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I'm struggling to understand the beginning of the solution to the following exercise:


Let $(X_n)_{n\geq 1}$ and $X$ be random variables. Prove that $X_n \to X$ almost surely if and only if for every $\epsilon>0$ $$P(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\})=0$$


Solution:

Notice that $\omega \in\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\}$ iff there exist a subsequence $(n_k)_{k\geq 1}$ such that $|X_{n_k}-X|\geq \epsilon$, so we have $$\{\limsup\limits_{n\to \infty}|X_n-X|> \epsilon\}\subset \limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\} \subset \{\limsup\limits_{n\to \infty}|X_n-X|\geq \epsilon\}\quad (*)$$

and so on..


I don't understand what this means.

  1. First what subsequence are we referring to? A subsequence that does converge? Why do we need this exactly?
  2. What's the difference between those 3 sets in line $(*)$. Why does it matter if the $\limsup$ is within the brackets and why does it matter whether $>$ or $\geq$? I really don't understand what the difference are.

I'm happy if someone could explain it to me.

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  • $\begingroup$ How does the proof continues? I am interested $\endgroup$
    – edamondo
    Oct 22, 2021 at 13:05

1 Answer 1

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Note the definition of lim sup for a sequence $(A_n)$ of sets: $$ \limsup_{n\to\infty} A_n=\bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n.$$ This translates to: $$ \omega\in\limsup_{n\to\infty} A_n \Leftrightarrow \forall k\, \exists n\ge k\colon \omega\in A_n.$$ Or, in everyday language: There are arbitrarily large $n$ for which $\omega\in A_n$. If you collect all such $n$ in a set and number them consecutively, you get an increasing sequence $(n_k)$ with $\omega\in A_{n_k}$ for all $k$.

That is the sequence mentioned in the proof, where $A_n=\{|X_n-X|\ge\epsilon\}$.

It matters whether lim sup is inside or outside the sets, because the two kinds of lim sup are entirely different beasts! But clearly, related.

The difference between sharp and non-sharp inequalities arise for $\omega$ such that $\limsup|X_n-X|=\epsilon$ exactly. Such an $\omega$ belongs in the set on the right end of $(*)$, but not in the left set. And it may or may not be in the middle set, depending on whether $|X_n-X|\ge\epsilon$ infinitely often or not. Both are possible, under the current assumption.

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  • $\begingroup$ Thank you very much for the answer. Unfortunately I'm still struggling to understand the difference between $\limsup\{|X_n-X|\geq \epsilon \}$ and $\{\limsup |X_n-X|\geq \epsilon \}$. Is the difference that in the first case the $\epsilon$ has influence on the $\limsup$ but in the later it has not? Is this related to Fatou's lemma? $\endgroup$
    – user185346
    Aug 27, 2016 at 10:54
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    $\begingroup$ The short answer is that $\omega\in\limsup\{|X_n-X|\ge\epsilon\}$ iff $|X_n-X|\ge\epsilon$ infinitely often (i.e., for infinitely many $n$), while if you untangle the other expression using the definition of the lim sup of a real sequence, you arrive at $\omega\in\{\limsup|X_n-X|\ge\epsilon\}$ iff for every $\eta<\epsilon$ it is true that $|X_n-X|\ge\eta$ infinitely often. The difference may seem subtle, but it cannot be neglected. $\endgroup$ Aug 27, 2016 at 16:01

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