- #1

marcusl

Science Advisor

Gold Member

- 2,762

- 412

[tex]p_{\Delta Y} = p_{(Y_1 - Y_2)},[/tex]where [itex]y = \sin \theta[/itex], and where [itex]\theta_1[/itex] and [itex]\theta_2[/itex] are random variables with the same uniform distribution [itex]p_{\theta}=\mathrm{rect}\left(\frac{\theta}{\pi}\right)[/itex]. This has support [itex]-\pi/2\leq\theta\leq \pi/2[/itex]. (Please let me know if I am misusing terminology, as Math is not my native language )

I can derive the distribution of

*y*[tex]p_Y=\frac{\mathrm{rect}\left(\frac{y}{2}\right)}{\pi \sqrt{1-y^2}}[/tex]by starting from the uniform distribution. This is nonzero over [itex]|y| \leq 1[/itex]. I get bogged down with the pdf of Δ

*y*, however. A direct approach involving the convolution[tex]p_{\Delta Y}(\Delta y)=p(y)*p(y)=\int_{-\infty}^{\infty}p_Y(\Delta y-u)p_Y(u)du[/tex]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 [itex]p_Y[/itex] is a Bessel function[tex]\varphi_Y(z)=\frac{1}{2\pi}J_0(z)[/tex][itex]p_{\Delta Y}[/itex] is then the inverse FT of the product of two of these CF's[tex]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}[/tex]but I cannot find the inverse FT of the square of the Bessel function. Can anyone help me finish this off?