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Let $X_1, X_2, ...$ be an independent sequence of random variables on $(\Omega, \mathscr{F}, \mathbb{P})$.

What I'm trying to prove is:

Prove that $X_1, X_2, ..., X_k$ is independent of $\liminf X_n$.

I've seen a proof of it already, but it was complicated. I am trying to prove it in another way.

This seems that this can be almost immediately just from the definition/s and can be extended to limsup and to any finite collection of the random variables of the sequence.

Based on this, I guess:

$\liminf X_n = \sup_{n \geq 1} \inf_{m \geq n} \ X_m = \sup_{n \geq k+1} \inf_{m \geq n} \ X_m$

I'm not quite sure how to say this precisely, but I think such shows that $\liminf X_n$ has nothing to do with $X_1, ..., X_k$. Perhaps we say something like $\sigma(\liminf X_n) \subseteq \sigma(X_{k+1}, X_{k+2}, ...)$, which is independent of $X_1, ..., X_k$ ?

Similarly, I guess that $\limsup X_n = \inf_{n \geq 1} \sup_{m \geq n} \ X_m = \inf_{n \geq k+1} \sup_{m \geq n} \ X_m$ and $\sigma(\limsup X_n) \subseteq \sigma(X_{k+1}, X_{k+2}, ...)$, which is independent of $X_1, ..., X_k$.

Extending to proving $\{X_k\}_{k \in K}$,where K is finite and $K \subseteq \mathbb{N}$, is independent of $\limsup X_n$ or $\liminf X_n$:

Pick the highest index , call it k* and then

$\liminf X_n = \sup_{n \geq 1} \inf_{m \geq n} \ X_m = \sup_{n \geq k^{*}+1} \inf_{m \geq n} \ X_m$

Pls do not present a different kind of proof if what I am attempting is wrong. Pls just state what is wrong with my attempt.

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    $\begingroup$ It seems correct to me. But could you please rephrase this as a question? (it is how mse works) $\endgroup$ – Siméon Jul 24 '15 at 16:35
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    $\begingroup$ @Siméon 'I'm not quite sure how to say this precisely. Perhaps we say something like σ(lim infXn)⊆σ(Xk+1,Xk+2,...), which is independent of X1,...,Xk ?' ? $\endgroup$ – BCLC Jul 24 '15 at 16:45
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    $\begingroup$ @Siméon Oh wait. I excluded something important. Will edit. $\endgroup$ – BCLC Jul 24 '15 at 16:46
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    $\begingroup$ @Siméon Edited 8th paragraph $\endgroup$ – BCLC Jul 24 '15 at 16:58
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Here is an example of how you can make your idea rigorous.

Let $X_1,X_2,X_3,\dots$ be a sequence of independent real random variables. Let us fix some integer $k$ and let $$ \mathcal F = \sigma(X_m ; m \leq k), \qquad \mathcal G = \sigma(X_m ; m \geq k+1). $$ For all $n \geq k+1$, the random variable $Y_n = \sup_{m \geq n} X_m$ is $\mathcal G$-measurable. Since a limit of $\mathcal G$-measurable functions is still $\mathcal G$-measurable, the random variable $\lim_{n\to \infty} Y_n$ is $\mathcal G$-measurable. But $\lim_{n\to\infty} Y_n = \limsup_{n\to\infty} X_n$ by definition of $\limsup$.

The random variable $(X_1,\dots,X_k)$ is $\mathcal F$-measurable and the random variable $\limsup X_n$ is $\mathcal G$-measurable. They are therefore independent since the $\sigma$-algebras $\mathcal F$ and $\mathcal G$ are independent $\because X_1,X_2,X_3,\dots$ are independent. In particular, the random variables $X_1,\dots,X_k$ are independent of $\limsup_{n\to\infty} X_n$.

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  • $\begingroup$ Thanks Siméon! How is the edit? $\endgroup$ – BCLC Sep 21 '15 at 15:55
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    $\begingroup$ @BCLC: Your edit was alright, but I prefer the current version. $\endgroup$ – Siméon Sep 22 '15 at 16:21

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