Let $X_1 , X_2,..$ be independent r.v's, defined on some probability space $(\Omega, \mathcal{F},P)$, s.t.
$$P(X_n = 0) = P(X_n = 1) =1/2 \text{ } \forall n$$ Let $U = \sum_{n=1}^{\infty}X_n 2^{-n}$.

(i) Show that the distribution of $U$ is the Lebesgue measure. (ii) Construct an independent sequence $U_1,U_2,..$ on $((0,1),\mathcal{B},\lambda)$ s.t. the distribution of each $U_i$ is $\lambda$ where $\lambda$ is Lebesgue.

My attempt: Let $E_n = \{ X_i = T_i | i=1,..,n \text{ } \& X_{n+1}<T_{n+1}\}$. To find the distribution of $U$, I need $P(X_1 X_2... < T_1 T_2...)$ where $X_i , T_i \in \{0,1\}$. Then $$\{X_1 X_2... < T_1 T_2...\} = \coprod_{n=1}^{n=\infty}\{E_n \cap \{T_{n+1}=1\}\}$$ where $\coprod$ is the disjoint union.
$$P\{E_n \cap \{T_{n+1}=1\}\} = 1_{\{T_{n+1}=1\}}.(\frac{1}{2})^{n+1}$$ and thus:
$$P\{X_1 X_2... < T_1 T_2...\} = \sum_{n=1}^{\infty}1_{\{T_{n+1}=1\}}.(\frac{1}{2})^{n+1}$$

Am I thinking in the right direction? and if yes, I'm not sure which set on $\mathbb{R}$ I should be looking at to take it's Lebesgue measure to be equal to the distribution obtained above? Thanks and appreciate a hint.

  • 1
    $\begingroup$ Show $P(U \in (a, b]) = b - a$ for $a$ and $b$ dyadic rationals and then use continuity of the probability $P(A_n) \to P(A)$ when $A_n$ monotonically converges to $A.$ $\endgroup$ – Will M. Mar 13 '19 at 15:40

As stated in the hint but using the fact that $(k/2^m , (k+1)/2^m]$ form a $\pi$ system, we can show that $P\{\sum_{n=1}^{\infty}\frac{X_n}{2^n}\in (k/2^m , (k+1)/2^m]\} = \frac{1}{2^m} = \lambda((k/2^m , (k+1)/2^m])$ and thus the two measures agree on $\mathcal{B}$. To see the above equality, there is only one way with positive probability s.t. $\sum_{n=1}^{\infty}\frac{X_n}{2^n}\in (k/2^m , (k+1)/2^m]$ where in $0's$ and $1's$ are assigned to $n=1,..,m$ and that happens with probability $1/2^m$.
$U$ comes from an infinite coin toss, so I think to construct independent $U_i$ whose distribution is Lebesgue, we can take independent infinite coin tosses.


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