Show $X$ is a Stationary Process Assume $\{Y_n : n \in \mathbb{Z_+} \}$ are iid with $P(Y_0=0)=P(Y_0=1)=0.5$. Let $X_n=Y_n+Y_{n+1}$ for all $n \in \mathbb{Z_+}$.
Show $X$ is a stationary process. 
I've showed that $X$ is not $(F_n^X)$-Markov Chain by showing that $P(X_2=2 |F_1^X)$ is not equal to $\phi(X_1)$ for any $\phi: \{0,1,2 \} \rightarrow [0,1]$ for the first part of the question. 
Now to show that it is not a stationary process, I want to show that $P((X_0,X_1,...) \in B) \neq P((X_1,X_2,...) \in B) $. I've tried to use the definition of $X$. But I couldn't get anywhere after that.
Please help.
Thank you! 
 A: $$
X_{t+i}=x_i\ \text{ for }\ i=0,1,\dots,n\ .
$$
if and only if
$$
Y_{t+i}=\sum_{j=0}^{i-1}(-1)^{i-j-1}x_j+(-1)^iY_t\ \text{ for }\ i=1,2,\dots,n+1
$$
Therefore, if $\ y_i=\sum_\limits{j=0}^{i-1}(-1)^{i+j-1}x_j\ $, then
\begin{align}
P\left(\bigcap_{i=0}^n\left\{X_{t+i}=x_i\right\}\right)&= P\left(\bigcap_{i=1}^{n+1}\left\{Y_{t+i}=y_i+(-1)^iY_t\right\}\right)\\
&= \frac{1}{2}P\left(\bigcap_{i=1}^{n+1}\left\{Y_{t+i}=y_i\right\} \,\big|Y_t=0\right)\\
&\ \ + \frac{1}{2}P\left(\bigcap_{i=1}^{n+1}\left\{Y_{t+i}=y_i +(-1)^i\right\} \,\big|Y_t=1\right)\\
&=\cases{\frac{1}{2^{n+1}}&if $\ y_i, y_i+(-1)^i\in\{0,1\}$\\
&for $\ i=1,2\dots,n+1$\\
\frac{1}{2^{n+2}}&if $\ y_i\in\{0,1\}\ $for $\ i=1,2\dots,n+1$\\
&but $\ y_k+(-1)^k\not\in\{0,1\}\ $for\\
&some $\ k\in\{1,2,\dots,n+1\}$\\
\frac{1}{2^{n+2}}&if $\ y_k\not\in\{0,1\}\ $for some $\ k\in\{1,2,\dots,n+1\}\ $\\
&but $\ y_i+(-1)^i\in\{0,1\}\ $\\
&for $\ i=1,2,\dots,n+1\ $\\
0&otherwise.}
\end{align}
Since this is independent of $\ t\ $, it follows that $\ \left\{X_i\right\}_{i=0}^\infty\ $ is stationary.
