# 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?

• 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. Commented Jun 22, 2015 at 20:24
• @Enredanrestos: What is your idea? Commented Jun 22, 2015 at 20:52

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.

• I think you are correct. I got a deeper understanding of this question. I will up vote your answer with my appreciation. Commented Jun 23, 2015 at 14:35
• you need to show the joint distribution is independent of time shift, that is $\mathbb{P}(X_{t_{1}+r}\leq x_{1}, X_{t_{2}+r}\leq x_{2}, \cdots X_{t_{n}+r}\leq x_{n})=\mathbb{P}(X_{t_{1}}\leq x_{1},X_{t_{2}}\leq x_{2},\cdots, X_{t_{n}}\leq x_{n})$ for all $r\in\mathbb{R}$, so why is it sufficient to only show $\mathbb{P}(z\sin(\omega t+\theta)\leq \xi)$ is independent of $t$? Commented Mar 4, 2020 at 21:33

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.

• Could you explain $\phi(dy)$? or $\phi^{-1}(dy)$? i'm a little confused. thanks Commented Jun 8, 2020 at 22:23
• @Bearandbunny sorry, I should use $dy$. Let me edit the answer. Commented Jun 8, 2020 at 22:32
• @Bearandbunny basically the $\mathbb{E}_{\phi}$ is the expectation with respect to $\phi(dx)$, the uniform distribution. but the density is $\frac{1}{2\pi}$, so $\phi(dx)=\frac{1}{2\pi}dx.$ Commented Jun 8, 2020 at 22:36
• I guess maybe it is appropriate to write as $\phi^{-1}(dx)$ because $\phi$ is a measurable function from $(\Omega, \mathcal{F}, P) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$. I think this proof is very smart and will upvote it. Commented Jun 8, 2020 at 22:50
• @Bearandbunny ah. You are right. thanks for your nice words :) Commented Jun 8, 2020 at 22:58

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 $z.$)

• 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? Commented Jun 20, 2015 at 22:39
• @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...
– zoli
Commented Jun 20, 2015 at 22:43
• @Bearandbunny: I studied stochastic processes from this book: books.google.hu/… -- thousands of years ago.
– zoli
Commented Jun 20, 2015 at 22:58
• 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. Commented Jun 23, 2015 at 16:21
• $\frac{d}{dt}\int_b^af(x+t)dx=f(t+b)-f(t+a)$ Commented Jun 23, 2015 at 18:03