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$X,Z,$ and $U$ are independent random variables. $X$ is arcsine distributed random variable with $$\frac{1}{\pi\sqrt{x(1-x)}}$$ pdf on $[0,1]$ ($0$ otherwise), $Z$ is a Bernoulli distributed random variable with $p=1/2,$ and $U$ is a uniform distributed random variable on $[0,1].$ How can I prove, that $$ UX+Z(1-X) $$ has the same distribution as $X?$

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Take a bounded function $f.$ So we have $$E[f(UX+Z(1-X))]=\int_{0}^{1}\frac{1}{2\pi\sqrt{x(1-x)}}(\int_{0}^1(f(ux)+f(ux+1-x))du)dx=\int_0^1\frac{1}{2x\pi\sqrt{x(1-x)}}(\int_0^xf(y)dy+\int_{1-x}^1f(y)dy)dx=\int_{0}^1f(y)(\int_y^1\frac{1}{2\pi x\sqrt{x(1-x)}}dx+\int_{1-y}^1\frac{1}{2\pi x\sqrt{x(1-x)}}dx)dy=\int_0^1f(y)\frac{1}{\pi\sqrt{y(1-y)}}(1-y+y)dy=\int_0^1f(y)\frac{1}{\pi\sqrt{y(1-y)}}dy.$$

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