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Prove: Let $X_1 ,X_2 , ... , X_n , ...$ be i.i.d. random variables with $\mathbb{E}[X_1^+]=\mathbb{E}[X_1^-]=+\infty$.

If $S_n=\sum_{i=1}^{n}{X_i}$, then $$\limsup_{n\rightarrow\infty}{\frac{S_n}{n}=+\infty}\text{ a.s., }\liminf_{n\rightarrow\infty}{\frac{S_n}{n}=-\infty}\text{ a.s.}$$

I have proven that $$\mathbb{P}\left(\left\{\omega :\limsup_{n\rightarrow\infty}{\left|\frac{S_n}{n}(\omega)\right|=+\infty}\right\}\right)=1,$$

so at least one of $$\mathbb{P}\left(\left\{\omega:\limsup_{n\rightarrow\infty}{\frac{S_n}{n}(\omega)=+\infty}\right\}\right)=1\text{ and }\mathbb{P}\left(\left\{\omega:\liminf_{n\rightarrow\infty}{\frac{S_n}{n}(\omega)=-\infty}\right\}\right)=1$$ is true, but I don't know how to prove both of them.

Update: According to the paper The strong law of large numbers when the mean is undefined (Erickson K B, 1973), this proposition is wrong.

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  • $\begingroup$ And I also wonder how can you prove that $\mathbb{P}(\{\Omega:\limsup_{n\to+\infty}|\frac{S_n}{n}(\omega)|=+\infty\})=1$. Is it proved by using Kolmogorov Zero-One law? $\endgroup$
    – lostlife
    Jun 19, 2016 at 5:13

1 Answer 1

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Corollary 3 (p. 1195) in [Kesten (1970)][1] states:

If $$\mathbb E[X_1^+]=\mathbb E[X_1^-]=\infty,$$ then one of the following three cases must prevail

(i) $\displaystyle\lim_{n\to\infty}\frac{S_n}n = \infty$ w.p. 1

(ii) $\displaystyle\lim_{n\to\infty}\frac{S_n}n = -\infty$ w.p. 1

(iii) $\displaystyle\liminf_{n\to\infty}\frac{S_n}n = -\infty$ and $\displaystyle\limsup_{n\to\infty}\frac{S_n}n = +\infty$ w.p. 1

From the equivalence of (b) and (c) in Theorem 6:

If $\mathbb E[X_1^+]=\infty$ then the following statements are equivalent

(b) $\displaystyle\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n>-\infty\right)=1$

(c) $\displaystyle\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n=+\infty\right)=1$

and the Hewitt-Savage zero-one law, if (ii) holds then $$\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n=\infty\right)=1. $$ If neither (i) nor (ii) hold, then similarly $$\mathbb P\left(\liminf_{n\to\infty}\frac{S_n}n=-\infty\right)=1,$$ from which the result follows.

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    $\begingroup$ I am still confused by the inequality $\frac{S_{m_{n_j}}}{m_{n_j}}>m$. How can it hold? $\endgroup$
    – lostlife
    Jun 19, 2016 at 2:16
  • $\begingroup$ @lostlife That part of the proof does need revision. I couldn't figure out exactly how to use $\mathbb P\left(\limsup_{n\to\infty} \{X_{n+1}>n\}\right)=1$ to show that $\limsup_{n\to\infty}\frac{S_n}n=\infty$ almost surely. $\endgroup$
    – Math1000
    Jun 19, 2016 at 4:03
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    $\begingroup$ By using this condition, I can only show that $\mathbb{P}(\lim_{n\to+\infty}\frac{S_n}{n}=+\infty)=1$ or $\mathbb{P}(\lim_{n\to+\infty}\frac{S_n}{n}=-\infty)=1$ or $\mathbb{P}(\limsup_{n\to+\infty}\frac{S_n}{n}=+\infty)=\mathbb{P}(\liminf_{n\to+\infty}\frac{S_n}{n}=-\infty)=1$, one of the three holds. But why not the first two, I have no idea. $\endgroup$
    – lostlife
    Jun 19, 2016 at 4:07
  • $\begingroup$ @lostlife Indeed, you are correct: ams.org/journals/tran/1973-185-00/S0002-9947-1973-0336806-5/… $\endgroup$
    – Math1000
    Jun 19, 2016 at 7:23
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    $\begingroup$ Thank you. I have read that paper. $\endgroup$
    – lostlife
    Jun 19, 2016 at 7:44

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