. Then, would the pdf also contain half of the variance as $\sigma^2_x/2$?


Let me try to establish the relation between the univariate PDF for a real Gaussian and the univariate PDF for a complex proper (i.e. circular symmetric) Gaussian.

You know that
$p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-x^2}{2\sigma^2})$ is the PDF of a real-valued Gaussian with variance $\sigma^2$. We write $x\sim\mathcal{N}(0,\sigma^2)$ to denote that $x$ is a random variable that follows a real-valued Gaussian with zero mean and variance $\sigma^2$.

Now, let $z=a+jb$ be a circular symmetric random variable with real part $a$ and imaginary part $b$. In a circularly symmetric Gaussian random variable, the real and imaginary part are i.i.d., i.e. $a\sim\mathcal{N}(0,\sigma^2)$ and $b\sim\mathcal{N}(0,\sigma^2)$. Since $a,b$ are independent, their joint PDF is the product of their PDFs.


which gives the PDF of the complex variable $z$ at the value $a+jb$. We can reformulate this to $$\begin{align}p_z(z=a+jb)&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-(a^2+b^2)}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-z^*z}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-\|z\|^2}{2\sigma^2}\right)\end{align}$$

We also write for this case that $z$ follows a circular Gaussian distribution and denote this by $z\sim\mathcal{CN}(0,2\sigma^2)$. Why is it suddenly $2\sigma^2$ (i.e. the double variance compared to the uni-variate, real case? Well, because $E[\|z\|^2]=2\sigma^2$ since it consists of a real and an imaginary part (which are independent and hence their variance adds up together).

So, to sum up:

  • if $x\sim\mathcal{N}(0,\sigma^2)$, then $p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$
  • if $z\sim\mathcal{CN}(0,\sigma^2)$, then $p_z(z)=\frac{1}{\pi\sigma^2}\exp\left(-\frac{z^*z}{\sigma^2}\right)$

Does this answer your confusions?

  • 1
    $\begingroup$ This answer is clear and well-written. I just would like to add a little thing that a complex random variable is Gaussian if its real and imaginary parts are jointly Gaussian, by definition. The circular symmetry assumption implies these two parts are uncorrelated and identically distributed. By combining these two things, real and imaginary parts of a circularly symmetric Gaussian random variable, the real and imaginary part are identically distributed, and independent. $\endgroup$
    – AlexTP
    Apr 18 '17 at 11:54
  • $\begingroup$ Thank you so much, cannot express my gratitude in words here.....just to clarify 2 things from you - (1) if I consider design where the r.v $z \sim CN(0,2\sigma^2)$ then the pdf has the 2 in the denominator, otherwise if $z \sim CN(0,\sigma^2)$ then there is no 2 in the pdf's denominator. So, either design is acceptable. Is my understanding right? (2) notations - in my question I wrote the $[.]^H$ complex conjugate transpose but you have written * which is only the conjugate. Why the $[.]^H$ is wrong? $\endgroup$
    – Ria George
    Apr 18 '17 at 15:33
  • 1
    $\begingroup$ sure, if $z\sim CN(0,2\sigma^2)$, then there is the two in the denominator, if its $z\sim CN(0,\sigma^2)$, then it's not there. Here, the two comes from the description of the variance (e.g. if we had $z\sim CN(0,1.23456\sigma^2)$, then in the denominator there would be $1.23456$). For a scalar variable $z$, $[]^H$ and $[]^*$ are equal. When it comes to vector-variables, you'd need $[]^H$. However, note that in this case, also the denominator changes to $\pi^k\sigma^{2k}$, since the PDF is the product of the $k$ PDFs of each component (k is the vector size); each component has var. $\sigma^2$. $\endgroup$ Apr 18 '17 at 17:58
  • $\begingroup$ This answer has the correct ideas but is marred badly by a confusion of the concepts of probability and probability density. The probabiity that a continuous random variable equals a given number is $0$, regardless of the choice of number. Indeed, the probability that $z$ equals $a+jb$ (where we choose $a=b=0$) is claimed to be $$P(z=0+j0)=\frac{1}{2\pi\sigma^2}$$ which has value greater than $1$ if $\sigma^2 < \frac{1}{2\pi}$. -1 pending clean-up of the answer to be correct. $\endgroup$ Apr 18 '17 at 22:26
  • $\begingroup$ @DilipSarwate I agree to your concern, and I have corrected my misleading wording. I've now only used PDF and not probability any more. $\endgroup$ Apr 18 '17 at 22:37

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