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Let $X$ be a continuous random variable with uniform distribution between $0$ and $1$. Compute the distribution of $Y = \sin(2\pi X)$.

$\sin(2\pi \cdot0)$ and $\sin(2\pi \cdot1) =0$. So, the inverse image of the function has multiple roots. How can I find the PDF of $Y$ then?

Tianlalu
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Christy
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  • Since $2\pi X\sim U(0,2\pi)$, see the old post https://math.stackexchange.com/questions/1026366/distribution-of-sine-of-uniform-random-variable-on-0-2-pi?rq=1. – StubbornAtom Oct 31 '18 at 16:42

2 Answers2

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Since $X$ lies between $0$ and $1$ you can explicitly know what the two values of $X$ are if $Y = y \in [-1, 1]$ and hence, determine the probability of $Y \in [-1, y].$

William M.
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  • could you please give the final answer since I am new to this subject. My apologies for not being able to understand the sub-steps involved. – Christy Oct 31 '18 at 16:33
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By symmetry, we can restrict $X\sim U(-0.25,0.25)$, since $X$ is uniform and the function sending it to $Y$ reflects on any half-integer multiple of 0.5. Then the inverse is well-defined: $$Y=\sin 2\pi X\implies X=\frac{\sin^{-1}Y}{2\pi}$$ $$J=\frac{\partial X}{\partial Y}=\frac1{2\pi\sqrt{1-Y^2}}$$ $$f_Y(y)=\frac1{\pi\sqrt{1-y^2}}\qquad y\in(-1,1)$$ where the multiplication by 2 in the last step is because the new $X$ has range 0.5. $Y$ is a scaled and shifted version of the arcsine distribution (or the arcsine distribution itself, depending on convention).

Parcly Taxel
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