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We have independent random variables such that

$$\mathbb{P}(X_n=n)=\mathbb{P}(X_n=-n)=\frac{1}{2(n+1)\ln(n+1)}$$ and

$$\mathbb{P}(X_n=0)=1-\frac{1}{(n+1)\ln(n+1)}$$

I am trying to show that $\frac{S_n}{n}$ does not converge to $0$ almost surely. I'm thinking about Borel-Cantelli Lemmas, so I'd like to show that $\sum_{n=0}^{\infty} \mathbb{P}(S_n=0)\lt\infty$.

I have tried coming up with a recurrence relation for $\mathbb{P}(S_n=0)$. I have also tried finding out whether $X_n$ could take the value $0$/$n$/$-n$ for infinitely many $n$'s. But these two did not help much.Do you see a hint you could give me?

Did
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Kika
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  • The convergence of the series $\sum P(S_n=0)$ seems quite unrelated to what you try to show. Why do you bring it to the fore? – Did Nov 24 '15 at 10:38
  • Infinitely often X_n = 0 so it seems –  Nov 24 '15 at 10:38
  • @Did Would $\mathbb{P}(\frac{S_n}{n}=0)$ be better? – Kika Nov 24 '15 at 10:40
  • @JackBauer I agree but $X_n \neq 0$ infinitely often too, right? – Kika Nov 24 '15 at 10:42
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    I'm fairly certain I've seen this exact same problem on MSE before, I'll try to find it.. – Math1000 Nov 24 '15 at 12:37
  • Kika: Sorry? Do you see any difference between $P(S_n=0)$ and $P(S_n/n=0)$? Then what are you talking about exactly? – Did Nov 24 '15 at 13:16
  • @JackBauer Rather, infinitely often $X_n\ne0$... – Did Nov 24 '15 at 13:17
  • "I have also tried finding out whether $X_n$ could take the value $0$/$n$/$-n$ for infinitely many $n$'s." That is much more interesting... How did it go? – Did Nov 24 '15 at 13:18
  • @Math1000 I couldn't find but if I've missed it, can you give the link? – Kika Nov 24 '15 at 13:43
  • @Did Sorry, that was me misunderstanding the statement of Borel-Cantelli Lemmas, don't we get that $X_n=0$ infinitely often and $X_n \neq 0$ infinitely often? Both series diverge. – Kika Nov 24 '15 at 13:49
  • I feel like I should be able to conclude knowing that $X_n \neq 0$ infiitely often... But I'm confused as how... – Kika Nov 24 '15 at 14:09
  • I didn't find the MSE link. But this problem is in Counterexamples in Probability by Stoyanov. – Math1000 Nov 24 '15 at 14:55
  • Did you fully read and understand the answer below before accepting it? – Did Nov 24 '15 at 15:10
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    I thought I 'unaccepted it', it's now done. What about using that $X_n=n$ for infinitely many $n$'s. Then if $X_{n+1}=n+1$, considering $\frac{S_{n+1}}{n+1}-\frac{S_n}{n}$, we can get that it is $\geq$ to $1/2$ using that $S_{n} \leq 1+...+n=\frac{n(n+1)}{2}$. That holds for infinitely many $n$'s so we're done? – Kika Nov 24 '15 at 15:23

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We have $$\sum_{n=1}^\infty \mathbb P(X_n\geqslant n) = \sum_{n=1}^\infty\frac1{2(n+1)\log (n+1)}=\infty $$ which implies that $$\mathbb P\left(\limsup_{n\to\infty} X_n \geqslant n\right)=1 $$ by the second Borel-Cantelli lemma. Now consider the difference in consecutive partial sums, as @Kika: $$\frac{S_{n+1}}{n+1}-\frac{S_n}n = \frac{X_{n+1}}{n+1} - \frac1{n(n+1)}\sum_{k=1}^n X_k.$$ Since $$\sum_{k=1}^n X_k \leqslant \sum_{k=1}^n k = \frac{n(n+1)}2, $$ it follows that $$\frac{S_{n+1}}{n+1} - \frac{S_n}n \geqslant 1 - \frac{n(n+1)}{2n(n+1)} = \frac12, $$ for infinitely many $n$, and hence $$\mathbb P\left(\limsup_{n\to\infty} \frac{S_n}n\geqslant\frac12\right) =1. $$ We conclude that $\frac{S_n}n$ does not converge to $0$ a.s.

Math1000
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    Actually, the implication is only that $$\limsup_{n\to\infty} |X_n| \geqslant n\implies\limsup_{n\to\infty} \frac{|S_n|}n\geqslant \frac12,$$ for the same conclusion. – Did Nov 24 '15 at 15:09
  • @Did Fair enough. The book simply states $$\mathbb P(|X_n|\geqslant n\ ; \mathrm{ i.o.}) \implies\mathbb P\left(\lim_{n\to\infty} S_n/n \ne 0\right) = 1.$$ – Math1000 Nov 24 '15 at 15:53
  • Which book is that? – Kika Nov 24 '15 at 16:24
  • 15.3 in Counterexamples in Probability by Stoyanov. – Math1000 Nov 24 '15 at 16:25
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    I understand your answer reproduces this book. Mentioning explicitely the fact in the answer seems to be requisite. – Did Nov 24 '15 at 17:32
  • @Did I added detail that the book does not include. Mentioning the book in the answer now seems redundant ;) – Math1000 Nov 24 '15 at 20:52
  • I disagree. You know, the ethics of proper attribution runs rather deep and strict amongst mathematicians... – Did Nov 24 '15 at 21:01
  • When I deem it necessary, I do provide sources, for example here: http://math.stackexchange.com/questions/1536488/when-superposition-of-two-renewal-processes-is-another-renewal-process/1536806

    But for a relatively straightforward exercise such as this, the "proper attribution" would be the prior MSE post where I last saw it. Which I was not able to find.

     – Math1000 Nov 24 '15 at 21:26
  • Finally found it: http://math.stackexchange.com/questions/1288502/sequence-satisfies-weak-law-of-large-numbers-but-doesnt-satisfy-strong-law-of-l/1299888 – Math1000 Nov 25 '15 at 01:22
  • Can you show why the event: $$\frac{S_{n+1}}{n+1} - \frac{S_n}n \geqslant \frac12 $$ infinitely often, implies the event: $$\limsup_{n\to\infty} \frac{S_n}{n} \geq \frac12 $$ In particular, the first event is about differences between partial sums, the second one is about their values. – yoshi Apr 27 '20 at 15:02
  • Heuristically, I think if the partial sums differ by some finite amount (here 1/2) infinitely often, then a limit can't exist. I'm having trouble formulating this precisely though. – yoshi Apr 27 '20 at 15:51