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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.

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    $\begingroup$ r u sure the $n^{-1}$ inside the series is supposed to be there? $\endgroup$ Sep 2, 2020 at 2:30
  • $\begingroup$ @mathworker21 yes very sure. $\endgroup$ Sep 2, 2020 at 15:08
  • $\begingroup$ Just curious, do you want to apply Borel Cantelli or why are you so focused on it? $\endgroup$
    – Diger
    Sep 5, 2020 at 11:43
  • $\begingroup$ @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. $\endgroup$ Sep 6, 2020 at 10:59

2 Answers 2

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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.

Note that we do not use the pairwise independence of the sequence in this step; this is certainly needed in order to establish the convergence of the series given in the assumption.

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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.

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  • $\begingroup$ 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. $\endgroup$ Sep 5, 2020 at 12:40
  • $\begingroup$ 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? $\endgroup$
    – Diger
    Sep 5, 2020 at 12:56
  • $\begingroup$ I mean the equation you wrote in the comment is true, but that it only gives the convergence in probability. $\endgroup$ Sep 5, 2020 at 13:07
  • $\begingroup$ Isn't that the same? With probability 1 (almost sure), we have that for $n\rightarrow \infty$ $$\max _{1 \leq m \leq n} n^{-1} \left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|<\varepsilon \, .$$ $\endgroup$
    – Diger
    Sep 5, 2020 at 13:09
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    $\begingroup$ 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. $\endgroup$ Sep 6, 2020 at 9:04

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