# Weak stationarity versus Strong stationarity for a stochastic process

I'd like a check in this exercise, taken from Beichelt's book:

Let $X(t) = sin(\Phi t)$, where $\Phi$ is uniformly distributed over the interval $[0,2\pi]$.

Verify:

(1) The discrete-time stochastic process $\{X(t), t= 1,2, \dots\}$ is weakly but not strongly stationary.

(2) The continuous-time $\{X(t), t \in (-\infty,+\infty)\}$ is neither weakly nor strongly stationary.

$\textit{Proof(s)}$:

First, I need to check that $X(t) \in L^2(\Omega)$, with $\Omega=[0,2\pi]$, and this is shown by the computation of $\int_{0}^{2\pi} sin^2(\Phi t)d\Phi= \pi$.

$\mathbb{E}[X(t)]= \frac{1}{2\pi}\int_{0}^{2 \pi}sin(\Phi t) d \Phi=\frac{\cos(2\pi t) - 1}{2\pi}$. If $T \subset N$, then $\mathbb{E}[X(t)]=0 \quad \forall t$.

Conversely, if $T \in \mathbb{R}$, then the trend function is not constant and thus the process is not stationary (dns this answers secondo question).

So from now I suppose $T \subset \mathbb{N}$.

$Cov(X(t),X(s))=\mathbb{E}[\sin(\Phi t) \sin(\Phi s)] = \frac{1}{2}\cdot \mathbb{E}[\cos(\Phi(t-s)) + \cos(\Phi(t+s))]$

The computation of this integral gives $\frac{1}{4\pi} [\frac{\sin(2\pi(s-t))}{s-t} + \frac{\sin(2\pi(s+t))}{s+t}]$. Since $t,s \in \mathbb{N}$, then $s+t$ and $s-t \in \mathbb{N}$ and the sum summand is identically zero, i. e., $Cov(X(t),X(s))=0$.

In order to show that $X(t)$ is not strongly stationary I need to check if $F_t(x)=F(x), t \in T$. So, for $x \in [0,1]$:

$F_t(x)=\mathbb{P}(sin(\Phi t)\leq x)=\mathbb{P}(\Phi \leq \frac{arcsin(x)}{t}) = \int_{0}^{\frac{\arcsin(x)}{t}} \frac{1}{2\pi} d \Phi=\frac{\arcsin(x)}{2\pi t }$.

Since $F_t(x)$ depends on $t$, this process is weakly but non strong stationary.

But I have a big doubt: is it correct to state that the covariance function depends on the time splitting $\tau=t-s$, in order to state that the process is weakly stationary?

I hope to have explained myself well