Is there any connection between these two distributions? I was playing with standard uniform distributions where I noticed a "weird" relation between two combinations and was wondering if there was an underlying reason for it (or if it is just coincidence).
Consider two variables $X \sim U(0,1)$ and $Y | X \sim U(0,x)$. The product distribution $Z = XY$ can be calculated and I find the result to be $f_Z(z) = \frac{1}{\sqrt z} -1$ when $z \in [0,1]$.
Consider now two standard uniform variables $U,V$ and a new variable $W = (U - V)^2$. This distribution can be computed analytically as well and I get $f_W(w) = \frac{1}{\sqrt w} - 1$ when $w \in [0,1]$.
Aside from this showing that two (different?) random variables ($W$ and $Z$) can share the same distribution, is there any connection between them that I am missing?
 A: There is a different representation. For $X \sim U(0,1)$ and $Y \sim U(0,1)$ we have that
$$Z = XY^2 \sim (X-Y)^2$$
with $f_Z(z) = \frac{1}{\sqrt{z}} - 1$
Below is a plot of isolines for the two cases $Z = XY^2$ (broken black lines) and $Z = (X-Y)^2$ (solid gray lines). The equations to draw these lines are $Y = \sqrt{Z/X}$ for black and $Y = X - \sqrt{Z}$ and $Y = X + - \sqrt{Z}$ for gray.

Let's focus on the complementary cumulative distribution $\bar{F}_Z(z) = 1 - {F}_Z(z)  = P(Z>z)$.
This probability is equal to the area bounded by those isolines.

*

*For the gray isolines, which are straight lines, we can compute it easily. It is the area bounded by a triangle of size $1-\sqrt(z)$ (and the isolines are on both sides so we have to multiply by two). $${F}_Z(z) = (1-\sqrt{z})^2$$


*For the black isolines, which have a bend, we can use an integral
$$\begin{array}{}
\bar{F}_Z(z) &=& \int_z^1 1- \frac{\sqrt{z}}{\sqrt{x}} dx \\ 
&=&  x - 2\sqrt{x}\sqrt{z} \,\,  \Biggr|_{z}^{1} \\
&=& (1 - 2\sqrt{1}\sqrt{z}) - (z - 2\sqrt{z}\sqrt{z}) \\
 &=& (1-\sqrt{z})^2
\end{array} $$
We see that the pattern is quite different but both of these have the same distribution.
I doubt that there might be an intuitive connection. One try that can be done is to see if some transformation from the points on the line $Z = XY^2$ to the line $Z = (X-Y)^2$ might be possible.
