# Pdf of Difference of Random Variables

• I
Gold Member
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 convolution$$p_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-\infty}^{\infty}p_Y(\Delta y-u)p_Y(u)du$$results 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's$$p_{\Delta Y}=F^{-1}(\varphi_{\Delta Y})=F^{-1}(\varphi_Y^2)=\frac{F^{-1}\left(J_0^2(z)\right)}{4\pi^2}$$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 convolution$$p_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-\infty}^{\infty}p_Y(\Delta y-u)p_Y(u)du$$......
$$p_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-1}^{1}p_Y(\Delta y+u)p_Y(u)du$$
$$\int_{-1-\min(\Delta y,0)}^{1-\max(\Delta y, 0)}p_Y(\Delta y+u)p_Y(u)du$$