Let $X\sim Uni[0,1]$. Then define $X_n= \frac{\lfloor 2^n X\rfloor}{2^n}$. I want to show that this is uniformly distributed on $A=\{\frac{k}{2^n}: 0\leq k<2^n\}$ for all $n\geq 1$.
My Idea was the following. Let me remark that $$\frac{2^nX-1}{2^n}< \frac{\lfloor 2^n X\rfloor}{2^n}\leq \frac{2^nX}{2^n}$$then as $n\rightarrow \infty$ we get $$\frac{\lfloor 2^n X\rfloor}{2^n}\rightarrow X$$ Then for all $x\in A$ we have that $F_{X_n}(x)=\Bbb{P}\left( \frac{\lfloor 2^n X\rfloor}{2^n}\leq x\right)\rightarrow \Bbb{P}(X\leq x)$ using Fubini/Tonelli. So by portmanteau theorem we get $X_n\rightarrow X$ in distribution and hence $X_n$ has the uniform distribution.
Is this true or what am I doing wrong?
Best Answer
This seems much harder than it needs to be though.
Let us define $$X_n = 2^n \times \frac{\lfloor 2^n X \rfloor}{2^n}.$$ Then $X_n$ is clearly integral; the numerator $2^n \times \lfloor X \rfloor$ is both integral and is clearly an integral multiple of the denominator $2^n$. Furthermore, $X_n$ must be nonnegative [lest $X$ is negative], and it must be less than $2^n$ with probability $1$ [lest $X \ge 1$, note that as $X$ is uniformly distributed on $[0,1]$, that $X=1$ with probability $0$, and $X$ cannot be larger than $1$.] So the support of $X_n$ is $A_n =\{0,1,2,\ldots, 2^n-1\}$.
So now we show that $X_n$ is uniformly distributed on $A_n$. It is clear [due to the fact that $X$ is uniformly distributed on $[0,1]$] that, for each integer $n$ [no need to take limits], and each $k \in A_n = \{0 \le k < 2^n\}$, the following equation holds:
$$\mathbb{P}[X_n = k] = \mathbb{P}\Big[X \in \Big[\frac{k}{2^n},\frac{k+1}{2^n}\Big)\Big]$$ $$= \frac{(k+1)-k}{2^n} = \frac{1}{2^n},$$ where the "$=$" on the second line follows from the fact that $X$ is uniformly distributed on $[0,1]$. Then this gives $X_n$ uniformly distributed on $A_n$; indeed, for all $k \in A_n$, the probability that $X_n$ is $k$ is the same, namely $\frac{1}{2^n}$.