The theorem 2.3.7 states the following.

If $X_1, X_2, \ldots$ are i.i.d with $E \left| X_i \right| = \infty$, then $P(\left|X_n\right| \geq n \;\textit{i.o.}) = 1$. So if $S_n = X_1 + \ldots + X_n$, then $P(\lim \frac{S_n}{n} \textit{exists} \in (-\infty, \infty)) = 0$.

where i.o. stands for infinitely often, or limsup of sequence of sets.

I do not understand the second part of the proof.

It lets $C = \{\omega \colon \lim \frac{S_n}{n} \text{exists} \in (-\infty, \infty) \}$ and states that on $C \cap \{\omega \colon \left|X_n\right| \geq n \}$, $\left| \frac{S_n}{n} - \frac{S_{n+1}}{n+1} \right| > \frac{2}{3}$ i.o.

I tried using the inequality $\left|a-b\right| \geq \left| \left|a\right| - \left|b\right| \right|$, but at best I could only replace $S_{n+1}/(n+1)$ with $1$, which I don't think helps much.

So if anyone could explain why $\left| \frac{S_n}{n} - \frac{S_{n+1}}{n+1} \right| > \frac{2}{3}$ i.o., I would appreciate that, thank you.

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1 Answer 1


By algebra, \begin{align*} \big|\frac{S_{n+1}}{n+1}-\frac{S_n}{n}\big|&=\big|\frac{X_{n+1}}{n+1}+S_n(\frac{1}{n+1}-\frac{1}{n})\big|\\ &=\big|\frac{X_{n+1}}{n+1}-\frac{S_n}{n(n+1)}\big|\\ &\ge \big|\frac{X_{n+1}}{n+1}\big|-\big|\frac{S_n}{n(n+1)}\big|. \end{align*} On $\{X_n\ge n\ \text{i.o.}\}$, the first term $\ge 1$ infinitely often, and the second term goes to zero, so in particular, for all $n$ large enough, $|\frac{S_n}{n(n+1)}|<1/3$. Hence $$ \big|\frac{X_{n+1}}{n+1}\big|-\big|\frac{S_n}{n(n+1)}\big| > 2/3\quad\text{infinitely often}. $$

  • $\begingroup$ the choice of $2/3$ is arbitrary? $\endgroup$
    – Phil
    Aug 11, 2022 at 22:43
  • $\begingroup$ @Phil: Yes, the choice of 2/3 is completely arbitrary, because the point of what is being shown is to show that $S_n/n$ is not a Cauchy sequence. Any fixed $\epsilon_0>0$ would work instead of $1/3$. $\endgroup$
    – Alex Ortiz
    Aug 11, 2022 at 22:47

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