Does there exist a **continuous** (differentiable) function $h:[0,1]\times [0,1] \to [0,1]$ such that if $\alpha,\beta\in [0,1]$ are independent and uniformly distributed on $[0,1]$, the random variable $h(\alpha,\beta)$ is uniformly distributed on $[0,1]$ **independent** of $\alpha,\beta$?


**Clarification:** By independent I mean pairwise independent, i.e.


$\mathbb{P}[h(\alpha,\beta)\leq x\mid \alpha]=x$ for all $x,\alpha\in[0,1]$ 


and


$\mathbb{P}[h(\alpha,\beta)\leq x\mid \beta]=x$ for all $x,\beta\in[0,1]$..


Thanks a lot!



Constructing an independent uniform random variable from two independent ones
I think this works for a continuous $h$.


Let $f : \mathbb{R} \to [0,1]$ be the "triangle wave" function given on $[0,1]$ by
$$f(u) = \begin{cases}1-2u, & 0 \le u \le \frac{1}{2} \\ 2u-1, & \frac{1}{2} \le u \le 1 \end{cases}$$
and extended periodically. Note that for any $t \in [0,1]$ we have $$m(\{x \in [0,1] : f(x) \le t\}) = m\left(\left[\frac{1-t}{2}, \frac{1+t}{2}\right]\right) = t.$$
By the translation invariance of $m$ and the 1-periodicity of $f$ it also follows that for any $y \in \mathbb{R}$ we have $m(\{x \in [0,1] : f(x-y) \le t\}) = t$. So take $h(x,y) = f(x-y)$. Then if $\alpha, \beta$ are iid $U(0,1)$ the above computation says precisely that $P(h(\alpha, \beta) \le t \mid \beta) = t$, i.e. $h(\alpha, \beta)$ is $U(0,1)$ and independent of $\beta$. Since $f$ is an even function we have $h(x,y) = h(y,x)$ and thus by symmetry $h$ is also independent of $\alpha$.


I'm not sure how to get a differentiable example, though.


