Convergence in distribution of the digits of uniform random variables to one Uniform(0,1)

Question

Let $X_i$ be a discrete uniform random variable $X_i\sim \text{Uniform}\{0,1,2,\dots,9\}$, for $i=0,1,2,\dots$ and let $$Y_n = 0.X_1X_2X_3X_4\cdots X_n$$ Is it true that $Y_n$ converge in distribution (weak convergence) to a uniform distribution on $(0,1)$?s I hope you can help me, because I don't know how to prove this. Have a great day

Answer 1

They are indeed the same distribution. Try comparing their CDFs. Let $F$ be the CDF of the uniform distribution on $[0,1]$ given by $$ F(x) = \begin{cases} 0 & \text{if } x < 0 \\ x & \text{if } 0 \leq x \leq 1 \\ 1 & \text{if } x > 1 \end{cases} $$ Let $F_n$ be the CDF of $Y_n$. For example, since $Y_1$ is a uniform distribution on $\{0,0.1,0.2,\dots,0.9\}$, its CDF is $$ F_1(x) = \begin{cases} 0 & \text{if } x < 0 \\ 0.1 & \text{if } 0 \leq x < 0.1 \\ 0.2 & \text{if } 0.1 \leq x < 0.2 \\ & \vdots \\ 0.9 & \text{if } 0.8 \leq x \leq 0.9 \\ 1 & \text{if } x > 0.9 \end{cases} $$ Then similarly, the distribution of $Y_2$ is the uniform distribution on $\{0,0.01,0.02,\dots,0.99\}$. Then by observation, or inspection, you could notice the general equation of $F_n$ is given by $$ F_n(x) = \begin{cases} 0 & \text{if } x < 0 \\ \frac{1}{10^n}\text{floor}(10^nx) & \text{if } 0 \leq x \leq 1 \\ 1 & \text{if } x > 1 \end{cases} $$ which is basically a step function on $[0,1]$ with $10^n$ steps, which is always within $10^{-n}$ of the line $y=x$. To prove $\lim_{n\rightarrow\infty}F_n(x)=x$ for all $x\in[0,1]$, you must show for all $\varepsilon>0$ there exists natural number $N$ such that if $n\geq N$ then $$|F_n(x)-F(x)|<\varepsilon$$ This represents the error between real numbers in $[0,1]$ and their $n$-digit approximation, so it should not be surprising that a choice of $N=-\log_{10}(\varepsilon)$ is sufficient.