Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let us consider the quadratic form

$$q_1 = \mathbf{x}_1^\mathrm{H} \mathbf{A}\,\mathbf{x}_1 $$

and the quadratic form $$ q_2= \mathbf{x}_2^\mathrm{H} \mathbf{A}\, \mathbf{x}_2 $$

where $\mathbf{x}_1\in\mathbb{C}^{N\times 1}$, $\mathbf{x}_2\in\mathbb{C}^{N\times 1}$ and the elements of both $\mathbf{x}_1$ and $\mathbf{x}_2$ are independent and identically distributed circularly-symmetric Gaussian random variables. The matrix $\mathbf{a}\in\mathbb{C}^{N\times N}$ is singular and idempotent. The superscript $(.)^H$ is the conjugate transpose.

I am confused about the independence of $q_1$ and $q_2$. For example if I apply Criag's theorem for the independence of quadratic forms, then $q_1$ and $q_2$ are not independent since $\mathbf{A}\mathbf{A}=\mathbf{A}\ne \mathbf{O}_N$, where $\mathbf{O}_N$ is the $N\times N$ null matrix. So, the question is are they independent?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Yes. $q_1$ and $q_2$ are independent because they are functions of independent random variables, respectively $x_1$ and $x_2$. You don't need Craig's theorem, which is about independence of different quadratic functions of the same random variable.

share|improve this answer
    
Thanks Jonathan –  Remy Oct 28 '12 at 15:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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