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?


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.

  • $\begingroup$ 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

The value of $\text{inf}_{k\ge n}\, \text{sup}_{m \ge k}\, x_m$ is independent of $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.