The Rademacher functions are an explicit iid sequence with Bernoulli law. Does it exist an explicit construction of an iid sequence with uniform law?
1 Answer
$\newcommand\N{\mathbb N}$Let $(r_k\colon k\in\N)$ be the sequence of the Rademacher functions. Re-enumerate this sequence into a two-way array $(r_{i,j}\colon(i,j)\in\N^2)$. So, the $r_{i,j}$'s are iid random variables (r.v.'s, defined on the standard probability space over the interval $[0,1]$) each uniformly distributed on the two-point set $\{-1,1\}$.
For each $i\in\N$, define the r.v. $U_i$ on the standard probability space over the interval $[0,1]$ by the formula $$U_i:=\sum_{j=1}^\infty\frac{r_{i,j}}{2^j}.$$ Then the $U_i$'s are iid r.v.'s each uniformly distributed on $[-1,1]$.
Moreover, if $F$ is any cumulative distribution function (cdf) and $X_i:=F^{-1}(\frac{1+U_i}2)$ for $i\in\N$, then the $X_i$'s are iid r.v.'s each with cdf $F$. Here, as usual, $$F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\}$$ for $u\in(0,1)$.
So, one can explicitly get the distribution of any sequence of iid real-valued r.v.'s from the single sequence of the Rademacher functions.
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$\begingroup$ yes, concerning your edit: that was precisely my purpose , to have a general construction for an arbitrary pdf, via composition $\endgroup$ Commented Aug 17, 2022 at 20:46
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$\begingroup$ @PieroD'Ancona : I am glad this was of help. $\endgroup$ Commented Aug 17, 2022 at 21:28