When $x$ is a zero mean random variable then

$$\sum_{n=1}^N x_n x_n^T = N \sigma^2_x\,\text,$$

where the variance is $\sigma^2_x$.

I'm considering Complex Normal Distributions where the real and imaginary part are uncorrelated.

https://en.wikipedia.org/wiki/Complex_normal_distribution explains about the form of the distribution for complex normal distribution.

I have a confusion because the denominator for the real case in the distribution has a 2 but for the complex that is not there.

Does this mean that if $x$ is a complex valued random variable, then the variance becomes half i.e., $$\sum_{n=1}^N x_n x_n^H = N \sigma^2_x/2\,\text,$$ where the variance is $\sigma^2_x/2$ because the variance gets equally distributed in the real and imaginary component?

I have this doubt because when implementing, if I need to generate a complex noise of variance 1, I would be doing (in Matlab)

noise = sqrt(1/2) * (randn(N,1) + 1j*randn(N,1))

Since each component (real and imaginary) needs to have variance 1/2, such that their sum becomes 1.

So, the variance $\sigma^2$ is half mathematically. Is my understanding correct?

UPDATE based on valuable information provided under the comments : I am considering at the circularly symmetric complex normal distribution, where real and imaginary part are completely uncorrelated.

  • $\begingroup$ (also, I have my slight problems with your initial statement. The CLT is a limit theorem, and hence, the $=$ is not generally true, unless you let $N\rightarrow\infty$). $\endgroup$ Apr 17, 2017 at 16:17
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    $\begingroup$ You are correct, in my case the real and imaginary part are uncorrelated. I don't know what the expression for the estimator of the variance or in general $\Sigma x_n x_n^H$ would be. Thank you for your followup. Appreciate your help. $\endgroup$
    – Ria George
    Apr 17, 2017 at 16:28
  • $\begingroup$ @MarcusMüller: Why did you delete your answer? Your answer was helpful, only thing I wanted to confirm in the comment which I wrote after your answer was if the expression $\Sigma x_n x_n ^H$ would evaluate to $N\sigma^2_x/2$ or not $\endgroup$
    – Ria George
    Apr 17, 2017 at 17:11
  • $\begingroup$ if the real and imaginary are uncorrelated $\endgroup$
    – Ria George
    Apr 17, 2017 at 17:16
  • $\begingroup$ Aside from the part that I got wrong, you can find any formula directly on Wikipedia's Normal Distribution article, so my answer has no value that you couldn't have found with minutes of reading, thus I deleted it :) $\endgroup$ Apr 17, 2017 at 17:34

1 Answer 1


I will focus on the reason of the factor $1/2$ and leave aside the estimation things.

The exact understanding should be : if a scalar Gaussian random variable (rv) is circular symmetric, its real and imaginary parts must be uncorrelated (this is equivalent to independence if they are assumed jointly Gaussian) and identically distributed with zero mean. Thus, your Matlab code is correct for a rv $\sim \mathcal{CN}(0,1)$.

The story behind is that a complex random variable (rv) is simply a vector of two real random variables. A vector of $n$ complex rv is indeed a vector of $2n$ real rv.

You are talking about the case $n=1$, 1 complex Gaussian rv $Z = Z_r + jZ_i$ or a vector of 2 real Gaussian rv $[Z_r,Z_i]^T$. As with real Gaussian rv which is described by its variance, a real vector $[Z_r,Z_i]^T$ must be described by its covariance matrix $$\mathbb{E}\left\lbrace [Z_r,Z_i]^T \times [Z_r,Z_i] \right\rbrace = \mathbb{E}\left\lbrace\begin{bmatrix} Z_r^2 & Z_rZ_i \\ Z_iZ_r & Z_i^2 \\ \end{bmatrix}\right\rbrace$$

Take a look at the variance $\mathbb{E}\left\lbrace ZZ^H \right\rbrace = \mathbb{E}\left\lbrace Z_r^2+Z_i^2 \right\rbrace$ and pseudo-variance $\mathbb{E}\left\lbrace ZZ^T \right\rbrace = \mathbb{E}\left\lbrace Z_r^2-Z_i^2+j2Z_rZ_i \right\rbrace$, the covariance matrix of real vector (or complex scalar) is fully described by the variance and pseudo-variance. Threfore, you need both variance and pseudo-variance to characterize a complex Gaussian rv (with the prior condition that its real and imaginary parts are jointly Gaussian).

Now we use the circular symmetry property : $e^{j\phi}Z$ has the same probability distribution as $Z$ for all real $\phi$. This leads to $\mathbb{E}\left\lbrace e^{j\phi}Z(e^{j\phi}Z)^T \right\rbrace = \mathbb{E}\left\lbrace e^{j2\phi}ZZ^T \right\rbrace = \mathbb{E}\left\lbrace ZZ^T \right\rbrace$ for all $\phi$ thus $E[ZZ^T] = 0$ and variance $\mathbb{E}\left\lbrace ZZ^H \right\rbrace$ is sufficient statistic for $Z$. Note that $\mathbb{E}\left\lbrace ZZ^T \right\rbrace = \mathbb{E}\left\lbrace Z_r^2-Z_i^2+j2Z_rZ_i \right\rbrace \implies \mathbb{E}\left\lbrace Z_r^2 \right\rbrace = \mathbb{E}\left\lbrace Z_i^2 \right\rbrace$ and $\mathbb{E}\left\lbrace Z_rZ_i \right\rbrace=0$, the real and imaginary parts are uncorrelated then independent (because they are Gaussian), with the same variance, this is the reason of the factor $1/2$.

To sum up, your code is correct because you are estimating circular symmetric complex Gaussian rv. The jointly Gaussian assumption between real and imaginary parts must be used. If this is not about circular symmetric rv (or real random vector with two elements), you must calculate the pseudo-variance.

For more details and to understand the formula of the wikipedia article, you can read this article Circular Symmetric Gaussian R.Gallager.

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    $\begingroup$ My answer is indeed a quick draft of the article rle.mit.edu/rgallager/documents/CircSymGauss.pdf that I find extremely intituitive to understand Gaussian random variable/vector/matrix. $\endgroup$
    – AlexTP
    Apr 17, 2017 at 20:41
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    $\begingroup$ Thumbs up for the reference $\endgroup$ Apr 17, 2017 at 20:47
  • $\begingroup$ Thank you for your reply. WOuld the pdf for the complex case be given by the pdf in the link en.wikipedia.org/wiki/… ? There the pdf has 2 in the denominator . To simplfy things, I was considering the bivariate case $\Sigma = [\sigma_1^2 , 0; 0, \sigma_2^2]$. So for $N$ samples, the joint pdf would turn out to be $P_x(x_1,x_2,...,x_N) = \prod_{n=1}^N\frac{1}{\pi \sigma^2_x} \exp \bigg(\frac{-{(({x}))}^H (({x}))}{\sigma^2_x} \bigg)$ Would there be a 2 in the denominator in the pdf expression? Is this correct? $\endgroup$
    – Ria George
    Apr 17, 2017 at 21:03
  • $\begingroup$ where $\sigma_1^2$ is the variance for real component and $\sigma_2^2$ of the complex rv $x$ $\endgroup$
    – Ria George
    Apr 17, 2017 at 21:04
  • $\begingroup$ I am sorry I think I don't get what you are talking. In the wikipedia article, it is stated clearly that the pdf is for the real vector : where is a real k-dimensional column vector*. In real vector, you have factor 2, in complex vector, pdf does not have factor 2. In the document I cited, the proof is clear and easy to follow. $\endgroup$
    – AlexTP
    Apr 18, 2017 at 6:14

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