continuous RV from discrete RV

Question

So I am reading some notes in stochastic processes and I don't really understand the solution of this problem:

Problem:

Let $(\Omega,F,\mathbb{P})$ be a probability space where $\Omega$ is the set of all infinite sequences from $\{0, 1\}$. In other words $\Omega = \{(x_1, x_2, \cdots): x_i \in \{0,1\} \} $. For each $\omega =(x_1, x_2, \cdots)\in \Omega $ set $X_n(\omega)=x_n$. Let us now define the following random variable : $$X(\omega) = \sum^{\infty}_{n=1} \frac{X_n(\omega)}{2^n}.$$ Prove that $X$ is uniformly distributed on $[0,1]$

The main point of the problem is to tell that we can come up with continuous random variables, as is $X$ in the question by using discrete random variables (which is pretty cool!). I know how to prove that $X$ is a random variable but I don't understand how to prove that $\mathbb{P}[X\leq a]=a$. Can someone help me with that? Here in the notes is says that if $a$ is a dyadic fraction than $X(\omega) \leq a$ iff $x_j \leq a_j \forall j$, and that from here it is easy to see that for $0 \leq a \leq 1$ it follows that $\mathbb{P}[X\leq a]=a$. I don't really understand how does that follow, any hints?

Answer 1

This is essentially the problem of identifying $(\Omega, F, \Bbb{P})$ with the space $([0, 1], \mathcal{B}, \mathrm{Leb})$ equipped with the Lebesbue measure on Borel sets. (So if you can take it granted, then there is nothing to prove.)

A direct approach is also possible, but we give a more neat approach using characteristic functions:

\begin{align*} \Bbb{E}\exp\left\{ it \sum_{n=1}^{\infty} \frac{X_{n}}{2^{n}} \right\} &= \prod_{n=1}^{\infty} \Bbb{E} \exp\left\{ \frac{it X_{n}}{2^{n}} \right\} = \prod_{n=1}^{\infty} \Bbb{E} \exp\left\{ \frac{it X_{n}}{2^{n}} \right\} \\ &= \prod_{n=1}^{\infty} \left( \frac{1 + \exp\{ 2^{-n}it \}}{2} \right). \end{align*}

By noting that

$$ (1 - \exp\{2^{-N}it\}) \prod_{n=1}^{N} \left( \frac{1 + \exp\{ 2^{-n}it \}}{2} \right) = \frac{1 - e^{it}}{2^{N}}, $$

it follows that

$$ \prod_{n=1}^{\infty} \left( \frac{1 + \exp\{ 2^{-n}it \}}{2} \right) = \lim_{N\to\infty} \frac{1 - e^{it}}{2^{N}(1 - \exp\{2^{-N}it\})} = \frac{e^{it} - 1}{it}, $$

which is exactly the characteristic function for the uniform distribution on $[0, 1]$. Therefore the proof is complete.

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Here is a more direct approach. Let $D_{m} = \{ k/2^{m} : 0 \leq k \leq 2^{m} \}$ and $D = \cup_{m=1}^{\infty} D_{m}$ be the set of dyadic numbers on $[0, 1]$. Also we put

$$ Y = \sum_{n=1}^{\infty} \frac{X_{n+1}}{2^{n}}. $$

Then by the right-continuity of $x \mapsto \Bbb{P}(X \leq x)$, it suffices to prove that $\Bbb{P}(X \leq x) = x$ for dyadic $x \in D$. We prove that this is true for $x \in D_{m}$ for all $m$ by induction.

1. When $m = 1$, only the non-trivial case arises when $x = 1/2$. But in this case, we have $$ \Bbb{P}(X \leq 1/2) = \Bbb{P}(X_{1} = 0) + \Bbb{P}(X_{1} = 1 \text{ & } X_{2} = X_{3} = \cdots = 0 ) = 1/2. $$
2. Assume that the claim holds for $x \in D_{m}$. Then for $x \in D_{m+1} \setminus D_{m}$, we have either $x \in (0, 1/2)$ or $x \in (1/2, 1)$. By noting that $X_{0} \! \perp\!\!\perp \! Y$ and $Y \stackrel{\text{law}}{=} X$, in the former case $$ \Bbb{P}(X \leq x) = \Bbb{P}(X_{1} = 0 \text{ & } Y \leq 2x) = \Bbb{P}(X_{1} = 0) \Bbb{P}(X \leq 2x) = \tfrac{1}{2}(2x) = x $$ and in the latter case \begin{align*} \Bbb{P}(X \leq x) &= \Bbb{P}(X_{1} = 0) + \Bbb{P}(X_{1} = 1 \text{ & } Y \leq 2x-1) \\ &= \tfrac{1}{2} + \Bbb{P}(X_{1} = 1)\Bbb{P}(X \leq 2x-1) \\ &= \tfrac{1}{2} + \tfrac{1}{2}(2x-1) = x. \end{align*}

Therefore the conclusion follows by mathematical induction.