What Is the PDF of a Circularly Symmetric Complex Gaussian Vector?

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The discussion centers on the probability distribution function (PDF) of a zero-mean circularly symmetric complex Gaussian vector. It clarifies that a complex Gaussian vector is considered circularly symmetric if its real and imaginary parts form a Gaussian distribution with a specific covariance structure. The PDF is derived as f_{z}(z) = (1/π^n det(R_z)) exp{-(z - z̄)^H R_z^{-1}(z - z̄)}. There is also a query about extending this concept to circularly symmetric complex Gaussian matrices, suggesting that matrices can be treated as vectors in this context. The conversation highlights the importance of understanding the covariance matrix and its implications for the distribution.
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Hi,

Suppose that an n-dimensional vector \mathbf{z}=\begin{pmatrix}z_1&z_2&\cdots & z_n\end{pmatrix}^T is characterized as a zero-mean circularly symmetric complex Gaussian random vector. What is the distribution (the probability distribution function PDF) of this vector in both: complex and real representations?

Thanks in advance
 
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What is the meaning of "circularly symmetric" in n dimensions ? Do you mean "spherically symmetric" in the sense that |z_1|^2 + |z_2|^2 + ...|z_n|^2 = c is a surface where the PDF is constant? And can we also assume that |RE(z_1)|^2 + |IM( z_1)|^2 + |RE(z_2)|^ + |IM( z_2)|^2 + ... |RE( z_n)|^2 + |IM (z_n)|^2 = c is a surface where the PDF is constant?

Is |z| the quantity that has a gaussian distribution?
 
Stephen Tashi said:
What is the meaning of "circularly symmetric" in n dimensions ? Do you mean "spherically symmetric" in the sense that |z_1|^2 + |z_2|^2 + ...|z_n|^2 = c is a surface where the PDF is constant? And can we also assume that |RE(z_1)|^2 + |IM( z_1)|^2 + |RE(z_2)|^ + |IM( z_2)|^2 + ... |RE( z_n)|^2 + |IM (z_n)|^2 = c is a surface where the PDF is constant?

Is |z| the quantity that has a gaussian distribution?

A complex random vector \mathbf{x}\in C^n is said to be Gaussian, if the real vector \mathbf{\hat{x}}\in R^{2n} consisting of the real and imaginary parts of \mathbf{x} as \mathbf{\hat{x}}=\begin{pmatrix}\text{Re}\{\mathbf{x}\}&\text{Im}\{\mathbf{x}\}\end{pmatrix}^T is Gaussian.

A complex Gaussian random vector \mathbf{x} is said to be circularly symmetric if the covariance of the corresponding \mathbf{\hat{x}} has the following structure:

E(\left(\mathbf{\hat{x}-\mu}\right)\left(\mathbf{\hat{x}-\mu}\right)^H)=\frac{1}{2}\begin{pmatrix}\text{Re}(Q)&-\text{Im}(Q)\\\text{Im}(Q)&\text{Re}(Q)\end{pmatrix}

where E(\mathbf{\hat{x}})=\mu and Q is some non negative matrix.
 
If I take a Gaussian distribution of the components of a vector to mean a multivariate Gaussian distribution and take the covariance matrix as given, is the question "What is the PDF of a multivariate Gaussian distribution?". I guess I still don't understand the question.
 
Stephen Tashi said:
If I take a Gaussian distribution of the components of a vector to mean a multivariate Gaussian distribution and take the covariance matrix as given, is the question "What is the PDF of a multivariate Gaussian distribution?". I guess I still don't understand the question.

That is right, but for complex Gaussian. Actually, I got the result, which is:

f_{\mathbf{z}}(\mathbf{z})=\frac{1}{\pi^n\text{det}(\mathbf{R}_z)}\text{exp}\left(-(\mathbf{z}-\overline{\mathbf{z}})^H\mathbf{R}_z^{-1}(\mathbf{z}-\overline{\mathbf{z}})\right)

Now the problem with me was that, I read in some paper that the distribution is given by:

f_{\mathbf{z}}(\mathbf{z})=\frac{1}{\text{det}(\pi\mathbf{R}_z)}\text{exp}\left(-(\mathbf{z}-\overline{\mathbf{z}})^H\mathbf{R}_z^{-1}(\mathbf{z}-\overline{\mathbf{z}})\right)

But knowing that:

\text{det}(cA)=c^n\text{det}(A)

solved the confusion.

Thanks
 
Last edited:
Ok, now what if Z is a circularly symmetric complex Gaussian matrix not vector, what then the PDF of Z?
 
Any suggestion?
 
From the appropriate point of view, matrices are vectors. What property would an nxn "circularly symmetric" matrix have that an n^2 dimensional circularly symmetric vector wouldn't?
 
Stephen Tashi said:
From the appropriate point of view, matrices are vectors. What property would an nxn "circularly symmetric" matrix have that an n^2 dimensional circularly symmetric vector wouldn't?

So, you are saying it is just like the vector case. But I have a formula in matrix form, and I am not sure how the authors got there. I mean it is like the following:

\frac{1}{\pi^{2NK}\text{det}^KQ}\text{exp}\left\{-\|Q^{-1/2}(Y-HX)\|_F^2\right\}

where Y is 2N-by-K matrix, Q is 2N-by-2N, H is 2N-by-M, and X is M-by-K. Any hint in this?
 

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