Weak stationarity for a stochastic process $\{X(t) \}$

Question

I've done this exercise from Beichelt, and I'd like a check in my proof.

Let $X(t)= A(t) sin(\omega t + \Phi)$, where $A(t)$ and $\Phi$ are independent, non-negative random variables for all $t$, and let $\Phi$ be uniformly distributed over $[0,2\pi]$.

Show that if $A(t)$ is a weakly stationary process, then $\{X(t), t \in \mathbb{R}\}$ is also a weakly stationary process.

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$\textit{Proof}$:

First of all, if a process is weakly stationary, it needs to be in $L^{2}(\mathbb{R}$): so $\int_{\mathbb{R}} |A(t)|^2 \ < +\infty$.

I need to check that:

• $X(t) \in L^2$
• $\mathbb{E}[X(t)]$, also called 'trend function' is constant

• $Cov(s,t)=Cov(s-t)$, also called 'covariance function' depends only on the 'time displacement' $t-s$

  1. $\int_R |X(t)|^2<\int_{\mathbb{R}}|A(t)|^2 |sin(\omega t+ \Phi)| \text{d} \Phi <\int_{\mathbb{R}} |A(t)|^2 < +\infty$, so $X(t)$ is in $L^2$.

  2. $\mathbb{E}[X(t)]=\mathbb{E}[A(t)]\cdot \mathbb{E}[sin(\omega t+\Phi)]=\mathbb{E}[A(t)]\cdot0=0$

  3. $Cov(s,t)=\mathbb{E}[X(s)X(t)]=\mathbb{E}[A(s)A(t) sin(\omega t+\Phi)sin(\omega s + \Phi)]=_\underbrace{\text{independence of A and $\Phi$ }}=\mathbb{E} [A(t)A(s)] \cdot \mathbb{E}[sin(\omega t+ \Phi)sin(\omega s + \Phi)]$

Using Werner's formulas, can be shown that $\mathbb{E}[sin(\omega t+ \Phi)sin(\omega s + \Phi)]=cos(\omega(t-s))$.

Since $A(t)$ is a weakly stationary process, then $Cov_{A}(t,s)=\mathbb{E}[A(t)A(s)] - m_s m_t=g(t-s)$, where $g$ is a function which depends only on one variable $t-s$.

So $Cov(s,t)=\mathbb{E}[A(s)A(t)]cos(w(t-s))=g(t-s) cos(w(t-s))$. So the covariance function is a function of the 'time displacement'.

- Edit

$\mathbb{E}[\sin(\omega t + \Phi)] = \frac{1}{2\pi} \int_{\mathbb{R}} \mathbb{1}_{[0,2\pi]} \sin(\omega t + \Phi) d \Phi=0$ because it's an integral of a periodic function over its period (otherwise I could have shown that its primitive $[\cos(\omega t + \Phi)]_{0}^{2\pi}=0$.

As written before, in order to justify that $\mathbb{E}[sin(\omega t+ \Phi)sin(\omega s + \Phi)]=cos(\omega(t-s))$, I'll use Werner's formulas:

$\mathbb{E}[sin(\omega t+ \Phi)sin(\omega s + \Phi)]=\frac{1}{2\pi}\int_{0}^{2 \pi} \cos(\omega (t-s)) - \cos(\omega(t+s) + 2\Phi) \text{d}\Phi= \cos(\omega (t-s))$

Hence, $\{ X(t) \}$ is a weakly stationary process.

Answer 1

The proof is OK in general, but some comments:

• The first point would be $\int |X(t)|^2=\int |A(t)|^2\sin^2(\omega t +\Phi)$, etc. (check that there's an equality instead of a strict inequality).
• Anyway, both these integrals (and the next one) are with respect to the measure of the probability space, that is, $\int_{\Omega} \ldots d\mathbb P$. And $\Omega=\mathbb R$ only if you know that the underlying probability space over which $A(t)$ and $X(t)$ (for every $t\in \mathbb R$) as well as $\Phi$ are defined is in fact $\mathbb R$ (be sure also that you understand that this has nothing to do with the fact that $t\in \mathbb R$). And probably this is not the case.
• Likewise, you need $X(t)\in L^2(\Omega)$. Think that $X(t)$ is a function with domain on $\Omega$ for each given $t\in \mathbb R$. It would actually look better the notation $X_t=X_t(w)$, where $w\in \Omega$ (I use $w$ because $\omega$ is already used in this example). So this integral is $$\int_\Omega |X_t(w)|^2 d\mathbb P,$$ or you can also find the notation $$\int_\Omega |X_t(w)|^2 d\mathbb P(w),$$ where $(\Omega,\mathcal E, \mathbb P)$ is the underlying measure space.
• The other two points are fine, although I don't know how much detail is expected when justifying that $\textrm{cov}\big(\sin(\omega t + \Phi),\sin(\omega s + \Phi)\big)=\cos\big(\omega(t-s)\big)$.
• I also think you should be a little more specific when you affirm that $\textrm E\big(\sin(\omega t + \Phi)\big)=0$.

But of course, that depends on which properties you're allowed to assume and which you should proof explicitly. The rest is fine. Well done!