Why is this random process stationary? I am studying stationary from wikipedia. (https://en.wikipedia.org/wiki/Stationary_process?wprov=sfii1)
In the examples section, the following is written.
let $Y$ have a uniform distribution on $(0,2π]$ and define the time series $\{ X_t \}$ by
$X_t=\cos (t+Y) \quad \text{ for } t \in \mathbb{R}$. 
Then $\{ X_t \}$ is strictly stationary. 
I don't understand why it is stationary.
Can someone prove this?
 A: Since the cosine is periodic with period 2$\pi$, $Y$ already covers all possible inputs within a whole period of the cosine, and so does $Y+t$.
A formal proof:
We need to show that $P(\cos(Y)\le x) = P(\cos(Y+t)\le x)$. We have:
$P(\cos(Y)\le x)2\pi = I_{]1, \infty]}(x)2\pi + I_{[-1,1]}(x)\mu(\{y\in]0,2\pi]|\cos(y)\le x\})$
and
$P(\cos(Y+t)\le x)2\pi = I_{]1, \infty]}(x)2\pi + I_{[-1,1]}(x)\mu(\{y\in]0,2\pi]|\cos(y+t)\le x\})$.
If $x \in [-1,1]$, then $\arccos(x)$ exists and we have:
$\mu(\{y\in]0,2\pi]|\cos(y)\le x\}) = I_{[-1,1]}(x)\mu(\{]0,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]0,t]\cup]t,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{(]t,2\pi]\cup]2\pi,t+2\pi])\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]t,t+2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]0,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi-t,2\pi-\arccos(x)+2k\pi-t]\}) = \mu(\{y\in]0,2\pi]|\cos(y+t)\le x\})$
We further need to show that $P(X_{t_1+\tau} \le x_1, \ldots, X_{t_n+\tau} \le x_n) = P(X_{t_1} \le x_1, \ldots, X_{t_n} \le x_n)$ for $n \ge 2$.
