If $\theta$ is uniformly distributed in $(0, 2\pi),$ then what is the distribution of $e^{i\theta},$ where $i = \sqrt{-1}?$ And what are the statistical properties of $\left[e^{i0\theta}\, e^{i1\theta}\, e^{i2\theta}\dots\, e^{i(N-1)\theta}\right],$ where $\theta$ is a single observation and $N$ some integer?


$e^{ik\theta}$, where $k$ is any non-zero integer, is uniformly distributed on the complex plane unit circle. The expected value of $e^{ik\theta}$ is $\displaystyle\frac{\int_0^{2\pi} e^{ik\theta} d\theta}{2\pi} = \frac{i - ie^{2\pi i k}}{k} = \frac{i - i}{k} = 0.$ The variance of $e^{ik\theta}$ is $1,$ because all values that may be observed are at a distance of $1$ from the expected value $0.$

$e^{i0\theta} = 1$ is constant.

If you by notation $(0, 2\pi)$ mean that values $0$ and $2\pi$ are never observed, then one may observe $e^{ik\theta} = 1$ only if $|k| > 1$, because the distribution wraps around passing the point $e^{ik\theta} = 1$ at $\theta = 2\pi/k.$

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    $\begingroup$ +1. This is one of those curious cases where the expected value of a random variable is actually a value that the variable can never actually take on. It follows from the symmetry of the unit circle about the origin. $\endgroup$ – Jason R Aug 10 '17 at 18:27
  • $\begingroup$ @JasonR For discrete random variables it's very common that the RV would never take its expected value. On the other hand for continuous random variables one might expect that RV takes every possible value including the expected one. That's probably the curious case. That would happen for example with a symmetric pdf with a gap around the part of its domain which includes the mean of the RV. $\endgroup$ – Fat32 Aug 10 '17 at 23:07
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    $\begingroup$ @Fat32 indeed, you're right. I was thinking of continuous-valued random variables. $\endgroup$ – Jason R Aug 10 '17 at 23:08

A more general solution can behad from the Von Mises Distribution $$ f(x\mid\mu,\kappa)=\frac{e^{\kappa\cos(x-\mu)}}{2\pi I_0(\kappa)} $$ When $\kappa=0$, it is equal to the uniform on $-\pi \le x \le \pi$

The circular moments $$ m_n = E\{ e^{\jmath n x} \}= \int_{-\pi}^{\pi} e^{\jmath n x}f(x\mid\mu,\kappa) dx =\frac{I_{|n|}(\kappa)}{I_0(\kappa)}e^{i n \mu} $$



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