Is $\{\sin(\omega n), n \geq 1\}$ a strictly stationary process? Let $X(t)=\sin(\omega t)$, where $\omega$ is is uniformly distributed R.V. on $[0,2π]$. Let $X_n=X(n)$, is $\{X_n,n \geq 1\}$ a strictly stationary process?
I've calculated that the distribution function of $X_n$ is
$f_{X_n}(x)=\frac{1}{\pi\sqrt{1-x^2}}$.
Can anybody help me then?
 A: The process is not stationary. For example, for every small $\varepsilon$, the event $[X_1\geqslant1-\varepsilon,|X_3|\leqslant\varepsilon]$ is empty while the event $[X_2\geqslant1-\varepsilon,|X_4|\leqslant\varepsilon]$ has positive probability.
To see this, note that $[X_1\geqslant1-\varepsilon]$ corresponds to $\omega$ close to $\frac\pi2$ and $[|X_3|\leqslant\varepsilon]$ corresponds to $\omega$ close to $n\frac\pi3$ for some integer $0\leqslant n\leqslant6$, hence $[X_1\geqslant1-\varepsilon]$ and $[|X_3|\leqslant\varepsilon]$ are not compatible when $\varepsilon$ is small, while $[X_2\geqslant1-\varepsilon,|X_4|\leqslant\varepsilon]$ is realized when $\omega$ is close to $\frac\pi4$.
A: A nice and simple argument as been provided by did. 
Here we use the fact that if $(Y_n,n\geqslant 1)$ is a strictly stationary sequence, and $\varphi\colon\mathbb R^\infty\to\mathbb R^\infty$ is measurable, then $\varphi((Y_n,n\geqslant 1))$ is a strictly stationary sequence. We use this with $\varphi((x_n,n\geqslant 1))=(x_nx_{n+1},n\geqslant 1)$. If $(X(n),n\geqslant 1)$ like in the OP was stationary, so would be the sequence $(\sin(nU)\sin((n+1)U),n\geqslant 1)$, that is, the sequence $(\cos U-\cos((2n+1)U),n\geqslant 1)$. Using this times $\varphi((x_n,n\geqslant 1)):=(x_
1,x_1+x_2,x_1+x_2+x_3,\dots)$, we would get the stationarity of $(y_n,n\geqslant 1)$ with $Y_n=\cos((2n-1)U)-\cos((2n+1)U)$. In this case, the random variables
$$\sum_{j=1}^nY_j=1-\cos((2n+1)U)\quad \mbox{and}\quad \sum_{j=n+1}^{2n}Y_j=\cos((2n+1)U)-\cos((4n+1)U)$$
should have the same distribution. But the first one is non-negative and the second one takes negative values with a positive probability.
