Let $X,Y$ be i.i.d random variable having a density function $f(x).$ I consider the $k$- bit quantized version of $X$, i.e $\lfloor X \rfloor_k= \frac{\lfloor 2^kX \rfloor}{2^k}=\frac{m}{2^k}$ if $\frac{m}{2^k} \leq X < \frac{m+1}{2^k}$. Thus $X=\lfloor X \rfloor_k +a_k, a_k \in [0,\frac{1}{2^k})$. Similarly $Y=\lfloor Y \rfloor _k +b_k.$
Let $Z=X+Y$. Clearly, $\lfloor Z \rfloor_k\begin{align}&=& \lfloor X \rfloor_k +\lfloor Y \rfloor _k \ ,\ \text{if} \ 0 \leq \ a_k+b_k < 1/2^k \\ &=& \lfloor X \rfloor_k +\lfloor Y \rfloor _k + 1/2^k \ ,\ \text{if} \ 1/2^k \leq \ a_k+b_k < 2/2^k\end{align}$.
I am trying to prove that as $k\rightarrow \infty$, $\lfloor X \rfloor_k+\lfloor Y \rfloor_k$ and $a_k+b_k$ become independent. Can anyone guide me in this please ?
I've checked this fact for $X$ a uniform random variable in $(0,1).$ This is because a uniform random variable is same as infinite independent coin flips of probability $1/2$