Infinite sum of random variables

Question

Let $\xi_i$ be independent discrete random variables such that $P[\xi_i = k] = {1 \over 10}$ for all $i \in \mathbb{N}$ and $k = 0,\dots,9$. Define

$$ X = \sum_{i = 1}^\infty \xi_i{1 \over 10^i}. $$

What's the distribution of $X$?

The sum converges as every summand is less than ${10 \over 10^i} = {1 \over 10^{i-1}}$ and $\sum_{i=1}^\infty {1 \over 10^i}$ converges.

I let $X_i = {\xi_i\over 10^i}$. Then I have $X = \sum_{i = 1}^\infty X_i$. From this formula it's easy to see that every $X_i \in \{0, {1\over 10^i}, {2\over 10^i}, \dots, {9\over 10^i}\}$, ie. the $i$-th random variable contributes only to the $i$-th decimal place of the sum $X$ and does not affect other decimal places.

This immediately gives $X \in [0, 1)$. Also I think that $P[X=x] = 0$ for any $x$ in $[0, 1)$, suggesting "continuous behaviour" of $X$.

Could anyone provide hint which way to go from here? Maybe the Central limit theorem or Law of large numbers could give something.

Thanks.

Answer 1

Simply we have $$x=0\cdot \xi_1\xi_2\xi_3\xi_4\cdots$$which means that $\xi_i$s in companion make $x$. For example for $x=1$ we have $\xi_1=\xi_2=\xi_3=\cdots =9$. Since all the real numbers in the interval $[0,1]$ can be made using this method and $\xi$s are independent, the probability distribution of $X$ is uniform over $[0,1]$.

Answer 2

For example, you can find CDF of $X$. Using $X_i$, one can write $X$ as $0,X_1X_2X_3\ldots$. Each $x\in(0,\,1)$ can be represented as $$x=\sum\limits_{i=1}^{\infty}x_i\dfrac{1}{10^i}=0,x_1x_2x_3\ldots,$$ and $$ F_X(x)=P(X\leq x) = P(0,X_1X_2X_3\ldots \leq 0,x_1x_2x_3\ldots)= $$ $$ =P(X_1 < x_1) + P(X_1=x_1,\ X_2 < x_2) + P(X_1=x_1,\ X_2 = x_2, \ X_3< x_3) + \ldots + P(X_1=x_1,\ X_2=x_2,\ \ldots). $$ Calculate all probabilities and get answer.

Answer 3

I think this is quite a fun exercise :) but be aware of

SPOILERS

I would say that $X$ has a uniform distribution on $[0,1].$ In fact all you have to do is prove that $$(*) \ \ \ \ \ \mathbb{P}(X \in A) = \lambda(A),$$ where $\lambda$ is the Lebesgue measure.

Even less: you can restrict yourself to the following $\pi \ -$ system: take the $A$'s of the form $A = (p,q),$ where $p$ and $q$ are real numbers with finite nonzero decimals in the decimal expansion. Why is it a $\pi \ - $ system? Does it generate the borel sets?

Now for example: $\mathbb{P}(X \in (0.1, 0.2) \ ) = \mathbb{P}(\xi_1 = 1 ) = \frac{1}{10} = \lambda( \ (0.1, \ 0.2) \ )$

Answer 4

Since you want some hints I would do the following. Show that the moment generating function of each $\xi_i$ is $$M(t)=\begin{cases}1, & t=0 \\ \frac{1}{10}\left(\frac{e^{10t}-1}{e^t-1}\right), & t\neq0\end{cases}$$ using the geometric sum formula. Use this to show that the moment generating function of $X_n=\sum_{i=1}^n10^{-i}\xi_i$ is $$\phi_n(t)=\begin{cases}1,& t=0 \\ 10^{-n}\left(\frac{e^t-1}{e^{10^{-n}t}-1}\right),& t\neq0\end{cases}$$ where you may need to recognize that a certain product is in telescoping form. Finally show that $\phi_n$ converges pointwise to $$\phi_{\infty}(t)=\begin{cases}1,& t=0 \\ \frac{e^t-1}{t},& t\neq0\end{cases}$$ which is the moment generating function of a uniform random variable.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

For a given $\ds{N \in \mathbb{N}_{\geq 1}}$, the $\ds{X}$-distribution is given by $\ds{\pars{~\mbox{note that, in this case,}\ 0 \leq X \leq 10 - {1 \over 10^{N}}~}}$:

\begin{align} &\left.\sum_{\xi_{1} = 0}^{9}{1 \over 10}\cdots\sum_{\xi_{N} = 0}^{9}{1 \over 10}\right\vert_{\ X\ =\ \sum_{i = 1}^{N}\xi_{i}/10^{i}} = {1 \over 10^{N}}\sum_{\xi_{1} = 1}^{9}\cdots\sum_{\xi_{N} = 1}^{9} \bracks{z^{X}}z^{\sum_{i = 1}^{N}\xi_{i}/10^{i}} \\[5mm] = &\ {1 \over 10^{N}}\sum_{\xi_{1} = 1}^{9}\cdots\sum_{\xi_{N} = 1}^{9} \bracks{z^{X}}\pars{z^{\xi_{1}/10}z^{\xi_{2}/100}\cdots z^{\xi_{N}/10^{N}}} \\[5mm] = &\ {1 \over 10^{N}}\bracks{z^{X}}\pars{% \sum_{\xi_{1} = 0}^{9}z^{\xi_{1}/10}\sum_{\xi_{2} = 0}^{9}z^{\xi_{2}/100} \ldots\sum_{\xi_{N} = 0}^{9}z^{\xi_{N}/10^{N}}} \\[5mm] = &\ {1 \over 10^{N}}\bracks{z^{X}}\pars{{z - 1 \over z^{1/10} - 1} {z^{1/10} - 1 \over z^{1/100} - 1}\ldots {z^{1/10^{N - 1}} - 1 \over z^{1/10^{N}} - 1}} = {1 \over 10^{N}}\bracks{z^{X}}\pars{1 - z \over 1 - z^{1/10^{N}}} \\[5mm] = &\ {1 \over 10^{N}}\bracks{z^{X}}\pars{1 \over 1 - z^{1/10^{N}}} - {1 \over 10^{N}}\bracks{z^{X - 1}}\pars{1 \over 1 - z^{1/10^{N}}} \\[5mm] = &\ {1 \over 10^{N}}\bracks{z^{X}}\sum_{k = 0}^{\infty}z^{k/10^{N}} - {1 \over 10^{N}}\bracks{z^{X - 1}}\sum_{k = 0}^{\infty}z^{k/10^{N}} = \mrm{f}_{N}\pars{X} - \,\mrm{f}_{N}\pars{X - 1} \\ &\ \bbx{\ds{\mbox{where}\ 0 \leq X \leq 10 - {1 \over 10^{N}}}} \end{align}

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$$ \mbox{and}\quad \mrm{f}_{N}\pars{x} \equiv \left\{\begin{array}{lcl} \ds{1 \over 10^{N}} & \mbox{if} & \ds{10^{N}x\ \in\ \mathbb{N}_{\geq 0}} \\[2mm] \ds{0} && \mbox{otherwise} \end{array}\right. $$