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I want to show the following:

Let $X_1, X_2,\ldots$ i.i.d. random variables on $(\Omega, \mathfrak A, P)$ with infinite expectation. Let $S_n = X_1+\ldots+X_n$. Then $$ P\left(\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R\right) = 0. $$

Using the second Borel-Cantelli-Lemma, I already showed that $$ P(|X_n| \ge n \text{ infinitely often}) = 1. $$

Now I'd like to show that $$ P\left(\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R\right) \le P(|X_n| \ge n \text{ at most finitely often}). \tag{1} $$ Here's what I did:

Suppose that there exists $\omega \in \Omega$ s.t. $X_n(\omega) \ge n$ infinitely often and $\frac{S_{n}(\omega)}{n} \to L \in \mathbb R$.

Note that $$ \lim\limits_{n \to \infty}\frac{S_n(\omega)}{n+1} = \lim\limits_{n \to \infty}\frac{n}{n+1}\frac{S_n(\omega)}{n} = L $$

Then $$ \begin{align} \underset{n\to\infty}{\lim\sup} \left| \frac{S_{n+1}(\omega)}{n+1} - L \right| &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} + \frac{S_{n}(\omega)}{n} - L \right| \\ &\ge \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \left| \frac{S_{n}(\omega)}{n+1} - L \right| \\ &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \underset{n\to\infty}{\lim\inf}\left| \frac{S_{n}(\omega)}{n+1} - L \right| \tag{2}\\ &= 1-0 = 1, \end{align} $$ which is a contradiction to $\lim\limits_{n \to \infty}\frac{S_{n}(\omega)}{n} = L$. Therefore $$ \left\{\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R \right\} \subset \left\{ |X_n| \ge n \text{ at most finitely often}\right\}, $$ which implies $(1)$ and therefore the claim.


Is this correct? I'm mostly unsure about equation $(2)$. Is it always true that the limsup of a difference of two positive sequences is the limsup of the first sequence minus the liminf of the second one?

Thank you for your help!

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  • $\begingroup$ It is not always true that the limsup of a difference of two positive sequences is the limsup of the first sequence minus the liminf of the second one. In fact, the inequality you need in your proof is also not valid. For example, consider the sequences {02,0,2,0,...} and {1,2,1,2,...}. $\endgroup$ – Kavi Rama Murthy Sep 25 '17 at 7:41
  • $\begingroup$ You're correct, thank you for your answer. $\endgroup$ – Epiousios Sep 29 '17 at 16:58
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As @KaviRamaMurthy correctly pointed out, $(2)$ is incorrect. However, the proof can be fixed since for any real sequences $a_n$ and $b_n$, if $b_n$ is convergent, it holds that $$ \underset{n\to\infty}{\lim\sup}\, (a_n + b_n) = \underset{n\to\infty}{\lim\sup}\, a_n + \lim\limits_{n\to\infty} b_n. $$

So I can write $$ \begin{align} \underset{n\to\infty}{\lim\sup} \left| \frac{S_{n+1}(\omega)}{n+1} - L \right| &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} + \frac{S_{n}(\omega)}{n} - L \right| \\ &\ge \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \left| \frac{S_{n}(\omega)}{n+1} - L \right| \\ &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \lim\limits_{n\to\infty}\left| \frac{S_{n}(\omega)}{n+1} - L \right| \tag{2´}\\ &= 1-0 = 1, \end{align} $$

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