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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.


$\textit{Proof}$:

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.

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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!

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  • $\begingroup$ The first three points of your comments are quite clear to me. Just edited my answer for the last two points: let me know if it's all clear ! $\endgroup$ – VoB Apr 20 '18 at 7:49
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    $\begingroup$ I'm quite interested in your third point. I know the essential stuff about measure theory in probability, but I'd like a good reference in order to improve this skill. for example, 'W.Rudin (real & complex analysis)' is a good choice? Or you suggest something else? $\endgroup$ – VoB Apr 20 '18 at 8:12
  • $\begingroup$ W. Rudin is a classical text and will do the job. It would also be helpful to read something more specifically oriented to probability theory: that includes most text books with titles similar to 'advanced probability theory', or even 'intermediate probability theory' in some cases. Books on 'measure and integration' or 'measure theory' with one or more chapter specially devoted to probability theory are a good alternative, too. $\endgroup$ – Alejandro Nasif Salum Apr 21 '18 at 22:13

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