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Let $X_n, n\geq 2$ be a random variable taking the values $n,0$ or $-n$ with probabilities $\frac{1}{2n \log (n)}, 1 - \frac{1}{n \log (n)}$ and $\frac{1}{2n \log (n)}$, respectively. Define the sum $S_n = \sum_{i = 1}^n X_i$. Suppose we know: $$\mathbb{P}\{|X_n| \geq n \,\,\text{infinitely often}\} = 1$$ Why does this imply: $$\mathbb{P}\left\{\lim_{n \rightarrow \infty} \frac{S_n}{n} \neq 0\right\} = 1$$

Context: I'm reading about an example of a random sequence that satisfies the weak LLN but not the strong LLN. This example is 15.3 in Stoyanov's Counterexamples in Probability. I have found more detail in this answer but not enough to figure it out myself.

yoshi
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1 Answers1

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We have that

$$X_{n}=S_{n}-S_{n-1},$$

therefore $$|X_{n}| \le |S_{n}|+|S_{n-1}|$$

Therefore, if $|X_n|=n$, or $|S_{n-1}| \ge n/2$ or $|S_n| \ge n/2$, otherwise $|X_n|<n$. That means

$$\{X_n=n\} \subseteq \left\{\frac{|S_n|}{n}\ge\frac{1}{2} \right\} \bigcup\left\{\frac{|S_{n-1}|}{n}\ge\frac{1}{2} \right\}.$$

By Borel-Cantelli, is easily seen that $$\mathbb{P}\{X_n=n \text{ infinitely often}\}=1,$$

then there are, with probability $1$, infinitely many $n$ for which either $\frac{|S_n|}{n}\ge \frac{1}{2}$ or $\frac{|S_{n-1}|}{n} \ge \frac{1}{2}$.But note that if $\frac{|S_{n-1}|}{n} \ge \frac{1}{2}$, then $\frac{|S_{n-1}|}{n-1}=\frac{n}{n-1} \frac{|S_{n-1}|}{n} \ge \frac{1}{2}.$

We see that, almost surely, we have infinite $n$'s for which $\frac{|S_n|}{n} \ge \frac{1}{2}$, therefore $S_n/n$ cannot converge to $0$.

Célio Augusto
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