Rademacher functions independent random Variables You call $r_n$=sign($sin^{2n}(\pi t))$ Rademacher functions. 
I read " the $r_n's$ are independant random variables"
Why is this true ? I don't get it...
 A: Read the Wikipedia Rademacher system article. I think you want
$\;r_n = \textrm{sign}(\sin(2^n 2\pi t))\;$ instead. The  $r_n$ are orthogonal as functions on $\;t\in[0,1]\;$ and and also the random
variables associated with them are independent.
I have no proof of that.
A: The question is almost $4$ years old, but apparently has still not received a correct answer. The reason why the Rademacher variables are independent is not because they are orthogonal, as was claimed in the accepted answer. (For functions whose mean is zero, being orthogonal is equivalent to being uncorrelated, and that in itself is not sufficient in general to guarantee independence). The reason is that whenever $m<n$, if we look at the set $P_m$ of points where $r_m=1$, then we can split it into two essentially disjoint subsets of equal measure $P_m^+$ and $P_m^-$, such that $f_n=1$ on $P_m^+$ and $f_n=-1$ on $P_m^-$. Similarly, if $N_m$ denotes the set where $f_m=-1$, we can split it into two subsets of equal measure $N_m^+$ and $N_m^-$ such that $f_n=1$ on $N_m^+$ and $f_n=-1$ on $N_m^-$. Now we can calculate the joint probabilities
$$\hbox{Prob}(f_m=1\wedge f_n=1)=\frac{1}{2}\hbox{Prob}P_m=\hbox{Prob}(f_n=1)\cdot\hbox{Prob}(f_m=1)$$
and similarly for the other possibilities
$$\hbox{Prob}(f_m=1\wedge f_n=-1)=\frac{1}{2}\hbox{Prob}P_m=\hbox{Prob}
(f_n=-1)\cdot\hbox{Prob}(f_m=1)$$
$$\hbox{Prob}(f_m=-1\wedge f_n=1)=\frac{1}{2}\hbox{Prob}N_m=\hbox{Prob}
(f_n=1)\cdot\hbox{Prob}(f_m=-1)$$
$$\hbox{Prob}(f_m=-1\wedge f_n=-1)=\frac{1}{2}\hbox{Prob}N_m=\hbox{Prob}
(f_n=-1)\cdot\hbox{Prob}(f_m=-1)$$
A: because the probability of its values are independent such that $p ( r_1=e_1 , r_2=e_2 , ... , r_n=e_n ) = p ( r_1=e_1 )  p ( r_2=e_2 ) .... p ( r_n=e_n )$ where $p$ is the probability measure and $e$ are borel subsets of the space , if this satisfied then $r_n$'s are independent random variables , you can take some specific values of n it would be much easier to prove it , GOOD LUCK.
