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Given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, let $\{X_n: n \ge 1\}$ be a sequence of i.i.d random variables with the common distribution $$\mathbb{P}(X_1 = 1) = p \text{ and } \mathbb{P}(X_1 = 0) = 1-p.$$ Define a mapping $D: \Omega \to [0,1]$ by $$D = \sum_{n = 1} ^ {\infty} \frac{X_n}{2^n}.$$ Obviously $D$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $\mu_p$ and $F_p$ the law and the distribution function of $D$ respectively. (therefore, $\mu_p$ is a probability measure on $([0, 1] , \mathcal{B} ([0, 1])))$.

(i) Prove that $F_p$ is a continuous function.

(ii) Prove that $$\mu_p \left( \left\{x \in [0, 1]: \lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n E_n(x) = p \right\} \right) = 1,$$ where, for each $x \in [0,1]$ and $n\ge 1$ $$E_n(x) := \begin{cases} 1, & \text{ if } 2^{n-1}x - [2^{n-1}x] \ge 0.5 \\ 0, & \text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5\end{cases}, $$ where $[s]$ denotes the integer part of $s\ge 0$.

From this argue that if $p_1 \neq p_2$, then $\mu_{p_1} \perp \mu_{p_2}$ , and in particular, if $p \neq \frac{1} {2}$ , $\mu_p \perp \lambda$ where $\lambda$ is the Lebesgue measure on $\mathbb{R}$.

For the first part, one can argue that $F_p$ is continuous iff $\mu_p(\{x\}) = 0, \forall x \in [0, 1]$. On the other hand, $\mu_p$ for each interval $\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right], k = 0, \ldots 2^n$ can be readily shown to be $$\mu_p\left(\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right]\right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $n_1(n, k)$ and $n_0(n, k)$ are obtained by writing $k$ in the binary basis, padded with zeros on the left until it reaches $n$ binary digits, and counting the number of 1s and 0s. For example $\mu_p([0.25, 0.375]) = \mu_p([\frac{2}{2^3}, \frac{2+1}{2^3}]) = \mu_p([0.010, 0.011]) = p(1-p)^2$. Now for each $x \in [0,1]$ there exists $\{k_{x,n}, n\ge 1\}$ such that $\{x\} = \bigcap_{n=1}^{\infty} [\frac{k_{x,n}}{2^n}, \frac{k_{x,n} +1}{2^n}]$, which implies $$\mu_p(\{x\}) = \lim_{n \to \infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$

As for the second part, I understand that $\{E_n(x)\}$ gives the binary representation of $x \in [0,1]$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $p \neq 1$, $F_p$ is a continuous distribution function but $\mu_p$ is singular to the Lebesgue measure."

Any hint is appreciated.

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  • $\begingroup$ I don't know if this help. Have you thought of the measure of the set $$\bigg\{x\in[0,1]:\liminf_n\frac{1}{n}\sum_{j=1}^nE_n(x)<\alpha<\beta<\limsup_n\frac{1}{n}\sum_{j=1}^nE_n(x)\bigg\}$$ where $\alpha$, $\beta$ are rationals such that $\alpha<p<\beta$? $\endgroup$ – ei2kpi Nov 15 '18 at 13:47

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