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

Question

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.

Answer 1

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.

Answer 2

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.