Sum and product of arcsine distributed random variables How can I prove if $X$ and $Y$ are independent arcsine distributed
random variables, then 
$$
X+Y-1\;\;\;\text{and}\;\;\;(2X-1)(2Y-1)
$$
 has the same distribution?
I realized the distribution of $X+Y-1$ is equivalent to the distribution
of 
$$
\frac{1}{2}\left[\sin\left(2\pi U_{X}\right)+\sin\left(2\pi U_{Y}\right)\right],
$$
 and the distribution of $(2X-1)(2Y-1)$ is equivalent to the distribution
of 
$$
\sin\left(2\pi U_{X}\right)\cdot\sin\left(2\pi U_{Y}\right),
$$
 where $U_{X}$ and $U_{Y}$ are independent random variables on $[0,1].$
I'm not convinced this is a right beginning to solve the problem,
but this is what I've got so far...
 A: This is a very nice problem (I will certainly borrow it), but a bit tricky. Nevertheless you have already done a very important step. There are few things left to mention: 


*

*If $U$ is uniformly distributed on $[0,1]$, then $\sin 2\pi U$, $\cos 2\pi U$, $-\cos 2\pi U$  have the same distribution. 

*If $U$ and $V$ are independent and uniformly distributed on $[0,1]$, then the fractional parts $(U+V)\mod 1$ and $(U-V) \mod 1$ are independent and uniformly distributed on $[0,1]$.

*$\sin x \sin y = \frac12 \big(\cos (x-y) - \cos(x+y)\big)$. 
Hence we have, with $\overset{\mathrm{d}}{=}$ denoting equality in distribution,
$$
\sin\left(2\pi U_{X}\right)\cdot\sin\left(2\pi U_{Y}\right) = \frac12 \big(\cos (2\pi(U_{X}-U_{Y})) - \cos(2\pi(U_{X}+U_{Y}))\big) \\
= \frac12 \big(\cos \big(2\pi((U_{X}-U_{Y})\mod 1)\big) - \cos \big(2\pi((U_{X}+U_{Y})\mod 1)\big)\big)\\
\overset{\mathrm{d}}{=} \frac12 \big(\cos (2\pi U_{X}) - \cos (2\pi U_{Y})\big)
\overset{\mathrm{d}}{=} \frac12 \big(\sin (2\pi U_{X}) + \sin (2\pi U_{Y})\big),
$$
as required.
