$U\sim \mathcal U(0,a)\overset{?}{\implies}U-\lfloor U \rfloor \sim \mathcal U (0,a-\lfloor a \rfloor$)

Question

Suppose $U\sim \mathcal U(0,a)$ for some $a>0$. Is it true that $U-\lfloor U \rfloor \sim \mathcal U (0,a-\lfloor a \rfloor$)? How can I prove this?

If $a\in \mathbb N$ then the following calculation suffices ($t\in [0,1)$). $$P[U-\lfloor U \rfloor \leq t]=P\left(U\in\bigcup_{i=0}^{a-1}[i,i+t]\right)=\sum_{i=0}^{a-1}t\frac 1{a}=t$$

I'm not sure how to generalize it to $a>0$...

Answer 1

I think it's not true, It is true for any integer, but consider $a = 1.5$, then $$P(A-[A]\leq 0.5) = \frac23$$ Although it should have actually be $1$ if $(A-[A]) \sim(0,0.5)$