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Let $([0, 1], B, m)$ be the given probability space where $B$ is the Borel σ-algebra and $m$ is the Lebesgue measure on $[0, 1]$. Construct a sequence of random variables ${\{X_i}\}_{i∈N}$ on the same probability space such that

a) for each $i, X_i$ takes values only $0$ and $1$ with equal probabilities,

b) for each pair $i \not= j, X_i$ and $X_j$ are independent random variables.

(c) Using ${X_i}$, construct a random variable $Y := \Sigma_{i≥1}(X_{2i}/2^i)$.

Find the distribution of Y .

The first two parts are easy. Define $X_i$ as $$X_i=\bigcup_{m=1}^{m=2^{i-1}}1_{[\frac{2m-1}{2^i},\frac{2m}{2^i}]} $$ but i don't know how to approach the third part of the problem. I know since these are i.i.d's I can write probability of any finite sum by considering whether each of the term in the sum will be $0$ or $1$. But how do i even begin to make a list of possible values of the random variable $Y$ since it will take values of the following series $$\Sigma_{i\geq1} \frac{x_i}{2^i}$$ with each $x_i$ possibly being $0$ or $1$.

confusedsam
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2 Answers2

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Let $X$ be a Bernoulli random variable of parameter $1/2$ and $Y$ be a uniform random variable supported on $[0,1]$. Let $\varphi_X (\xi)=\mathbb{E}[e^{i\xi X}]$ and $\varphi_Y(\xi)=\mathbb{E}[e^{i\xi Y}]$ be their characteristic functions.

Using the fact that $X_1, X_2,\dots$ is an i.i.d. sequence of Bernoulli random variables of parameter $1/2$, we have that the characteristic function of $Z:=\sum_{k=1}^\infty 2^{-k}X_k$ is \begin{align*} \varphi_{Z}(\xi) &= \mathbb{E}\bigg[\exp\bigg(i \xi \sum_{k=1}^\infty 2^{-k}X_k\bigg)\bigg] = \mathbb{E}\bigg[\prod_{k = 1}^\infty e^{i \xi 2^{-k} X_k}\bigg] = \prod_{k = 1}^\infty\mathbb{E}\big[ e^{i \xi 2^{-k} X_k}\big] = \prod_{k = 1}^\infty \varphi_X(2^{-k}\xi) \\ &= \prod_{k = 1}^\infty \Bigg( \exp \bigg(\frac{i \xi}{2^{k+1} }\bigg) \cdot \cos \bigg( \frac{\xi}{ 2^{k+1}}\bigg) \Bigg) = \exp \bigg( \sum_{k=1}^\infty \frac{i \xi}{2^{k+1}} \bigg) \cdot \prod_{k = 1}^\infty\cos \bigg( \frac{\xi}{ 2^{k+1}}\bigg) \\ &= e^{i \xi / 2 } \cdot \prod_{k = 1}^\infty\cos \bigg( \frac{\xi}{ 2^{k+1}}\bigg) = e^{i \xi / 2 }\cdot \frac{2 \sin(\xi/2)}{\xi} = \frac{e^{i \xi }-1}{i \xi} = \varphi_Y(\xi)\;, \end{align*} where the equality $\prod_{k = 1}^\infty\cos \Big( \frac{\xi}{ 2^{k+1}}\Big) = \frac{2 \sin(\xi/2)}{\xi}$ can be deduced as shown here.

Since $Z$ and $Y$ share the same characteristic function, they also share the same distribution. It follows that $Z$ is a uniform random variable supported on $[0,1]$.

Bob
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Consider for each $M \in \mathbb{N}$ and each $m\in\{1,\dots,2^M\}$ the half-open interval $I_{M,m} :=\big[\frac{m-1}{2^M},\frac{m}{2^M} \big)$. For each $n \in \mathbb{N}$, define $Z_n := \sum_{k=1}^n 2^{-k}X_k$ and $Z:= \sum_{k=1}^{\infty} 2^{-k}X_k$. Furthermore, let $U$ be a uniform random variable supported in $[0,1]$.

Notice that, since $\mathbb{P}[Z\in(0,1)] = 1 = \mathbb{P}[U\in(0,1)]$, to show that $Z$ has the same distribution of $U$, it will be enough to show that for every Borel measurable $A \subset (0,1)$ we have that $\mathbb{P}[Z\in A] = \mathbb{P}[U\in A]$.

Notice that, for each $M \in \mathbb{N}$, each $m \in \{1,\dots, 2^{M}\}$, and each $n \ge M$, the symmetric difference between the two events $\{Z_n \in I_{M,m}\}$ and $\{Z \in I_{M,m}\}$ is an event of probability zero, and so: \begin{equation*} \mathbb{P}[Z \in I_{M,m}] = \mathbb{P}[Z_n \in I_{M,m}] =\frac{1}{2^M} = \mathbb{P}[U \in I_{M,m}]\;, \end{equation*} where, to show that $\mathbb{P}[Z_n \in I_{M,m}] =\frac{1}{2^M}$, we have used the fact that $Z_n$ is a discrete uniform random variable on $\{0,\frac{1}{2^n},\dots,\frac{2^{n}-1}{2^{n}}\}$.

Since every open interval $(a,b) \subset (0,1)$ is a countable disjoint union of sets of the form $I_{M,m}$, we can deduce that, for each open interval $I \subset (0,1)$, it holds that $\mathbb{P}[Z \in I] = \mathbb{P}[U \in I]$. Since the $\sigma$-algebra generated by the open intervals contained in $(0,1)$ coincides with the Borel $\sigma$-algebra of $(0,1)$, it follows the for each Borel subset $A$ of $(0,1)$ it holds that $\mathbb{P}[Z\in A] = \mathbb{P}[U\in A]$.

Bob
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