2
$\begingroup$

I'm trying to understand the concept of complex-valued random variables, but I'm struggling. If you consider two real-valued random variables $U$ and $V$ with values $u$ and $v$ and the joint random variable $UV$ with values $(u,v)$ then under the following transformation of random variable $UV$ to random variable $ZW$ $$z=u+iv$$ $$w=u$$ the probability density function of $ZW$ is (Jacobian determinant =1) $$p_{ZW}(z,w)=p_{UV}(z-iw,w)$$ and the marginal probability density function $p_{Z}(z)$ is then given by $$p_{Z}(z)=\int_{-\infty}^{\infty}{p_{UV}(z-iw,w)}dw$$ Is this the probability density function of complex-valued random variable $Z$?

$\endgroup$
3
  • 1
    $\begingroup$ Without taking into account anything like a Jacobian at all, just begin by first defining what you mean by the probability density function of a complex-valued random variable $Z$ and perhaps relate it to the cumulative probability distribution function $F_Z(z)$. If $Z$ were a real random variable, then $$F_Z(z) = P\{Z \leq z\}.$$ What is the corresponding definition when $Z$ is complex? What is the meaning of $Z \leq z$ for complex numbers? $\endgroup$ Jul 19, 2012 at 18:52
  • $\begingroup$ This book tells me that complex-valued random variable $Z$ is considered to be the joint random variable of $U$ and $V$ that takes values $z=u+iv$. On the other hand they define $$F_{Z}(z)=P\{U\leq u,V\leq v\}$$ $\endgroup$
    – Wox
    Jul 20, 2012 at 8:42
  • $\begingroup$ This would be the definition of the cumulative probability distribution function of any joint random variable. However, this is a particular joint variable because $z=u+iv$. Hence I was considering the transformation of random variable $UV$ to include this particularity, but I'm sure this is wrong as I didn't find anything like it in the books I've read. $\endgroup$
    – Wox
    Jul 20, 2012 at 8:56

1 Answer 1

4
$\begingroup$

Complex number is treated as 2-d real vector here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .