How to judge second-order stationary? Given that a series {$x_t$} where $x_t=\sin(t+U)$, $U$ has uniform distrubution on $(0,2\pi)$. Is {$x_t$} second-order stationary?
I know here we need to show $E(x_t)$ is constant and $cov(x_{t+s},x_{t})$ only depends on $s$. Now $E(U)=\pi$, $\mathrm{Var}(U)=\frac{\pi^2}{3}$. My problem is how to find the expection value and covariance of $\sin$ function? Indeed $\mathrm{Cov}(x_{t+s},x_{t})$ = $E(x_{t+s}x_{t})-E(x_{t+s})E(x_{t})$, so the probelm is expected value..
Here I think it is hard to find the p.d.f of this $\sin$ function. Can anyone help me with that?
Thank you in advance!
 A: $$\mathbb{E}[x_t] = \int_{0}^{2\pi}\sin(t+u) \cdot \dfrac{1}{2\pi}\text{ d}u = \dfrac{1}{2\pi}\int_{0}^{2\pi}\sin(t+u)\text{ d}u\text{.}$$
Let $v = t + u$, then $\text{d}v = \text{d}u$ and
$$\begin{align}
\mathbb{E}[x_t] &= \dfrac{1}{2\pi}\int_{t}^{t + 2\pi}\sin(v)\text{ d}v \\
&= \dfrac{1}{2\pi}\left.[-\cos(v)]\right|^{v=t+2\pi}_{v=t} \\
&= \dfrac{-1}{2\pi}[\cos(t+2\pi)-\cos(t)] \\
&= \dfrac{\cos(t)-\cos(t+2\pi)}{2\pi}\text{.}\end{align}$$
Because $\cos$ is $2\pi$-periodic, we have $\cos(t) = \cos(t + 2\pi)$, hence $\mathbb{E}[x_t] = 0$.
You must also demonstrate that the variance of $x_t$ is constant with respect to $t$, and finite.
We have similarly that
$$\mathbb{E}[x_t^2] = \dfrac{1}{2\pi}\int_{0}^{2\pi}\sin^2(t+u)\text{ d}u = \dfrac{1}{2\pi}\int_{t}^{t+2\pi}\sin^2(v)\text{ d}v$$
and you will need to use the power-reducing identity $$\sin^2(v) = \dfrac{1-\cos(2v)}{2}\text{.}$$
I will leave this to you as an exercise.
For the covariance, we have from the above work that
$$\begin{align}
\text{Cov}(x_{t+s}, x_t) &= \mathbb{E}[x_{t+s}x_t] \\
&= \mathbb{E}[\sin(t+s+U)\sin(t+U)] \\
&= \dfrac{1}{2\pi}\int_{0}^{2\pi}\sin(t+s+u)\sin(t+u)\text{ d}u\text{.} 
\end{align}$$
You will want to make use of the product-to-sum identity
$$\sin(t+s+u)\sin(t+u) = \dfrac{1}{2}\left[\cos(s) - \cos(2t+2u+s) \right]\text{.}$$
I will leave this to you as an exercise.
