Proving a distribution function is continuous at irrational points

Question

Let $\{x_1, x_2, \ldots\}$ be a collection of rational points from the interval $[0, 1]$. A random variable $X$ takes on $x_n$ with probability $1/2^{n}$. Prove the distribution function $F_{X}(x)$ of $X$ is continuous at every irrational point $x$.

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I am pretty sure this type of question will involve a density argument by using the fact that the irrationals are dense in $\mathbb{R}$ (I'm not completely sure -- this is just my guess from other problems).

I'm really not too sure how to proceed though because they also gave us a collection of rational points as well. I don't really see how to combine both of these facts to show continuity.

Also, some other problems I've solved use inequalities like Markov's and Chebyshev's, but I can't really do that here since there's no expectation.

I would greatly appreciate your help. I am trying to get better at these sort of problems.

Answer 1

Pick any irrational point and look at its neighborhood containing none of the $x_i$ (why must this exist). $F_X$ is constant in this neighborhood and hence is continuous

Edit: This also means $F_X$ is continuous at every point except the $x_i$

Answer 2

The argument here is actually analysis, not probability at all.

$$F_X(x):= P(X \leq x)=\sum_{n \geq 1}{2^{-n}1(x \geq x_n)}.$$

The series is normally convergent, and, if $x$ is irrational, each term is continuous at $x$. So the sum of the series is also continuous at $x$.

Answer 3

$F_X$ is an increasing right-continuous function. So it is continuous at a point $x$ iff $F_X(x)=F_X(x-)$ where $F_X(x-)$ is the left hand limit at $x$. Also, $P(X=x)=\lim P(X-\frac 1 n < X \leq x)=\lim [F_X(x)-F_X(x-\frac 1 n)]=F_X(x)=F_X(x-)$. Hence $F_X$ is continuous at $x$ iff $P(X=x)=0$. For $x$ irrational $P(X=x)=0$ so $F_X$ is continuous at $x$.

Answer 4

Let $A=\{r_{n}\mid n\in\mathbb{N}\}$ be a subset of $\mathbb{Q}\cap[0,1]$. Let $F:\mathbb{R}\rightarrow[0,1]$ be the cumulative distribution function of $X$. Note that $F(x)$ is explicitly given by $$ F(x)=\sum\left\{ \frac{1}{2^{n}}\mid r_{n}\leq x\right\} . $$ Let $x_{0}\in[0,1]\setminus\mathbb{Q}$. Let $\varepsilon>0$ be arbitrary. Choose $N\in\mathbb{N}$ such that $$ \sum_{n=N+1}^{\infty}\frac{1}{2^{n}}<\varepsilon. $$ Let $\delta=\min\{|r_{n}-x_{0}|\mid1\leq n\leq N\}>0$, then $(x_{0}-\delta,x_{0}+\delta)\cap\{r_{n}\mid1\leq n\leq N\}=\emptyset$. For any $x_{1},x_{2}\in(x_{0}-\delta,x_{0}+\delta)$ with $x_{1}<x_{2}$, we have that \begin{eqnarray*} 0 & \leq & F(x_{2})-F(x_{1})\\ & = & \sum\{\frac{1}{2^{n}}\mid r_{n}\in(x_{1},x_{2}]\}\\ & \leq & \sum\{\frac{1}{2^{n}}\mid r_{n}\in(x_{0}-\delta,x_{0}+\delta)\}\\ & \leq & \sum_{n=N+1}^{\infty}\frac{1}{2^{n}}\\ & < & \varepsilon. \end{eqnarray*} It follows that $F$ is continuous at $x_{0}$.