PDF of $X\sqrt{Y(1-Y)}$ if $X\sim \mathrm{Arcsine}[-2,2]$ and $Y\sim \mathrm{uniform}[0,1]$ are independent Let $X\sim \mathrm{Arcsine}[-2,2]$ and $Y\sim \mathrm{uniform}[0,1]$ be independent random variables on some probability space. I want to find the distribution of the random variable $Z=X\sqrt{Y(1-Y)}$. I know that due to independence, I know that in principle, I can write an integral to find the CDF of $Z$ and then differentiate to find the associated PDF but is there a simpler way in general and for this case in particular?
 A: 
Indeed, staying at the level of PDFs is often simpler. To deduce the PDF of $Z$ from the joint PDF of $(X,Y)$, the key tool is of course a change of variables formula but, to use it without pain, the simplest approach is to add the intermediate step of computing the joint PDF of $(Z,T)$ with $T=Y$, say (other choices for $T$ are possible, naturally), by the usual "Jacobian" technique, and then to deduce the PDF of $Z$ as a marginal.

More in details, the change of variable $$z=x\sqrt{y(1-y)}\qquad t=y$$ yields $$dzdt=\sqrt{y(1-y)}dxdy$$ hence $$dxdy=\frac{dzdt}{\sqrt{t(1-t)}}$$ and $$x=\frac{z}{\sqrt{t(1-t)}}\qquad y=t$$ hence $$f_{Z,T}(z,t)=f_X\left(\frac{z}{\sqrt{t(1-t)}}\right)f_Y(t)\frac{1}{\sqrt{t(1-t)}}$$ and, integrating this, $$f_Z(z)=\int_\mathbb Rf_X\left(\frac{z}{\sqrt{t(1-t)}}\right)f_Y(t)\frac{dt}{\sqrt{t(1-t)}}$$
Beware that the correct formulas for $f_X$ and $f_Y$ should include indicator functions, that is, $$f_X(x)=\frac{\mathbf 1_{|x|<2}}{\pi\sqrt{4-x^2}}$$ and $$f_Y(y)=\mathbf 1_{0<y<1}$$ Thus, $$f_Z(z)=\int_0^1\mathbf 1_{|z|<2\sqrt{t(1-t)}}\frac{1}{\pi\sqrt{4-z^2/(t(1-t))}}\frac{dt}{\sqrt{t(1-t)}}$$ that is,
$$f_Z(z)=\int_0^1\mathbf 1_{4t(1-t)>z^2}\frac{dt}{\pi\sqrt{4t(1-t)-z^2}}$$
The interval of integration is $4t^2-4t+z^2<0$, that is, $(2t-1)^2<1-z^2$, that is, $$\frac12(1-\sqrt{1-z^2})<t<\frac12(1+\sqrt{1-z^2})$$ This suggests the change of variable $t\to s$ defined by $$2t-1=s\sqrt{1-z^2}$$ which yields $$f_Z(z)=\mathbf 1_{|z|<1}\int_{-1}^1\frac{\frac12\sqrt{1-z^2}ds}{\pi\sqrt{1-z^2}\sqrt{1-s^2}}=\mathbf 1_{|z|<1}\int_{-1}^1\frac{ds}{2\pi\sqrt{1-s^2}}$$ which simplifies into $$f_Z(z)=\frac12\mathbf 1_{|z|<1}$$
