Let $\left(X_i\right)_{i\ge1}$ be independent identically distributed random variables such that $\mathrm{P}\left( X_i = -1 \right)=\mathrm{P}\left( X_i = 1 \right)=\frac{1}{2}$.
Let $X(t)=X_{\left \lfloor t \right \rfloor}$ be a stochastic process such that the initial time follows $U\left( 0,1 \right)$.
Prove the process is weakly stationary.
EDIT: to completely change the previous version of the definition which was bogus: $X(t)=X_{\left \lfloor t_0+t \right \rfloor}$ where $t_0\in\mathrm{U}\left(0,1\right)$
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So far I have proved that:
$m_X(t)=0$
$\mathrm{E}\left( X(t)X(t') \right) = \left\{\begin{array}{c} 1\quad\text{if $\left \lfloor t \right \rfloor = \left \lfloor t' \right \rfloor$ }\\0\quad \text{otherwise} \end{array} \right.$
So for the moment it is not weakly stationary since the autocorrelation depends not only on $t'-t$, but also on whether there is an integer between $t$ and $t'$. I can't find a way to use the uniform distribution.
I have tried the following:
If $|\tau|=\left| t'-t\right|>1$, then $\left \lfloor t \right \rfloor \neq \left \lfloor t' \right \rfloor$. So $\mathrm{E}\left( X(t)X(t') \right) =0$
If $|\tau|=\left| t'-t\right|\leq1$, and assuming, without loss of generality, that $ t=t_0+t_1 \leq t_0+t_2=t'$, I want to prove that $\mathrm{P}\left(\left \lfloor t \right \rfloor \neq \left \lfloor t' \right \rfloor\right)$ is a function of $\tau$.
$\mathrm{P}\left(\left \lfloor t \right \rfloor \neq \left \lfloor t' \right \rfloor\right) = \mathrm{P}\left( \exists k\in\mathbb{N}/k\in\left[t,t'\right] \right) = \mathrm{P}\left( \exists k\in\mathbb{N}/ t_0\in \left[k-t_2,k-t_1 \right] \right)$
Here I have an issue because I don't know if such probability is a function of $\tau$.