I don't know if the title is accurate, but I have this problem:

I have iid RVs $Y_k$ that has a value from {0,1,...,9} with equal probability. I need to show that $$ X_n = \sum_{k=1}^{n}Y_k10^{-k} $$ converges to a uniform distribution between [0,1], as $n \rightarrow\infty$.

Can anyone help about this? Thanks in advance.


I've tried to use characteristic functions. I thought that maybe I could show that $$\Phi_{X_n}(\omega) = \frac{1}{j\omega}(e^{j\omega}-1)=\Phi_Y(\omega) $$ as $n\rightarrow\infty$, which is the characteristic function of a uniform RV.

Since $$ X_n = \frac{Y_1}{10^1}+\frac{Y_2}{10^2}+\frac{Y_3}{10^3}+...+\frac{Y_n}{10^n} $$ I know that pdf of $X_n$ can be found by convolution. Therefore, by multiplicating the characteristic functions of $Y_k$'s.

I calculated $\Phi_{Y_k}(\omega)= E[e^{j\omega\frac{Y_k}{10^k}}]$, which is actually $\Phi_Y(\frac{\omega}{10^k})$.

Finally, I get $$ \Phi_{X_n}(\omega)=\Phi_Y(\frac{\omega}{10})\Phi_Y(\frac{\omega}{10^2})\Phi_Y(\frac{\omega}{10^3})...\Phi_Y(\frac{\omega}{10^n}) $$

But I stuck here...

Do you thinks this is a valid way to go on or a dead end?

  • 1
    $\begingroup$ This is the base-10 version of Method 5 in my answer at stats.stackexchange.com/a/117711/919. I believe the demonstration is straightforward if you begin with a definition of a uniform distribution. Apply that definition to the distribution of $X_n.$ $\endgroup$ – whuber May 29 '18 at 18:14

You need to show that the c.d.f., $F_{X_n}(x)$ converges to $F(x)=x$ (the c.d.f. of a uniform distribution) as $n\rightarrow \infty$.

A way to prove it is to show that $F_{X_n}(x) = Pr(X_n \le x) = \lceil 10^nx \rceil/10^n$. Then you just need to argue that $\lceil 10^nx \rceil/10^n \rightarrow x$ as $n \rightarrow \infty$.

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