Distribution of a fractional part of the sum of uniform RVs I had a question in class not long ago which I couldn't solve. I've been digging into it for a few hours now but I can't find the right direction. So the question is: 

Let $ U_1,..,U_n$ be I.I.D random variables with uniform distribution : $ U_i \sim U[0,1] $. I need to prove that the fractional part of their sum - $\sum_{i=1}^n U_i $ , is also uniformly distributed. 

I know that this sum has the Irwin-Hall distribution, but not sure how it helps me. 
Any ideas on this one?
Thanks.
 A: A simple argument by induction will also work.  If we define $\{x\} = x - \lfloor x \rfloor$ as the fractional part of $x$, then proving that $\{U_1 + U_2\} \sim U \sim \operatorname{Uniform}(0,1)$ will then allow you to inductively show the main result, because $$\{U_1 + \cdots + U_n\} = \{\{U_1 + \cdots + U_{n-1} \} + U_n \}$$ and each $U_1, \ldots, U_n$ are iid.  So all that remains is to prove the base case, which is quite straightforward as an explicit integral.  First, the distribution of $S_2 = U_1 + U_2$ is easily calculated:  $$f_{S_2}(s) = \begin{cases} s, & 0 \le s \le 1 \\ 2-s, & 1 < s \le 2. \end{cases}.$$  Then the CDF of $\{S_2\}$ is obtained by noting $$ F_{\{S_2\}}(s) = \Pr[\{S_2\} \le s] = \int_{u=0}^s f_{S_2}(u) \, du + \int_{u=1}^{1+s} f_{S_2}(u) \, du = s.$$ 
A: Here is a rough and sloppy sketch of a proof. It should not be difficult to fill in any gaps.
Let $\{x\}$ refer to the fractional part of $x$. 
First, we need the following fact $\{a_1+a_2\} = \{a_1\} + \{a_2\} - \lfloor \{a_1\}+\{a_2\}  \rfloor $.
Let us start with $\{\sum_{i=1}^n U_i\}$. The $U_i$'s are i.i.d uniformly distributed in $[0,1]$.
\begin{equation}
\{\sum_{i=1}^n U_i\} = \{U_1+U_2\} + \{\sum_{i=3}^n U_i\} - \lfloor \{U_1+U_2\} + \{\sum_{i=3}^n U_i\} \rfloor  \\
\end{equation}
Consider $\{U_1+U_2\}$. You can prove that $\{U_1+U_2\}$ is also uniformly distributed in $[0,1]$. We will refer to this term as $\{V\}$ ($V$ and $\{V\}$ are interchangeable here). Therefore, we have
\begin{equation}
\begin{split}
\{\sum_{i=1}^n U_i\} &= \{V\} + \{\sum_{i=3}^n U_i\} - \lfloor \{V\} + \{\sum_{i=3}^n U_i\} \rfloor  \\
&=\{V + \sum_{i=3}^n U_i\}
\end{split}
\end{equation}
Now, the RHS in the above equation is a sum of uniform random variables, so you can apply the above process again until you are left with only a single uniformly distributed random variable.
A: The characteristic function of the sum ($S_n$) of i.i.d. $U[0,1]$ random variables
$$\varphi_{S_n}(2\pi h)=\left(\frac{e^{2\pi h i}-1}{2\pi h i}\right)^n=0$$
for any $h\in\mathbb{Z}\text{\\} \{0\}$. Here is some reference.
