Consider $X,Y,Z \in [0,1]$ be three independent uniform random variables.
It's easy to show that $A=max\{X,Y\}$ and $B=\sqrt{Z}$ have same distribution.
But I want a direct proof that completely couples these two random variables without computing their distributions...
Has any one know any way to do this?
Consider sampling points in the unit square $ X \times Y$.
Then $A=\max\{X,Y\}$ is formed by the rule of taking the $y$-coordinate for points above the $y=x$ line and the $x$-coordinate for points below it.
Now consider the rule where we reject points in the upper triangle and consider only $x$-coordinates from the lower triangle. The (conditional) probability of getting an $x$ less than a given $a$ is ratio of the area of the triangle $[(0,0), (a,0), (a,a)]$ to the entire lower triangle $[(0,0), (1,0), (1,1)]$, which is simply $a^2$. The same is true of $y$ by symmetry for the upper triangle. But this is the same as the distribution for $\sqrt{Z}$:
$$F_{\sqrt{Z}}(a) = P(\sqrt{Z} < a) = P(Z < a^2) = a^2$$
Figure: Lines illustrate the rule for $A = \max\{X,Y\}$. Shading illustrates calculation of probability for $\sqrt{Z}$.