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$X_1,X_2,\ldots$ is a sequence of independent random variables and

$$P\{X_n = n^2 - 1\} = 1 - P\{X_n = -1\} = n^{-2}$$

Clearly, $E[X_n] = 0$. However, $\frac{1}{n}S_n \to -1$ almost surely where $S_n = \sum_{i=1}^nX_i$.

My proof of this assertion is as follows and I would appreciate it if someone could check its correctness as well as the precision of my reasoning.

$(X_n)_n$ being a collection of independent random variables, the event $\left\{\omega: \lim_{n\to\infty}\frac{1}{n}S_n(\omega) = -1\right\}$ belongs to the tail $\sigma$-algebra and hence has a probability of either zero or one. If I can show that the probability of this event is strictly positive, then I am done.

I note that $$\{X_2 = -1, X_3 = -1, \ldots\} \subset \left\{\lim_{n\to\infty}\frac{1}{n}S_n = -1\right\}$$

By monotone convergence $$P\{X_2 = -1, X_3 = -1, \ldots\} = \lim_{n\to\infty}P\{X_2 = -1, \ldots, X_n = -1\}$$ This yields $P\{X_2 = -1, X_3 = -1, \ldots\} = \frac{1}{2}$. Therefore, $P\left\{\lim_{n\to\infty}\frac{1}{n}S_n = -1\right\} \geq \frac{1}{2}$, which implies $\frac{1}{n}S_n \to -1$ almost surely.

I would also appreciate seeing different proofs of this claim, in particular ones that do not invoke Kolmogorov's $0-1$ law.

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    $\begingroup$ Borel Cantelli shows that, almost surely, there exists $N$ such that $X_n=-1$ for every $n\geqslant N$, hence you are done. $\endgroup$
    – Did
    Commented Aug 29, 2018 at 15:03
  • $\begingroup$ @Did Thanks a lot. $\endgroup$
    – Calculon
    Commented Aug 29, 2018 at 15:04
  • $\begingroup$ How did you get $P(X_2=-1,X_3=-1,...)=\frac{1}{2})$? My gut feeling says it should $=0$. $\endgroup$ Commented Aug 29, 2018 at 15:05
  • $\begingroup$ @herbsteinberg $P\{X_2 = -1, \ldots, X_n = -1\} = \frac{n+1}{2n}$ (just factorize each term in the resulting product and the pattern of cancellation becomes obvious) $\endgroup$
    – Calculon
    Commented Aug 29, 2018 at 15:08
  • $\begingroup$ I presume you noticed that the (weak) law of large numbers doesn't apply since the variance becomes infinite with n. $\endgroup$ Commented Aug 29, 2018 at 17:14

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