4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.