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Let $\{X_n\}_{n \geq 1}$ be an independent sequence of random variables on $(\Omega, \mathcal{F}, \mathbb{P})$. Fix $n \geq 1$. I want to prove that $X_1, \ldots, X_n$ is independent of $\limsup X_n$.

This result is incredibly obvious, at least intuitively: the limsup does not depend on the first $n$ elements of the sequence. But I'm having trouble showing this rigorously.

The issue I'm having is in working with a $\sigma$-field generated by a finite sequence (i.e. $X_1, \ldots, X_n$) versus a $\sigma$-field generated by the infinite remainder of the sequence (i.e. $\limsup X_n$). The definitions are straightforward enough, but showing independence between the two $\sigma$-fields is confusing me. I know that I have to use continuity of $\mathbb{P}$ somewhere, but for some reason it's eluding me. Can anyone point me in the right direction?

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This is a consequence of Dynkin's $\pi$-$\lambda$ theorem.

To show that $\mathcal F_n=\sigma(X_k;k\leqslant n)$ and $\mathcal G_n=\sigma(X_k;k\geqslant n+1)$ are independent sigma-algebras, consider the class $\mathcal C_n$ of the events in $\mathcal G_n$ that depend on a finite number of $X_k$, $k\geqslant n+1$, and the class $\mathcal M_n$ of events in $\mathcal G_n$ independent of $\mathcal F_n$. Then $\mathcal C_n$ is a $\pi$-system, $\mathcal M_n$ is a $\lambda$-system, and $\mathcal C_n\subseteq\mathcal M_n$ (note that this inclusion is concerned with independence between events depending on $X_k$ for a finite number of $k$ only). Hence $\sigma(\mathcal C_n)\subseteq\mathcal M_n$. Since $\sigma(\mathcal C_n)=\mathcal G_n$, this proves that $\mathcal F_n$ and $\mathcal G_n$ are independent.

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  • $\begingroup$ Hi Did. Sorry for being slow, but you claim $\sigma(\mathcal{C} _n) = \mathcal{F}_n$. Did you mean $\mathcal{G}_n$? And if so, is this direct from a definition or does it follow from another fact? Thanks so much for your patience. $\endgroup$ – gogurt Jul 21 '13 at 14:24
  • $\begingroup$ Yes $\mathcal G_n$. And yes this is direct since $\mathcal C_n$ contains $\sigma(X_k)$ for every $k\geqslant n+1$. $\endgroup$ – Did Jul 21 '13 at 19:03
  • $\begingroup$ Thanks! And also just to be complete, just after you mention that $\mathcal{C}_n$ is a $\pi$-system and $\mathcal{M}_n$ is a $\lambda$-system, you mean that $\mathcal{C}_n \subset \mathcal{M}_n$, correct? Then the $\pi-\lambda$ theorem gives $\sigma(\mathcal{C}_n) \subset \mathcal{M}_n$. $\endgroup$ – gogurt Jul 21 '13 at 19:33
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The value of $\text{inf}_{k\ge n}\, \text{sup}_{m \ge k}\, x_m$ is independent of $n$.

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