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In this paper (page 6) I'm reading, the author has a uniform random variable $\zeta$ which takes on values between 0 and 1. He computes

$$ 2 \arccos( \sqrt{ 1 - \zeta } ) $$

But isn't that the same as computing $2 \arccos( \sqrt{ \zeta } )$, since $\zeta$ is uniform on [0..1]? Or is there a reason he may be doing that I'm missing?

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    They both have the same probability distribution. But if in a particular case $\zeta$ happens to be equal to $0.25$, then $1-\zeta$ will not be equal to $0.25$, so you can't say they're equal, but only that their distributions are equal.2011-08-09
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    Let $\eta = 1- \zeta$. Then $\eta$ and $\zeta$ will have the same probability distribution. But the correlation between $\zeta$ and $\eta$ is different from the correlation between $\zeta$ and $\zeta$. So it really depends on what the author goes on to use the expression for.2011-08-11
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    @Willie Wong: +1 I thought of an expectation sign there (though, I see that there were no reasons for it). On the other hand my answer is still correct besides the phrase "You're right".2011-08-11

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You are right, if you consider $\eta = 1-\zeta$ then the distribution of $\eta$ is also uniform on $[0,1].$