Distribution of a fractional part of the sum of uniform RVs

Solution 1:

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 $$ F_{\{S_2\}}(s) = \Pr[\{S_2\} \le s] = \int_{u=0}^s f_{S_2}(u) \, du + \int_{u=1}^{1+s} f_{S_2}(u) \, du = s.$$

Solution 2:

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.

Solution 3:

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.