Let $X$ be a continuous random variable having uniform distribution on $[0, 2\pi]$. What distribution has the random variable $Y=\sin X$ ? I think, it is also uniform. Am I right?
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Davide Giraudo
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Nikita Martynov
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1No, it is not. Loosely speking: Since the sine is "slow" near its extremes, if is more likely that $Y\approx 1$ than that $Y\approx 0$. – Hagen von Eitzen Nov 17 '14 at 18:14

https://math.stackexchange.com/q/118108/321264 – StubbornAtom Oct 10 '20 at 15:00
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By the Jacobian formula, $$f_Y(y)=\frac{\mathbf 1_{y\lt1}}{\pi\sqrt{1y^2}}.$$ This is the socalled Arcsine distribution, which famously appears in probability theory and in number theory, as shown by two mathematical giants from the 20th century:
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The support of the distribution is of course $[1,1]$. For $y\in[0,1]$, we have \begin{align} \Pr(Y\le y) & = \Pr(0\le X\le\arcsin y\text{ or }2\pi\ge X\ge \pi\arcsin y) \\[10pt] & = \frac{\arcsin y + (2\pi(\pi\arcsin y))}{2\pi}. \end{align} The density is the derivative of that, and that is not a constant function, so it's not uniformly distributed.
Michael Hardy
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