• $k \in \mathbb{N}$ is fixed
  • $(X_n)_{n \geq 1}$ are all independent and follow an uniform law on $[0,k]$
  • We define $f(x)=x -\lfloor x \rfloor$
  • $S_n= \sum_{i=1}^{n} X_i$
  • $Z_n= f(S_n)$
  • We want to show that $\forall n \geq 1, S_n -\lfloor S_n \rfloor \sim U[0,1]$

Here are the steps :

  1. I have found a density of $S_2$
  2. Show that $Z_2 \sim U[0,1]$
    3.(a) Express $f(f(S_n) + X_{n+1})$ with $Z_{n+1}$
    3.(b) Deduce that $Z_n \sim U[0,1]$

My attempt:
1. $f_{S_2}(s)= \begin{cases} \frac{1}{k^2} s \quad \text{si} \quad 0 \leq s\leq k \\ \frac{1}{k} (2-\frac{s}{k}) \quad \text{si} \quad k \leq s \leq 2k\\ \end{cases} $

$F_{S_2}(s)= \begin{cases} \frac{s^2}{2 k^2} \quad \text{si} \quad 0 \leq s\leq k \\ 2\frac{s}{k}-\frac{s^2}{2 k^2} -1 \quad \text{si} \quad k \leq s \leq 2k\\ \end{cases} $

  1. For this question, let $Z=Z_2$

$0\leq Z \leq 1 $
For $a \leq 1$
$0\leq Z \leq a \iff Z \in \bigcup_{j=0}^{j=k-1} [j,j+a]$

$F_Z(a)= \sum_{j=0}^{j=2k-1} F(j+a)-F(j)$

$ \begin{align*} f_Z(a) &= \sum_{j=0}^{j=2k-1} f_S(a+j) \\ &= \sum_{j=0}^{j=k-1} f_S(a+j) + \sum_{j=k}^{j=2k-1} f_S(a+j) \\ &= \sum_{j=0}^{k-1} \big( \frac{a}{k^2} + \frac{j}{k^2} \big) + \sum_{j=k}^{2k-1} \big( \frac{2}{k} - \frac{a}{k^2} - \frac{j}{k^2}) \\ &= \big( \sum_{j=0}^{k-1} \frac{a}{k^2} - \sum_{j=k}^{2k-1}\frac{a}{k^2} \big) + \sum_{j=0}^{k-1} \frac{j}{k^2} - \sum_{j=0}^{k-1} \frac{j+k}{k^2} + \sum_{j=k}^{2k-1} \frac{2}{k} \\ &= -1 +\sum_{j=k}^{2k-1} \ \frac{2}{k} \\ &=-1+2=1\\ \end{align*} $

3.$f ( f(S_n) + X_{n+1})= f( S_n - \lfloor S_n \rfloor + X_{n+1} )$
Let $Z_n= S_n - \lfloor S_n\rfloor $
$S_{n+1} = S_n+ X_{n+1} = Z_n + \lfloor S_n\rfloor + X_{n+1}$
$ S_{n+1} - \lfloor S_{n+1}\rfloor = f( Z_n + X_{n+1} )$
because $f(x+p)=f(x)$ for all integer $p$ so : $f ( f(S_n) + X_{n+1}) = Z_{n+1}$


2 Answers 2


This is one of those things that's annoying to compute directly, but becomes easy if you use modular arithmetic. In this case, you should work with real numbers modulo $1$. Then the claim follows directly from the fact that the uniform measure on $\Bbb R/\Bbb Z$ is invariant under convolution (i.e., independent sums).

Here is the argument drawn out in full detail, in case it is helpful.

The idea is to work on $\Bbb R/\Bbb Z$ instead of $\Bbb R$. Let $\pi: \Bbb R \to \Bbb R/\Bbb Z$ be the projection map $x \mapsto x \pmod 1$. Henceforth, whenever I refer to a "sum" it will be with respect to the additive group structure on $\Bbb R/\Bbb Z$.

Let $Y_i=\pi(X_i)$. Note that the $Y_i$ are uniformly distributed on $\Bbb R/\Bbb Z$ (i.e., they are distributed according to arclength measure if you view $\Bbb R/\Bbb Z$ as a circle, or Haar measure if you view it as a topological group).

Also note that any finite sum of independent uniformly distributed variables in $\Bbb R/\Bbb Z$ is still uniformly distributed on $\Bbb R/\Bbb Z$ (i.e., the arclength measure on the circle is invariant under convolution of measures).

Note also that $\pi$ is a group homomorphism, so we have that $\pi(S_n) = \sum_1^n Y_i$. We thus conclude from the previous paragraph that $\pi(S_n)$ is uniformly distributed on $\Bbb R/\Bbb Z$.

The final step is to note that $\pi$ is invariant under $f$, i.e., $\pi \circ f = \pi$. Thus $\pi(f(S_n))$ has a uniform distribution on $\Bbb R / \Bbb Z$. But $\pi$ is invertible if we restrict its domain to $[0,1)$. Moreover, $f(S_n)$ takes values in $[0,1)$.

The pushforward by $\pi^{-1}$ of the uniform measure on $\Bbb R/\Bbb Z$ is the uniform measure on $[0,1)$, so we conclude that $f(S_n)$ is uniformly distributed on $[0,1)$.

  • $\begingroup$ Thank you so much $\endgroup$
    – zestiria
    Aug 21, 2020 at 7:46
  • 1
    $\begingroup$ I think this is the first time I have seen a purely algebraic proof of a probabilistic problem. Very elegant! $\endgroup$
    – Math1000
    Aug 21, 2020 at 11:07

Here is a more elementary rendition of @shalop's answer. The point is that it all boils down to showing the following two claims:


  1. If $U \sim \mathcal{U}[0,k]$ for some $k\in\mathbb{N}$, then $f(U) \sim \mathcal{U}[0,1]$.
  2. If $U \sim \mathcal{U}[0,1]$ and $a \in \mathbb{R}$, then $f(a+U) \sim \mathcal{U}[0,1]$.

Using this claim, we know that

$$f(X_n+a) = f(f(X_n) + a) \sim \mathcal{U}[0,1]$$

whenever $a \in \mathbb{R}$ and $X_n \sim \mathcal{U}[0,k]$ for some $k\in\mathbb{N}$. Then for any $r \in [0, 1)$, by the independence of $X_n$ and $S_{n-1}$,

\begin{align*} \mathbb{P}(Z_{n} \leq r) = \mathbb{E}[\mathbb{P}(f(X_{n}+S_{n-1}) \leq r \mid S_{n-1})] = \mathbb{E}[r] = r. \end{align*}

Therefore the desired conclusion follows.

Proof of Claim. In the first part, it is clear that $f(U)$ takes values only in $[0,1)$. Now for any $r \in [0,1]$, we have

$$ P(f(U) \leq r) = \sum_{q=0}^{k-1} P(q \leq U \leq q+r) = \sum_{q=0}^{k-1} \frac{r}{k} = r, $$

and therefore $f(U)$ has the desired distribution. In the second part, write $a = \lfloor a \rfloor + \langle a \rangle$, where $\langle a \rangle$ denotes the fractional part of $a$. Then for any $r \in [0,1)$,

\begin{align*} P(f(a+U) \leq r) &= P(\{ 0 \leq U < 1 - \langle a \rangle \} \cap \{ U+\langle a \rangle \leq r \}) \\ &\quad + P( \{ 1 - \langle a \rangle \leq U < 1 \} \cap \{ U+\langle a \rangle - 1 \leq r \}). \end{align*}

Considering the cases $r < \langle a \rangle$ and $r \geq \langle a \rangle$ separately, this can be easily computed as $r$, again proving that $f(a+U) \sim \mathcal{U}[0,1]$. $\square$


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