How to show Rademacher functions are independent.

Question

Rademacher functions are defined on [0,1] in the following way. Let $x= 0.d_1 d_2 d_3...$ , the binary expansion of x in [0,1], where the expansion is non terminating (so 1= 0.111111111.... and not 1). Then the kth radamacher functions is defined as $R_k(x)= -1$ if $d_k=0$ and $R_k(x)=1$ if $d_k=1$. The measure used is borel measure on [0,1]. We say $f$ and $g$ are independent if $m(f^{-1}(A) \cap g^{-1}(B))= m(f^{-1}(A)) m(g^{-1}(B))$ for all borel sets A,B.

I have tried to show any two Rademacher functions are independent but have not succeeded. I thought of splitting the cases where each function equals 1 and -1 and so on but it gets more and more complicated. Can anyone help me out? Thanks.

Answer 1

Note that $R_k$ only takes values $\pm 1$.

Note that for any $k$ and $v \in \{-1,1\}$ we have $m\{x \in [0,1] | R_k(x) = v \} = {1 \over 2}$.

Suppose $k\neq l$ and $v_k,v_l \in \{-1,1\}$, then note that $m\{x \in [0,1] | R_k(x) = v_k, R_l(x)= v_l \} = {1 \over 4}$ (this is the essence of the solution).

Since $R_k^{-1}(A) = R_k^{-1}(A \cap \{-1,1\})$, you only need to consider $A$ of the form $\emptyset, \{-1\}, \{1\}, \{-1,1\}$, and similarly for $B$. The first and last of these are straightforward, and the other cases can be addressed using the previous two remarks.

Answer 2

Here is a proof based binary expansion of numbers in the unit intervals and building a sequence of Bernoulli iid random variables.

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As in the posting of the OP states, every $x\in[0,1]$ has a unique binary expansion $x=\sum_{n\geq1}d_n(x)/2^n$ where $d_n(x)\in\{0,1\}$, and $\sum_{n\geq1}d_n(x)=\infty$ for $x>0$. Notice that for each $n\in\mathbb{N}$, the $n$--th bit map $x\mapsto d_n(x)$ defines a measurable function from $([0,1],\mathscr{B}([0,1]))$ to $(\{0,1\},\mathscr{P}(\{0,1\}))$. Therefore, the map $\beta:[0,1]\rightarrow\{0,1\}^{\mathbb{N}}$ given by $x\mapsto(d_n(x))$ is measurable. The next result is a mathematical formulation of tossing a fair coin.

Lemma A: Suppose $\theta\sim U[0,1]$, and for each $n\in\mathbb{N}$ let $X_n=d_n\circ\theta$. Then, $(X_n:n\in\mathbb{N})$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$. Conversely, if $(X_n)$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$, then $\theta=\sum_{n\geq1}2^{-n}X_n\sim U[0,1]$.

Here I present a proof of this Lemma. Suppose that $\theta\sim U(0,1)$. For any $N\in\mathbb{N}$ and $k_1,\ldots,k_N\in\{0,1\}$, \begin{align} \bigcap^N_{j=1}\{x\in(0,1]:d_j(x)=k_j\}&=(\sum^N_{j=1}\tfrac{k_j}{2^j}, \sum^N_{j=1}\tfrac{k_j}{2^j}+\tfrac{1}{2^N}]\\ \{x\in(0,1]: d_N(x)=0\}&=\bigcup^{2^{N-1}-1}_{j=0}(\tfrac{2j}{2^N},\tfrac{2j+1}{2^N}]\\ \{x\in(0,1]:d_N(x)=1\}&=\bigcup^{2^{N-1}-1}_{j=0} (\tfrac{2j+1}{2^N},\tfrac{2(j+1)}{2^N}] \end{align} It follows immediately that $ \mathbb{P}\big[\bigcap^N_{j=1}\{X_j=k_j\}\big]=\tfrac{1}{2^N}=\prod^N_{j=1}\mathbb{P}[X_j=k_j]$. Hence $\{X_n\}$ is a Bernoulli sequence with rate $\tfrac12$.

Conversely, suppose $\{X_n:n\geq1\}$ is a Bernoulli sequence with rate $\tfrac12$. If $\widetilde{\theta}\sim U(0,1)$ then, the first part shows that the sequence of bits $\{\widetilde{X}_n\}\stackrel{law}{=}\{X_n\}$. Therefore, \begin{align} \theta:=\sum_{n\geq1}2^{-n}X_n\stackrel{law}{=} \sum_{n\geq1}2^{-n}\widetilde{X}_n=\widetilde{\theta} \end{align} since $\theta$ is a measurable function of $\{X_n\}$. This concludes the proof of the Lemma.

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To conclude, consider in particular the probability space $(\Omega,\mathscr{F},\mathbb{P})=([0,1],\mathscr{B}([0,1]),m)$, where $m$ is Lebesgue measure in the unit interval. The Lemma above shows that the bits $d_n:[0,1]\rightarrow\{0,1\}$ in the binary expansion of numbers in $[0,1]$ form an i.i.d. sequence of Bernoulli $0-1$ random variables with parameter $p=1/2$. Hence, the sequence $$R_n(x)=2d_n(x)-1$$ forms an i.i.d. sequence of Bernoulli $\pm1$ random variables with parameter $1/2$.

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Observation: For each $n\in\mathbb{N}$, $R_n(x)=\operatorname{sign}(\sin(2^n\pi x))$ a.s. This is typically how the Rademacher sequence is defined in many Probability and Analysis textbooks.