Equidistributed Random Variables[No Fancy Stuff]

Question

Let me first set up the background. It might be tedious but not fancy I promise.

Let the probability space be the interval $[0,1]$ of the real line. Let the sigma algebra be the sigma algebra generated by Lebesgue measurable sets on $[0,1]$. The probability measure is the Lebesgue measure on $[0,1]$.

Note that for every $\omega \in [0,1]$, we have the binary expansion as follows: $$\omega=\sum_{n=1}^\infty \beta_n 2^{-n},\,\text{where } \beta_n=\beta_n(\omega) \text{ which is either 0 or 1}.$$

Just by convention, we require one more condition that $\sum \beta_n = \infty$ except for $\omega=0$. This condition basically means that in the expansion there are infinitely many $1$'s.

Therefore, we can view $\beta_n$ as a function from $[0,1]$ to $\{0,1\}$, which is a random variable in our probability space.

For every fixed $j\in \mathbb N$, we may further define $$\omega_j=\omega_j(\omega)=\sum_{n\in \mathbb N} \beta_{m(n,j)}2^{-n},\text{ where }m \text{ is an arbitrary but fixed bijection between } \mathbb N^2 \mbox{ and } \mathbb N.$$

The author then wrote that, each of $\omega_j$ is equidistibuted on $[0,1]$, i.e., given a subinterval $I$ of $[0,1]$, the probability of the $\omega$-set where $\omega_j(\omega)\in I$ is the length of $I$.

My question is, why the probability of the $\omega$-set where $\omega_j(\omega)\in I$ is the length of $I$?

---

My thoughts: Given a fixed $\omega\in I$, we have its binary expansion, which determines all $\beta_n$'s. Fix $j\in \mathbb N$, $\{m(n,j)| n\in \mathbb N\}$ is a proper subset of $\mathbb N$, which means that, $\omega_j$ is defined by only using some of $\beta_n$'s. For those $n$ which are not used in defining $\omega_j$, even if we alter the values of the corresponding $\beta_n$ resulting in new $\tilde{\omega}$, we still have $\omega_j(\omega)=\omega_j(\tilde{\omega})$. Therefore, I guess we should have $|\omega_j^{-1}(I)|>|I|$, where $|\cdot|$ means Lebesgue measure rather than cardinality. (I know my reasoning is all about cardinality and it is different from measure but they are also somehow related, especially speaking of Lebesgue measure. My argument may not be sufficient, but I still somehow believe my guess that $|\omega_j^{-1}(I)|>|I|$ shall be correct).

---

Any comment or insight is welcome. Thanks!

Answer 1

Claim: $\beta_n$s are i.i.d with $P(\beta_n = 0) = P(\beta_n = 1) = 1/2$. Actually it seems slightly easier to show the converse, i.e if $(X_n)_{n \geq 1}$ is an i.i.d sequence uniformly distributed on $\{0, 1\}$, then $\sum_{n = 1}^{\infty}X_n2^{-n}$ is uniform on $[0, 1]$. Here it is easy to argue that the probabilities match for dyadic intervals of the form $[0, j2^{-n}]$, and then the $\pi-\lambda$ theorem implies that the probabilities match for all Borel subsets of $[0, 1]$.

Then the author's claim follows because $(\beta_{m(n, j)})_{n,j\in\mathbb{N}}$ are i.i.d.

Answer 2

First, we notice that $\beta_n$ is measurable. For this purpose let $q(b)=\sum_{k=1}^nb_k2^{-k}$ be the number for $b\in\{0,1\}^n$, and notice that $\{\omega\in[0,1]:\beta_n(\omega)=0\}=\bigcup_{b\in\{0,1\}^{n-1}}[q(b,0),q(b,1))$ is measurable. Further, notice that since $q(b,1)=q(b,0)+2^{-n}$, each interval has length $2^{-n}$ and hence $\mathbb P(\beta_n=0)=2^{n-1}\cdot 2^{-n}=0.5$. Next, we show that the distribution of $(\beta_n)_n$ is the product measure. Using the Ionescu-Tulcea Theorem, it is sufficient to show that $\beta_n$ is independent of $(\beta_k)_{k<n}$ for all $n$. But we have just seen that $\mathbb P(\beta_n=0|\beta_{[n-1]}=b)=0.5$ for all $b\in\{0,1\}^{n-1}$, where $\beta_{[n-1]}=(\beta_k)_{k\in[n-1]}$. Hence, $\beta_n$ does not depend on its predecessors and the joint distribution of $(\beta_k)_k$ is the product measure.

To see that $(\beta_{m(n,j)})_{n,j}$ has the same distribution, we may proceed accordingly. We could show this by using the order given by $n,j$. However, it is more instructive to observe that the order does not matter. For this purpose notice that $\mathbb P(\beta_2=b_2,\beta_1=b_1)=\mathbb P(\beta_1=b_1,\beta_2=b_2)=0.5\cdot 0.5$ and this clearly extends to any rearrangement of $\beta_{[n]}$. The fact that $(\beta_i)_{i\in\mathcal N}$ for a subset $\mathcal N\subseteq[n]$ is still i.i.d. Bernoulli with success probability $0.5$ can be seen by expanding the complete joint distribution and aggregating, say \begin{aligned} \mathbb P(\beta_1=b_1,\beta_3=b_3) &=\mathbb P(\beta_1=b_1,\beta_2=0,\beta_3=b_3)+\mathbb P(\beta_1=b_1,\beta_2=1,\beta_3=b_3)\\ &=0.125+0.125=0.25, \end{aligned} so, beyond all doubt, no matter what selection $\beta_{\mathcal N}$, they are i.i.d. Bernoulli with success probability $0.5$.

So, in particular any selection $(\beta_{m(n,j)})_{(n,j)\in\mathcal S}$, for any finite $\mathcal S\subseteq\mathbb N^2$, is also i.i.d. Bernoulli. And this means that the distribution of $(\beta_{m(n,j)})_{n}$ for fixed $j$ ist still the product measure given by the theorem above.

Next, we show that $\omega_j$ is measurable, which is all but obvious. But notice that $\lim_{n\rightarrow\infty}q((\beta_{m(k,j)})_{k\in[n]})=\omega_j$ from below, and the $q$'s are clearly measurable, being just linear combinations, and non-negative, so $\omega_j$ is measurable because it is both the pointwise limit and the supremum, and we obtain the distribution of $\omega_j$ using the monotone convergence theorem (since probabilities are also just integrals), yielding \begin{align*} \mathbb P(\omega_j\in\mathcal E)=\lim_{n\rightarrow\infty}\mathbb P(q((\beta_{m(k,j)})_{k\in[n]})\in\mathcal E). \end{align*} Now, we can use Theorem 1.23 with Remark 1.24 and Lemma 1.42 in Klenke's Probability Theory, i.e. it is sufficient to show that $\mathbb P(\omega_j\in(-\infty,a])=\int_{(0,1)\cap(-\infty,a]}\mathrm dx$. For $a<0$ and $a>1$ this holds by the above, so let $a\in[0,1]$. The sequence $P_n=\mathbb P(q((\beta_{m(k,j)})_{k\in[n]})\in(-\infty,a])$ gives weight $2^{-n}$ to each $b\in\{0,1\}^{n}$ with $q(b)\le a$. How many are there? Let $a=\sum_{k=1}^\infty b^*_k$, then we count from including $0$ to $(b^*_k)_{k\in[n]}$ in binary numbers, meaning that we count $\sum_{k=1}^n2^{n-k}b^*_k+1$ numbers, yielding $P_n=\sum_{k=1}^n2^{-k}b^*_k+n^{-1}$, which converges to $a$ by the above. This shows that $\omega_j$ is uniform.

Finally, we also want to see that $\omega_i,\omega_j$ are independent for $i\neq j$. For this purpose we consider the joint distribution $\mathbb P(\omega_i\in\mathcal E,\omega_j\in\mathcal F)$ and approximate it by $\mathbb P(q((\beta_{m(i,k)})_{k\in[n]}\in\mathcal E,q((\beta_{m(j,k)})_{k\in[n]}\in\mathcal F)$. Since the $\beta$'s are independent, this factorizes, and using the generator $(-\infty,a]\times(-\infty,b]$ we obtain the result analogously. This can be further extended to any finite selection $\omega_{\mathcal J}$, thus this result shows how we can create countably many i.i.d. copies of $\omega$.

EDIT In its essence, this construction is nothing but a countably infinite version of this construction (the first item).