# Stronger version of strong law of large numbers

Let $(X_i)_{i\in\mathbb{N}}$ be pairwise independent random variables where $E\left[X_i\right]=0$ for all $i\in\mathbb{N}$ and $\sup_{n}E\left[X_n^2\right]\lt\infty$. Then for $S_n=\sum_{i=1}^n X_i$ and $t>\dfrac{3}{4}$ we have $\dfrac{S_n}{n^t}\to 0$ almost surely.

I looked at different proofs of the SLLN, but I can't see where we could be using such a $t$. Given the assumptions I was able to see we have finite variance, but I didn't make much ground in that direction. Can anyone give me any hints to get me started?

• Kolomorogov's condition for a sequence of independent r.v.s is that the sum of the $\frac{X_n}{b_n^2} < \infty,\;\; b_n>0, b_n\to \infty$. However, its the pairwise independence that is difficult to see how to accomodate, as it implies that the variance of the sum is the sum of the variance, but there may be correlations among groups of random variables. – user76844 Dec 17 '14 at 18:15
• Also, with independent variables, you can take $t=\frac{1}{2}$, so something about lack of true independence is decreasing the convergence rate. Sorry, that's all that came to my mind. You'll need someone with more theoretical background. – user76844 Dec 17 '14 at 18:17
• One note is that you get a weak law immediately under these assumptions, since the variance of $S_n/n^t$ goes like $n^{1-2t}$. – Nate Eldredge Oct 27 '17 at 19:37

It seems that the wanted result can be improved in two directions.

• The random variables do not need to be pairwise independent: uncorrelatedness is sufficient.
• We have the result for $t\gt 1 /2$ and not only $3/4$.

Using the fact that the random variables are uncorrelated, we derive easily that $$\mathbb E\left[S_n^2\right]\leqslant n\sup_{i\geqslant 1}\mathbb E \left[X_i^2\right].$$ Now, we shall show the following:

Lemma. Let $\left(X_i\right)_{i\geqslant 1}$ be a centered uncorrelated sequence such that $\sup_{i\geqslant 1}\mathbb E \left[X_i^2\right]$ is finite. Then there exists a constant $C$ such that for any $n\geqslant 1$, $$\sup_{n\geqslant 1}\frac 1{2^nn^2} \mathbb E \left[\max_{1\leqslant j\leqslant 2^n} S_{j} ^2\right]\leqslant C.$$ This is contained in a paper by Stout (1974).

Let $s:=\sup_{i\geqslant 1}\mathbb E\left[X_i^2\right]$. We shall show by induction on $n$ that for any uncorrelated sequence $\left(X_i\right)$ with partial sums $S_n$,
$$\frac 1{2^nn^2} \mathbb E \left[\max_{1\leqslant j\leqslant 2^n} S_{j} ^2\right]\leqslant \max\left\{2s , 2\mathbb E\left[\max\left\{X_1^2,S_2^2\right\}\right] \right\} =:A .$$ This is satisfied for $n=1$ by assumption. Now, assume that it is true for some $n$. Denote $M_n:=\max_{1\leqslant j\leqslant 2^n}\left\lvert S_j\right\rvert$. Then $$M_{n+1}^2\leqslant M_n^2+\max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}+S_{2^n}\right\rvert^2$$
hence $$\mathbb E\left[M_{n+1}^2 \right] \leqslant \mathbb E\left[M_{n}^2 \right] +\left( \left\lVert \max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}\right\rvert\right\rVert _2 +\left\lVert S_{2^n}\right\rVert_2 \right)^2.$$ Using the induction assumption, we obtain that $\mathbb E\left[M_{n}^2 \right]\leqslant n^22^nA$; using again the induction assumption with the sequence $\left(X_{i+2^n}\right)_{i\geqslant 0}$ instead of the original sequence, we obtain that $\left\lVert \max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}\right\rvert\right\rVert _2^2\leqslant n^22^nA$. Consequently, $$\mathbb E\left[M_{n+1}^2 \right] \leqslant n^22^nA +\left( \sqrt{ n^22^nA} +2^{n/2}s^{1/2} \right)^2=2^{n+1}n^2A+2^{n+1}\sqrt As^{1/2} +2^ns.$$ We have to show that
$$2^{n+1}n^2A+2^{n+1}\sqrt As^{1/2} +2^ns\leqslant 2^{n+1}\left(n+1\right)^2A$$
which is equivalent to $$2\sqrt As^{1/2} +s\leqslant 2 A\left(2n+1\right).$$ It is thus sufficient to prove that $2\sqrt As^{1/2} +s\leqslant 2 A$, which holds since $A\geqslant 2s$.

Now, to conclude, observe that for any $t\gt 1/2$, the expectation of $$\sum_{n=1}^N2^{-2nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert^2$$ can be bounded independently of $N$, hence the random variable $\sum_{n=1}^{+\infty} 2^{-2nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert^2$ is almost surely finite hence $2^{-nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert$ goes to zero almost surely.