Assume we are trying to classify a sample $X$ as coming from one of two distributions: $$ \mathcal{CN}(\mu, \sigma^2) \\ \mathcal{CN}(\nu, \sigma^2), $$ where $\mathcal{CN}$ denotes a (circularly symmetric) complex Gaussian distribution with total noise variance $\sigma^2$, meaning it has $\sigma^2/2$ variance across the real and imaginary components.

In the real-valued case, the classification performance depends only on the SNR of the problem, also called the detectability i.e. $$d' = \frac{\lVert \mu - \nu \rVert_2^2}{\sigma^2}.$$

Intuitively, this should be identical in the complex case as the total noise power is $\sigma^2$. However, when I do the likelihood ratio calculations treating the complex Gaussian as a bi-variate Gausian with covariance matrix $\pmatrix{\sigma^2/2 &0 \\0 & \sigma^2/2}$, the result comes out to be $$d' = \frac{2\lVert \mu - \nu \rVert_2^2}{\sigma^2}.$$

Now, I think something's wrong here. The total variance is $\sigma^2$ since the complex-valued Gaussian is not just a multivariate Gaussian and the real and imaginary part of the signal are "joined" in one complex number. Why does this factor $2$ pop up and how can I get rid of it? Is there a special way to calculate the detectability for a complex-valued problem where I have to use the trace of the covariance matrix in the denominator? If yes, can someone maybe point me to some relevant literature on the topic?

  • $\begingroup$ Didn't you ask essentially the same question in dsp.stackexchange.com/q/84242/235? $\endgroup$ Commented Aug 29, 2022 at 3:43
  • $\begingroup$ I would not say they are essentially the same when one deals with the KL divergence and the other with detectability and their only similarity is the factor 2 issue, but if you/the community feel strongly about it, I am happy to merge the questions. $\endgroup$
    – Sami
    Commented Aug 29, 2022 at 4:26

1 Answer 1


Maybe I'm missing something here but given that a circularly symmetric complex normal RV X can be broken down into two independent real normal RVs (I+jQ), the complex RV classification problem can be split up into two separate real classification problems, as you showed above - except that each now has variance $\sigma_2/2$. Thus the detectability for each problem is trivially scaled by 2.

However I guess that is still not the same as the detectability for the overall problem. But, given that the two problems are independent, I would think you can show something like:

$P(\text{correctly classify }X) = P(\text{correctly classify } I \text{ AND correctly classify } Q) = P(\text{correctly classify } I) \cdot P(\text{correctly classify } Q)$


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