Stronger version of strong law of large numbers

Question

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?

Answer 1

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.