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I have the following exercise:

Let $\Omega=\left\{0,1\right\}^{\infty}$ and $\mathcal{F}$ the cylynders sigma algebra on $\Omega$. The function $\varphi\left(\omega_1,\omega_2,\ldots\right)=\left(\omega_2,\omega_3,\ldots\right)$ defined on $\Omega$ is called a shift. An event $A\in\mathcal{F}$ is said to be shift invariant if $\varphi^{-1}\left(A\right)=A$. Let $\mathcal{C}$ be the class of shift-invariant events in $\mathcal{F}$. Show $\mathcal{C}$ is a sigma algebra. Let $A_n=\left\{\omega\in\Omega:\omega_n=1\right\}$. Show that $\mathcal{C}$ is not included in the tail sigma-algebra of $\left(A_n\right)_n$.

Having been able to show that $\mathcal{C}$ is indeed a sigma-algebra and that it contains the events $\liminf_n A_n$ and $\limsup_n A_n$, I was not able to show the last point, i.e. I can't exhibit an event which is shift invariant but does not belong to the tail sigma-algebra.

I would really appreciate a hint in order to visualize the problem better!

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    $\begingroup$ Try an event that depends on all the coordinates. $\endgroup$
    – user940
    Nov 15 '15 at 13:32
  • $\begingroup$ That was my initial strategy. But can an event that depends on all coordinates be transition invariant ?? If I add a random first coordinate to every singleton, the property of such singleton will change if it depends on all coordinates, no? $\endgroup$ Nov 15 '15 at 14:19
  • $\begingroup$ $A$ above is a singleton, and it I think it isn't shift invariant. Indeed $\varphi^{-1}\left(\omega\right)=\left\{0,1\right\}\times\left\{1\right\}^{\infty}$ $ \neq\omega$ $\endgroup$ Nov 15 '15 at 14:26
  • $\begingroup$ Yes.. you are defining shift-invariant based on the PRE-IMAGE by the shift operator, that is, $\varphi^{-1}\left(A\right)=A$. On the other hand, the "usual definition" of shift-invariant is $\varphi\left(A\right)=A$. $\endgroup$
    – Ramiro
    Nov 15 '15 at 14:40
  • $\begingroup$ Unfortunately this is the exercise I was given :) Could you give me a reference for the definition of "shift-invariant"? $\endgroup$ Nov 15 '15 at 14:56
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Either the statement or the conclusion of this exercise is incorrect.

With your definition of a shift-invariant set $A$, you have $\omega\in A$ if and only if $\varphi(\omega)\in A$. Thus, by induction, for all $n\geq 1$ you have $\omega\in A$ if and only if $\varphi^n(\omega)\in A$. This gives $$A=\{\omega: \varphi^n(\omega)\in A\}\in\sigma(A_n, A_{n+1},\dots).$$ Taking the intersection over $n$ shows that $A$ belongs to the tail $\sigma$-algebra.

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  • $\begingroup$ Yes, I agree: either the statement or the conclusion of this exercise is incorrect. $\endgroup$
    – Ramiro
    Nov 15 '15 at 17:10

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