Let $X_n \sim$ Uniform discrete on $\{0,9\}$, find the convergence in law of $Y_n = \sum_{ i = 1}^n 10^{-i} X_i$

Question

Let $X_n \sim$ Uniform discrete on $\{0,9\}$ all independent. I want to find the convergence in law of $Y_n = \sum_{j = 1}^n 10^{-j} X_j$

Now, I want to use characteristic functions, so:

$$\Phi_{X_n}(t) = \frac{1}{10} \frac{e^{9it}-1}{9it}$$

$$\Phi_{Y_n}(t) = \Phi_{\sum_{i = j}^n 10^{-j} X_j} (t) = ... .$$

By independence:

$$ = \prod_{j = 1}^n \Phi_{X_j 10^{-j}}(t) = \prod_{j = 1}^n \frac{1}{10} \frac{e^{9i\frac{t}{10^j}}-1}{9i\frac{t}{10^j}}. $$

But now I am stuck as I do not know how to go on.

EDIT: Thanks to the comment I changed my CF and I got here:

$$ = \prod_{j = 1}^n \Phi_{X_j 10^{-j}}(t) = $$

$$ = \prod_{j = 1}^n \frac{1- e^{\frac{it}{10^{j-1}}}}{1-e^{\frac{it}{10^j}}} $$

But still I do not know how to go on

Answer 1

We find that $$\begin{eqnarray} \prod_{j = 1}^n \Phi_{10^{-j}\cdot X_j }(t) &=&\prod_{j = 1}^n \frac{1}{10}\frac{\exp\left(10^{-j+1} i t\right)-1}{\exp\left(10^{-j} i t\right)-1}\\&=& \frac{1}{10^n}\frac{\exp\left(i t\right)-1}{\exp\left(10^{-n} i t\right)-1}\\&\stackrel{n\to\infty}\longrightarrow&\frac{e^{it}-1}{it}. \end{eqnarray}$$ It remains to check $$ \Bbb E[e^{it U}]=\int_0^1 e^{itx}\mathrm{d}x =(1/it)(e^{it}-1). $$ Thus, by Levy's continuity theorem, $\sum_{j=1}^\infty 10^{-j}X_j \to_d U$ where $U$ is a uniform r.v. on $[0,1]$.

Answer 2

Notice, that $\{Y_n\}$ converges almost surely since the series $\sum\limits_{n=1}^\infty \frac{X_n}{10^n}$ is convergent everywhere on $\Omega$. Let $Y:\Omega \to [0,1]$ be the sum of the series. We show that $Y\sim U[0,1]$ (the sequence $(X_1,X_2,...)$ is simply the decimal expansion of $Y$, we exploit this fact below).

Take any $n\in \mathbb{N}$ and an integer $0\leq k\leq 10^n-1$, then $$ \mathbb{P}\left(Y\in \left[ \frac{k}{10^n}, \frac{k+1}{10^n} \right)\right) = \frac{1}{10^n}. \tag{1} $$ Indeed, we need to compute the probability of $k \leq 10^n Y (\omega) < k +1$, where $k$ and $n$ are fixed. By definition $$ 10^nY(\omega) = 10^n \sum\limits_{j = 1}^n 10^{n-j} X_j + 10^n\sum\limits_{j=n+1}^\infty \frac{X_j}{10^j} := A + B. $$ We have $B \leq 10^n \sum\limits_{j=n+1}^\infty \frac{9}{10^j} = 1$, but observe that $B<1$ unless all $\{X_j\}_{j=k+1}^\infty$ equal to $9$. The latter is a probability $0$ event, hence $\mathbb{P}(0\leq B<1) = 1$. We conclude that $$ A(\omega) \leq 10^n Y \leq A(\omega) + 1 $$ holds with probability $1$. But $A(\omega)$ is an integer, and for it to be equal to the fixed $k$, we need all $X_1,...,X_n$ to coincide with the decimal expansion of the integer $k$. In view of the independence condition, this has probability $10^{-n}$, hence $(1)$ follows.

Now, $(1)$ shows that for $F_Y(x)$, the distribution function of $Y$ we have $$ F_Y(x) = x \qquad \text{ if } x = \frac{k}{10^n} \in [0,1]. \tag{2} $$ Since $F_Y$ is right-continuous, and the set of points $\{k10^{-n}: n\in\mathbb{N}, k=0,1,...,10^{n}-1\}$ is dense in $[0,1]$ (these are points with finite decimal expansion) it follows from $(2)$ that $F_Y(x) = x$ for all $x\in[0,1]$, hence $Y\sim U[0,1]$.