Showing that $S_n -\lfloor S_n \rfloor \sim U[0,1]$

ceiling-and-floor-functionsprobabilityprobability distributionsuniform distribution

  • $k \in \mathbb{N}$ 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 :
$f ( f(S_n) + X_{n+1}) = Z_{n+1}$

Best Answer

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 $[0,1)$.

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