I am wondering what the distribution of
$$ y = exp(i\phi)$$
where $\phi$ is Rayleigh distributed and $i=\sqrt{-1}$.
My thought process to solve was to use LOTUS but I am unsure how to handle the $i$ as the arg into the Rayleigh distribution.
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What is $i$? (Integer? Imaginary unit?) – angryavian Feb 21 '19 at 21:07
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LOTUS ? Do you refer to a yoga position in order to get inspiration ? – Jean Marie Feb 21 '19 at 21:21
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@angryavian Yes. Updated question. – Avedis Feb 21 '19 at 21:26
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@JeanMarie LOTUS = https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician – Avedis Feb 21 '19 at 21:26
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Thanks. Not lost my day... My second thought was that LOTUS was a new software I wasn't aware of. – Jean Marie Feb 21 '19 at 21:50
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More seriously, there is a problem because your definition would imply that Rayleigh distribution (https://en.wikipedia.org/wiki/Rayleigh_distribution) has a $2 \pi$ period... Truncating it to $[0,2\pi]$ wouldn't fix the problem because by doing that, you would loose all the tail of the distribution... – Jean Marie Feb 22 '19 at 07:26
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@JeanMarie No, the phase is [0, inf] just like a rayliegh. – Avedis Feb 22 '19 at 14:11
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I understand now... – Jean Marie Feb 22 '19 at 21:38