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Let $X(t)=\sin(\omega t)$, where $\omega$ is is uniformly distributed R.V. on $[0,2π]$. Let $X_n=X(n)$, is $\{X_n,n \geq 1\}$ a strictly stationary process?

I've calculated that the distribution function of $X_n$ is

$f_{X_n}(x)=\frac{1}{\pi\sqrt{1-x^2}}$.

Can anybody help me then?

Davide Giraudo
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DANG Fan
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2 Answers2

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The process is not stationary. For example, for every small $\varepsilon$, the event $[X_1\geqslant1-\varepsilon,|X_3|\leqslant\varepsilon]$ is empty while the event $[X_2\geqslant1-\varepsilon,|X_4|\leqslant\varepsilon]$ has positive probability.

To see this, note that $[X_1\geqslant1-\varepsilon]$ corresponds to $\omega$ close to $\frac\pi2$ and $[|X_3|\leqslant\varepsilon]$ corresponds to $\omega$ close to $n\frac\pi3$ for some integer $0\leqslant n\leqslant6$, hence $[X_1\geqslant1-\varepsilon]$ and $[|X_3|\leqslant\varepsilon]$ are not compatible when $\varepsilon$ is small, while $[X_2\geqslant1-\varepsilon,|X_4|\leqslant\varepsilon]$ is realized when $\omega$ is close to $\frac\pi4$.

Did
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  • Your example is really simple and nice. But could you please give a bit more explanations on the definition of stationary process? According to my calculation, the PDF of ${X_n}$ is $\left(\frac{1}{\pi\sqrt{1-x^2}}\right)^k$, which is not the function of time. – DANG Fan Oct 07 '13 at 12:40
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    As you should know since you posted the question, a process $(X_t)t$ is stationary if, for every index set $T$, the distribution of $(X{s+t})_{t\in T}$ is independent of $s$. You checked this property for $|T|=1$, I disproved it for $|T|=2$. – Did Oct 07 '13 at 13:21
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A nice and simple argument as been provided by did.

Here we use the fact that if $(Y_n,n\geqslant 1)$ is a strictly stationary sequence, and $\varphi\colon\mathbb R^\infty\to\mathbb R^\infty$ is measurable, then $\varphi((Y_n,n\geqslant 1))$ is a strictly stationary sequence. We use this with $\varphi((x_n,n\geqslant 1))=(x_nx_{n+1},n\geqslant 1)$. If $(X(n),n\geqslant 1)$ like in the OP was stationary, so would be the sequence $(\sin(nU)\sin((n+1)U),n\geqslant 1)$, that is, the sequence $(\cos U-\cos((2n+1)U),n\geqslant 1)$. Using this times $\varphi((x_n,n\geqslant 1)):=(x_ 1,x_1+x_2,x_1+x_2+x_3,\dots)$, we would get the stationarity of $(y_n,n\geqslant 1)$ with $Y_n=\cos((2n-1)U)-\cos((2n+1)U)$. In this case, the random variables $$\sum_{j=1}^nY_j=1-\cos((2n+1)U)\quad \mbox{and}\quad \sum_{j=n+1}^{2n}Y_j=\cos((2n+1)U)-\cos((4n+1)U)$$ should have the same distribution. But the first one is non-negative and the second one takes negative values with a positive probability.

Davide Giraudo
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    $P(X_1 \ge 0, x_2 \ge 0)=P(\omega \in [0, \pi / 2])=1/4$, and $P(X_2 \ge 0, X_3 \ge 0)=P(\omega \in [0, \pi / 3] \cup [4\pi / 3, 3\pi / 2])=1/4$. They equals. – DANG Fan Oct 07 '13 at 01:21
  • @upvoters Why the upvote? – Did Oct 07 '13 at 08:11
  • @did: you are right, the previous answer was not correct. I have an other argument, but it's far more complicated than what Did gave. – Davide Giraudo Oct 07 '13 at 11:33