$X(\omega)=\sum_{n=1}^{\infty} X_n(\omega)2^{-n}$ is a Uniform Random Variable?

Question

I am working on a problem on the infinite coin-tossing space and I'm having trouble making any meaningful progress.

Let $(\Omega, \mathcal F, P)$, where $\Omega=\{0,1\}^{\mathbb N}$, $F$ is the $\sigma$-field generated by the finite-dimensional sets, and $P$ is the coin(a fair coin) tossing probability. Finite dimensional sets are defined as sets $A=\{\omega\in\Omega\lvert(\omega_1,\omega_2,...,\omega_n)\in B\}$ for some $n\in\mathbb N$, $B\subset\{0,1\}^n$.

Define $X_n(\omega)=\omega_n \in\{0,1\}$ is the result of the nth toss and define $X(\omega)=\sum_{n=1}^{\infty} X_n(\omega)2^{-n}$.

• Show that the above series converges for every $\omega \in \Omega$ and defines a random variable taking values in $[0,1]$.
• Show that $X$ has uniform distribution on $[0,1]$.

My attempt:

• I believe the series converges (absolutely) by using the comparison test with $\sum_{i=1}^\infty 2^{-n}$, as $X_n\leq 1$, and $\lvert X_n2^{-n} \rvert\leq\lvert2^{-n}\rvert$. Now I'm hoping to show that $$E=\{\omega :X(\omega)\in (a,b)\}\in\sigma(F)$$ as if that holds, I can conclude that any Borel set will be in $\sigma(F)$ as well, since $\{(a,b):a,b\in[0,1]\}$ generates all the Borel sets in $[0,1]$. Now I can certainly find $\omega$'s that will be in $E$, since $X$ is a binary expansion of some real number in $[0,1]$, but I'm a bit stuck at this point.

• I was given a hint to find $P(X\in[i2^{-m},(i+1)2^{-m}])$ for $i=0,...,2^{m}-1$. I'm not sure how to find the probabilities, and how it can help to show that $X$ has a uniform distribution.

Any help on both questions would be greatly appreciated. Thank you.

Answer 1

The point is to use the connection with binary expansions.

Fix $n \in \mathbb{N}$ and consider the sum $A: = \sum\limits_{k=1}^n \frac{1}{2^k} X_k(\omega) = \frac{1}{2^n} \sum\limits_{k=1}^n 2^{n-k} X_k(\omega) $. Observe that the last sum is a binary representation of some integer $0\leq i \leq 2^n-1$, regardless of the outcomes $X_k(\omega)$. Due to the independence of $\{X_n\}$, for each fixed integer $0\leq i \leq 2^n - 1 $ we get $\mathbb{P}( A = \frac{i}{2^n}) = 2^{-n} $ since all bits $X_1(\omega),...,X_n(\omega)$ must coincide with the bits of the given integer $i$. On the other hand, for the rest of the sum in $X(\omega)$ we have $$ 0\leq \sum\limits_{k = n+1}^\infty \frac{1}{2^k} X_k(\omega) \leq \sum\limits_{k = n+1}^\infty \frac{1}{2^k} = \frac{1}{2^n}. $$ It follows that $\mathbb{P}(X(\omega) \in [\frac{i}{2^n}, \frac{i+1}{2^n}] ) = 2^{-n}$. From here we get that $\mathbb{P}(X \leq \frac{i}{2^n} ) = \frac{i}{2^n}$.

Now if $F(x)$ is the cdf (distribution function) of $X$, we have $F(x) = 0 $ for $x\leq 0$, $F(x) = 1$ for $x\geq 1$ by trivial estimates, and $F(x) = x$ if $x \in [0,1] $ and is of the form $\frac{i}{2^n}$ (dyadic rational) for some $n\in \mathbb{N}$ and $0\leq i \leq 2^n-1$. Finally, using the right continuity of $F$ and density of dyadic rationals in $[0,1]$ we conclude that $F (x) = x$ for all $x\in [0,1]$ and hence $X$ has uniform distribution in $[0,1]$.

Answer 2

The intuition behind the proof is as follows. Let us start from $P(X\in[0,2^{-1}])$; the event of interest can occur in 2 ways:

• $X_1=0$;
• $X_1=1$, and all $X_i=0$, $i>1$.

The 2nd way has zero probability and we ignore it, $$P(X\in[0,2^{-1}])=P(X_1=0)=2^{-1}$$

Arguing this way, we prove that $$P(X\in[i2^{-m},(i+1)2^{-m}])=2^{-m}$$ because the event of interest is a unique combination of the values of random variables $X_1,X_2,\ldots,X_m$. For example, $$P(X\in[2\cdot 2^{-2}, 3\cdot 2^{-2}])=P(X_1=1,X_2=0)=2^{-2}$$

Answer 3

my contribution in providing histogram of a 10,000 runs of 12 flips

Answer 4

We can use the Central Limit Theorem. Notice that if $U\sim \text{Unif}[0,1]$: $$\phi_U(t)=\int_0^1 e^{ixt}dx=\begin{cases}\frac{e^{it}+1}{it} & t\not=0\\ 1&t=0\end{cases}$$

For $t\not=0$, one has that: $$\phi_{\sum_{k=1}^n \frac{X_k}{2^k}}(t)=\prod_{k=1}^n\mathbb{E}e^{i \frac{X_k}{2^k} t}=\frac{1}{2^n}\prod_{k=1}^n (e^{i \frac{t}{2^k}}+1)=\frac{1}{2^n}\prod_{k=1}^n\frac{e^{i\frac{t}{2^{k-1}}}-1}{e^{i \frac{t}{2^k}}-1}=\frac{e^{it}-1}{2^n (e^{i\frac{t}{2^n}}-1)}$$

Notice that the denominator may be rewritten as: $$ t\left(\frac{\cos(t/2^n)-1}{t/2^n}+i\frac{\sin(t/2^n)}{t/2^n}\right)\xrightarrow{n\rightarrow \infty}it$$ When $t=0$, convergence is even easier as $\phi_{\sum_{k=1}^n}(0)=1=\phi_U(0)$. Therefore, if $Y_n:= \sum_{k=1}^n X_k/2^k $, we have proved that:

$$Y_n\Rightarrow U\quad \text{and}\quad Y_n\rightarrow X \text{ pointwise}$$

It is a general fact that pointwise convergence and convergence in distribution implies $\mathbb{P}_X=\mathbb{P}_U$.

Let us prove this. If $g$ is bounded and continuous, we have that:

$$\mathbb{E}g(X)=\mathbb{E}g(\lim Y_n)=\mathbb{E}\lim g(Y_n)=\lim \mathbb{E}g(Y_n)=\mathbb{E}g(U)$$

Where the equalities are respectively due to $Y_n$ converging to $X$ pointwise, continuity of $g$, the dominated convergence theorem, as $\|g\|_{\infty}<\infty$, and finally by convergence in distribution.

That $\mathbb{E}g(X)=\mathbb{E}g(U)$ for every bounded continuous function is enough to conclude they have the same distribution by this fact.