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Given the random variable $\nu(t)$ and given the function $$y(t)=sin \left(\pi\nu(t) \right)^2$$ how can I find the distribution of the $y(t)$ knowing that $\nu(t)$ is a gaussian white noise? Thanks

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If $Y=\sin^2(\pi X)$, then, for every $y$ in $(0,1)$, $$ [Y\leqslant y]=\bigcup\limits_{n\in\mathbb Z}[n-x(y)\leqslant X\leqslant n+x(y)], $$ for some $x(y)$, which solves $\sin^2(\pi x)=y$. Hence, $$ \mathrm P(Y\leqslant y)=\sum\limits_{n\in\mathbb Z}\int_{n-x(y)}^{n+x(y)}f_X(s)\mathrm ds, $$ where $f_X$ denotes the density of $X$, hence $f_X$ is standard gaussian in your case. More precisely, $x(y)=\pi^{-1}\arcsin(\sqrt{y})$ and $2\pi\sin(\pi x)\cos(\pi x)\mathrm dx=\mathrm dy$, hence the density $f_Y$ of $Y$ is defined on $(0,1)$ by the series $$ f_Y(y)=\frac1{2\pi\sqrt{y(1-y)}}\sum\limits_{n\in\mathbb Z}f_X(n+\pi^{-1}\arcsin(\sqrt{y}))-f_X(n-\pi^{-1}\arcsin(\sqrt{y})). $$

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