How to prove it is a strictly stationary process? $ξ(t) = z*sin(ωt + θ)$ where $z$ is a random variable and its distribution is unknown and $θ$ is another random variable that is independent of $z$ and $θ$ is uniformly distributed on $(0, 2\pi)$. Besides, $ω$ is a constant greater than $0$. I've been asked to show $ξ(t)$ is a strictly stationary stochastic process using characteristic function or say $E(e^{jvξ(t)})$. 
I've tried but it seems that $E(e^{jvξ(t)})$ depends on the $t$ I choose, which means it is not a strictly stationary stochastic process. I think my answer can be wrong and how to prove it?
One more question: I've got quite confused why a characteristic function of a stochastic process can be used to prove property of strictly stationary?
The definition of strictly stationary is $F_ξ(x_1, x_2, x_3,..., x_n; t_1, t_2, t_3,...,t_n) = F_ξ(x_1, x_2, x_3,..., x_n; t_1 + τ, t_2 + τ, t_3 + τ,...,t_n + τ)$ where capital $F$ denotes the probability distribution function(PDF) of ξ(t). My book never told me anything about relationship between characteristic function of a stochastic process and its PDF. So when this problem appeared, I think they want me to show $E(e^{jvξ(t)})$ does not depend on $t$ while forget to tell me why not depending on $t$ imply its strictly stationary?
 A: I am going to write a demonstration that the process is stationary, but I am aware this is not what the original poster asked since it is not based on calculating the characteristic function. This post is in response to a comment I made.
Strict sense. It suffices to show that
$$\mathbb{P}(z\sin(\omega t+\theta)\le\xi)$$ is independent of $t$.
But this is clear since  $s=\sin(\omega t+\theta)$ distributes in $(-1,1)$ with density $f_s=1/(\pi\sqrt{1-s^2})$ independent of $t$.
Wide sense. This becomes a bit unnecessary, but anyways. We need to show that the autocorrelation depends only on the temporal displacement, not the coordinate, that is, $\mathbb{E}[\xi(t)\xi(t+\tau)]$ is function of $\tau$ and not $t$. We have that
$$ 
\begin{multline}
\mathbb{E}[\xi(t)\xi(t+\tau)]=\cos(\omega t)\cos(\omega (t+\tau))\mathbb{E}[z^2\sin^2\theta]+\sin(\omega t)\sin(\omega (t+\tau))\mathbb{E}[z^2\cos^2\theta]+\\
\sin(\omega(2t+\tau))\mathbb{E}[z^2\sin\theta\cos\theta]~~.\\
\end{multline}$$
But $\mathbb{E}[z^2\sin^2\theta]=\mathbb{E}[z^2\cos^2\theta]=\mathbb{E}[z^2]/2=\mu_z^2/2$ (assuming it exists). We can separate the variables in the expectation since thet are independent.
Also $\mathbb{E}[z^2\sin\theta\cos\theta]=0$. We conclude
$$\mathbb{E}(\xi(t)\xi(t+\tau))=\frac{\mu_z^2}{2}\cos(\omega\tau)~~,$$
independent of t. Therefore the process is wide sense stationary.
A: I know this is a post from 4 years ago, and you got two answers. However, it seems that the accepted answer did not give you a complete proof, and I cannot agree with the second answer in terms of strict stationarity.
To prove the strict stationarity, you need to show $$\mathbb{P}(X_{t_{1}+h}\leq x_{1},\cdots, X_{t_{n}+h}\leq x_{n})=\mathbb{P}(X_{t_{1}}\leq x_{1},\cdots, X_{t_{n}}\leq x_{n})$$ so only arguing the first dimensional distribution (as suggested by the second answer) may not be sufficient. Since that answer did not give you why it is sufficient, I think we need to be cautious here. 
Here is an alternative but still straightforward proof (and I don't think the char function can give you any result of the time shift of the f.d.d, it only gives you if the f.d.d. is consistent).

So you have $X_{t}:=A\cos(\eta t+\phi)$ where $A,\eta\geq 0$ and is independent of $\phi$ which is uniform on $[0,2\pi]$. Let $h\in \mathbb{R}$.
Firstly, if $X_{t}:=\cos(t+\phi)$, then the proof is straightforward 
\begin{align*}
\mathbb{P}(X_{t_{1}+h}\leq x_{1},\cdots, X_{t_{n}+h}\leq x_{n})&=\mathbb{P}(\cos(t_{1}+h+\phi)\leq x_{1},\cdots, \cos(t_{n}+h+\phi)\leq x_{n})\\
&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+h+y)\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+h+y)\leq x_{n}}dy\\
&=\int_{h}^{2\pi+h}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+\tilde{y})\leq x_{n}}d\tilde{y}\\
&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+\tilde{y})\leq x_{n}}d\tilde{y}\\
&=\mathbb{P}(X_{t_{1}}\leq x_{1},\cdots, X_{t_{n}}\leq x_{n}),
\end{align*}
where the third equality was obtained by a change of variable $\tilde{y}:=y+h$, and the fourth equality is obtained by the observation that the integrand is a $2\pi-$periodic function (since $\cos$ is ) and as such it has the same value of the integral over any interval of length $2\pi.$
Now, with $A\geq 0$ and $\eta\geq 0$,  we use the same spirit, but apply Fubini as follows:
\begin{align*}
\mathbb{P}(X_{t_{1}+h}\leq x_{1}, \cdots, X_{t_{n}+h}\leq x_{n})&=\mathbb{P}\Big(A\cos(\eta(t_{1}+h)+\phi)\leq x_{1},\cdots, A\cos(\eta(t_{n}+h)+\phi)\leq x_{n}\Big)\\
&= \mathbb{E}_{A,\eta}\mathbb{E}_{\phi}\Big(\mathbb{1}_{A\cos(\eta(t_{1}+h)+\phi)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta(t_{n}+h)+\phi)\leq x_{n}}\Big),
\end{align*}
then using the independence for the change of variable (so you can just treat $\eta$ as constant in the integral of uniform), and the hypothesis of $\eta\geq 0$ (so the integral length is still $2\pi$),  we go back to a similar case: 
\begin{align*}
\mathbb{E}_{\phi}\Big(\mathbb{1}_{A\cos(\eta(t_{1}+h)+\phi)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta(t_{n}+h)+\phi)\leq x_{n}}\Big)&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{A\cos(\eta(t_{1}+h)+y)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta(t_{1}+h)+y)\leq x_{n}}dy\\
&=\int_{\eta h}^{2\pi+\eta h}\dfrac{1}{2\pi}\mathbb{1}_{\cos(\eta t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(\eta t_{n}+\tilde{y})\leq x_{n}}d\tilde{y}\\
&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\cos(\eta t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(\eta t_{n}+\tilde{y})\leq x_{n}}d\tilde{y}\\
&=\mathbb{E}_{\phi}\Big(\mathbb{1}_{A\cos(\eta t_{1}+\phi)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta t_{n}+\phi)\leq x_{n}}\Big),
\end{align*}
where the second and the third equality were obtained due to the same reasons.
This implies that 
\begin{align*}
\mathbb{P}(X_{t_{1}+h}\leq x_{1}, \cdots, X_{t_{n}+h}\leq x_{n})&=\mathbb{E}_{A,\eta}\mathbb{E}_{\phi}\Big(\mathbb{1}_{A\cos(\eta(t_{1}+h)+\phi)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta(t_{n}+h)+\phi)\leq x_{n}}\Big)\\
&=\mathbb{E}_{A,\eta}\mathbb{E}_{\phi}\Big(\mathbb{1}_{A\cos(\eta t_{1}+\phi)\leq x_{1}}\cdots\mathbb{1}_{A\cos(\eta t_{n}+\phi)\leq x_{n}}\Big)\\
&=\mathbb{P}(X_{t_{1}}\leq x_{1},\cdots, X_{t_{n}}\leq x_{n}).
\end{align*}
Since this holds for all $h\in\mathbb{R}$, we are able to conclude that $X_{t}$ is strictly stationary. 

This is basically an application of Fubini. 
A: HINT
To compute the characteristic function we can use the so called tower rule:
$$E\left[e^{j\times v \times ξ(t)}\right]=E\left[E\left[e^{j\times v\times Z \times \sin(ωt + θ)} \mid Z\right]\right].$$
Since $Z$ and $\theta$ are independent we can tell that
$$E\left[e^{j\times v\times Z\times \sin(ωt + θ)}\mid Z=z\right]=E\left[e^{j\times v \times z\times  \sin(ωt + θ)}\right]=\frac{1}{2\pi}\int_0^{2\pi}e^{j\times  v \times z\times \sin(ωt + x)}dx.$$
I cannot formally prove but I am convinced that 
$$\int_0^{2\pi}e^{j\times  v\times  z \times \sin(ωt + x)}dx$$
will not depend on $t$.
This statement is intuitively clear if we take into account that the integral above has a real part and an imaginary part (For the sake of simplicity, let $v=z=\omega=1$):
$$\int_0^{2\pi}e^{j\times \sin(t + x)}dx=$$
$$=\int_0^{2\pi}\cos(\sin(t + x))dx+j \int_0^{2\pi}\sin(\sin(t + x))dx.$$
In the figure below I depicted the two integrands for $t=0$:

The blue curve is the integrand of the real part and the red curve is the integrand of the imaginary part. If we chose another $t$ then we would get the same curves shifted accordingly. The result of the integrals would be $0$ for the imaginary part and a fixed positive number for the real part independently of $t$. (This fact is independent of $\omega, v$, and $z.$)
