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I have the following sequence of independent r.v. $$\mathbb{P}(\{X=n\})=\big(\frac{1}{n^2}\big), \mathbb{P}(\{X=0\})=\big(1-\frac{1}{n^2}\big)$$ and also, $S_n = \sum_{i=1}^{n} X_i$.

I am trying to find if the sequence $\frac{S_n}{n}$ converges to a constant $c_1$ in probability.

Now, not being iid I can not use the laws of large numbers. I was tryint to use Chebyshev inequality:

Since

$$E(S_n)=\sum_{i=1}^{n}\frac{1}{i}=H_n, Var(S_n)=n-\sum_{i=1}^{n}\frac{1}{i^2}=K_n, E(S^2_n)=n-K_n+H^2_n$$

and that $$\mathbb{P}(|S_n-nc_1|>n\epsilon)\leq \frac{E((S_n-nc_1)^2)}{n^2\epsilon^2} = \frac{n-K_n+H^2_n+n^2c_1^2-2nc_1H_n}{n^2\epsilon^2}\to \frac{c_1^2}{\epsilon^2}$$

Now, since I don't think that calculating higher moments can be of any help, what do you think? Chebyshev is just an upper bound, so does not provide any mean to say that it does not converge to any constant in Probability or am I missing something? I wanted to use the second Borel-Cantelli Lemma, but once again I would only get that $\mathbb{P}(A_n) \leq + \infty$ so I can not say that $|\frac{S_n}{n}-c_1|>\epsilon$ infinitely often.

Also, what could I say about the almost sure convergence of $\frac{S_n}{nH_n}$ to a constant $c_2$?

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  • $\begingroup$ how can $\mathbb P\{X=0\}=1-\frac{1}{n^2}$ ? $\endgroup$ – Surb Apr 14 '18 at 10:47
  • $\begingroup$ Use characteristic functions. My calculations show that $Ee^{itS_n} \to 1$ so $S_n \to 1$ in probability. Of course, this implies $S_n /n$ and $S_n /ES_N$ both $\to 0$ in probability. $\endgroup$ – Kavi Rama Murthy Apr 14 '18 at 11:53
  • $\begingroup$ Use the Borel Cantelli lemma to show that $\{n \in \mathbb{N}; X_n(\omega) \neq 0\}$ is for each fixed $\omega$ a finite set. $\endgroup$ – saz Apr 14 '18 at 18:12
  • $\begingroup$ @saz Why did you delete your answer? $\endgroup$ – Davide Giraudo Dec 18 '18 at 21:47
  • $\begingroup$ @DavideGiraudo I don't remember why I deleted it... perhaps because my comment is saying (almost) as much as my answer. $\endgroup$ – saz Dec 19 '18 at 7:29

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