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I want to know the distribution of $z = \exp(j\varphi)$, with $\varphi \sim \mathcal{U}[-\pi;+\pi]$.

From the book "Probability, Random Variables and Stochastic Processes" by Papoulis and Pillai I figured the following: If $y=g(x)$ is a function of random variable $x$ with the density function $f_x(x)$, one can retrieve the density function $f_y(y)$ of $y$ by $f_y(y)=f_x(x)/|g'(x)|$. Therefore, if $g(x)=y=exp(x)$, this yields $f_y(y)=f_x(ln(y))/|exp(x)|$. In my case with $f_x(x)$ being a uniform distribution.

Unfortunately, I haven't figured out yet how to include this uniform distribution into the equation and even more important how to transfer the solution to complex variables.

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You are confusing real and complex exponentials, and the Jacobian formula you cite applies to real exponentials. Here $z$ is a complex-valued random variable and whether $z$ has a density or not depends on the target space one considers and the reference measure one puts on it. As a random variable with values in $\mathbb C\sim\mathbb R^2$ whose Borel sigma-algebra is endowed with the two-dimensional Lebesgue measure $\lambda_2$, $z$ has no density since $z\in S^1$ almost surely where the unit circle $S^1$ is such that $\lambda_2(S^1)=0$. But one can also consider $z$ as a random variable with values in $S^1$, whose Borel sigma-algebra is endowed with its own one-dimensional uniform measure $\sigma_1$. Then $z$ has density $1$ with respect to $\sigma_1$.

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  • $\begingroup$ Unfortunately I am absolutely not familiar with the concepts of Borel sigma-algebra or Lebesgue measures. Could you point me in the right direction here? $\endgroup$ – Michael Jan 30 '14 at 9:50
  • $\begingroup$ Distributions without measures? Well... $\endgroup$ – Did Jan 30 '14 at 10:32
  • $\begingroup$ That was very helpful, thank you. $\endgroup$ – Michael Jan 30 '14 at 11:11

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