# #StackBounty: #normal-distribution #multivariate-normal-distribution #linear-algebra #complex-numbers Density of a degenerate complex n…

### Bounty: 50

I’m looking for the properties of a degenerate complex normal distribution. Wikipedia has a section on degeneracy in the real multivariate normal case here, stating that the density (in a subspace of $$mathbb R^k$$ where the distribution is supported) looks like:

$$f(mathbf{x}) = (text{det}^{*}(2piSigma))^{-1/2} e^{-1/2 (x-mu)^TSigma^+(x-mu)}$$

… where $$text{det}^*$$ is a pseudo-determinant (product of non-zero eigenvalues) and $$Sigma^+$$ is the pseudo-inverse.

I have a couple of questions relating to whether or not such a formulation exists for the complex case.

Question 1: If $$Z sim mathcal{C}N(mu, Gamma, C)$$ where the support is $$mathbb C^2$$ (for example), then multiplication of $$Z$$ with a matrix $$Ain mathbb{C}_{3,2}$$ would yeild: $$AZ sim mathcal{C}N(Amu, AGamma A^H, ACA^T)$$ where $$AZ in mathbb{C}^3$$ and could, in general, be degenerate.

Assuming that $$A$$ is unknown, how would one find the subspace of $$mathbb{C}^{3}$$ where the density would have a support, and in this subspace, what would the density be? Would be be similar to the real case wherein the determinants and inverses are replaced by their pseudo-counterparts?

Question 2: On the wikipedia article, it is shown that the density of a complex normal can be written as:

$$f(z) = tfrac{sqrt{detleft(overline{P^{-1}}-R^{ast} P^{-1}Rright)det(P^{-1})}}{pi^n}, e^{ -(z-mu)^astoverline{P^{-1}}(z-mu) + operatorname{Re}left((z-mu)^intercal R^intercaloverline{P^{-1}}(z-mu)right)}$$

…where $$P, R$$ are functions of $$Gamma, C$$. I assume that this density is only valid if, for example, $$Gamma$$ is invertible. In a specific problem setting, I’ve encountered a likelihood that can be written in this form, but $$detleft(overline{P^{-1}}-R^{ast} P^{-1}Rright) = 0$$. Does this imply that the density is degenerate? Or is it the case that my likelihood isn’t a likelihood at all?

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