Pdf of Difference of Random Variables

[marcusl]

I want to find the probability density function (pdf) of the difference of two RV's,
p_{\Delta Y} = p_{(Y_1 - Y_2)},where y = \sin \theta, and where \theta_1 and \theta_2 are random variables with the same uniform distribution p_{\theta}=\mathrm{rect}\left(\frac{\theta}{\pi}\right). This has support -\pi/2\leq\theta\leq \pi/2. (Please let me know if I am misusing terminology, as Math is not my native language )

I can derive the distribution of y p_Y=\frac{\mathrm{rect}\left(\frac{y}{2}\right)}{\pi \sqrt{1-y^2}}by starting from the uniform distribution. This is nonzero over |y| \leq 1. I get bogged down with the pdf of Δy, however. A direct approach involving the convolutionp_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-\infty}^{\infty}p_Y(\Delta y-u)p_Y(u)duresults in the square root of a quartic in the denominator, which seems like a dead end.

I instead tried characteristic functions. From tables of Fourier transforms (FT's), I find that the CF of p_Y is a Bessel function\varphi_Y(z)=\frac{1}{2\pi}J_0(z)p_{\Delta Y} is then the inverse FT of the product of two of these CF'sp_{\Delta Y}=F^{-1}(\varphi_{\Delta Y})=F^{-1}(\varphi_Y^2)=\frac{F^{-1}\left(J_0^2(z)\right)}{4\pi^2}but I cannot find the inverse FT of the square of the Bessel function. Can anyone help me finish this off?

 

[andrewkirk]

A direct approach involving the convolutionp_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-\infty}^{\infty}p_Y(\Delta y-u)p_Y(u)du...

That doesn't look quite right to me. I think it should be:
$$
p_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-1}^{1}p_Y(\Delta y+u)p_Y(u)du
$$
which is equal to:
$$
\int_{-1-\min(\Delta y,0)}^{1-\max(\Delta y, 0)}p_Y(\Delta y+u)p_Y(u)du
$$
I would be inclined to stop there and just use numerical integration. Expressing the result as a Bessel function may look more compact, but will still involve numerical integration to evaluate the Gamma function inside the Bessel sum.