# Confusion regarding pdf of circularly symmetric complex gaussian rv

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.

$$p_z(z=a+jb)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-a^2}{2\sigma^2}\right)\cdot\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-b^2}{2\sigma^2}\right)$$

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 $p_z(z)=\frac{1}{\pi\sigma^2}\exp\left(-\frac{z^*z}{\sigma^2}\right)$

• 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? – Ria George Apr 18 '17 at 15:33
• 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. $\sigma^2$. – Maximilian Matthé Apr 18 '17 at 17:58
• 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. – Dilip Sarwate Apr 18 '17 at 22:26