Prove that if the series is convergent then the law of large numbers hold.

Let $$(X_n)_{n \geq 1}$$ be a sequence of pairwise independent random variables such that :

$$\sum_{n=1}^{\infty} n^{-1} P\left\{\max _{1 \leq m \leq n}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon n\right\}<\infty$$

show that $$n^{-1} \sum_{k=1}^{n}\left(X_{k}-E X_{k}\right) \rightarrow 0$$ almost surely.

I'm fairly certain Borel Cantelli Lemma for pairwise independent random variables is to be used here but I dont know how to get rid of the $$n^{-1}$$ inside the series.

• r u sure the $n^{-1}$ inside the series is supposed to be there? Sep 2, 2020 at 2:30
• @mathworker21 yes very sure. Sep 2, 2020 at 15:08
• Just curious, do you want to apply Borel Cantelli or why are you so focused on it? Sep 5, 2020 at 11:43
• @Diger this is part of paper I was reading, in the beginning the borel cantelli lemma is stated so I thought it had to be there for a reason. Sep 6, 2020 at 10:59

Denote $$p_{n,\varepsilon}:= P\left\{\max _{1 \leqslant m \leqslant n}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon n\right\}.$$ If $$2^N\leqslant n\leqslant 2^{N+1}-1$$, then $$p_{2^N,2\varepsilon}\leqslant p_{n,\varepsilon}\leqslant P\left\{\max _{1 \leqslant m \leqslant 2^{N+1}-1}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon 2^N\right\}\leqslant p_{2^{N+1},\varepsilon/2}.$$ Therefore, by splitting the series into a series of indexed between two consecutive dyadic numbers, one derives that the initial assumption is equivalent to $$\sum_{N\geqslant 0}p_{2^N,\varepsilon}<+\infty$$ for all $$\varepsilon$$. By an application of the Borel-Cantelli lemma, one derives that $$\frac 1{2^N}\max_{1\leqslant m\leqslant 2^N}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|\to 0\, a.s.$$ from which it follows that $$n^{-1} \sum_{k=1}^{n}\left(X_{k}-E X_{k}\right) \rightarrow 0$$ almost surely.
Using the same notation as DG $$p_{n,\varepsilon}:= P\left\{\max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon \right\}$$ we have that $$\sum_{n=1}^\infty \frac{p_{n,\epsilon}}{n}<\infty$$ for all $$\epsilon>0$$. Since $$\sum_{n=1}^\infty \frac{1}{n} = \infty$$ it must also be true that $$\lim_{n\rightarrow \infty} p_{n,\epsilon} = 0$$ otherwise the series would diverge. This is equivalent to $$\lim_{n\rightarrow\infty}P\left\{\max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|<\varepsilon \right\} = 1$$ which is true $$\forall \epsilon>0$$. Therefore almost surely $$\lim_{n\rightarrow \infty} n^{-1} \left|\sum_{k=1}^{n}\left(X_{k}-E X_{k}\right)\right| \leq \lim_{n\rightarrow \infty} \max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|<\varepsilon$$ $$\forall \epsilon>0$$, which is what we need.
• It seems that your penultimate equation would only give the convergence in probability to $0$, not the almost sure convergence. This is the best we can deduce from the fact that $p_{n,\varepsilon}\to 0$. For the almost sure convergence, one really has to exploit the convergence of the series. Sep 5, 2020 at 12:40
• What do you mean? $$\lim_{n\rightarrow\infty}P\left\{\max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|<\varepsilon \right\} = 1$$ is not true? Sep 5, 2020 at 12:56
• Isn't that the same? With probability 1 (almost sure), we have that for $n\rightarrow \infty$ $$\max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|<\varepsilon \, .$$ Sep 5, 2020 at 13:09
• If we follow your reasoning, you deduce from $p_{n,\varepsilon}\to 0$ the almost sure convergence. This cannot be correct, as $p_{n,\varepsilon}\to 0$ for each positive $\varepsilon$ is equivalent to the convergence in probability. Here stats.stackexchange.com/questions/2230/… is a discussion on the difference between the two modes of convergence. Sep 6, 2020 at 9:04