$ξ(t) = z*sin(ωt + θ)$ 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?

  • $\begingroup$ The process is stationary (wide sense and strictly I think), but calculating the characteristic function do not seem to be the simplest way to prove it. $\endgroup$ – Enredanrestos Jun 22 '15 at 20:24
  • $\begingroup$ @Enredanrestos: What is your idea? $\endgroup$ – Bear and bunny Jun 22 '15 at 20:52


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 $t=0$:

enter image description here

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 $z.$)

  • $\begingroup$ Thanks. I didn't got answer notification until I got in just now. I have one more question do you know why does characteristic function not depending on t imply its strictly stationary? $\endgroup$ – Bear and bunny Jun 20 '15 at 22:39
  • $\begingroup$ @Bearandbunny: I know only that the time independence of the characteristic function implies the time independence of all the moments. To be honest, I don't know the formal proof that then the multidimensional distribution functions will be time independent. But I am still convinced... $\endgroup$ – zoli Jun 20 '15 at 22:43
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    $\begingroup$ @Bearandbunny: I studied stochastic processes from this book: books.google.hu/… -- thousands of years ago. $\endgroup$ – zoli Jun 20 '15 at 22:58
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    $\begingroup$ I see. You are right then. By the way, to demonstrate that these integrals do not depend on t, you could derive on $t$ and use $\frac{d}{dt}\sin\cos(\omega t +x)=\omega \frac{d}{dx} \sin\cos(\omega t +x)$ and then do the integrals and see that you get cero. $\endgroup$ – Enredanrestos Jun 23 '15 at 16:21
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    $\begingroup$ $\frac{d}{dt}\int_b^af(x+t)dx=f(t+b)-f(t+a)$ $\endgroup$ – Enredanrestos Jun 23 '15 at 18:03

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.

  • $\begingroup$ I think you are correct. I got a deeper understanding of this question. I will up vote your answer with my appreciation. $\endgroup$ – Bear and bunny Jun 23 '15 at 14:35

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