- $k \in \mathbb{N}$ 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 :
- I have found a density of $S_2$
- 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}
$
- 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.