Product Convolution

If we have a random variable Z = XY, and we know the distributions for X and Y, then the distribution for Z is given by P_z(t) = ^\infty {s})}{|s|} \; ds it then seems that if X and Y are uniformly distributed between −1 and 1, that this results in the normalised distribution for Z being given as P_z(t) = {2} for two distributions between 0 and 1, then it seems that P_z(t) = -\log(t) and between 0 and r, P_z(t) = - {r} If we keep convolving the uniform distribution on the range −1 to 1, we get further distributions -{2}\\ {4}\\ -{12} \\ {48}\\ we can conclude that the product convolution of n uniformly distributed variables in this range has distribution p_z(t) = (-1)^{n+1}}{2 (n-1)!} now we might consider the determinant of a matrix A whose elements are in the range −1 to 1. Being of the form det(A)=ad − bc, we see that the random variable is given by d = AD - BC the distributions of AD and BC are the same, given by $-{2}$, so the total variable will be a convolution of these two P_d(t) = ^{1} P_{AD}(x)P_{BC}\left({-1}\right) \; dx P_d(t) = ^{1} P_{AD}(x)P_{BC}\left(x-t\right) \; dx P_d(t) = ^{1} {2}{2} \; dx this gives a piecewise distribution {4} \left( {cc} \{ & {cc} 2-{6} & y=-1\lor y=1 \\ \log (-y) (\log (y+1)-i \pi ) y-_2\left(1+{y}\right) y+_2\left(-{y}\right) y+2 y-2 (y+1) \log (y+1)+4 & -1